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n v = rω has units length We need to talk about units here. The units of v are length time , the units of r are length only, and the units of ω are radians time , whereas time = length·radians the right hand side has units length · radians . The supposed contradiction in units is resolved by remembering that radians are a dimensionless quantity and angles in radian measure are identiﬁed with real numbers so that the units length·radians time . We are long overdue for an example. reduce to the units length time time Example 10.1.5. Assuming that the surface of the Earth is a sphere, any point on the Earth can be thought of as an object traveling on a circle which completes one revolution in (approximately) 24 hours. The path traced out by the point during this 24 hour period is the Latitude of that point. Lakeland Community College is at 41.628◦ north latitude, and it can be shown19 that the radius of the earth at this Latitude is approximately 2960 miles. Find the linear velocity, in miles per hour, of Lakeland Community College as the world turns. Solution. To use the formula v = rω, we ﬁrst need to compute the angular velocity ω. The earth π makes one revolution in 24 hours, and one revolution is 2π radians, so ω = 2π radians 12 hours , 24 hours = 17You guessed it, using Calculus . . . 18See the discussion on Page 161 for more details on the idea of an ‘instantaneous’ rate of change. 19We will discuss how we arrived at this approximation in Example 10.2.6. 708 Foundations of Trigonometry where, once again, we are using the fact that radians are real numbers and are dimensionless. (For simplicity’s sake, we are also assuming that we are viewing the rotation of the earth as counterclockwise so ω > 0.) Hence, the linear velocity is v = 2960 miles · π 12 hours ≈ 775 miles hour It is worth noting that the quantity 1 revolution in Example 10.1.5 is called the ordinary frequency 24 hours of the motion and is usually denoted by the variable f . The ordinary frequency is a measure of how often an object makes a complete cycle of the motion. The fact that ω = 2πf suggests that ω is also a frequency. Indeed, it is called the angular frequency of the motion. On a related note, the quantity T = is called the period of the motion and is the amount of time it takes for the 1 f object to complete one cycle of the motion. In the scenario of Example 10.1.5, the period of the motion is 24 hours, or one day. The concepts of frequency and period help frame the equation v = rω in a new light. That is, if ω is ﬁxed, points which are farther from the center of rotation need to travel faster to maintain the same angular frequency since they have farther to travel to make one revolution in one period’s time. The distance of the object to the center of rotation is the radius of the circle, r, and is the ‘magniﬁcation factor’ which relates ω and v. We will have more to say about frequencies and periods in Section 11.1. While we have exhaustively discussed velocities associated with circular motion, we have yet to discuss a more natural question: if an object is moving on a circular path of radius r with a ﬁxed angular velocity (frequency) ω, what is the position of the object at time t? The answer to this question is the very heart of Trigonometry and is answered in the next section. 10.1 Angles and their Measure 709 10.1.2 Exercises In Exercises 1 - 4, convert the angles into the DMS system. Round each of your answers to the nearest second. 1. 63.75◦ 2. 200.325◦ 3. −317.06◦ 4. 179.999◦ In Exercises 5 - 8, convert the angles into decimal degrees. Round each of your answers to three decimal places. 5. 125◦50 6. −32◦1012 7. 502◦35 8. 237◦5843 In Exercises 9 - 28, graph the oriented angle in standard position. Classify each angle according to where its terminal side lies and then give two coterminal angles, one of which is positive and the other negative. 9. 330◦ 10. −135◦ 13. −270◦ 17. 3π 4 21. − π 2 25. −2π 14. 5π 6 18. − π 3 22. 7π 6 26. − π 4 11. 120◦ 15. − 11π 3 19. 7π 2 23. − 5π 3 27. 15π 4 12. 405◦ 16. 20. 5π 4 π 4 24. 3π 28. − 13π 6 In Exercises 29 - 36, convert the angle from degree measure into radian measure, giving the exact value in terms of π. 29. 0◦ 33. −315◦ 30. 240◦ 34. 150◦ 31. 135◦ 35. 45◦ 32. −270◦ 36. −225◦ In Exercises 37 - 44, convert the angle from radian measure into degree measure. 37. π 41. π 3 38. − 2π 3 42. 5π 3 39. 7π 6 43. − π 6 40. 44. 11π 6 π 2 710 Foundations of Trigonometry In Exercises 45 - 49, sketch the oriented arc on the Unit Circle which corresponds to the given real number. 45. t = 5π 6 46. t = −π 47. t = 6 48. t = −2 49. t = 12 50. A yo-yo which is 2.25 inches in diameter spins at a rate of 4500 revolutions per minute. How fast is the edge of the yo-yo spinning in miles per hour? Round your answer to two decimal places. 51. How many revolutions per minute would the yo-yo in exercise 50 have to complete if the edge of the yo-yo is to be spinning at a rate of 42 miles per hour? Round your answer to two decimal places. 52. In the yo-yo trick ‘Around the World,’ the performer throws the yo-yo so it sweeps out a vertical circle whose radius is the yo-yo string. If the yo-yo string is 28 inches long and the yo-yo takes 3 seconds to complete one revolution of the circle, compute the speed of the yo-yo in miles per hour. Round your answer to two decimal places. 53. A computer hard drive contains a circular disk with diameter 2.5 inches and spins at a rate of 7200 RPM (revolutions per minute). Find the linear speed of a point on the edge of the disk in miles per hour. 54. A rock got stuck in the tread of my tire and when I was driving 70 miles per hour, the rock came loose and hit the inside of the wheel well of the car. How fast, in miles per hour, was the rock traveling when it came out of the tread? (The tire has a diameter of 23 inches.) 55. The Giant Wheel at Cedar Point is a circle with diameter 128 feet which sits on an 8 foot tall platform making its overall height is 136 feet. (Remember this from Exercise 17 in Section 7.2?) It completes two revolutions in 2 minutes and 7 seconds.20 Assuming the riders are at the edge of the circle, how fast are they traveling in miles per hour? 56. Consider the circle of radius r pictured below with central angle θ, measured in radians, and subtended arc of length s. Prove that the area of the shaded sector is A = 1 (Hint: Use the proportion s circumference of the circle .) A area of the circle = 2 r2θ. r s θ r 20Source: Cedar Point’s webpage. 10.1 Angles and their Measure 711 In Exercises 57 - 62, use the result of Exercise 56 to compute the areas of the circular sectors with the given central angles and radii. 57. θ = π 6 , r = 12 60. θ = π, r = 1 58. θ = 5π 4 , r = 100 61. θ = 240◦, r = 5 59. θ = 330◦, r = 9.3 62. θ = 1◦, r = 117 63. Imagine a rope tied around the Earth at the equator. Show that you need to add only 2π feet of length to the rope in order to lift it one foot above the ground around the entire equator. (You do NOT need to know the radius of the Earth to show this.) 64. With the help of your classmates, look for a proof that π is indeed a constant. 712 Foundations of Trigonometry 10.1.3 Answers 1. 63◦45 5. 125.833◦ 2. 200◦1930 3. −317◦336 4. 179◦5956 6. −32.17◦ 7. 502.583◦ 8. 237.979◦ 9. 330◦ is a Quadrant IV angle coterminal with 690◦ and −30◦ 10. −135◦ is a Quadrant III angle coterminal with 225◦ and −4954 −3 −2 −1 −1 1 2 3 4 x −4 −3 −2 −1 −1 1 2 3 4 x −2 −3 −4 −2 −3 −4 11. 120◦ is a Quadrant II angle coterminal with 480◦ and −240◦ 12. 405◦ is a Quadrant I angle coterminal with 45◦ and −3154 −3 −2 −1 −1 1 2 3 4 x −4 −3 −2 −1 −1 1 2 3 4 x −2 −3 −4 −2 −3 −4 13. −270◦ lies on the positive y-axis 14. coterminal with 90◦ and −630◦ is a Quadrant II angle 5π 6 coterminal with 17π 6 and − 7π 4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x −4 −3 −2 −1 −1 −2 −3 −4 10.1 Angles and their Measure 713 15. − 11π 3 is a Quadrant I angle 5π π 3 3 and − coterminal with y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x 17. 19. is a Quadrant II angle 3π 4 coterminal with 11π 4 and − 5π 4 −3 −2 −1 −1 −2 −3 −4 7π 2 coterminal with 3π 2 and − π 2 y 4 3 2 1 16. is a Quadrant III angle 5π 4 coterminal with 13π 4 and − 3π 4 −3 −2 −1 −1 −2 −3 −4 18. − π 3 is a Quadrant IV angle 7π 3 and − 5π 3 coterminal with y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x π 4 coterminal with 9π 4 and − 7π 4 y 4 3 2 1 lies on the negative y-axis 20. is a Quadrant I angle −4 −3 −2 −1 −1 1 2 3 4 x −4 −3 −2 −1 −1 1 2 3 4 x −2 −3 −4 −2 −3 −4 714 21. − π 2 lies on the negative y-axis 5π 3π 2 2 and − coterminal with y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x Foundations of Trigonometry 22. is a Quadrant III angle 7π 6 coterminal with 19π 6 and − 5π 4 −3 −2 −1 −1 −2 −3 −4 23. − 5π 3 is a Quadrant I angle 24. 3π lies on the negative x-axis coterminal with y 4 3 2 1 π 3 and − 11π 3 coterminal with π and −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x −4 −3 −2 −1 −1 −2 −3 −4 25. −2π lies on the positive x-axis 26. − π 4 is a Quadrant IV angle coterminal with 2π and −4π coterminal with 7π 4 and − 9π 4 −4 −3 −2 −1 −1 1 2 3 4 x −4 −3 −2 −1 −1 1 2 3 4 x −2 −3 −4 −2 −3 −4 10.1 Angles and their Measure 715 27. 15π 4 is a Quadrant IV angle π 4 and − 7π 4 coterminal with y 28. − 13π 6 is a Quadrant IV angle π 11π 6 6 and − coterminal with y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x 29. 0 33. − 7π 4 37. 180◦ 41. 60◦ 45. t = 5π 6 47. t = 6 30. 34. 4π 3 5π 6 38. −120◦ 42. 300◦ 1 x y 1 y 1 31. 35. 3π 4 π 4 39. 210◦ 43. −30◦ 46. t = −π 48. t = −2 32. − 36. − 3π 2 5π 4 40. 330◦ 44. 90 716 Foundations of Trigonometry 49. t = 12 (between 1 and 2 revolutions) y 1 1 x 50. About 30.12 miles per hour 51. About 6274.52 revolutions per minute 52. About 3.33 miles per hour 53. About 53.55 miles per hour 54. 70 miles per hour 55. About 4.32 miles per hour 57. 12π square units 59. 79.2825π ≈ 249.07 square units 61. 50π 3 square units 58. 6250π square units 60. π 2 square units 62. 38.025π ≈ 119.46 square units 10.2 The Unit Circle: Cosi
ne and Sine 717 10.2 The Unit Circle: Cosine and Sine In Section 10.1.1, we introduced circular motion and derived a formula which describes the linear velocity of an object moving on a circular path at a constant angular velocity. One of the goals of this section is describe the position of such an object. To that end, consider an angle θ in standard position and let P denote the point where the terminal side of θ intersects the Unit Circle. By associating the point P with the angle θ, we are assigning a position on the Unit Circle to the angle θ. The x-coordinate of P is called the cosine of θ, written cos(θ), while the y-coordinate of P is called the sine of θ, written sin(θ).1 The reader is encouraged to verify that these rules used to match an angle with its cosine and sine do, in fact, satisfy the deﬁnition of a function. That is, for each angle θ, there is only one associated value of cos(θ) and only one associated value of sin(θ). y 1 θ 1 x y 1 P (cos(θ), sin(θ)) θ 1 x Example 10.2.1. Find the cosine and sine of the following angles. 1. θ = 270◦ 2. θ = −π 3. θ = 45◦ 4. θ = π 6 5. θ = 60◦ Solution. 1. To ﬁnd cos (270◦) and sin (270◦), we plot the angle θ = 270◦ in standard position and ﬁnd the point on the terminal side of θ which lies on the Unit Circle. Since 270◦ represents 3 4 of a counter-clockwise revolution, the terminal side of θ lies along the negative y-axis. Hence, the point we seek is (0, −1) so that cos (270◦) = 0 and sin (270◦) = −1. 2. The angle θ = −π represents one half of a clockwise revolution so its terminal side lies on the negative x-axis. The point on the Unit Circle that lies on the negative x-axis is (−1, 0) which means cos(−π) = −1 and sin(−π) = 0. 1The etymology of the name ‘sine’ is quite colorful, and the interested reader is invited to research it; the ‘co’ in ‘cosine’ is explained in Section 10.4. 718 Foundations of Trigonometry y 1 θ = 270◦ P (0, −1) y 1 P (−1, 0) 1 x 1 x θ = −π Finding cos (270◦) and sin (270◦) Finding cos (−π) and sin (−π) 3. When we sketch θ = 45◦ in standard position, we see that its terminal does not lie along any of the coordinate axes which makes our job of ﬁnding the cosine and sine values a bit more diﬃcult. Let P (x, y) denote the point on the terminal side of θ which lies on the Unit Circle. By deﬁnition, x = cos (45◦) and y = sin (45◦). If we drop a perpendicular line segment from P to the x-axis, we obtain a 45◦ − 45◦ − 90◦ right triangle whose legs have lengths x and y units. From Geometry,2 we get y = x. Since P (x, y) lies on the Unit Circle, we have √ 2 x2 + y2 = 1. Substituting y = x into this equation yields 2x2 = 1, or x = ± 2 . √ 2 Since P (x, y) lies in the ﬁrst quadrant, x > 0, so x = cos (45◦) = 2 and with y = x we have y = sin (45◦) = = 45◦ P (x, y) x 1 P (x, y) 45◦ y θ = 45◦ x 2Can you show this? 10.2 The Unit Circle: Cosine and Sine 719 4. As before, the terminal side of θ = π 6 does not lie on any of the coordinate axes, so we proceed using a triangle approach. Letting P (x, y) denote the point on the terminal side of θ which lies on the Unit Circle, we drop a perpendicular line segment from P to the x-axis to form = 1 a 30◦ − 60◦ − 90◦ right triangle. After a bit of Geometry3 we ﬁnd y = 1 2 . Since P (x, y) lies on the Unit Circle, we substitute y = 1 4 , or x = ± 2 into x2 + y2 = 1 to get x2 = 3 2 . Here, x > 0 so x = cos π 2 so sin x, y) x 1 P (x, y) 60◦ y θ = π 6 = 30◦ x 5. Plotting θ = 60◦ in standard position, we ﬁnd it is not a quadrantal angle and set about using a triangle approach. Once again, we get a 30◦ − 60◦ − 90◦ right triangle and, after the usual computations, ﬁnd x = cos (60◦) = 1 2 and y = sin (60◦) = √ 3 2 . y 1 P (x, y) θ = 60◦ x 1 P (x, y) 30◦ y θ = 60◦ x 3Again, can you show this? 720 Foundations of Trigonometry In Example 10.2.1, it was quite easy to ﬁnd the cosine and sine of the quadrantal angles, but for In these latter cases, we made good non-quadrantal angles, the task was much more involved. use of the fact that the point P (x, y) = (cos(θ), sin(θ)) lies on the Unit Circle, x2 + y2 = 1. If we substitute x = cos(θ) and y = sin(θ) into x2 + y2 = 1, we get (cos(θ))2 + (sin(θ))2 = 1. An unfortunate4 convention, which the authors are compelled to perpetuate, is to write (cos(θ))2 as cos2(θ) and (sin(θ))2 as sin2(θ). Rewriting the identity using this convention results in the following theorem, which is without a doubt one of the most important results in Trigonometry. Theorem 10.1. The Pythagorean Identity: For any angle θ, cos2(θ) + sin2(θ) = 1. The moniker ‘Pythagorean’ brings to mind the Pythagorean Theorem, from which both the Distance Formula and the equation for a circle are ultimately derived.5 The word ‘Identity’ reminds us that, regardless of the angle θ, the equation in Theorem 10.1 is always true. If one of cos(θ) or sin(θ) is known, Theorem 10.1 can be used to determine the other, up to a (±) sign. If, in addition, we know where the terminal side of θ lies when in standard position, then we can remove the ambiguity of the (±) and completely determine the missing value as the next example illustrates. Example 10.2.2. Using the given information about θ, ﬁnd the indicated value. 1. If θ is a Quadrant II angle with sin(θ) = 3 5 , ﬁnd cos(θ). 2. If π < θ < 3π 2 with cos(θ) = − √ 5 5 , ﬁnd sin(θ). 3. If sin(θ) = 1, ﬁnd cos(θ). Solution. 1. When we substitute sin(θ) = 3 5 into The Pythagorean Identity, cos2(θ) + sin2(θ) = 1, we obtain cos2(θ) + 9 5 . Since θ is a Quadrant II angle, its terminal side, when plotted in standard position, lies in Quadrant II. Since the x-coordinates are negative in Quadrant II, cos(θ) is too. Hence, cos(θ) = − 4 5 . 25 = 1. Solving, we ﬁnd cos(θ) = ± 4 √ √ 5 5 2. Substituting cos(θ) = − 5 . Since we 5 are given that π < θ < 3π 2 , we know θ is a Quadrant III angle. Hence both its sine and cosine are negative and we conclude sin(θ) = − 2 into cos2(θ) + sin2(θ) = 1 gives sin(θ) = ± 2. When we substitute sin(θ) = 1 into cos2(θ) + sin2(θ) = 1, we ﬁnd cos(θ) = 0. Another tool which helps immensely in determining cosines and sines of angles is the symmetry inherent in the Unit Circle. Suppose, for instance, we wish to know the cosine and sine of θ = 5π 6 . We plot θ in standard position below and, as usual, let P (x, y) denote the point on the terminal side of θ which lies on the Unit Circle. Note that the terminal side of θ lies π 6 radians short of one half revolution. In Example 10.2.1, we determined that cos π 2 . This means 6 2 and sin π = 1 = √ 6 3 4This is unfortunate from a ‘function notation’ perspective. See Section 10.6. 5See Sections 1.1 and 7.2 for details. 10.2 The Unit Circle: Cosine and Sine 721 that the point on the terminal side of the angle π . From the ﬁgure below, it is clear that the point P (x, y) we seek can be obtained by reﬂecting that point about the y-axis. Hence, cos 5π 6 6 , when plotted in standard position, is 2 and sin 5π = x, y) θ = 5π = 5π is called the reference angle for the angle 5π In the above scenario, the angle π 6 . In general, for a non-quadrantal angle θ, the reference angle for θ (usually denoted α) is the acute angle made between the terminal side of θ and the x-axis. If θ is a Quadrant I or IV angle, α is the angle between the terminal side of θ and the positive x-axis; if θ is a Quadrant II or III angle, α is the angle between the terminal side of θ and the negative x-axis. If we let P denote the point (cos(θ), sin(θ)), then P lies on the Unit Circle. Since the Unit Circle possesses symmetry with respect to the x-axis, y-axis and origin, regardless of where the terminal side of θ lies, there is a point Q symmetric with P which determines θ’s reference angle, α as seen below Reference angle α for a Quadrant I angle Reference angle α for a Quadrant II angle 722 Foundations of Trigonometry Reference angle α for a Quadrant III angle Reference angle α for a Quadrant IV angle We have just outlined the proof of the following theorem. Theorem 10.2. Reference Angle Theorem. Suppose α is the reference angle for θ. Then cos(θ) = ± cos(α) and sin(θ) = ± sin(α), where the choice of the (±) depends on the quadrant in which the terminal side of θ lies. In light of Theorem 10.2, it pays to know the cosine and sine values for certain common angles. In the table below, we summarize the values which we consider essential and must be memorized. Cosine and Sine Values of Common Angles θ(degrees) 0◦ 30◦ 45◦ 60◦ 90◦ θ(radians) cos(θ) sin(θ Example 10.2.3. Find the cosine and sine of the following angles. 1. θ = 225◦ 2. θ = 11π 6 3. θ = − 5π 4 4. θ = 7π 3 Solution. 1. We begin by plotting θ = 225◦ in standard position and ﬁnd its terminal side overshoots the negative x-axis to land in Quadrant III. Hence, we obtain θ’s reference angle α by subtracting: α = θ − 180◦ = 225◦ − 180◦ = 45◦. Since θ is a Quadrant III angle, both cos(θ) < 0 and 10.2 The Unit Circle: Cosine and Sine sin(θ) < 0. The Reference Angle Theorem yields: cos (225◦) = − cos (45◦) = − sin (225◦) = − sin (45◦) = − √ 2 2 . 723 √ 2 2 and 2. The terminal side of θ = 11π 6 , when plotted in standard position, lies in Quadrant IV, just shy of the positive x-axis. To ﬁnd θ’s reference angle α, we subtract: α = 2π − θ = 2π − 11π 6 = π 6 . Since θ is a Quadrant IV angle, cos(θ) > 0 and sin(θ) < 0, so the Reference Angle Theorem gives: cos 11π 6 2 and sin 11π = − sin π 6 = cos = 225◦ 45 = 11π 6 Finding cos (225◦) and sin (225◦) Finding cos 11π 6 and sin 11π 6 3. To plot θ = − 5π 4 , we rotate clockwise an angle of 5π 4 from the positive x-axis. The terminal side of θ, therefore, lies in Quadrant II making an angle of α = 5π 4 − π = π 4 radians with respect to the negative x-axis. Since θ is a Quadrant II angle, the Reference Angle Theorem gives: cos − 5π 4 = − cos π 4 = sin . Since the angle θ = 7π one full revolution followed by an additional α = 7π coterminal, cos 7π 3 2 and sin 7π = cos π 3 = 1 = sin π 3 3 3 , we ﬁnd the terminal side of θ by rotating 3 − 2π = π 3 radians. Since θ and α ar
e √ = 3 2 . 2 = − 2 and sin − 5π 3 measures more than 2π = 6π y 1 π 4 1 x θ = − 5π 4 y 1 θ = 7π 3 π 3 1 x Finding cos − 5π 4 and sin − 5π 4 Finding cos 7π 3 and sin 7π 3 724 Foundations of Trigonometry 6 as a reference angle, those with a denominator of 4 have π The reader may have noticed that when expressed in radian measure, the reference angle for a non-quadrantal angle is easy to spot. Reduced fraction multiples of π with a denominator of 6 have π 4 as their reference angle, and those with a denominator of 3 have π 3 as their reference angle.6 The Reference Angle Theorem in conjunction with the table of cosine and sine values on Page 722 can be used to generate the following ﬁgure, which the authors feel should be committed to memory. y (0, 1 5π 6 2π 3 3π (−1, 0) 0, 2π (1, 0) x 7π 6 11π 5π 4 4π 3 − 1 2 , − √ 3 2 7π 4 5π 3π 2 (0, −1) Important Points on the Unit Circle 6For once, we have something convenient about using radian measure in contrast to the abstract theoretical nonsense about using them as a ‘natural’ way to match oriented angles with real numbers! 10.2 The Unit Circle: Cosine and Sine 725 The next example summarizes all of the important ideas discussed thus far in the section. Example 10.2.4. Suppose α is an acute angle with cos(α) = 5 13 . 1. Find sin(α) and use this to plot α in standard position. 2. Find the sine and cosine of the following angles: (a) θ = π + α (b) θ = 2π − α (c) θ = 3π − α (d) θ = π 2 + α Solution. 1. Proceeding as in Example 10.2.2, we substitute cos(α) = 5 ﬁnd sin(α) = ± 12 Hence, sin(α) = 12 x-axis to the ray which contains the point (cos(α), sin(α)) = 5 13 into cos2(α) + sin2(α) = 1 and 13 . Since α is an acute (and therefore Quadrant I) angle, sin(α) is positive. 13 . To plot α in standard position, we begin our rotation on the positive . 13 , 12 13 y 1 5 13 , 12 13 α 1 x Sketching α 2. (a) To ﬁnd the cosine and sine of θ = π + α, we ﬁrst plot θ in standard position. We can imagine the sum of the angles π+α as a sequence of two rotations: a rotation of π radians followed by a rotation of α radians.7 We see that α is the reference angle for θ, so by 13 and sin(θ) = ± sin(α) = ± 12 The Reference Angle Theorem, cos(θ) = ± cos(α) = ± 5 13 . Since the terminal side of θ falls in Quadrant III, both cos(θ) and sin(θ) are negative, hence, cos(θ) = − 5 13 and sin(θ) = − 12 13 . 7Since π + α = α + π, θ may be plotted by reversing the order of rotations given here. You should do this. 726 Foundations of Trigonometry Visualizing θ = π + α θ has reference angle α (b) Rewriting θ = 2π − α as θ = 2π + (−α), we can plot θ by visualizing one complete revolution counter-clockwise followed by a clockwise revolution, or ‘backing up,’ of α radians. We see that α is θ’s reference angle, and since θ is a Quadrant IV angle, the Reference Angle Theorem gives: cos(θ) = 5 13 and sin(θ) = − 12 13 . y 1 y 1 θ θ 2π −α 1 x 1 x α Visualizing θ = 2π − α θ has reference angle α (c) Taking a cue from the previous problem, we rewrite θ = 3π − α as θ = 3π + (−α). The angle 3π represents one and a half revolutions counter-clockwise, so that when we ‘back up’ α radians, we end up in Quadrant II. Using the Reference Angle Theorem, we get cos(θ) = − 5 13 and sin(θ) = 12 13 . 10.2 The Unit Circle: Cosine and Sine 727 y 1 −α 3π y 1 θ α 1 x 1 x Visualizing 3π − α θ has reference angle α (d) To plot θ = π 2 + α, we ﬁrst rotate π 2 radians and follow up with α radians. The reference angle here is not α, so The Reference Angle Theorem is not immediately applicable. (It’s important that you see why this is the case. Take a moment to think about this before reading on.) Let Q(x, y) be the point on the terminal side of θ which lies on the Unit Circle so that x = cos(θ) and y = sin(θ). Once we graph α in standard position, we use the fact that equal angles subtend equal chords to show that the dotted lines in the ﬁgure below are equal. Hence, x = cos(θ) = − 12 13 . Similarly, we ﬁnd y = sin(θ) = 5 13 . 13 , 12 13 Q (x, y) α α 1 x 1 x Visualizing θ = π 2 + α Using symmetry to determine Q(x, y) 728 Foundations of Trigonometry Our next example asks us to solve some very basic trigonometric equations.8 Example 10.2.5. Find all of the angles which satisfy the given equation. 1. cos(θ) = 1 2 2. sin(θ) = − 1 2 3. cos(θ) = 0. Solution. Since there is no context in the problem to indicate whether to use degrees or radians, we will default to using radian measure in our answers to each of these problems. This choice will be justiﬁed later in the text when we study what is known as Analytic Trigonometry. In those sections to come, radian measure will be the only appropriate angle measure so it is worth the time to become “ﬂuent in radians” now. 1. If cos(θ) = 1 Unit Circle at x = 1 2 , then the terminal side of θ, when plotted in standard position, intersects the 2 . This means θ is a Quadrant I or IV angle with reference angle One solution in Quadrant I is θ = π coterminal with π IV case, we ﬁnd the solution to cos(θ) = 1 θ = 5π 3 , we ﬁnd θ = π 3 + 2πk for integers k. 3 , and since all other Quadrant I solutions must be 3 + 2πk for integers k.9 Proceeding similarly for the Quadrant 3 , so our answer in this Quadrant is 2 here is 5π 2. If sin(θ) = − 1 2 , then when θ is plotted in standard position, its terminal side intersects the Unit Circle at y = − 1 2 . From this, we determine θ is a Quadrant III or Quadrant IV angle with reference angle π 6 . 8We will study trigonometric equations more formally in Section 10.7. Enjoy these relatively straightforward exercises while they last! 9Recall in Section 10.1, two angles in radian measure are coterminal if and only if they diﬀer by an integer multiple of 2π. Hence to describe all angles coterminal with a given angle, we add 2πk for integers k = 0, ±1, ±2, . . . . 10.2 The Unit Circle: Cosine and Sine 729 In Quadrant III, one solution is 7π multiples of 2π: θ = 7π are of the form θ = 11π 6 , so we capture all Quadrant III solutions by adding integer so all the solutions here 6 + 2πk. In Quadrant IV, one solution is 11π 6 + 2πk for integers k. 6 3. The angles with cos(θ) = 0 are quadrantal angles whose terminal sides, when plotted in standard position, lie along the y-axis While, technically speaking, π answers. If we follow the procedure set forth in the previous examples, we ﬁnd θ = π and θ = 3π θ = π 2 isn’t a reference angle we can nonetheless use it to ﬁnd our 2 + 2πk 2 + 2πk for integers, k. While this solution is correct, it can be shortened to 2 + πk for integers k. (Can you see why this works from the diagram?) One of the key items to take from Example 10.2.5 is that, in general, solutions to trigonometric equations consist of inﬁnitely many answers. To get a feel for these answers, the reader is encouraged to follow our mantra from Chapter 9 - that is, ‘When in doubt, write it out!’ This is especially important when checking answers to the exercises. For example, another Quadrant IV solution to 2 is θ = − π sin(θ) = − 1 6 . Hence, the family of Quadrant IV answers to number 2 above could just have easily been written θ = − π 6 + 2πk for integers k. While on the surface, this family may look 730 Foundations of Trigonometry diﬀerent than the stated solution of θ = 11π they represent the same list of angles. 6 + 2πk for integers k, we leave it to the reader to show 10.2.1 Beyond the Unit Circle We began the section with a quest to describe the position of a particle experiencing circular motion. In deﬁning the cosine and sine functions, we assigned to each angle a position on the Unit Circle. In this subsection, we broaden our scope to include circles of radius r centered at the origin. Consider for the moment the acute angle θ drawn below in standard position. Let Q(x, y) be the point on the terminal side of θ which lies on the circle x2 + y2 = r2, and let P (x, y) be the point on the terminal side of θ which lies on the Unit Circle. Now consider dropping perpendiculars from P and Q to create two right triangles, ∆OP A and ∆OQB. These triangles are similar,10 thus it follows that x 1 = r, so x = rx and, similarly, we ﬁnd y = ry. Since, by deﬁnition, x = cos(θ) and y = sin(θ), we get the coordinates of Q to be x = r cos(θ) and y = r sin(θ). By reﬂecting these points through the x-axis, y-axis and origin, we obtain the result for all non-quadrantal angles θ, and we leave it to the reader to verify these formulas hold for the quadrantal angles. x = r y r 1 Q (x, y) P (x, y) θ y 1 1 x r P (x, y) Q(x, y) = (r cos(θ), r sin(θ)) θ O A(x, 0) B(x, 0) x Not only can we describe the coordinates of Q in terms of cos(θ) and sin(θ) but since the radius of x2 + y2, we can also express cos(θ) and sin(θ) in terms of the coordinates of Q. the circle is r = These results are summarized in the following theorem. Theorem 10.3. If Q(x, y) is the point on the terminal side of an angle θ, plotted in standard position, which lies on the circle x2 + y2 = r2 then x = r cos(θ) and y = r sin(θ). Moreover, cos(θ) = x r = x x2 + y2 and sin(θ) = y r = y x2 + y2 10Do you remember why? 10.2 The Unit Circle: Cosine and Sine 731 Note that in the case of the Unit Circle we have r = our deﬁnitions of cos(θ) and sin(θ). x2 + y2 = 1, so Theorem 10.3 reduces to Example 10.2.6. 1. Suppose that the terminal side of an angle θ, when plotted in standard position, contains the point Q(4, −2). Find sin(θ) and cos(θ). 2. In Example 10.1.5 in Section 10.1, we approximated the radius of the earth at 41.628◦ north latitude to be 2960 miles. Justify this approximation if the radius of the Earth at the Equator is approximately 3960 miles. Solution. 1. Using Theorem 10.3 with x = 4 and y = −2, we ﬁnd r = √ √ 5 5 5 . 5 and sin(θ) = y that cos(θ) = x r = −4)2 + (−2)2 = √ 20 = 2 √ 5 so 2. Assuming the Earth is a sphere, a cross-section through the poles produces a circle of radius 3960 miles. Viewing the Equator as the x-axis, the value we seek is the x-coordinate of the point Q(x, y) indicated in the ﬁgure below. y 4 2 −4 −2 2
4 x Q(4, −2) −2 −4 y 3960 Q (x, y) 41.628◦ 3960 x The terminal side of θ contains Q(4, −2) A point on the Earth at 41.628◦N Using Theorem 10.3, we get x = 3960 cos (41.628◦). Using a calculator in ‘degree’ mode, we ﬁnd 3960 cos (41.628◦) ≈ 2960. Hence, the radius of the Earth at North Latitude 41.628◦ is approximately 2960 miles. 732 Foundations of Trigonometry Theorem 10.3 gives us what we need to describe the position of an object traveling in a circular path of radius r with constant angular velocity ω. Suppose that at time t, the object has swept out an angle measuring θ radians. If we assume that the object is at the point (r, 0) when t = 0, the angle θ is in standard position. By deﬁnition, ω = θ t which we rewrite as θ = ωt. According to Theorem 10.3, the location of the object Q(x, y) on the circle is found using the equations x = r cos(θ) = r cos(ωt) and y = r sin(θ) = r sin(ωt). Hence, at time t, the object is at the point (r cos(ωt), r sin(ωt)). We have just argued the following. Equation 10.3. Suppose an object is traveling in a circular path of radius r centered at the origin with constant angular velocity ω. If t = 0 corresponds to the point (r, 0), then the x and y coordinates of the object are functions of t and are given by x = r cos(ωt) and y = r sin(ωt). Here, ω > 0 indicates a counter-clockwise direction and ω < 0 indicates a clockwise direction. y r 1 Q (x, y) = (r cos(ωt), r sin(ωt)) θ = ωt 1 x r Equations for Circular Motion Example 10.2.7. Suppose we are in the situation of Example 10.1.5. Find the equations of motion of Lakeland Community College as the earth rotates. Solution. From Example 10.1.5, we take r = 2960 miles and and ω = π of motion are x = r cos(ωt) = 2960 cos π measured in miles and t is measured in hours. 12 hours . Hence, the equations 12 t, where x and y are 12 t and y = r sin(ωt) = 2960 sin π In addition to circular motion, Theorem 10.3 is also the key to developing what is usually called ‘right triangle’ trigonometry.11 As we shall see in the sections to come, many applications in trigonometry involve ﬁnding the measures of the angles in, and lengths of the sides of, right triangles. Indeed, we made good use of some properties of right triangles to ﬁnd the exact values of the cosine and sine of many of the angles in Example 10.2.1, so the following development shouldn’t be that much of a surprise. Consider the generic right triangle below with corresponding acute angle θ. The side with length a is called the side of the triangle adjacent to θ; the side with length b is called the side of the triangle opposite θ; and the remaining side of length c (the side opposite the 11You may have been exposed to this in High School. 10.2 The Unit Circle: Cosine and Sine 733 right angle) is called the hypotenuse. We now imagine drawing this triangle in Quadrant I so that the angle θ is in standard position with the adjacent side to θ lying along the positive x-axis. y c θ P (a, b) x c b c a θ According to the Pythagorean Theorem, a2 + b2 = c2, so that the point P (a, b) lies on a circle of radius c. Theorem 10.3 tells us that cos(θ) = a c , so we have determined the cosine and sine of θ in terms of the lengths of the sides of the right triangle. Thus we have the following theorem. c and sin(θ) = b Theorem 10.4. Suppose θ is an acute angle residing in a right triangle. If the length of the side adjacent to θ is a, the length of the side opposite θ is b, and the length of the hypotenuse is c, then cos(θ) = and sin(θ) = a c b c . Example 10.2.8. Find the measure of the missing angle and the lengths of the missing sides of: 30◦ 7 Solution. The ﬁrst and easiest task is to ﬁnd the measure of the missing angle. Since the sum of angles of a triangle is 180◦, we know that the missing angle has measure 180◦ − 30◦ − 90◦ = 60◦. We now proceed to ﬁnd the lengths of the remaining two sides of the triangle. Let c denote the 7 length of the hypotenuse of the triangle. By Theorem 10.4, we have cos (30◦) = 7 cos(30◦) . 2 , we have, after the usual fraction gymnastics, c = 14 Since cos (30◦) = . At this point, we have two ways to proceed to ﬁnd the length of the side opposite the 30◦ angle, which we’ll denote b. We know the length of the adjacent side is 7 and the length of the hypotenuse is 14 , so we 3 c , or c = √ √ √ 3 3 3 3 734 Foundations of Trigonometry could use the Pythagorean Theorem to ﬁnd the missing side and solve (7)2 + b2 = Alternatively, we could use Theorem 10.4, namely that sin (30◦) = b b = c sin (30◦) = 14 3 √ 3 3 . The triangle with all of its data is recorded below. 2 = 7 · 1 √ 3 c . Choosing the latter, we ﬁnd 2 √ 14 3 3 for b. √ 3 c = 14 3 60◦ b = 7 √ 3 3 30◦ 7 We close this section by noting that we can easily extend the functions cosine and sine to real numbers by identifying a real number t with the angle θ = t radians. Using this identiﬁcation, we deﬁne cos(t) = cos(θ) and sin(t) = sin(θ). In practice this means expressions like cos(π) and sin(2) can be found by regarding the inputs as angles in radian measure or real numbers; the choice is the reader’s. If we trace the identiﬁcation of real numbers t with angles θ in radian measure to its roots on page 704, we can spell out this correspondence more precisely. For each real number t, we associate an oriented arc t units in length with initial point (1, 0) and endpoint P (cos(t), sin(t)). cos(t), sin(t)) θ = t 1 x In the same way we studied polynomial, rational, exponential, and logarithmic functions, we will study the trigonometric functions f (t) = cos(t) and g(t) = sin(t). The ﬁrst order of business is to ﬁnd the domains and ranges of these functions. Whether we think of identifying the real number t with the angle θ = t radians, or think of wrapping an oriented arc around the Unit Circle to ﬁnd coordinates on the Unit Circle, it should be clear that both the cosine and sine functions are deﬁned for all real numbers t. In other words, the domain of f (t) = cos(t) and of g(t) = sin(t) is (−∞, ∞). Since cos(t) and sin(t) represent x- and y-coordinates, respectively, of points on the Unit Circle, they both take on all of the values between −1 an 1, inclusive. In other words, the range of f (t) = cos(t) and of g(t) = sin(t) is the interval [−1, 1]. To summarize: 10.2 The Unit Circle: Cosine and Sine 735 Theorem 10.5. Domain and Range of the Cosine and Sine Functions: • The function f (t) = cos(t) • The function g(t) = sin(t) – has domain (−∞, ∞) – has range [−1, 1] – has domain (−∞, ∞) – has range [−1, 1] Suppose, as in the Exercises, we are asked to solve an equation such as sin(t) = − 1 2 . As we have already mentioned, the distinction between t as a real number and as an angle θ = t radians is often blurred. Indeed, we solve sin(t) = − 1 2 in the exact same manner12 as we did in Example 10.2.5 number 2. Our solution is only cosmetically diﬀerent in that the variable used is t rather than θ: t = 7π 6 + 2πk for integers, k. We will study the cosine and sine functions in greater detail in Section 10.5. Until then, keep in mind that any properties of cosine and sine developed in the following sections which regard them as functions of angles in radian measure apply equally well if the inputs are regarded as real numbers. 6 + 2πk or t = 11π 12Well, to be pedantic, we would be technically using ‘reference numbers’ or ‘reference arcs’ instead of ‘reference angles’ – but the idea is the same. 736 Foundations of Trigonometry 10.2.2 Exercises In Exercises 1 - 20, ﬁnd the exact value of the cosine and sine of the given angle. 1. θ = 0 5. θ = 9. θ = 13. θ = 2π 3 5π 4 7π 4 17. θ = − 3π 4 2. θ = 6. θ = 10. θ = 14. θ = π 4 3π 4 4π 3 23π 6 18. θ = − π 6 3. θ = π 3 7. θ = π 11. θ = 3π 2 15. θ = − 13π 2 19. θ = 10π 3 4. θ = 8. θ = 12. θ = π 2 7π 6 5π 3 16. θ = − 43π 6 20. θ = 117π In Exercises 21 - 30, use the results developed throughout the section to ﬁnd the requested value. 21. If sin(θ) = − 7 25 with θ in Quadrant IV, what is cos(θ)? 22. If cos(θ) = 23. If sin(θ) = 4 9 5 13 with θ in Quadrant I, what is sin(θ)? with θ in Quadrant II, what is cos(θ)? 24. If cos(θ) = − 2 11 with θ in Quadrant III, what is sin(θ)? 25. If sin(θ) = − 2 3 with θ in Quadrant III, what is cos(θ)? 26. If cos(θ) = 27. If sin(θ) = 28. If cos(θ) = 28 53 √ 2 5 √ 10 10 with θ in Quadrant IV, what is sin(θ)? 5 and π 2 < θ < π, what is cos(θ)? and 2π < θ < 5π 2 3π 2 , what is sin(θ)? , what is cos(θ)? 29. If sin(θ) = −0.42 and π < θ < 30. If cos(θ) = −0.98 and π 2 < θ < π, what is sin(θ)? 10.2 The Unit Circle: Cosine and Sine 737 In Exercises 31 - 39, ﬁnd all of the angles which satisfy the given equation. 31. sin(θ) = 34. cos(θ) = 1 2 √ 2 2 32. cos(θ) = − √ 3 2 35. sin(θ) = √ 3 2 √ 3 2 33. sin(θ) = 0 36. cos(θ) = −1 39. cos(θ) = −1.001 37. sin(θ) = −1 38. cos(θ) = In Exercises 40 - 48, solve the equation for t. (See the comments following Theorem 10.5.) 40. cos(t) = 0 41. sin(t) = − 43. sin(t) = − 1 2 44. cos(t) = 1 2 √ 2 2 42. cos(t) = 3 45. sin(t) = −2 46. cos(t) = 1 47. sin(t) = 1 48. cos(t) = − √ 2 2 In Exercises 49 - 54, use your calculator to approximate the given value to three decimal places. Make sure your calculator is in the proper angle measurement mode! 49. sin(78.95◦) 50. cos(−2.01) 51. sin(392.994) 52. cos(207◦) 53. sin (π◦) 54. cos(e) In Exercises 55 - 58, ﬁnd the measurement of the missing angle and the lengths of the missing sides. (See Example 10.2.8) 55. Find θ, b, and c. 56. Find θ, a, and c. c θ b 30◦ 1 45◦ c a θ 3 738 Foundations of Trigonometry 57. Find α, a, and b. 58. Find β, a, and c. b α 8 a 33◦ a 48◦ 6 c β In Exercises 59 - 64, assume that θ is an acute angle in a right triangle and use Theorem 10.4 to ﬁnd the requested side. 59. If θ = 12◦ and the side adjacent to θ has length 4, how long is the hypotenuse? 60. If θ = 78.123◦ and the hypotenuse has length 5280, how long is the side adjacent to θ? 61. If θ = 59◦ and the side opposite θ has length 117.42, how long is the hypotenuse? 62. If θ = 5◦ and the hypotenuse has length 10, how lo
ng is the side opposite θ? 63. If θ = 5◦ and the hypotenuse has length 10, how long is the side adjacent to θ? 64. If θ = 37.5◦ and the side opposite θ has length 306, how long is the side adjacent to θ? In Exercises 65 - 68, let θ be the angle in standard position whose terminal side contains the given point then compute cos(θ) and sin(θ). 65. P (−7, 24) 66. Q(3, 4) 67. R(5, −9) 68. T (−2, −11) In Exercises 69 - 72, ﬁnd the equations of motion for the given scenario. Assume that the center of the motion is the origin, the motion is counter-clockwise and that t = 0 corresponds to a position along the positive x-axis. (See Equation 10.3 and Example 10.1.5.) 69. A point on the edge of the spinning yo-yo in Exercise 50 from Section 10.1. Recall: The diameter of the yo-yo is 2.25 inches and it spins at 4500 revolutions per minute. 70. The yo-yo in exercise 52 from Section 10.1. Recall: The radius of the circle is 28 inches and it completes one revolution in 3 seconds. 71. A point on the edge of the hard drive in Exercise 53 from Section 10.1. Recall: The diameter of the hard disk is 2.5 inches and it spins at 7200 revolutions per minute. 10.2 The Unit Circle: Cosine and Sine 739 72. A passenger on the Big Wheel in Exercise 55 from Section 10.1. Recall: The diameter is 128 feet and completes 2 revolutions in 2 minutes, 7 seconds. 73. Consider the numbers: 0, 1, 2, 3, 4. Take the square root of each of these numbers, then divide each by 2. The resulting numbers should look hauntingly familiar. (See the values in the table on 722.) 74. Let α and β be the two acute angles of a right triangle. (Thus α and β are complementary angles.) Show that sin(α) = cos(β) and sin(β) = cos(α). The fact that co-functions of complementary angles are equal in this case is not an accident and a more general result will be given in Section 10.4. 75. In the scenario of Equation 10.3, we assumed that at t = 0, the object was at the point (r, 0). If this is not the case, we can adjust the equations of motion by introducing a ‘time delay.’ If t0 > 0 is the ﬁrst time the object passes through the point (r, 0), show, with the help of your classmates, the equations of motion are x = r cos(ω(t − t0)) and y = r sin(ω(t − t0)). 740 Foundations of Trigonometry 10.2.3 Answers 1. cos(0) = 1, sin(0) = 0 1 2 = 3. cos 5. cos π 3 2π 3 , sin π 3 = = − 1 2 , sin 2π 3 √ 3 2 √ 3 2 = 7. cos(π) = −1, sin(π) = 0 9. cos 11. cos 13. cos 5π 4 3π 2 7π 4 √ 2 2 = − , sin 5π 4 = − √ 2 2 = 0, sin 3π 2 = −1 √ 2 2 = , sin 7π 4 = − √ 2 2 2. cos 4. cos 6. cos 8. cos 10. cos 12. cos π 4 π 2 3π 4 7π 6 4π 3 5π 3 √ 2 2 = , sin π 4 = √ 2 2 = 0, sin , sin , sin 3π 4 7π , sin 4π 3 = − √ 3 2 = 1 2 , sin 5π 3 = − √ 3 2 14. cos 23π 6 √ 3 2 , sin 23π 6 = − 1 2 = 15. cos − 13π 2 = 0, sin − 13π 2 = −1 16. cos − 43π 6 √ 3 2 , sin − 43π 6 = 1 2 = − 17. cos 19. cos − 3π 4 10π 3 = − √ 2 2 , sin − 3π 4 √ 2 2 = − 18. cos − π 6 = √ 3 2 , sin − , sin 10π 3 = − √ 3 2 20. cos(117π) = −1, sin(117π) = 0 21. If sin(θ) = − 7 25 22. If cos(θ) = 23. If sin(θ) = 4 9 5 13 24. If cos(θ) = − 2 11 25. If sin(θ) = − 2 3 26. If cos(θ) = 28 53 with θ in Quadrant IV, then cos(θ) = with θ in Quadrant I, then sin(θ) = with θ in Quadrant II, then cos(θ) = − 12 13 . with θ in Quadrant III, then sin(θ) = − 24 25 . . √ 65 9 √ 117 11 . √ 5 3 . with θ in Quadrant III, then cos(θ) = − with θ in Quadrant IV, then sin(θ) = − 45 53 . 10.2 The Unit Circle: Cosine and Sine 741 27. If sin(θ) = 28. If cos(θ) = √ 2 5 5 √ 10 10 and π 2 < θ < π, then cos(θ) = − and 2π < θ < , then sin(θ) = √ . 5 5 √ 3 10 . 10 √ 5π 2 3π 2 29. If sin(θ) = −0.42 and π < θ < , then cos(θ) = − 0.8236 ≈ −0.9075. 30. If cos(θ) = −0.98 and < θ < π, then sin(θ) = √ 0.0396 ≈ 0.1990. π 2 31. sin(θ) = 1 2 32. cos(θ) = − √ 3 2 when θ = π 6 + 2πk or θ = 5π 6 + 2πk for any integer k. when θ = 5π 6 + 2πk or θ = 7π 6 + 2πk for any integer k. 33. sin(θ) = 0 when θ = πk for any integer k. 34. cos(θ) = 35. sin(θ) = √ 2 2 √ 3 2 when θ = when θ = π 4 π 3 + 2πk or θ = + 2πk or θ = 7π 4 2π 3 + 2πk for any integer k. + 2πk for any integer k. 36. cos(θ) = −1 when θ = (2k + 1)π for any integer k. 37. sin(θ) = −1 when θ = 38. cos(θ) = √ 3 2 when θ = 3π 2 π 6 + 2πk for any integer k. + 2πk or θ = 11π 6 + 2πk for any integer k. 39. cos(θ) = −1.001 never happens 40. cos(t) = 0 when t = π 2 + πk for any integer k. 41. sin(t) = − √ 2 2 when t = 5π 4 + 2πk or t = 7π 4 + 2πk for any integer k. 42. cos(t) = 3 never happens. 43. sin(t) = − 1 2 when t = 7π 6 + 2πk or t = 11π 6 + 2πk for any integer k. 44. cos(t) = 1 2 when t = π 3 + 2πk or t = 5π 3 + 2πk for any integer k. 45. sin(t) = −2 never happens 46. cos(t) = 1 when t = 2πk for any integer k. 742 Foundations of Trigonometry π 2 + 2πk for any integer k. 47. sin(t) = 1 when t = 48. cos(t) = − √ 2 2 when t = 3π 4 + 2πk or t = 5π 4 + 2πk for any integer k. 49. sin(78.95◦) ≈ 0.981 50. cos(−2.01) ≈ −0.425 51. sin(392.994) ≈ −0.291 52. cos(207◦) ≈ −0.891 53. sin (π◦) ≈ 0.055 54. cos(e) ≈ −0.912 55. θ = 60◦, b = √ 3 3 , c = 56. θ = 45◦, a = 3 57. α = 57◦, a = 8 cos(33◦) ≈ 6.709, b = 8 sin(33◦) ≈ 4.357 58. β = 42◦, c = 6 sin(48◦) ≈ 8.074, a = √ c2 − 62 ≈ 5.402 59. The hypotenuse has length 4 cos(12◦) ≈ 4.089. 60. The side adjacent to θ has length 5280 cos(78.123◦) ≈ 1086.68. 61. The hypotenuse has length 117.42 sin(59◦) ≈ 136.99. 62. The side opposite θ has length 10 sin(5◦) ≈ 0.872. 63. The side adjacent to θ has length 10 cos(5◦) ≈ 9.962. 64. The hypotenuse has length c = √ c2 − 3062 ≈ 398.797. 306 sin(37.5◦) 65. cos(θ) = − 7 25 , sin(θ) = 24 25 ≈ 502.660, so the side adjacent to θ has length , sin(θ) = 4 5 66. cos(θ) = 67. cos(θ) = 3 5 √ 5 , sin(θ) = − 68. cos(θ) = − , sin(θ) = − 106 106 √ 5 2 25 √ 106 9 106 √ 5 11 25 69. r = 1.125 inches, ω = 9000π radians minute , x = 1.125 cos(9000π t), y = 1.125 sin(9000π t). Here x and y are measured in inches and t is measured in minutes. 10.2 The Unit Circle: Cosine and Sine 743 70. r = 28 inches, ω = 2π 3 radians second , x = 28 cos 2π 3 t, y = 28 sin 2π 3 t. Here x and y are measured in inches and t is measured in seconds. 71. r = 1.25 inches, ω = 14400π radians minute , x = 1.25 cos(14400π t), y = 1.25 sin(14400π t). Here x and y are measured in inches and t is measured in minutes. 127 t, y = 64 sin 4π second , x = 64 cos 4π 72. r = 64 feet, ω = 4π 127 in feet and t is measured in seconds radians 127 t. Here x and y are measured 744 Foundations of Trigonometry 10.3 The Six Circular Functions and Fundamental Identities In section 10.2, we deﬁned cos(θ) and sin(θ) for angles θ using the coordinate values of points on the Unit Circle. As such, these functions earn the moniker circular functions.1 It turns out that cosine and sine are just two of the six commonly used circular functions which we deﬁne below. Deﬁnition 10.2. The Circular Functions: Suppose θ is an angle plotted in standard position and P (x, y) is the point on the terminal side of θ which lies on the Unit Circle. The cosine of θ, denoted cos(θ), is deﬁned by cos(θ) = x. The sine of θ, denoted sin(θ), is deﬁned by sin(θ) = y. The secant of θ, denoted sec(θ), is deﬁned by sec(θ) = 1 x , provided x = 0. The cosecant of θ, denoted csc(θ), is deﬁned by csc(θ) = The tangent of θ, denoted tan(θ), is deﬁned by tan(θ) = 1 y y x The cotangent of θ, denoted cot(θ), is deﬁned by cot(θ) = , provided y = 0. , provided x = 0. x y , provided y = 0. While we left the history of the name ‘sine’ as an interesting research project in Section 10.2, the names ‘tangent’ and ‘secant’ can be explained using the diagram below. Consider the acute angle θ below in standard position. Let P (x, y) denote, as usual, the point on the terminal side of θ which lies on the Unit Circle and let Q(1, y) denote the point on the terminal side of θ which lies on the vertical line x = 1. Q(1, y) = (1, tan(θ)) P (x, y) y 1 θ O A(x, 0) B(1, 0) x 1In Theorem 10.4 we also showed cosine and sine to be functions of an angle residing in a right triangle so we could just as easily call them trigonometric functions. In later sections, you will ﬁnd that we do indeed use the phrase ‘trigonometric function’ interchangeably with the term ‘circular function’. 10.3 The Six Circular Functions and Fundamental Identities 745 y = 1 x which gives y = y The word ‘tangent’ comes from the Latin meaning ‘to touch,’ and for this reason, the line x = 1 is called a tangent line to the Unit Circle since it intersects, or ‘touches’, the circle at only one point, namely (1, 0). Dropping perpendiculars from P and Q creates a pair of similar triangles ∆OP A and ∆OQB. Thus y x = tan(θ), where this last equality comes from applying Deﬁnition 10.2. We have just shown that for acute angles θ, tan(θ) is the y-coordinate of the point on the terminal side of θ which lies on the line x = 1 which is tangent to the Unit Circle. Now the word ‘secant’ means ‘to cut’, so a secant line is any line that ‘cuts through’ a circle at two points.2 The line containing the terminal side of θ is a secant line since it intersects the Unit Circle in Quadrants I and III. With the point P lying on the Unit Circle, the length of the hypotenuse of ∆OP A is 1. If we let h denote the length of the hypotenuse of ∆OQB, we have from similar triangles that h x = sec(θ). Hence for an acute angle θ, sec(θ) is the length of the line segment which lies on the secant line determined by the terminal side of θ and ‘cuts oﬀ’ the tangent line x = 1. Not only do these observations help explain the names of these functions, they serve as the basis for a fundamental inequality needed for Calculus which we’ll explore in the Exercises. x , or h = 1 1 = 1 Of the six circular functions, only cosine and sine are deﬁned for all angles. Since cos(θ) = x and sin(θ) = y in Deﬁnition 10.2, it is customary to rephrase the remaining four circular functions in terms of cosine and sine. The following theorem is a result of simply replacing x with cos(θ) and y with sin(θ) in Deﬁnition 10.2. Theorem 10.6. Reciprocal and Quotient Identities: sec(θ) = 1 cos(θ) ,
provided cos(θ) = 0; if cos(θ) = 0, sec(θ) is undeﬁned. csc(θ) = 1 sin(θ) , provided sin(θ) = 0; if sin(θ) = 0, csc(θ) is undeﬁned. tan(θ) = sin(θ) cos(θ) , provided cos(θ) = 0; if cos(θ) = 0, tan(θ) is undeﬁned. cot(θ) = cos(θ) sin(θ) , provided sin(θ) = 0; if sin(θ) = 0, cot(θ) is undeﬁned. It is high time for an example. Example 10.3.1. Find the indicated value, if it exists. 1. sec (60◦) 2. csc 7π 4 3. cot(3) 4. tan (θ), where θ is any angle coterminal with 3π 2 . 5. cos (θ), where csc(θ) = − √ 5 and θ is a Quadrant IV angle. 6. sin (θ), where tan(θ) = 3 and π < θ < 3π 2 . 2Compare this with the deﬁnition given in Section 2.1. 746 Solution. 1. According to Theorem 10.6, sec (60◦) = Foundations of Trigonometry 1 cos(60◦) . Hence, sec (60◦) = 1 (1/2) = 2. 2. Since sin 7π 4 = − √ 2 2 , csc 7π 4 = 1 sin( 7π 4 ) = 1 √ − 2/2 = − 2√ 2 √ = − 2. 3. Since θ = 3 radians is not one of the ‘common angles’ from Section 10.2, we resort to the calculator for a decimal approximation. Ensuring that the calculator is in radian mode, we ﬁnd cot(3) = cos(3) sin(3) ≈ −7.015. = 0 and sin(θ) = sin 3π 2 = −1. Attempting 4. If θ is coterminal with 3π to compute tan(θ) = sin(θ) 2 , then cos(θ) = cos 3π cos(θ) results in −1 √ 2 0 , so tan(θ) is undeﬁned. 5. We are given that csc(θ) = 1 √ 5 5 so sin(θ) = − 1√ 5 . As we saw in Section 10.2, 5 we can use the Pythagorean Identity, cos2(θ) + sin2(θ) = 1, to ﬁnd cos(θ) by knowing sin(θ). Substituting, we get cos2(θ) + θ is a Quadrant IV angle, cos(θ) > 0, so cos(θ) = 2 = 1, which gives cos2(θ) = 4 √ 5 5 . 5 , or cos(θ) = ± 2 √ 5 5 . Since sin(θ. If tan(θ) = 3, then sin(θ) cos(θ) = 3. Be careful - this does NOT mean we can take sin(θ) = 3 and cos(θ) = 1. Instead, from sin(θ) cos(θ) = 3 we get: sin(θ) = 3 cos(θ). To relate cos(θ) and sin(θ), we once again employ the Pythagorean Identity, cos2(θ) + sin2(θ) = 1. Solving sin(θ) = 3 cos(θ) for cos(θ), we ﬁnd cos(θ) = 1 3 sin(θ). Substituting this into the Pythagorean Identity, we ﬁnd sin2(θ) + 1 3 sin(θ)2 10 so sin(θ) = ± 3 10 . Since π < θ < 3π 2 , θ is a Quadrant III angle. This means sin(θ) < 0, so our ﬁnal answer is sin(θ) = − 3 = 1. Solving, we get sin2(θ) = 9 10 √ √ 10 10 . While the Reciprocal and Quotient Identities presented in Theorem 10.6 allow us to always reduce problems involving secant, cosecant, tangent and cotangent to problems involving cosine and sine, it is not always convenient to do so.3 It is worth taking the time to memorize the tangent and cotangent values of the common angles summarized below. 3As we shall see shortly, when solving equations involving secant and cosecant, we usually convert back to cosines and sines. However, when solving for tangent or cotangent, we usually stick with what we’re dealt. 10.3 The Six Circular Functions and Fundamental Identities 747 Tangent and Cotangent Values of Common Angles θ(degrees) 0◦ 30◦ 45◦ 60◦ 90◦ θ(radians) tan(θ) cot(θ undeﬁned undeﬁned √ 3 1 √ 3 3 0 Coupling Theorem 10.6 with the Reference Angle Theorem, Theorem 10.2, we get the following. Theorem 10.7. Generalized Reference Angle Theorem. The values of the circular functions of an angle, if they exist, are the same, up to a sign, of the corresponding circular functions of its reference angle. More speciﬁcally, if α is the reference angle for θ, then: cos(θ) = ± cos(α), sin(θ) = ± sin(α), sec(θ) = ± sec(α), csc(θ) = ± csc(α), tan(θ) = ± tan(α) and cot(θ) = ± cot(α). The choice of the (±) depends on the quadrant in which the terminal side of θ lies. We put Theorem 10.7 to good use in the following example. Example 10.3.2. Find all angles which satisfy the given equation. 1. sec(θ) = 2 Solution. 2. tan(θ) = √ 3 3. cot(θ) = −1. 1. To solve sec(θ) = 2, we convert to cosines and get 1 cos(θ) = 2 or cos(θ) = 1 same equation we solved in Example 10.2.5, number 1, so we know the answer is: θ = π or θ = 5π 3 + 2πk for integers k. 2 . This is the exact 3 + 2πk √ √ = 2. From the table of common values, we see tan π 3 3 must, therefore, have a reference angle of π 3. According to Theorem 10.7, we know 3 . Our next task is the solutions to tan(θ) = to determine in which quadrants the solutions to this equation lie. Since tangent is deﬁned as the ratio y x of points (x, y) on the Unit Circle with x = 0, tangent is positive when x and y have the same sign (i.e., when they are both positive or both negative.) This happens in Quadrants I and III. In Quadrant I, we get the solutions: θ = π 3 + 2πk for integers k, and for Quadrant III, we get θ = 4π 3 + 2πk for integers k. While these descriptions of the solutions are correct, they can be combined into one list as θ = π 3 + πk for integers k. The latter form of the solution is best understood looking at the geometry of the situation in the diagram below.4 4See Example 10.2.5 number 3 in Section 10.2 for another example of this kind of simpliﬁcation of the solution. 748 Foundations of Trigonometry . From the table of common values, we see that π 4 has a cotangent of 1, which means the solutions to cot(θ) = −1 have a reference angle of π 4 . To ﬁnd the quadrants in which our solutions lie, we note that cot(θ) = x y for a point (x, y) on the Unit Circle where y = 0. If cot(θ) is negative, then x and y must have diﬀerent signs (i.e., one positive and one negative.) Hence, our solutions lie in Quadrants II and IV. Our Quadrant II solution is θ = 3π 4 + 2πk, and for Quadrant IV, we get θ = 7π 4 +2πk for integers k. Can these lists be combined? Indeed they can - one such way to capture all the solutions is: θ = 3π 4 + πk for integers k We have already seen the importance of identities in trigonometry. Our next task is to use use the Reciprocal and Quotient Identities found in Theorem 10.6 coupled with the Pythagorean Identity found in Theorem 10.1 to derive new Pythagorean-like identities for the remaining four circular functions. Assuming cos(θ) = 0, we may start with cos2(θ) + sin2(θ) = 1 and divide both sides by cos2(θ) to obtain 1 + sin2(θ) cos2(θ) . Using properties of exponents along with the Reciprocal and Quotient Identities, this reduces to 1 + tan2(θ) = sec2(θ). If sin(θ) = 0, we can divide both sides of the identity cos2(θ) + sin2(θ) = 1 by sin2(θ), apply Theorem 10.6 once again, and obtain cot2(θ) + 1 = csc2(θ). These three Pythagorean Identities are worth memorizing and they, along with some of their other common forms, are summarized in the following theorem. cos2(θ) = 1 10.3 The Six Circular Functions and Fundamental Identities 749 Theorem 10.8. The Pythagorean Identities: 1. cos2(θ) + sin2(θ) = 1. Common Alternate Forms: 1 − sin2(θ) = cos2(θ) 1 − cos2(θ) = sin2(θ) 2. 1 + tan2(θ) = sec2(θ), provided cos(θ) = 0. Common Alternate Forms: sec2(θ) − tan2(θ) = 1 sec2(θ) − 1 = tan2(θ) 3. 1 + cot2(θ) = csc2(θ), provided sin(θ) = 0. Common Alternate Forms: csc2(θ) − cot2(θ) = 1 csc2(θ) − 1 = cot2(θ) Trigonometric identities play an important role in not just Trigonometry, but in Calculus as well. We’ll use them in this book to ﬁnd the values of the circular functions of an angle and solve equations and inequalities. In Calculus, they are needed to simplify otherwise complicated expressions. In the next example, we make good use of the Theorems 10.6 and 10.8. Example 10.3.3. Verify the following identities. Assume that all quantities are deﬁned. 1. 1 csc(θ) = sin(θ) 3. (sec(θ) − tan(θ))(sec(θ) + tan(θ)) = 1 5. 6 sec(θ) tan(θ) = 3 1 − sin(θ) − 3 1 + sin(θ) 2. tan(θ) = sin(θ) sec(θ) 4. 6. sec(θ) 1 − tan(θ) = 1 cos(θ) − sin(θ) sin(θ) 1 − cos(θ) = 1 + cos(θ) sin(θ) Solution. In verifying identities, we typically start with the more complicated side of the equation and use known identities to transform it into the other side of the equation. 1. To verify 1 csc(θ) = sin(θ), we start with the left side. Using csc(θ) = 1 sin(θ) , we get: 1 csc(θ) = 1 1 sin(θ) = sin(θ), which is what we were trying to prove. 750 Foundations of Trigonometry 2. Starting with the right hand side of tan(θ) = sin(θ) sec(θ), we use sec(θ) = 1 cos(θ) and ﬁnd: sin(θ) sec(θ) = sin(θ) 1 cos(θ) = sin(θ) cos(θ) = tan(θ), where the last equality is courtesy of Theorem 10.6. 3. Expanding the left hand side of the equation gives: (sec(θ) − tan(θ))(sec(θ) + tan(θ)) = sec2(θ) − tan2(θ). According to Theorem 10.8, sec2(θ) − tan2(θ) = 1. Putting it all together, (sec(θ) − tan(θ))(sec(θ) + tan(θ)) = sec2(θ) − tan2(θ) = 1. 4. While both sides of our last identity contain fractions, the left side aﬀords us more opportu- nities to use our identities.5 Substituting sec(θ) = 1 cos(θ) and tan(θ) = sin(θ) cos(θ) , we get: sec(θ) 1 − tan(θ) = 1 cos(θ) 1 − sin(θ) cos(θ) = 1 cos(θ) 1 − sin(θ) cos(θ) · cos(θ) cos(θ) = 1 cos(θ) 1 − sin(θ) cos(θ) (cos(θ)) = (cos(θ)) (1)(cos(θ)) − 1 sin(θ) cos(θ) (cos(θ)) = 1 cos(θ) − sin(θ) , which is exactly what we had set out to show. 5. The right hand side of the equation seems to hold more promise. We get common denomina- tors and add: 3 1 − sin(θ) − 3 1 + sin(θ) = = = = 3(1 + sin(θ)) (1 − sin(θ))(1 + sin(θ)) − 3(1 − sin(θ)) (1 + sin(θ))(1 − sin(θ)) 3 + 3 sin(θ) 1 − sin2(θ) − 3 − 3 sin(θ) 1 − sin2(θ) (3 + 3 sin(θ)) − (3 − 3 sin(θ)) 1 − sin2(θ) 6 sin(θ) 1 − sin2(θ) 5Or, to put to another way, earn more partial credit if this were an exam question! 10.3 The Six Circular Functions and Fundamental Identities 751 At this point, it is worth pausing to remind ourselves of our goal. We wish to transform this expression into 6 sec(θ) tan(θ). Using a reciprocal and quotient identity, we ﬁnd 6 sec(θ) tan(θ) = 6 . In other words, we need to get cosines in our denominator. Theorem 10.8 tells us 1 − sin2(θ) = cos2(θ) so we get: sin(θ) cos(θ) 1 cos(θ) 3 1 − sin(θ) − 3 1 + sin(θ) = = 6 sin(θ) cos2(θ) 6 sin(θ) 1 − sin2(θ) 1 cos(θ) = 6 sin(θ) cos(θ) = 6 sec(θ) tan(θ) 6. It is debatable which side of the identity is more complicated. One thing which stands out is that the denominator on the left hand side is 1 − cos(θ), while the numerator of the right hand side is 1 + cos(θ). This suggests the strategy of starting with the left han
d side and multiplying the numerator and denominator by the quantity 1 + cos(θ): sin(θ) 1 − cos(θ) = = = sin(θ) (1 − cos(θ)) · (1 + cos(θ)) (1 + cos(θ)) = sin(θ)(1 + cos(θ)) (1 − cos(θ))(1 + cos(θ)) sin(θ)(1 + cos(θ)) 1 − cos2(θ) sin(θ)(1 + cos(θ)) sin(θ) sin(θ) = = sin(θ)(1 + cos(θ)) sin2(θ) 1 + cos(θ) sin(θ) In Example 10.3.3 number 6 above, we see that multiplying 1 − cos(θ) by 1 + cos(θ) produces a diﬀerence of squares that can be simpliﬁed to one term using Theorem 10.8. This is exactly the √ same kind of phenomenon that occurs when we multiply expressions such as 1 − 2 or 3 − 4i by 3 + 4i. (Can you recall instances from Algebra where we did such things?) For this reason, the quantities (1 − cos(θ)) and (1 + cos(θ)) are called ‘Pythagorean Conjugates.’ Below is a list of other common Pythagorean Conjugates. 2 by 1 + √ Pythagorean Conjugates 1 − cos(θ) and 1 + cos(θ): (1 − cos(θ))(1 + cos(θ)) = 1 − cos2(θ) = sin2(θ) 1 − sin(θ) and 1 + sin(θ): (1 − sin(θ))(1 + sin(θ)) = 1 − sin2(θ) = cos2(θ) sec(θ) − 1 and sec(θ) + 1: (sec(θ) − 1)(sec(θ) + 1) = sec2(θ) − 1 = tan2(θ) sec(θ)−tan(θ) and sec(θ)+tan(θ): (sec(θ)−tan(θ))(sec(θ)+tan(θ)) = sec2(θ)−tan2(θ) = 1 csc(θ) − 1 and csc(θ) + 1: (csc(θ) − 1)(csc(θ) + 1) = csc2(θ) − 1 = cot2(θ) csc(θ) − cot(θ) and csc(θ) + cot(θ): (csc(θ) − cot(θ))(csc(θ) + cot(θ)) = csc2(θ) − cot2(θ) = 1 752 Foundations of Trigonometry Verifying trigonometric identities requires a healthy mix of tenacity and inspiration. You will need to spend many hours struggling with them just to become proﬁcient in the basics. Like many things in life, there is no short-cut here – there is no complete algorithm for verifying identities. Nevertheless, a summary of some strategies which may be helpful (depending on the situation) is provided below and ample practice is provided for you in the Exercises. Try working on the more complicated side of the identity. Strategies for Verifying Identities Use the Reciprocal and Quotient Identities in Theorem 10.6 to write functions on one side of the identity in terms of the functions on the other side of the identity. Simplify the resulting complex fractions. Add rational expressions with unlike denominators by obtaining common denominators. Use the Pythagorean Identities in Theorem 10.8 to ‘exchange’ sines and cosines, secants and tangents, cosecants and cotangents, and simplify sums or diﬀerences of squares to one term. Multiply numerator and denominator by Pythagorean Conjugates in order to take advan- tage of the Pythagorean Identities in Theorem 10.8. If you ﬁnd yourself stuck working with one side of the identity, try starting with the other side of the identity and see if you can ﬁnd a way to bridge the two parts of your work. 10.3.1 Beyond the Unit Circle In Section 10.2, we generalized the cosine and sine functions from coordinates on the Unit Circle to coordinates on circles of radius r. Using Theorem 10.3 in conjunction with Theorem 10.8, we generalize the remaining circular functions in kind. Theorem 10.9. Suppose Q(x, y) is the point on the terminal side of an angle θ (plotted in standard position) which lies on the circle of radius r, x2 + y2 = r2. Then: r x r y = = x2 + y2 x x2 + y2 y , provided x = 0. , provided y = 0. , provided x = 0. , provided y = 0. sec(θ) = csc(θ) = tan(θ) = cot(θ) = y x x y 10.3 The Six Circular Functions and Fundamental Identities 753 Example 10.3.4. 1. Suppose the terminal side of θ, when plotted in standard position, contains the point Q(3, −4). Find the values of the six circular functions of θ. 2. Suppose θ is a Quadrant IV angle with cot(θ) = −4. Find the values of the ﬁve remaining circular functions of θ. Solution. 1. Since x = 3 and y = −4, r = x2 + y2 = (3)2 + (−4)2 = √ cos(θ) = 3 5 , sin(θ) = − 4 5 , sec(θ) = 5 3 , csc(θ) = − 5 4 , tan(θ) = − 4 25 = 5. Theorem 10.9 tells us 3 and cot(θ) = − 3 4 . 2. In order to use Theorem 10.9, we need to ﬁnd a point Q(x, y) which lies on the terminal side y , and since θ is a −1 , we may choose6 x = 4 17. Applying Theorem 10.9 once √ √ 17 17 4 , csc(θ) = − 17 17 , sec(θ) = of θ, when θ is plotted in standard position. We have that cot(θ) = −4 = x Quadrant IV angle, we also know x > 0 and y < 0. Viewing −4 = 4 x2 + y2 = and y = −1 so that r = √ = 4 more, we ﬁnd cos(θ) = 4√ 17 and tan(θ) = − 1 4 . (4)2 + (−1)2 = = − 17 , sin(θ) = − 1√ √ √ 17 17 We may also specialize Theorem 10.9 to the case of acute angles θ which reside in a right triangle, as visualized below. b c a θ Theorem 10.10. Suppose θ is an acute angle residing in a right triangle. If the length of the side adjacent to θ is a, the length of the side opposite θ is b, and the length of the hypotenuse is c, then tan(θ) = sec(θ) = csc(θ) = cot(θ The following example uses Theorem 10.10 as well as the concept of an ‘angle of inclination.’ The angle of inclination (or angle of elevation) of an object refers to the angle whose initial side is some kind of base-line (say, the ground), and whose terminal side is the line-of-sight to an object above the base-line. This is represented schematically below. 6We may choose any values x and y so long as x > 0, y < 0 and x y = −4. For example, we could choose x = 8 and y = −2. The fact that all such points lie on the terminal side of θ is a consequence of the fact that the terminal side of θ is the portion of the line with slope − 1 4 which extends from the origin into Quadrant IV. 754 Foundations of Trigonometry object θ ‘base line’ The angle of inclination from the base line to the object is θ Example 10.3.5. 1. The angle of inclination from a point on the ground 30 feet away to the top of Lakeland’s Armington Clocktower7 is 60◦. Find the height of the Clocktower to the nearest foot. 2. In order to determine the height of a California Redwood tree, two sightings from the ground, one 200 feet directly behind the other, are made. If the angles of inclination were 45◦ and 30◦, respectively, how tall is the tree to the nearest foot? Solution. 1. We can represent the problem situation using a right triangle as shown below. If we let h 30 . From this we get denote the height of the tower, then Theorem 10.10 gives tan (60◦) = h h = 30 tan (60◦) = 30 3 ≈ 51.96. Hence, the Clocktower is approximately 52 feet tall. √ h ft. 60◦ 30 ft. Finding the height of the Clocktower 2. Sketching the problem situation below, we ﬁnd ourselves with two unknowns: the height h of the tree and the distance x from the base of the tree to the ﬁrst observation point. 7Named in honor of Raymond Q. Armington, Lakeland’s Clocktower has been a part of campus since 1972. 10.3 The Six Circular Functions and Fundamental Identities 755 h ft. 30◦ 200 ft. 45◦ x ft. Finding the height of a California Redwood Using Theorem 10.10, we get a pair of equations: tan (45◦) = h x+200 . Since tan (45◦) = 1, the ﬁrst equation gives h x = 1, or x = h. Substituting this into the second equation gives 3. The result is a linear equation for h, so we proceed to expand the right hand side and gather all the terms involving h to one side. √ 3 3 . Clearing fractions, we get 3h = (h + 200) x and tan (30◦) = h h+200 = tan (30◦) = √ h 3h = (h + 200) √ 3h = h √ 3 + 200 √ 3 = 200 3 √ 3 √ 3 3h − h √ (3 − 3)h = 200 √ 3 √ 3 √ 3 ≈ 273.20 Hence, the tree is approximately 273 feet tall. h = 200 3 − As we did in Section 10.2.1, we may consider all six circular functions as functions of real numbers. At this stage, there are three equivalent ways to deﬁne the functions sec(t), csc(t), tan(t) and cot(t) for real numbers t. First, we could go through the formality of the wrapping function on page 704 and deﬁne these functions as the appropriate ratios of x and y coordinates of points on the Unit Circle; second, we could deﬁne them by associating the real number t with the angle θ = t radians so that the value of the trigonometric function of t coincides with that of θ; lastly, we could simply deﬁne them using the Reciprocal and Quotient Identities as combinations of the functions f (t) = cos(t) and g(t) = sin(t). Presently, we adopt the last approach. We now set about determining the domains and ranges of the remaining four circular functions. Consider the function F (t) = sec(t) deﬁned as F (t) = sec(t) = 1 cos(t) . We know F is undeﬁned whenever cos(t) = 0. From Example 10.2.5 number 3, we know cos(t) = 0 whenever t = π 2 + πk for integers k. Hence, our domain for F (t) = sec(t), in set builder notation is {t : t = π 2 + πk, for integers k}. To get a better understanding what set of real numbers we’re dealing with, it pays to write out and graph this set. Running through a few values of k, we ﬁnd the domain to be {t : t = ± π 2 , . . .}. Graphing this set on the number line we get 2 , ± 5π 2 , ± 3π 756 Foundations of Trigonometry − 5π 2 − 3π 2 − π 2 0 π 2 3π 2 5π 2 Using interval notation to describe this set, we get . . . ∪ − 5π 2 , − 3π 2 ∪ − 3π , 3π 2 ∪ 3π 2 , 5π 2 ∪ . . . This is cumbersome, to say the least! In order to write this in a more compact way, we note that from the set-builder description of the domain, the kth point excluded from the domain, which we’ll call xk, can be found by the formula xk = π 2 +πk. (We are using sequence notation from Chapter 9.) Getting a common denominator and factoring out the π in the numerator, we get xk = (2k+1)π . The . domain consists of the intervals determined by successive points xk: (xk, xk + 1) = In order to capture all of the intervals in the domain, k must run through all of the integers, that is, k = 0, ±1, ±2, . . . . The way we denote taking the union of inﬁnitely many intervals like this is to use what we call in this text extended interval notation. The domain of F (t) = sec(t) can now be written as 2 , (2k+3)π 2 (2k+1)π 2 ∞ k=−∞ (2k + 1)π 2 , (2k + 3)π 2 The reader should compare this notation with summation notation introduced in Section 9.2, in particular the notation used to describe geometric series in Theorem 9.2. In the same way the index k in the series ∞ k=1 ark−1 can never equal the upper l
imit ∞, but rather, ranges through all of the natural numbers, the index k in the union ∞ (2k + 1)π 2 , (2k + 3)π 2 k=−∞ can never actually be ∞ or −∞, but rather, this conveys the idea that k ranges through all of the integers. Now that we have painstakingly determined the domain of F (t) = sec(t), it is time to discuss the range. Once again, we appeal to the deﬁnition F (t) = sec(t) = 1 cos(t) . The range of f (t) = cos(t) is [−1, 1], and since F (t) = sec(t) is undeﬁned when cos(t) = 0, we split our discussion into two cases: when 0 < cos(t) ≤ 1 and when −1 ≤ cos(t) < 0. If 0 < cos(t) ≤ 1, then we can divide the inequality cos(t) ≤ 1 by cos(t) to obtain sec(t) = 1 cos(t) ≥ 1. Moreover, using the notation introduced in Section 4.2, we have that as cos(t) → 0+, sec(t) = 1 very small (+) ≈ very big (+). In other words, as cos(t) → 0+, sec(t) → ∞. If, on the other hand, if −1 ≤ cos(t) < 0, then dividing by cos(t) causes a reversal of the inequality so that sec(t) = 1 sec(t) ≤ −1. In this case, as cos(t) → 0−, sec(t) = 1 very small (−) ≈ very big (−), so that as cos(t) → 0−, we get sec(t) → −∞. Since cos(t) ≈ cos(t) ≈ 1 1 10.3 The Six Circular Functions and Fundamental Identities 757 f (t) = cos(t) admits all of the values in [−1, 1], the function F (t) = sec(t) admits all of the values in (−∞, −1] ∪ [1, ∞). Using set-builder notation, the range of F (t) = sec(t) can be written as {u : u ≤ −1 or u ≥ 1}, or, more succinctly,8 as {u : \|u\| ≥ 1}.9 Similar arguments can be used to determine the domains and ranges of the remaining three circular functions: csc(t), tan(t) and cot(t). The reader is encouraged to do so. (See the Exercises.) For now, we gather these facts into the theorem below. Theorem 10.11. Domains and Ranges of the Circular Functions • The function f (t) = cos(t) • The function g(t) = sin(t) – has domain (−∞, ∞) – has range [−1, 1] – has domain (−∞, ∞) – has range [−1, 1] • The function F (t) = sec(t) = 1 cos(t) – has domain {t : t = π 2 + πk, for integers k} = – has range {u : \|u\| ≥ 1} = (−∞, −1] ∪ [1, ∞) ∞ k=−∞ (2k + 1)π 2 , (2k + 3)π 2 • The function G(t) = csc(t) = 1 sin(t) – has domain {t : t = πk, for integers k} = ∞ k=−∞ (kπ, (k + 1)π) – has range {u : \|u\| ≥ 1} = (−∞, −1] ∪ [1, ∞) • The function J(t) = tan(t) = sin(t) cos(t) – has domain {t : t = π 2 + πk, for integers k} = ∞ k=−∞ (2k + 1)π 2 , (2k + 3)π 2 – has range (−∞, ∞) • The function K(t) = cot(t) = cos(t) sin(t) – has domain {t : t = πk, for integers k} = – has range (−∞, ∞) ∞ k=−∞ (kπ, (k + 1)π) 8Using Theorem 2.4 from Section 2.4. 9Notice we have used the variable ‘u’ as the ‘dummy variable’ to describe the range elements. While there is no mathematical reason to do this (we are describing a set of real numbers, and, as such, could use t again) we choose u to help solidify the idea that these real numbers are the outputs from the inputs, which we have been calling t. 758 Foundations of Trigonometry We close this section with a few notes about solving equations which involve the circular functions. First, the discussion on page 735 in Section 10.2.1 concerning solving equations applies to all six circular functions, not just f (t) = cos(t) and g(t) = sin(t). In particular, to solve the equation cot(t) = −1 for real numbers t, we can use the same thought process we used in Example 10.3.2, number 3 to solve cot(θ) = −1 for angles θ in radian measure – we just need to remember to write our answers using the variable t as opposed to θ. Next, it is critical that you know the domains and ranges of the six circular functions so that you know which equations have no solutions. For example, sec(t) = 1 2 is not in the range of secant. Finally, you will need to review the notions of reference angles and coterminal angles so that you can see why csc(t) = −42 has an inﬁnite set of solutions in Quadrant III and another inﬁnite set of solutions in Quadrant IV. 2 has no solution because 1 10.3 The Six Circular Functions and Fundamental Identities 759 10.3.2 Exercises In Exercises 1 - 20, ﬁnd the exact value or state that it is undeﬁned. 1. tan π 4 5. tan − 11π 6 9. tan (117π) 13. tan 17. tan 31π 2 2π 3 2. sec π 6 6. sec − 10. sec − 3π 2 5π 3 14. sec π 4 18. sec (−7π) 3. csc 7. csc 5π 6 − π 3 4. cot 8. cot 4π 3 13π 2 11. csc (3π) 12. cot (−5π) 15. csc − 7π 4 19. csc π 2 16. cot 20. cot 7π 6 3π 4 In Exercises 21 - 34, use the given the information to ﬁnd the exact values of the remaining circular functions of θ. 21. sin(θ) = 3 5 with θ in Quadrant II 22. tan(θ) = 12 5 with θ in Quadrant III with θ in Quadrant I 24. sec(θ) = 7 with θ in Quadrant IV 25. csc(θ) = − with θ in Quadrant III 26. cot(θ) = −23 with θ in Quadrant II 27. tan(θ) = −2 with θ in Quadrant IV. 28. sec(θ) = −4 with θ in Quadrant II. 29. cot(θ) = √ 5 with θ in Quadrant III. 30. cos(θ) = 1 3 with θ in Quadrant I. 31. cot(θ) = 2 with 0 < θ < π 2 . 33. tan(θ) = √ 10 with π < θ < 3π 2 . 32. csc(θ) = 5 with π 2 < θ < π. 34. sec(θ) = 2 √ 5 with 3π 2 < θ < 2π. In Exercises 35 - 42, use your calculator to approximate the given value to three decimal places. Make sure your calculator is in the proper angle measurement mode! 35. csc(78.95◦) 36. tan(−2.01) 37. cot(392.994) 38. sec(207◦) 39. csc(5.902) 40. tan(39.672◦) 41. cot(3◦) 42. sec(0.45) 23. csc(θ) = 25 24 10 91 √ 91 760 Foundations of Trigonometry In Exercises 43 - 57, ﬁnd all of the angles which satisfy the equation. 43. tan(θ) = √ 3 44. sec(θ) = 2 45. csc(θ) = −1 46. cot(θ) = √ 3 3 47. tan(θ) = 0 48. sec(θ) = 1 49. csc(θ) = 2 50. cot(θ) = 0 51. tan(θ) = −1 52. sec(θ) = 0 53. csc(θ) = − 55. tan(θ) = − √ 3 56. csc(θ) = −2 57. cot(θ) = −1 1 2 54. sec(θ) = −1 In Exercises 58 - 65, solve the equation for t. Give exact values. 58. cot(t) = 1 59. tan(t) = 62. cot(t) = − √ 3 63. tan(t 60. sec(t) = − 64. sec(t) = √ 2 3 61. csc(t) = 0 65. csc(t) = √ 2 3 3 In Exercises 66 - 69, use Theorem 10.10 to ﬁnd the requested quantities. 66. Find θ, a, and c. 67. Find α, b, and c. 60◦ a c θ 9 b α c 12 34◦ 68. Find θ, a, and c. 69. Find β, b, and c. 47◦ c a θ 6 b β c 2.5 50◦ 10.3 The Six Circular Functions and Fundamental Identities 761 In Exercises 70 - 75, use Theorem 10.10 to answer the question. Assume that θ is an angle in a right triangle. 70. If θ = 30◦ and the side opposite θ has length 4, how long is the side adjacent to θ? 71. If θ = 15◦ and the hypotenuse has length 10, how long is the side opposite θ? 72. If θ = 87◦ and the side adjacent to θ has length 2, how long is the side opposite θ? 73. If θ = 38.2◦ and the side opposite θ has lengh 14, how long is the hypoteneuse? 74. If θ = 2.05◦ and the hypotenuse has length 3.98, how long is the side adjacent to θ? 75. If θ = 42◦ and the side adjacent to θ has length 31, how long is the side opposite θ? 76. A tree standing vertically on level ground casts a 120 foot long shadow. The angle of elevation from the end of the shadow to the top of the tree is 21.4◦. Find the height of the tree to the nearest foot. With the help of your classmates, research the term umbra versa and see what it has to do with the shadow in this problem. 77. The broadcast tower for radio station WSAZ (Home of “Algebra in the Morning with Carl and Jeﬀ”) has two enormous ﬂashing red lights on it: one at the very top and one a few feet below the top. From a point 5000 feet away from the base of the tower on level ground the angle of elevation to the top light is 7.970◦ and to the second light is 7.125◦. Find the distance between the lights to the nearest foot. 78. On page 753 we deﬁned the angle of inclination (also known as the angle of elevation) and in this exercise we introduce a related angle - the angle of depression (also known as the angle of declination). The angle of depression of an object refers to the angle whose initial side is a horizontal line above the object and whose terminal side is the line-of-sight to the object below the horizontal. This is represented schematically below. horizontal θ observer object The angle of depression from the horizontal to the object is θ (a) Show that if the horizontal is above and parallel to level ground then the angle of depression (from observer to object) and the angle of inclination (from object to observer) will be congruent because they are alternate interior angles. 762 Foundations of Trigonometry (b) From a ﬁretower 200 feet above level ground in the Sasquatch National Forest, a ranger spots a ﬁre oﬀ in the distance. The angle of depression to the ﬁre is 2.5◦. How far away from the base of the tower is the ﬁre? (c) The ranger in part 78b sees a Sasquatch running directly from the ﬁre towards the ﬁretower. The ranger takes two sightings. At the ﬁrst sighting, the angle of depression from the tower to the Sasquatch is 6◦. The second sighting, taken just 10 seconds later, gives the the angle of depression as 6.5◦. How far did the Saquatch travel in those 10 seconds? Round your answer to the nearest foot. How fast is it running in miles per hour? Round your answer to the nearest mile per hour. If the Sasquatch keeps up this pace, how long will it take for the Sasquatch to reach the ﬁretower from his location at the second sighting? Round your answer to the nearest minute. 79. When I stand 30 feet away from a tree at home, the angle of elevation to the top of the tree is 50◦ and the angle of depression to the base of the tree is 10◦. What is the height of the tree? Round your answer to the nearest foot. 80. From the observation deck of the lighthouse at Sasquatch Point 50 feet above the surface of Lake Ippizuti, a lifeguard spots a boat out on the lake sailing directly toward the lighthouse. The ﬁrst sighting had an angle of depression of 8.2◦ and the second sighting had an angle of depression of 25.9◦. How far had the boat traveled between the sightings? 81. A guy wire 1000 feet long is attached to the top of a tower. When pulled taut it makes a 43◦ angle with the ground. How tall is the tower? How far away from the base of the tower does the wire hit the ground? In Exercises 82
- 128, verify the identity. Assume that all quantities are deﬁned. 82. cos(θ) sec(θ) = 1 84. sin(θ) csc(θ) = 1 86. csc(θ) cos(θ) = cot(θ) 88. 90. 92. cos(θ) sin2(θ) = csc(θ) cot(θ) 1 − cos(θ) sin(θ) = csc(θ) − cot(θ) sin(θ) 1 − cos2(θ) = csc(θ) 83. tan(θ) cos(θ) = sin(θ) 85. tan(θ) cot(θ) = 1 87. 89. 91. 93. sin(θ) cos2(θ) = sec(θ) tan(θ) 1 + sin(θ) cos(θ) = sec(θ) + tan(θ) cos(θ) 1 − sin2(θ) sec(θ) 1 + tan2(θ) = sec(θ) = cos(θ) 10.3 The Six Circular Functions and Fundamental Identities 763 94. 96. csc(θ) 1 + cot2(θ) cot(θ) csc2(θ) − 1 = sin(θ) = tan(θ) 95. tan(θ) sec2(θ) − 1 = cot(θ) 97. 4 cos2(θ) + 4 sin2(θ) = 4 98. 9 − cos2(θ) − sin2(θ) = 8 99. tan3(θ) = tan(θ) sec2(θ) − tan(θ) 100. sin5(θ) = 1 − cos2(θ)2 sin(θ) 101. sec10(θ) = 1 + tan2(θ)4 sec2(θ) 102. cos2(θ) tan3(θ) = tan(θ) − sin(θ) cos(θ) 103. sec4(θ) − sec2(θ) = tan2(θ) + tan4(θ) 104. 106. cos(θ) + 1 cos(θ) − 1 1 − cot(θ) 1 + cot(θ) = = 1 + sec(θ) 1 − sec(θ) tan(θ) − 1 tan(θ) + 1 105. sin(θ) + 1 sin(θ) − 1 = 1 + csc(θ) 1 − csc(θ) 107. 1 − tan(θ) 1 + tan(θ) = cos(θ) − sin(θ) cos(θ) + sin(θ) 108. tan(θ) + cot(θ) = sec(θ) csc(θ) 109. csc(θ) − sin(θ) = cot(θ) cos(θ) 110. cos(θ) − sec(θ) = − tan(θ) sin(θ) 111. cos(θ)(tan(θ) + cot(θ)) = csc(θ) 112. sin(θ)(tan(θ) + cot(θ)) = sec(θ) 113. 1 1 − cos(θ) + 1 1 + cos(θ) = 2 csc2(θ) 114. 1 sec(θ) + 1 + 1 sec(θ) − 1 = 2 csc(θ) cot(θ) 115. 1 csc(θ) + 1 + 1 csc(θ) − 1 = 2 sec(θ) tan(θ) 116. 1 csc(θ) − cot(θ) − 1 csc(θ) + cot(θ) = 2 cot(θ) 117. cos(θ) 1 − tan(θ) + sin(θ) 1 − cot(θ) = sin(θ) + cos(θ) 118. 120. 122. 124. 126. 128. 1 sec(θ) + tan(θ) 1 csc(θ) − cot(θ) = sec(θ) − tan(θ) = csc(θ) + cot(θ) 1 1 − sin(θ) 1 1 − cos(θ) = sec2(θ) + sec(θ) tan(θ) = csc2(θ) + csc(θ) cot(θ) cos(θ) 1 + sin(θ) = 1 − sin(θ) cos(θ) 1 − sin(θ) 1 + sin(θ) = (sec(θ) − tan(θ))2 119. 121. 123. 125. 1 sec(θ) − tan(θ) 1 csc(θ) + cot(θ) = sec(θ) + tan(θ) = csc(θ) − cot(θ) 1 1 + sin(θ) 1 1 + cos(θ) = sec2(θ) − sec(θ) tan(θ) = csc2(θ) − csc(θ) cot(θ) 127. csc(θ) − cot(θ) = sin(θ) 1 + cos(θ) 764 Foundations of Trigonometry In Exercises 129 - 132, verify the identity. You may need to consult Sections 2.2 and 6.2 for a review of the properties of absolute value and logarithms before proceeding. 129. ln \| sec(θ)\| = − ln \| cos(θ)\| 130. − ln \| csc(θ)\| = ln \| sin(θ)\| 131. − ln \| sec(θ) − tan(θ)\| = ln \| sec(θ) + tan(θ)\| 132. − ln \| csc(θ) + cot(θ)\| = ln \| csc(θ) − cot(θ)\| 133. Verify the domains and ranges of the tangent, cosecant and cotangent functions as presented in Theorem 10.11. 134. As we did in Exercise 74 in Section 10.2, let α and β be the two acute angles of a right triangle. (Thus α and β are complementary angles.) Show that sec(α) = csc(β) and tan(α) = cot(β). The fact that co-functions of complementary angles are equal in this case is not an accident and a more general result will be given in Section 10.4. 135. We wish to establish the inequality cos(θ) < < 1 for 0 < θ < . Use the diagram from the beginning of the section, partially reproduced below, to answer the following. sin(θ(1, 0) x (a) Show that triangle OP B has area 1 2 sin(θ). (b) Show that the circular sector OP B with central angle θ has area (c) Show that triangle OQB has area 1 2 tan(θ). (d) Comparing areas, show that sin(θ) < θ < tan(θ) for 0 < θ < π 2 . (e) Use the inequality sin(θ) < θ to show that sin(θ) θ < 1 for 0 < θ < 1 2 θ. π 2 . (f) Use the inequality θ < tan(θ) to show that cos(θ) < with the previous part to complete the proof. sin(θ) θ for 0 < θ < π 2 . Combine this 10.3 The Six Circular Functions and Fundamental Identities 765 136. Show that cos(θ) < sin(θ) θ < 1 also holds for − π 2 < θ < 0. 137. Explain why the fact that tan(θ) = 3 = 3 solution to number 6 in Example 10.3.1.) 1 does not mean sin(θ) = 3 and cos(θ) = 1? (See the 766 Foundations of Trigonometry 10.3.3 Answers 1. tan 4. cot π 4 4π 3 = 1 = 7. csc − π 3 = − = 2 √ 3 = 10. sec 13. tan 16. cot 19. csc − 5π 3 31π 2 7π 6 π 2 = 1 2. sec 5. tan 8. cot 3 = π 6 − 11π 6 13π . csc 5π 6 = 2 6. sec − 3π 2 is undeﬁned 9. tan (117π) = 0 11. csc (3π) is undeﬁned 12. cot (−5π) is undeﬁned √ 3 3 √ 2 3 is undeﬁned 14. sec 15. csc − √ 2 = 7π 4 18. sec (−7π) = −1 = π 4 2π 3 3π 4 √ 2 √ = − 3 = −1 17. tan 20. cot 21. sin(θ) = 3 5 , cos(θ) = − 4 5 , tan(θ) = − 3 4 , csc(θ) = 5 3 , sec(θ) = − 5 4 , cot(θ) = − 4 3 22. sin(θ) = − 12 13 , cos(θ) = − 5 13 , tan(θ) = 12 5 , csc(θ) = − 13 12 , sec(θ) = − 13 5 , cot(θ) = 5 12 23. sin(θ) = 24 25 , cos(θ) = 7 25 , tan(θ) = 24 24 , sec(θ) = 25 7 , cot(θ) = 7 24 7 , csc(θ) = 25 √ 3 , cos(θ) = 1 7 , tan(θ) = −4 3, csc(θ) = − 7 √ 12 , sec(θ) = 7, cot(θ) = − 3 √ 3 12 √ 24. sin(θ) = −4 7 √ 25. sin(θ) = − √ 26. sin(θ) = 91 10 , cos(θ) = − 3 530 , cos(θ) = − 23 530 10 , tan(θ) = √ 91 3 , csc(θ) = − 10 91 √ 530 , tan(θ) = − 1 530 23 , csc(θ) = 3 , cot(θ) = 3 , sec(θ) = − 10 √ 530 23 , cot(θ) = −23 91 91 √ √ 91 √ 27. sin(θ) = − 2 √ 28. sin(θ) = √ 5 5 , cos(θ) = 4 , cos(θ) = − 1 15 29. sin(θ) = − √ 30. sin(θ) = 2 √ 31. sin(θ) = 32. sin(θ) = 1 2 √ 6 6 , cos(θ) = − 3 , cos(θ) = 1 5 , cos(θ) = 2 5 , cos(θ) = − 2 5 √ 5 5 , tan(θ) = −2, csc(θ) = − √ 4 , tan(θ) = − √ 30 6 , tan(θ) = √ 15, csc(θ) = 4 √ 5 5 , csc(θ) = − 530, sec(θ) = − √ 5, cot(θ) = − 1 2 √ 5 2 , sec(θ) = √ 15 15 , sec(θ) = −4, cot(θ) = − √ √ 30 5 , cot(θ) = 6, sec(θ) = − √ 15 15 √ 5 3 , tan(θ) = 2 √ 5 , tan(θ) = 1 5 2, csc(θ) = 3 √ 2 , csc(θ) = √ 2 4 , sec(θ) = 3, cot(θ) = √ 5 2 , cot(θ) = 2 5, sec(θ) = √ 2 4 √ 6 5 , tan(θ) = − √ 6 12 , csc(θ) = 5, sec(θ) = − 5 √ 12 , cot(θ) = −2 6 √ 6 33. sin(θ) = − 34. sin(θ) = − √ 110 11 , cos(θ) = − √ √ 95 5 10 , tan(θ) = − 10 , cos(θ) = √ 11 11 , tan(θ) = √ √ 10, csc(θ) = − √ 95 19 , sec(θ) = 2 √ 110 10 , sec(θ) = − √ 19, csc(θ) = − 2 √ 11, cot(θ) = √ 5, cot(θ) = − 19 19 √ 10 10 10.3 The Six Circular Functions and Fundamental Identities 767 35. csc(78.95◦) ≈ 1.019 37. cot(392.994) ≈ 3.292 39. csc(5.902) ≈ −2.688 41. cot(3◦) ≈ 19.081 43. tan(θ) = √ 3 when θ = π 3 36. tan(−2.01) ≈ 2.129 38. sec(207◦) ≈ −1.122 40. tan(39.672◦) ≈ 0.829 42. sec(0.45) ≈ 1.111 + πk for any integer k 44. sec(θ) = 2 when θ = π 3 + 2πk or θ = 5π 3 + 2πk for any integer k 45. csc(θ) = −1 when θ = 46. cot(θ) = √ 3 3 when θ = 3π 2 π 3 + 2πk for any integer k. + πk for any integer k 47. tan(θ) = 0 when θ = πk for any integer k 48. sec(θ) = 1 when θ = 2πk for any integer k 49. csc(θ) = 2 when θ = 50. cot(θ) = 0 when θ = π 6 π 2 51. tan(θ) = −1 when θ = + 2πk or θ = 5π 6 + 2πk for any integer k. + πk for any integer k 3π 4 + πk for any integer k 52. sec(θ) = 0 never happens 53. csc(θ) = − 1 2 never happens 54. sec(θ) = −1 when θ = π + 2πk = (2k + 1)π for any integer k 55. tan(θ) = − √ 3 when θ = 2π 3 + πk for any integer k 56. csc(θ) = −2 when θ = 57. cot(θ) = −1 when θ = 7π 6 3π 4 + 2πk or θ = 11π 6 + 2πk for any integer k + πk for any integer k 58. cot(t) = 1 when t = π 4 + πk for any integer k 59. tan(t) = √ 3 3 when t = π 6 + πk for any integer k 768 Foundations of Trigonometry 60. sec(t) = − √ 2 3 3 when t = 5π 6 + 2πk or t = 7π 6 61. csc(t) = 0 never happens + 2πk for any integer k 62. cot(t) = − √ 3 when t = 5π 6 5π 6 + πk for any integer k + πk for any integer k when t = √ 3 3 63. tan(t) = − 64. sec(t) = 65. csc(t when t = π 6 + 2πk or t = 11π 6 + 2πk for any integer k when t = + 2πk or t = 2π 3 + 2πk for any integer k π 3 √ 66. θ = 30◦, a = 3 √ 3, c = √ 108 = 6 3 67. α = 56◦, b = 12 tan(34◦) = 8.094, c = 12 sec(34◦) = 12 cos(34◦) ≈ 14.475 68. θ = 43◦, a = 6 cot(47◦) = 6 tan(47◦) ≈ 5.595, c = 6 csc(47◦) = 6 sin(47◦) ≈ 8.204 69. β = 40◦, b = 2.5 tan(50◦) ≈ 2.979, c = 2.5 sec(50◦) = 2.5 cos(50◦) ≈ 3.889 70. The side adjacent to θ has length 4 √ 3 ≈ 6.928 71. The side opposite θ has length 10 sin(15◦) ≈ 2.588 72. The side opposite θ is 2 tan(87◦) ≈ 38.162 73. The hypoteneuse has length 14 csc(38.2◦) = 14 sin(38.2◦) ≈ 22.639 74. The side adjacent to θ has length 3.98 cos(2.05◦) ≈ 3.977 75. The side opposite θ has length 31 tan(42◦) ≈ 27.912 76. The tree is about 47 feet tall. 77. The lights are about 75 feet apart. 78. (b) The ﬁre is about 4581 feet from the base of the tower. (c) The Sasquatch ran 200 cot(6◦) − 200 cot(6.5◦) ≈ 147 feet in those 10 seconds. This translates to ≈ 10 miles per hour. At the scene of the second sighting, the Sasquatch was ≈ 1755 feet from the tower, which means, if it keeps up this pace, it will reach the tower in about 2 minutes. 10.3 The Six Circular Functions and Fundamental Identities 769 79. The tree is about 41 feet tall. 80. The boat has traveled about 244 feet. 81. The tower is about 682 feet tall. The guy wire hits the ground about 731 feet away from the base of the tower. 770 Foundations of Trigonometry 10.4 Trigonometric Identities In Section 10.3, we saw the utility of the Pythagorean Identities in Theorem 10.8 along with the Quotient and Reciprocal Identities in Theorem 10.6. Not only did these identities help us compute the values of the circular functions for angles, they were also useful in simplifying expressions involving the circular functions. In this section, we introduce several collections of identities which have uses in this course and beyond. Our ﬁrst set of identities is the ‘Even / Odd’ identities.1 Theorem 10.12. Even / Odd Identities: For all applicable angles θ, cos(−θ) = cos(θ) sin(−θ) = − sin(θ) tan(−θ) = − tan(θ) sec(−θ) = sec(θ) csc(−θ) = − csc(θ) cot(−θ) = − cot(θ) In light of the Quotient and Reciprocal Identities, Theorem 10.6, it suﬃces to show cos(−θ) = cos(θ) and sin(−θ) = − sin(θ). The remaining four circular functions can be expressed in terms of cos(θ) and sin(θ) so the proofs of their Even / Odd Identities are left as exercises. Consider an angle θ plotted in standard position. Let θ0 be the angle coterminal with θ with 0 ≤ θ0 < 2π. (We can construct the angle θ0 by rotating counter-clockwise from the positive x-axis to the terminal side of θ as pictured below.) Since θ and θ0 are coterminal, cos(θ) = cos(θ0) and sin(θ) = sin(θ0). y 1 θ0 y 1 θ0 P (cos(θ0), sin(θ0)) 1 x θ Q(cos(−θ0), sin(−θ0)) 1 x −θ0 We now consider the angles −θ and −θ0. Since θ is cot
erminal with θ0, there is some integer k so that θ = θ0 + 2π · k. Therefore, −θ = −θ0 − 2π · k = −θ0 + 2π · (−k). Since k is an integer, so is (−k), which means −θ is coterminal with −θ0. Hence, cos(−θ) = cos(−θ0) and sin(−θ) = sin(−θ0). Let P and Q denote the points on the terminal sides of θ0 and −θ0, respectively, which lie on the Unit Circle. By deﬁnition, the coordinates of P are (cos(θ0), sin(θ0)) and the coordinates of Q are (cos(−θ0), sin(−θ0)). Since θ0 and −θ0 sweep out congruent central sectors of the Unit Circle, it 1As mentioned at the end of Section 10.2, properties of the circular functions when thought of as functions of angles in radian measure hold equally well if we view these functions as functions of real numbers. Not surprisingly, the Even / Odd properties of the circular functions are so named because they identify cosine and secant as even functions, while the remaining four circular functions are odd. (See Section 1.6.) 10.4 Trigonometric Identities 771 follows that the points P and Q are symmetric about the x-axis. Thus, cos(−θ0) = cos(θ0) and sin(−θ0) = − sin(θ0). Since the cosines and sines of θ0 and −θ0 are the same as those for θ and −θ, respectively, we get cos(−θ) = cos(θ) and sin(−θ) = − sin(θ), as required. The Even / Odd Identities are readily demonstrated using any of the ‘common angles’ noted in Section 10.2. Their true utility, however, lies not in computation, but in simplifying expressions involving the circular functions. In fact, our next batch of identities makes heavy use of the Even / Odd Identities. Theorem 10.13. Sum and Diﬀerence Identities for Cosine: For all angles α and β, cos(α + β) = cos(α) cos(β) − sin(α) sin(β) cos(α − β) = cos(α) cos(β) + sin(α) sin(β) We ﬁrst prove the result for diﬀerences. As in the proof of the Even / Odd Identities, we can reduce the proof for general angles α and β to angles α0 and β0, coterminal with α and β, respectively, each of which measure between 0 and 2π radians. Since α and α0 are coterminal, as are β and β0, it follows that α − β is coterminal with α0 − β0. Consider the case below where α0 ≥ β0. y P (cos(α0), sin(α0)) α0 − β0 Q(cos(β0), sin(β0)) y 1 A(cos(α0 − β0), sin(α0 − β0)) α0 β0 α0 − β0 O 1 x O B(1, 0) x Since the angles P OQ and AOB are congruent, the distance between P and Q is equal to the distance between A and B.2 The distance formula, Equation 1.1, yields (cos(α0) − cos(β0))2 + (sin(α0) − sin(β0))2 = (cos(α0 − β0) − 1)2 + (sin(α0 − β0) − 0)2 Squaring both sides, we expand the left hand side of this equation as (cos(α0) − cos(β0))2 + (sin(α0) − sin(β0))2 = cos2(α0) − 2 cos(α0) cos(β0) + cos2(β0) + sin2(α0) − 2 sin(α0) sin(β0) + sin2(β0) = cos2(α0) + sin2(α0) + cos2(β0) + sin2(β0) −2 cos(α0) cos(β0) − 2 sin(α0) sin(β0) 2In the picture we’ve drawn, the triangles P OQ and AOB are congruent, which is even better. However, α0 − β0 could be 0 or it could be π, neither of which makes a triangle. It could also be larger than π, which makes a triangle, just not the one we’ve drawn. You should think about those three cases. 772 Foundations of Trigonometry From the Pythagorean Identities, cos2(α0) + sin2(α0) = 1 and cos2(β0) + sin2(β0) = 1, so (cos(α0) − cos(β0))2 + (sin(α0) − sin(β0))2 = 2 − 2 cos(α0) cos(β0) − 2 sin(α0) sin(β0) Turning our attention to the right hand side of our equation, we ﬁnd (cos(α0 − β0) − 1)2 + (sin(α0 − β0) − 0)2 = cos2(α0 − β0) − 2 cos(α0 − β0) + 1 + sin2(α0 − β0) = 1 + cos2(α0 − β0) + sin2(α0 − β0) − 2 cos(α0 − β0) Once again, we simplify cos2(α0 − β0) + sin2(α0 − β0) = 1, so that (cos(α0 − β0) − 1)2 + (sin(α0 − β0) − 0)2 = 2 − 2 cos(α0 − β0) Putting it all together, we get 2 − 2 cos(α0) cos(β0) − 2 sin(α0) sin(β0) = 2 − 2 cos(α0 − β0), which simpliﬁes to: cos(α0 − β0) = cos(α0) cos(β0) + sin(α0) sin(β0). Since α and α0, β and β0 and α − β and α0 − β0 are all coterminal pairs of angles, we have cos(α − β) = cos(α) cos(β) + sin(α) sin(β). For the case where α0 ≤ β0, we can apply the above argument to the angle β0 − α0 to obtain the identity cos(β0 − α0) = cos(β0) cos(α0) + sin(β0) sin(α0). Applying the Even Identity of cosine, we get cos(β0 − α0) = cos(−(α0 − β0)) = cos(α0 − β0), and we get the identity in this case, too. To get the sum identity for cosine, we use the diﬀerence formula along with the Even/Odd Identities cos(α + β) = cos(α − (−β)) = cos(α) cos(−β) + sin(α) sin(−β) = cos(α) cos(β) − sin(α) sin(β) We put these newfound identities to good use in the following example. Example 10.4.1. 1. Find the exact value of cos (15◦). 2. Verify the identity: cos π 2 − θ = sin(θ). Solution. 1. In order to use Theorem 10.13 to ﬁnd cos (15◦), we need to write 15◦ as a sum or diﬀerence of angles whose cosines and sines we know. One way to do so is to write 15◦ = 45◦ − 30◦. cos (15◦) = cos (45◦ − 30◦) = cos (45◦) cos (30◦) + sin (45◦) sin (30◦) √ √ √ 10.4 Trigonometric Identities 773 2. In a straightforward application of Theorem 10.13, we ﬁnd cos − θ π 2 = cos π 2 π 2 = (0) (cos(θ)) + (1) (sin(θ)) cos (θ) + sin sin (θ) = sin(θ) The identity veriﬁed in Example 10.4.1, namely, cos π 2 − θ = sin(θ), is the ﬁrst of the celebrated ‘cofunction’ identities. These identities were ﬁrst hinted at in Exercise 74 in Section 10.2. From sin(θ) = cos π 2 − θ, we get: π 2 which says, in words, that the ‘co’sine of an angle is the sine of its ‘co’mplement. Now that these identities have been established for cosine and sine, the remaining circular functions follow suit. The remaining proofs are left as exercises. = cos(θ), π 2 π 2 = cos − θ − θ sin − Theorem 10.14. Cofunction Identities: For all applicable angles θ, cos sin − θ − θ π 2 π 2 = sin(θ) = cos(θ) sec csc − θ − θ π 2 π 2 = csc(θ) = sec(θ) tan cot − θ − θ π 2 π 2 = cot(θ) = tan(θ) With the Cofunction Identities in place, we are now in the position to derive the sum and diﬀerence formulas for sine. To derive the sum formula for sine, we convert to cosines using a cofunction identity, then expand using the diﬀerence formula for cosine sin(α + β) = cos = cos − (α + β = cos π 2 = sin(α) cos(β) + cos(α) sin(β) cos(β) + sin − α − α sin(β) We can derive the diﬀerence formula for sine by rewriting sin(α − β) as sin(α + (−β)) and using the sum formula and the Even / Odd Identities. Again, we leave the details to the reader. Theorem 10.15. Sum and Diﬀerence Identities for Sine: For all angles α and β, sin(α + β) = sin(α) cos(β) + cos(α) sin(β) sin(α − β) = sin(α) cos(β) − cos(α) sin(β) 774 Example 10.4.2. 1. Find the exact value of sin 19π 12 Foundations of Trigonometry 2. If α is a Quadrant II angle with sin(α) = 5 13 , and β is a Quadrant III angle with tan(β) = 2, ﬁnd sin(α − β). 3. Derive a formula for tan(α + β) in terms of tan(α) and tan(β). Solution. 1. As in Example 10.4.1, we need to write the angle 19π 12 as a sum or diﬀerence of common angles. The denominator of 12 suggests a combination of angles with denominators 3 and 4. One such combination is 19π 4 . Applying Theorem 10.15, we get 12 = 4π 3 + π 19π 12 sin = sin 4π 3 4π 3 √ 3 2 π 4 + cos √ π 4 2 2 √ 2 6 − 4 = sin − √ − = = π 4 + cos + − 1 2 4π 3 √ 2 2 sin 13 2 = 1, or cos(α) = ± 12 2. In order to ﬁnd sin(α − β) using Theorem 10.15, we need to ﬁnd cos(α) and both cos(β) and sin(β). To ﬁnd cos(α), we use the Pythagorean Identity cos2(α) + sin2(α) = 1. Since 13 , we have cos2(α) + 5 sin(α) = 5 13 . Since α is a Quadrant II angle, cos(α) = − 12 13 . We now set about ﬁnding cos(β) and sin(β). We have several ways to proceed, but the Pythagorean Identity 1 + tan2(β) = sec2(β) is a quick way to get sec(β), and hence, √ cos(β). With tan(β) = 2, we get 1 + 22 = sec2(β) so that sec(β) = ± 5. Since β is a √ 5 = − Quadrant III angle, we choose sec(β) = − 5 . We now need to determine sin(β). We could use The Pythagorean Identity cos2(β) + sin2(β) = 1, but we opt instead to use a quotient identity. From tan(β) = sin(β) cos(β) , we have sin(β) = tan(β) cos(β) √ 5 5 . We now have all the pieces needed to ﬁnd sin(α − β): so we get sin(β) = (2) 5 so cos(β) = 1 sec(β sin(α − β) = sin(α) cos(β) − cos(α) sin(β) √ 5 5 − − 12 13 √ 2 5 5 − = 5 13 29 − 5 √ = − 65 10.4 Trigonometric Identities 775 3. We can start expanding tan(α + β) using a quotient identity and our sum formulas tan(α + β) = = sin(α + β) cos(α + β) sin(α) cos(β) + cos(α) sin(β) cos(α) cos(β) − sin(α) sin(β) Since tan(α) = sin(α) cos(α) and tan(β) = sin(β) denominator by cos(α) cos(β) we will have what we want cos(β) , it looks as though if we divide both numerator and tan(α + β) = sin(α) cos(β) + cos(α) sin(β) cos(α) cos(β) − sin(α) sin(β) · 1 cos(α) cos(β) 1 cos(α) cos(β) sin(α) cos(β) cos(α) cos(β) cos(α) cos(β) cos(α) cos(β) sin(α) cos(β) cos(α) cos(β) cos(α) cos(β) cos(α) cos(β) + − + − cos(α) sin(β) cos(α) cos(β) sin(α) sin(β) cos(α) cos(β) cos(α) sin(β) cos(α) cos(β) sin(α) sin(β) cos(α) cos(β) tan(α) + tan(β) 1 − tan(α) tan(β) = = = Naturally, this formula is limited to those cases where all of the tangents are deﬁned. The formula developed in Exercise 10.4.2 for tan(α +β) can be used to ﬁnd a formula for tan(α −β) by rewriting the diﬀerence as a sum, tan(α+(−β)), and the reader is encouraged to ﬁll in the details. Below we summarize all of the sum and diﬀerence formulas for cosine, sine and tangent. Theorem 10.16. Sum and Diﬀerence Identities: For all applicable angles α and β, cos(α ± β) = cos(α) cos(β) ∓ sin(α) sin(β) sin(α ± β) = sin(α) cos(β) ± cos(α) sin(β) tan(α ± β) = tan(α) ± tan(β) 1 ∓ tan(α) tan(β) In the statement of Theorem 10.16, we have combined the cases for the sum ‘+’ and diﬀerence ‘−’ of angles into one formula. The convention here is that if you want the formula for the sum ‘+’ of 776 Foundations of Trigonometry two angles, you use the top sign in the formula; for the diﬀerence, ‘−’, use the bottom sign. For example, tan(α − β) = tan(α) − tan(β) 1 + tan(α) tan(β) If we specialize the sum formulas in Theorem 10.16 to the case when α = β, we obtain the following ‘Double Angle’ Identities. Theorem 10.17. Double Angle Identities: For all a
pplicable angles θ, cos(2θ) =    cos2(θ) − sin2(θ) 2 cos2(θ) − 1 1 − 2 sin2(θ) sin(2θ) = 2 sin(θ) cos(θ) tan(2θ) = 2 tan(θ) 1 − tan2(θ) The three diﬀerent forms for cos(2θ) can be explained by our ability to ‘exchange’ squares of cosine and sine via the Pythagorean Identity cos2(θ) + sin2(θ) = 1 and we leave the details to the reader. It is interesting to note that to determine the value of cos(2θ), only one piece of information is required: either cos(θ) or sin(θ). To determine sin(2θ), however, it appears that we must know both sin(θ) and cos(θ). In the next example, we show how we can ﬁnd sin(2θ) knowing just one piece of information, namely tan(θ). Example 10.4.3. 1. Suppose P (−3, 4) lies on the terminal side of θ when θ is plotted in standard position. Find cos(2θ) and sin(2θ) and determine the quadrant in which the terminal side of the angle 2θ lies when it is plotted in standard position. 2. If sin(θ) = x for − π 2 ≤ θ ≤ π 2 , ﬁnd an expression for sin(2θ) in terms of x. 3. Verify the identity: sin(2θ) = 2 tan(θ) 1 + tan2(θ) . 4. Express cos(3θ) as a polynomial in terms of cos(θ). Solution. 1. Using Theorem 10.3 from Section 10.2 with x = −3 and y = 4, we ﬁnd r = x2 + y2 = 5. Hence, cos(θ) = − 3 5 . Applying Theorem 10.17, we get cos(2θ) = cos2(θ) − sin2(θ) = − 3 25 , and sin(2θ) = 2 sin(θ) cos(θ) = 2 4 2 − 4 25 . Since both 5 5 cosine and sine of 2θ are negative, the terminal side of 2θ, when plotted in standard position, lies in Quadrant III. 5 and sin(θ) = 4 = − 7 = − 24 − 3 5 2 5 10.4 Trigonometric Identities 777 2. If your ﬁrst reaction to ‘sin(θ) = x’ is ‘No it’s not, cos(θ) = x!’ then you have indeed learned something, and we take comfort in that. However, context is everything. Here, ‘x’ is just a variable - it does not necessarily represent the x-coordinate of the point on The Unit Circle which lies on the terminal side of θ, assuming θ is drawn in standard position. Here, x represents the quantity sin(θ), and what we wish to know is how to express sin(2θ) in terms of x. We will see more of this kind of thing in Section 10.6, and, as usual, this is something we need for Calculus. Since sin(2θ) = 2 sin(θ) cos(θ), we need to write cos(θ) in terms of x to ﬁnish the problem. We substitute x = sin(θ) into the Pythagorean Identity, cos2(θ) + sin2(θ) = 1, 2 ≤ θ ≤ π to get cos2(θ) + x2 = 1, or cos(θ) = ± 2 , cos(θ) ≥ 0, and thus √ 1 − x2. cos(θ) = 1 − x2. Our ﬁnal answer is sin(2θ) = 2 sin(θ) cos(θ) = 2x 1 − x2. Since − π √ √ 3. We start with the right hand side of the identity and note that 1 + tan2(θ) = sec2(θ). From this point, we use the Reciprocal and Quotient Identities to rewrite tan(θ) and sec(θ) in terms of cos(θ) and sin(θ): 2 tan(θ) 1 + tan2(θ) = 2 tan(θ) sec2(θ) = = 2 sin(θ) cos(θ) 2 sin(θ) cos(θ) 1 cos2(θ) = 2 sin(θ) cos(θ) cos2(θ) cos(θ) cos(θ) = 2 sin(θ) cos(θ) = sin(2θ) 4. In Theorem 10.17, one of the formulas for cos(2θ), namely cos(2θ) = 2 cos2(θ) − 1, expresses cos(2θ) as a polynomial in terms of cos(θ). We are now asked to ﬁnd such an identity for cos(3θ). Using the sum formula for cosine, we begin with cos(3θ) = cos(2θ + θ) = cos(2θ) cos(θ) − sin(2θ) sin(θ) Our ultimate goal is to express the right hand side in terms of cos(θ) only. We substitute cos(2θ) = 2 cos2(θ) − 1 and sin(2θ) = 2 sin(θ) cos(θ) which yields cos(3θ) = cos(2θ) cos(θ) − sin(2θ) sin(θ) = 2 cos2(θ) − 1 cos(θ) − (2 sin(θ) cos(θ)) sin(θ) = 2 cos3(θ) − cos(θ) − 2 sin2(θ) cos(θ) Finally, we exchange sin2(θ) for 1 − cos2(θ) courtesy of the Pythagorean Identity, and get cos(3θ) = 2 cos3(θ) − cos(θ) − 2 sin2(θ) cos(θ) = 2 cos3(θ) − cos(θ) − 2 1 − cos2(θ) cos(θ) = 2 cos3(θ) − cos(θ) − 2 cos(θ) + 2 cos3(θ) = 4 cos3(θ) − 3 cos(θ) and we are done. 778 Foundations of Trigonometry In the last problem in Example 10.4.3, we saw how we could rewrite cos(3θ) as sums of powers of cos(θ). In Calculus, we have occasion to do the reverse; that is, reduce the power of cosine and sine. Solving the identity cos(2θ) = 2 cos2(θ) − 1 for cos2(θ) and the identity cos(2θ) = 1 − 2 sin2(θ) for sin2(θ) results in the aptly-named ‘Power Reduction’ formulas below. Theorem 10.18. Power Reduction Formulas: For all angles θ, cos2(θ) = sin2(θ) = 1 + cos(2θ) 2 1 − cos(2θ) 2 Example 10.4.4. Rewrite sin2(θ) cos2(θ) as a sum and diﬀerence of cosines to the ﬁrst power. Solution. We begin with a straightforward application of Theorem 10.18 1 − cos(2θ) 2 1 − cos2(2θ) 1 + cos(2θ) 2 sin2(θ) cos2(θ cos2(2θ) Next, we apply the power reduction formula to cos2(2θ) to ﬁnish the reduction sin2(θ) cos2(θ cos2(2θ) 1 + cos(2(2θ)) 2 − 1 8 cos(4θ) cos(4θ) Another application of the Power Reduction Formulas is the Half Angle Formulas. To start, we apply the Power Reduction Formula to cos2 θ 2 cos2 θ 2 We can obtain a formula for cos θ by extracting square roots. In a similar fashion, we may obtain 2 a half angle formula for sine, and by using a quotient formula, obtain a half angle formula for tangent. We summarize these formulas below. 1 + cos(θ) 2 1 + cos 2 θ 2 2 = = . 10.4 Trigonometric Identities 779 Theorem 10.19. Half Angle Formulas: For all applicable angles θ, cos sin θ 2 θ 2 = ± 1 + cos(θ) 2 = ± 1 − cos(θ) 2 tan = ± θ 2 1 − cos(θ) 1 + cos(θ) where the choice of ± depends on the quadrant in which the terminal side of θ 2 lies. Example 10.4.5. 1. Use a half angle formula to ﬁnd the exact value of cos (15◦). 2. Suppose −π ≤ θ ≤ 0 with cos(θ) = − 3 5 . Find sin θ 2 . 3. Use the identity given in number 3 of Example 10.4.3 to derive the identity tan θ 2 = sin(θ) 1 + cos(θ) Solution. 1. To use the half angle formula, we note that 15◦ = 30◦ 2 and since 15◦ is a Quadrant I angle, its cosine is positive. Thus we have cos (15◦) = + 1 + cos (30◦) = = Back in Example 10.4.1, we found cos (15◦) by using the diﬀerence formula for cosine. In that case, we determined cos (15◦) = . The reader is encouraged to prove that these two expressions are equal. 6+ 4 √ √ 2 2. If −π ≤ θ ≤ 0, then − π < 0. Theorem 10.19 gives 2 sin 2 ≤ 0, which means sin θ 2 ≤ θ θ 1 − cos (θ 10 √ 2 5 5 780 Foundations of Trigonometry 3. Instead of our usual approach to verifying identities, namely starting with one side of the equation and trying to transform it into the other, we will start with the identity we proved in number 3 of Example 10.4.3 and manipulate it into the identity we are asked to prove. The identity we are asked to start with is sin(2θ) = 2 tan(θ) 1+tan2(θ) . If we are to use this to derive an identity for tan θ , it seems reasonable to proceed by replacing each occurrence of θ with θ 2 2 2 tan θ 2 1 + tan2 θ 2 2 tan θ 2 1 + tan2 θ 2 sin 2 θ 2 sin(θ) = = We now have the sin(θ) we need, but we somehow need to get a factor of 1 + cos(θ) involved. . We continue to manipulate our To get cosines involved, recall that 1 + tan2 θ 2 given identity by converting secants to cosines and using a power reduction formula = sec2 θ 2 sin(θ) = 2 tan θ 2 1 + tan2 θ 2 2 tan θ 2 sec2 θ 2 sin(θ) = 2 tan θ 2 sin(θ) = sin(θ) = 2 tan θ 2 sin(θ) = tan θ 2 θ sin(θ) 1 + cos(θ) 2 = tan cos2 θ 2 1 + cos 2 θ 2 2 (1 + cos(θ)) Our next batch of identities, the Product to Sum Formulas,3 are easily veriﬁed by expanding each of the right hand sides in accordance with Theorem 10.16 and as you should expect by now we leave the details as exercises. They are of particular use in Calculus, and we list them here for reference. Theorem 10.20. Product to Sum Formulas: For all angles α and β, cos(α) cos(β) = 1 2 [cos(α − β) + cos(α + β)] sin(α) sin(β) = 1 2 [cos(α − β) − cos(α + β)] sin(α) cos(β) = 1 2 [sin(α − β) + sin(α + β)] 3These are also known as the Prosthaphaeresis Formulas and have a rich history. The authors recommend that you conduct some research on them as your schedule allows. 10.4 Trigonometric Identities 781 Related to the Product to Sum Formulas are the Sum to Product Formulas, which we will have need of in Section 10.7. These are easily veriﬁed using the Product to Sum Formulas, and as such, their proofs are left as exercises. Theorem 10.21. Sum to Product Formulas: For all angles α and β, cos(α) + cos(β) = 2 cos cos(α) − cos(β) = −2 sin sin(α) ± sin(β) = 2 sin Example 10.4.6 cos sin cos 1. Write cos(2θ) cos(6θ) as a sum. 2. Write sin(θ) − sin(3θ) as a product. Solution. 1. Identifying α = 2θ and β = 6θ, we ﬁnd cos(2θ) cos(6θ) = 1 = 1 = 1 2 [cos(2θ − 6θ) + cos(2θ + 6θ)] 2 cos(−4θ) + 1 2 cos(4θ) + 1 2 cos(8θ) 2 cos(8θ), where the last equality is courtesy of the even identity for cosine, cos(−4θ) = cos(4θ). 2. Identifying α = θ and β = 3θ yields sin(θ) − sin(3θ) = 2 sin θ − 3θ 2 cos θ + 3θ 2 = 2 sin (−θ) cos (2θ) = −2 sin (θ) cos (2θ) , where the last equality is courtesy of the odd identity for sine, sin(−θ) = − sin(θ). The reader is reminded that all of the identities presented in this section which regard the circular functions as functions of angles (in radian measure) apply equally well to the circular (trigonometric) functions regarded as functions of real numbers. In Exercises 38 - 43 in Section 10.5, we see how some of these identities manifest themselves geometrically as we study the graphs of the these functions. In the upcoming Exercises, however, you need to do all of your work analytically without graphs. 782 Foundations of Trigonometry 10.4.1 Exercises In Exercises 1 - 6, use the Even / Odd Identities to verify the identity. Assume all quantities are deﬁned. 1. sin(3π − 2θ) = − sin(2θ − 3π) 2. cos − π 4 − 5t = cos 5t + π 4 3. tan(−t2 + 1) = − tan(t2 − 1) 4. csc(−θ − 5) = − csc(θ + 5) 5. sec(−6t) = sec(6t) 6. cot(9 − 7θ) = − cot(7θ − 9) In Exercises 7 - 21, use the Sum and Diﬀerence Identities to ﬁnd the exact value. You may have need of the Quotient, Reciprocal or Even / Odd Identities as well. 7. cos(75◦) 10. csc(195◦) 8. sec(165◦) 11. cot(255◦) 13. cos 16. cos 19. cot 13π 12 7π 12 11π 12 14. sin 17. tan 20. csc 11π 12 17π 12 5π 12 9. sin(105◦) 12. tan(375◦) 15. tan 13π 12 18. sin π 12 21. sec − π 12 22. If α is a Quadrant IV angle with cos(α) = √ 5 5 , and sin(β) = √ 10 10 , where π 2 < β < π
, ﬁnd (a) cos(α + β) (b) sin(α + β) (c) tan(α + β) (d) cos(α − β) (e) sin(α − β) (f) tan(α − β) 23. If csc(α) = 3, where 0 < α < π 2 , and β is a Quadrant II angle with tan(β) = −7, ﬁnd (a) cos(α + β) (b) sin(α + β) (c) tan(α + β) (d) cos(α − β) (e) sin(α − β) (f) tan(α − β) 24. If sin(α) = 3 5 , where 0 < α < π 2 , and cos(β) = 12 13 where 3π 2 < β < 2π, ﬁnd (a) sin(α + β) (b) cos(α − β) (c) tan(α − β) 10.4 Trigonometric Identities 783 25. If sec(α) = − 5 3 , where π 2 < α < π, and tan(β) = 24 7 , where π < β < 3π 2 , ﬁnd (a) csc(α − β) (b) sec(α + β) (c) cot(α + β) In Exercises 26 - 38, verify the identity. 26. cos(θ − π) = − cos(θ) 27. sin(π − θ) = sin(θ) 28. tan θ + π 2 = − cot(θ) 29. sin(α + β) + sin(α − β) = 2 sin(α) cos(β) 30. sin(α + β) − sin(α − β) = 2 cos(α) sin(β) 31. cos(α + β) + cos(α − β) = 2 cos(α) cos(β) 32. cos(α + β) − cos(α − β) = −2 sin(α) sin(β) 33. sin(α + β) sin(α − β) = 1 + cot(α) tan(β) 1 − cot(α) tan(β) 34. 36. 37. cos(α + β) cos(α − β) = 1 − tan(α) tan(β) 1 + tan(α) tan(β) 35. tan(α + β) tan(α − β) = sin(α) cos(α) + sin(β) cos(β) sin(α) cos(α) − sin(β) cos(β) sin(t + h) − sin(t) h = cos(t) sin(h) h + sin(t) cos(h) − 1 h cos(t + h) − cos(t) h = cos(t) cos(h) − 1 h − sin(t) sin(h) h 38. tan(t + h) − tan(t) h = tan(h) h sec2(t) 1 − tan(t) tan(h) In Exercises 39 - 48, use the Half Angle Formulas to ﬁnd the exact value. You may have need of the Quotient, Reciprocal or Even / Odd Identities as well. 39. cos(75◦) (compare with Exercise 7) 40. sin(105◦) (compare with Exercise 9) 41. cos(67.5◦) 43. tan(112.5◦) 45. sin 47. sin π 12 5π 8 42. sin(157.5◦) 44. cos (compare with Exercise 16) 7π 12 π 8 7π 8 (compare with Exercise 18) 46. cos 48. tan 784 Foundations of Trigonometry In Exercises 49 - 58, use the given information about θ to ﬁnd the exact values of sin(2θ) θ 2 sin cos(2θ) θ 2 cos tan(2θ) θ 2 tan 49. sin(θ) = − 7 25 where 3π 2 < θ < 2π 50. cos(θ) = 28 53 where 0 < θ < π 2 51. tan(θ) = 53. cos(θ) = 55. cos(θ) = 57. sec(θ) = 12 5 3 5 12 13 √ where π < θ < 3π 2 where 0 < θ < π 2 52. csc(θ) = 4 where π 2 < θ < π 54. sin(θ) = − 4 5 where π < θ < 3π 2 where 5 where 3π 2 3π 2 < θ < 2π 56. sin(θ) = 5 13 where < θ < 2π 58. tan(θ) = −2 where In Exercises 59 - 73, verify the identity. Assume all quantities are deﬁned. 59. (cos(θ) + sin(θ))2 = 1 + sin(2θ) 60. (cos(θ) − sin(θ))2 = 1 − sin(2θ) 61. tan(2θ) = 1 1 − tan(θ) − 1 1 + tan(θ) 62. csc(2θ) = cot(θ) + tan(θ) 2 63. 8 sin4(θ) = cos(4θ) − 4 cos(2θ) + 3 64. 8 cos4(θ) = cos(4θ) + 4 cos(2θ) + 3 65. sin(3θ) = 3 sin(θ) − 4 sin3(θ) 66. sin(4θ) = 4 sin(θ) cos3(θ) − 4 sin3(θ) cos(θ) 67. 32 sin2(θ) cos4(θ) = 2 + cos(2θ) − 2 cos(4θ) − cos(6θ) 68. 32 sin4(θ) cos2(θ) = 2 − cos(2θ) − 2 cos(4θ) + cos(6θ) 69. cos(4θ) = 8 cos4(θ) − 8 cos2(θ) + 1 70. cos(8θ) = 128 cos8(θ) − 256 cos6(θ) + 160 cos4(θ) − 32 cos2(θ) + 1 (HINT: Use the result to 69.) 71. sec(2θ) = cos(θ) cos(θ) + sin(θ) + sin(θ) cos(θ) − sin(θ) 72. 73. 1 cos(θ) − sin(θ) 1 cos(θ) − sin(θ) + − 1 cos(θ) + sin(θ) 1 cos(θ) + sin(θ) = = 2 cos(θ) cos(2θ) 2 sin(θ) cos(2θ) 10.4 Trigonometric Identities 785 In Exercises 74 - 79, write the given product as a sum. You may need to use an Even/Odd Identity. 74. cos(3θ) cos(5θ) 75. sin(2θ) sin(7θ) 76. sin(9θ) cos(θ) 77. cos(2θ) cos(6θ) 78. sin(3θ) sin(2θ) 79. cos(θ) sin(3θ) In Exercises 80 - 85, write the given sum as a product. You may need to use an Even/Odd or Cofunction Identity. 80. cos(3θ) + cos(5θ) 81. sin(2θ) − sin(7θ) 82. cos(5θ) − cos(6θ) 83. sin(9θ) − sin(−θ) 84. sin(θ) + cos(θ) 85. cos(θ) − sin(θ) 86. Suppose θ is a Quadrant I angle with sin(θ) = x. Verify the following formulas (a) cos(θ) = √ 1 − x2 (b) sin(2θ) = 2x √ 1 − x2 (c) cos(2θ) = 1 − 2x2 87. Discuss with your classmates how each of the formulas, if any, in Exercise 86 change if we change assume θ is a Quadrant II, III, or IV angle. 88. Suppose θ is a Quadrant I angle with tan(θ) = x. Verify the following formulas (a) cos(θ) = √ 1 x2 + 1 (c) sin(2θ) = 2x x2 + 1 (b) sin(θ) = √ x x2 + 1 (d) cos(2θ) = 1 − x2 x2 + 1 89. Discuss with your classmates how each of the formulas, if any, in Exercise 88 change if we change assume θ is a Quadrant II, III, or IV angle. for − for − 90. If sin(θ) = 91. If tan(θ) = 92. If sec(θ) = x 2 x 7 x 4 , ﬁnd an expression for cos(2θ) in terms of x. , ﬁnd an expression for sin(2θ) in terms of x. for 0 < θ < π 2 , ﬁnd an expression for ln \| sec(θ) + tan(θ)\| in terms of x. 93. Show that cos2(θ) − sin2(θ) = 2 cos2(θ) − 1 = 1 − 2 sin2(θ) for all θ. 94. Let θ be a Quadrant III angle with cos(θ) = − . Show that this is not enough information to 1 5 by ﬁrst assuming 3π < θ < 7π 2 and then assuming π < θ < 3π 2 determine the sign of sin θ 2 and computing sin in both cases. θ 2 786 Foundations of Trigonometry 95. Without using your calculator, show that 96. In part 4 of Example 10.4.3, we wrote cos(3θ) as a polynomial in terms of cos(θ). In Exercise 69, we had you verify an identity which expresses cos(4θ) as a polynomial in terms of cos(θ). Can you ﬁnd a polynomial in terms of cos(θ) for cos(5θ)? cos(6θ)? Can you ﬁnd a pattern so that cos(nθ) could be written as a polynomial in cosine for any natural number n? 97. In Exercise 65, we has you verify an identity which expresses sin(3θ) as a polynomial in terms of sin(θ). Can you do the same for sin(5θ)? What about for sin(4θ)? If not, what goes wrong? 98. Verify the Even / Odd Identities for tangent, secant, cosecant and cotangent. 99. Verify the Cofunction Identities for tangent, secant, cosecant and cotangent. 100. Verify the Diﬀerence Identities for sine and tangent. 101. Verify the Product to Sum Identities. 102. Verify the Sum to Product Identities. 10.4 Trigonometric Identities 787 = 2 − √ 3 12. tan(375◦) = 10.4.2 Answers 7. cos(75◦) = √ 2 √ 6 − 4 9. sin(105◦) = √ 2 √ 6 + 4 11. cot(255◦) = √ √ 13. cos 15. tan 17. tan 19. cot 21. sec 13π 12 13π 12 17π 12 11π 12 − π 12 = 2 + √ 3 = −(2 + √ 3) √ √ 2 6 − = 18. sin 20. csc π 12 5π 12 8. sec(165◦) = − √ 4 2 + √ 6 √ √ 6 2 − = 10. csc(195◦) = √ √ √ = −( √ 6 + √ 14. sin 11π 12 = 16. cos √ 7π 12 = √ − = 22. (a) cos(α + β) = − √ 2 10 (b) sin(α + β) = √ 2 7 10 √ 2 2 (c) tan(α + β) = −7 (d) cos(α − β) = − (e) sin(α − β) = √ 2 2 23. (a) cos(α + β) = − √ 4 + 7 30 2 (c) tan(α + β) = (e) sin(α − β28 + 4 + 7 28 + 30 (f) tan(α − β) = −1 63 − 100 √ 2 41 = (b) sin(α + β) = (d) cos(α − β) = (f) tan(α − β) = √ 2 √ 28 − 30 2 −4 + 7 30 √ √ 28 + 4 − 7 2 2 = − 24. (a) sin(α + β) = 16 65 (b) cos(α − β) = 33 65 (c) tan(α − β) = 63 + 100 √ 2 41 56 33 788 Foundations of Trigonometry 25. (a) csc(α − β) = − 5 4 (b) sec(α + β) = 125 117 (c) cot(α + β) = 117 44 39. cos(75◦) = 41. cos(67.5◦) = √ 40. sin(105◦) = 42. sin(157.5◦) = √ 1 − √ 2 44. cos 7π 12 = − √ 3 2 − 2 46. cos 48. tan 2 + 2 = − π 8 = 7π 43. tan(112.5◦) = − 45. sin 47. sin π 12 5π 49. sin(2θ) = − √ sin θ 2 = 50. sin(2θ) = sin θ 2 = 51. sin(2θ) = sin θ 2 = 52. sin(2θ) = − 336 625 2 10 2520 2809 √ 5 106 106 120 169 √ 3 13 13 √ 15 8 cos(2θ) = cos θ 2 = − cos(2θ) = − cos θ 2 = 9 527 625 √ 2 7 10 1241 2809 √ 106 106 119 169 √ 2 13 13 cos(2θ) = − cos θ 2 = − cos(2θ) = 7 8 sin θ 2 = √ 15 8 + 2 4 cos θ 2 = √ 15 8 − 2 4 tan(2θ) = − tan θ 2 = − tan(2θ) = − tan θ 2 = 5 9 tan(2θ) = − tan θ 2 = − 336 527 1 7 2520 1241 120 119 3 2 √ 15 7 tan(2θ) = − tan θ 2 tan √ 15 √ √ 15 15 53. sin(2θ) = sin θ 2 = 24 25 √ 5 5 cos(2θ) = − cos θ 2 = 2 7 25 √ 5 5 24 7 tan(2θ) = − tan θ 2 = 1 2 10.4 Trigonometric Identities 789 54. sin(2θ) = sin θ 2 = 24 25 √ 2 5 sin(2θ) = − √ sin θ 2 = sin(2θ) = − sin θ 2 = 5 5 120 169 26 26 120 169 √ 26 26 4 5 55. 56. 57. cos(2θ) = − cos θ 2 = − cos(2θ) = cos θ 2 = − 26 26 cos(2θ) = cos θ 2 = 7 25 √ 5 5 119 169 5 √ 119 169 √ 26 26 3 5 tan(2θ) = − 24 7 = −2 tan θ 2 120 119 1 5 120 119 tan(2θ) = − tan θ 2 = − tan(2θ) = − tan θ 2 = 5 tan(2θ) = 4 3 tan θ 2 = − sin(2θ) = − cos(2θ) = − sin θ 2 = 50 − 10 10 √ 5 cos θ 2 = − √ 5 50 + 10 10 58. sin(2θ) = − 4 5 cos(2θ) = − 3 5 sin θ 2 = √ 5 50 + 10 10 cos θ 2 = 50 − 10 10 √ tan θ 2 = tan(2θ) = 5 tan θ 2 = tan 10 10 74. 77. cos(2θ) + cos(8θ) 2 cos(4θ) + cos(8θ) 2 80. 2 cos(4θ) cos(θ) 83. 2 cos(4θ) sin(5θ) 90. 1 − x2 2 75. 78. cos(5θ) − cos(9θ) 2 cos(θ) − cos(5θ) 2 9 2 sin θ 81. −2 cos θ 5 2 76. 79. sin(8θ) + sin(10θ) 2 sin(2θ) + sin(4θ) 2 11 2 sin θ 1 2 θ 82. 2 sin √ θ − 2 cos 84. π 4 91. 14x x2 + 49 √ 85. − θ − 2 sin π 4 √ 92. ln \|x + x2 + 16\| − ln(4) 790 Foundations of Trigonometry 10.5 Graphs of the Trigonometric Functions In this section, we return to our discussion of the circular (trigonometric) functions as functions of real numbers and pick up where we left oﬀ in Sections 10.2.1 and 10.3.1. As usual, we begin our study with the functions f (t) = cos(t) and g(t) = sin(t). 10.5.1 Graphs of the Cosine and Sine Functions From Theorem 10.5 in Section 10.2.1, we know that the domain of f (t) = cos(t) and of g(t) = sin(t) is all real numbers, (−∞, ∞), and the range of both functions is [−1, 1]. The Even / Odd Identities in Theorem 10.12 tell us cos(−t) = cos(t) for all real numbers t and sin(−t) = − sin(t) for all real numbers t. This means f (t) = cos(t) is an even function, while g(t) = sin(t) is an odd function.1 Another important property of these functions is that for coterminal angles α and β, cos(α) = cos(β) and sin(α) = sin(β). Said diﬀerently, cos(t+2πk) = cos(t) and sin(t+2πk) = sin(t) for all real numbers t and any integer k. This last property is given a special name. Deﬁnition 10.3. Periodic Functions: A function f is said to be periodic if there is a real number c so that f (t + c) = f (t) for all real numbers t in the domain of f . The smallest positive number p for which f (t + p) = f (t) for all real numbers t in the domain of f , if it exists, is called the period of f . We have already seen a family of periodic functions in Section 2.1: the constant functions. However, despite being periodic a constant function has no period. (We’ll leave that odd gem as an exercise for you.) Returning to the circular functions, we see that by Deﬁnition 10.3, f (t) = cos(t) is periodic, since cos(t + 2πk)
= cos(t) for any integer k. To determine the period of f , we need to ﬁnd the smallest real number p so that f (t + p) = f (t) for all real numbers t or, said diﬀerently, the smallest positive real number p such that cos(t + p) = cos(t) for all real numbers t. We know that cos(t + 2π) = cos(t) for all real numbers t but the question remains if any smaller real number will do the trick. Suppose p > 0 and cos(t + p) = cos(t) for all real numbers t. Then, in particular, cos(0 + p) = cos(0) so that cos(p) = 1. From this we know p is a multiple of 2π and, since the smallest positive multiple of 2π is 2π itself, we have the result. Similarly, we can show g(t) = sin(t) is also periodic with 2π as its period.2 Having period 2π essentially means that we can completely understand everything about the functions f (t) = cos(t) and g(t) = sin(t) by studying one interval of length 2π, say [0, 2π].3 One last property of the functions f (t) = cos(t) and g(t) = sin(t) is worth pointing out: both of these functions are continuous and smooth. Recall from Section 3.1 that geometrically this means the graphs of the cosine and sine functions have no jumps, gaps, holes in the graph, asymptotes, 1See section 1.6 for a review of these concepts. 2Alternatively, we can use the Cofunction Identities in Theorem 10.14 to show that g(t) = sin(t) is periodic with period 2π since g(t) = sin(t) = cos . 3Technically, we should study the interval [0, 2π),4since whatever happens at t = 2π is the same as what happens at t = 0. As we will see shortly, t = 2π gives us an extra ‘check’ when we go to graph these functions. 4In some advanced texts, the interval of choice is [−π, π). 10.5 Graphs of the Trigonometric Functions 791 corners or cusps. As we shall see, the graphs of both f (t) = cos(t) and g(t) = sin(t) meander nicely and don’t cause any trouble. We summarize these facts in the following theorem. Theorem 10.22. Properties of the Cosine and Sine Functions • The function f (x) = cos(x) • The function g(x) = sin(x) – has domain (−∞, ∞) – has range [−1, 1] – has domain (−∞, ∞) – has range [−1, 1] – is continuous and smooth – is continuous and smooth – is even – has period 2π – is odd – has period 2π In the chart above, we followed the convention established in Section 1.6 and used x as the independent variable and y as the dependent variable.5 This allows us to turn our attention to graphing the cosine and sine functions in the Cartesian Plane. To graph y = cos(x), we make a table as we did in Section 1.6 using some of the ‘common values’ of x in the interval [0, 2π]. This generates a portion of the cosine graph, which we call the ‘fundamental cycle’ of y = cos(x). x 0 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π cos(xx, cos(x)) (0, 1 √ 3π 4 , − 2 2 (π, −1 3π √ 7π 4 , (2π, 1) 2 2 5π y 1 −1 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π x The ‘fundamental cycle’ of y = cos(x). A few things about the graph above are worth mentioning. First, this graph represents only part of the graph of y = cos(x). To get the entire graph, we imagine ‘copying and pasting’ this graph end to end inﬁnitely in both directions (left and right) on the x-axis. Secondly, the vertical scale here has been greatly exaggerated for clarity and aesthetics. Below is an accurate-to-scale graph of y = cos(x) showing several cycles with the ‘fundamental cycle’ plotted thicker than the others. The 5The use of x and y in this context is not to be confused with the x- and y-coordinates of points on the Unit Circle which deﬁne cosine and sine. Using the term ‘trigonometric function’ as opposed to ‘circular function’ can help with that, but one could then ask, “Hey, where’s the triangle?” 792 Foundations of Trigonometry graph of y = cos(x) is usually described as ‘wavelike’ – indeed, many of the applications involving the cosine and sine functions feature modeling wavelike phenomena. y x An accurately scaled graph of y = cos(x). We can plot the fundamental cycle of the graph of y = sin(x) similarly, with similar results. x 0 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π sin(xx, sin(x)) (0, 0 √ 3π 4 , 2 2 0 √ 2 2 −1 √ 2 2 − − 5π 7π (π, 01 3π √ 4 , − 2 2 y 1 −1 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π x 0 (2π, 0) The ‘fundamental cycle’ of y = sin(x). As with the graph of y = cos(x), we provide an accurately scaled graph of y = sin(x) below with the fundamental cycle highlighted. y x An accurately scaled graph of y = sin(x). It is no accident that the graphs of y = cos(x) and y = sin(x) are so similar. Using a cofunction identity along with the even property of cosine, we have sin(x) = cos π 2 − x = cos − x − = cos x − π 2 π 2 Recalling Section 1.7, we see from this formula that the graph of y = sin(x) is the result of shifting the graph of y = cos(x) to the right π Now that we know the basic shapes of the graphs of y = cos(x) and y = sin(x), we can use Theorem 1.7 in Section 1.7 to graph more complicated curves. To do so, we need to keep track of 2 units. A visual inspection conﬁrms this. 10.5 Graphs of the Trigonometric Functions 793 the movement of some key points on the original graphs. We choose to track the values x = 0, π 2 , π, 3π 2 and 2π. These ‘quarter marks’ correspond to quadrantal angles, and as such, mark the location of the zeros and the local extrema of these functions over exactly one period. Before we begin our next example, we need to review the concept of the ‘argument’ of a function as ﬁrst introduced in Section 1.4. For the function f (x) = 1 − 5 cos(2x − π), the argument of f is x. We shall have occasion, however, to refer to the argument of the cosine, which in this case is 2x − π. Loosely stated, the argument of a trigonometric function is the expression ‘inside’ the function. Example 10.5.1. Graph one cycle of the following functions. State the period of each. 1. f (x) = 3 cos πx−π 2 + 1 Solution. 1. We set the argument of the cosine, πx−π 2 2. g(x) = 1 2 sin(π − 2x) + 3 2 , equal to each of the values: 0, π 2 , π, 3π 2 , 2π and solve for x. We summarize the results below. πx−π a πx−π πx− = 3π 4 2 2 = 2π 5 0 π 2 π 3π 2 2π πx−π πx−π πx−π 2 Next, we substitute each of these x values into f (x) = 3 cos πx−π corresponding y-values and connect the dots in a pleasing wavelike fashion. 2 + 1 to determine the x 1 2 3 4 5 f (x) (x, f (x)) 4 1 (1, 4) (2, 1) −2 (3, −2) 1 4 (4, 1) (5, 4) y 4 3 2 1 −1 −2 1 2 3 4 5 x One cycle is graphed on [1, 5] so the period is the length of that interval which is 4. One cycle of y = f (x). 2. Proceeding as above, we set the argument of the sine, π − 2x, equal to each of our quarter marks and solve for x. 794 Foundations of Trigonometry a π − 2x = a x π π − 2x = 0 0 2 π π π − 2x = π 2 4 2 π − 2x = π π 0 3π π − 2x = 3π 2 − π 2 4 2π π − 2x = 2π − π 2 We now ﬁnd the corresponding y-values on the graph by substituting each of these x-values into g(x) = 1 2 . Once again, we connect the dots in a wavelike fashion. 2 sin(π − 2x(xx, g(x)) π 2 , 3 π 4 , 2 0 One cycle of y = g(x). One cycle was graphed on the interval − π 2 , π 2 so the period is π 2 − − π 2 = π. The functions in Example 10.5.1 are examples of sinusoids. Roughly speaking, a sinusoid is the result of taking the basic graph of f (x) = cos(x) or g(x) = sin(x) and performing any of the transformations6 mentioned in Section 1.7. Sinusoids can be characterized by four properties: period, amplitude, phase shift and vertical shift. We have already discussed period, that is, how long it takes for the sinusoid to complete one cycle. The standard period of both f (x) = cos(x) and g(x) = sin(x) is 2π, but horizontal scalings will change the period of the resulting sinusoid. The amplitude of the sinusoid is a measure of how ‘tall’ the wave is, as indicated in the ﬁgure below. The amplitude of the standard cosine and sine functions is 1, but vertical scalings can alter this. 6We have already seen how the Even/Odd and Cofunction Identities can be used to rewrite g(x) = sin(x) as a transformed version of f (x) = cos(x), so of course, the reverse is true: f (x) = cos(x) can be written as a transformed version of g(x) = sin(x). The authors have seen some instances where sinusoids are always converted to cosine functions while in other disciplines, the sinusoids are always written in terms of sine functions. We will discuss the applications of sinusoids in greater detail in Chapter 11. Until then, we will keep our options open. 10.5 Graphs of the Trigonometric Functions 795 amplitude baseline period = sin(x). As the reader can verify, a phase shift of π The phase shift of the sinusoid is the horizontal shift experienced by the fundamental cycle. We have seen that a phase (horizontal) shift of π 2 to the right takes f (x) = cos(x) to g(x) = sin(x) since cos x − π 2 to the left takes g(x) = sin(x) to 2 f (x) = cos(x). The vertical shift of a sinusoid is exactly the same as the vertical shifts in Section 1.7. In most contexts, the vertical shift of a sinusoid is assumed to be 0, but we state the more general case below. The following theorem, which is reminiscent of Theorem 1.7 in Section 1.7, shows how to ﬁnd these four fundamental quantities from the formula of the given sinusoid. Theorem 10.23. For ω > 0, the functions C(x) = A cos(ωx + φ) + B and S(x) = A sin(ωx + φ) + B have period 2π ω have amplitude \|A\| have phase shift − φ ω have vertical shift B We note that in some scientiﬁc and engineering circles, the quantity φ mentioned in Theorem 10.23 is called the phase of the sinusoid. Since our interest in this book is primarily with graphing sinusoids, we focus our attention on the horizontal shift − φ The proof of Theorem 10.23 is a direct application of Theorem 1.7 in Section 1.7 and is left to the reader. The parameter ω, which is stipulated to be positive, is called the (angular) frequency of the sinusoid and is the number of cycles the sinusoid completes over a 2π interval. We can always ensure ω > 0 using the Even/Odd Identities.7 We now test out Theorem 10.23 using the functions f and g featured in Example 10.5.1. First, we write f (x) in
the form prescribed in Theorem 10.23, ω induced by φ. f (x) = 3 cos πx − π 2 + 1 = 3 cos π 2 x + − π 2 + 1, 7Try using the formulas in Theorem 10.23 applied to C(x) = cos(−x + π) to see why we need ω > 0. 796 Foundations of Trigonometry 2 , φ = − π π/2 = 4, the amplitude is \|A\| = \|3\| = 3, the phase shift is − φ so that A = 3, ω = π 2 and B = 1. According to Theorem 10.23, the period of f is ω = 2π 2π π/2 = 1 (indicating a shift to the right 1 unit) and the vertical shift is B = 1 (indicating a shift up 1 unit.) All of these match with our graph of y = f (x). Moreover, if we start with the basic shape of the cosine graph, shift it 1 unit to the right, 1 unit up, stretch the amplitude to 3 and shrink the period to 4, we will have reconstructed one period of the graph of y = f (x). In other words, instead of tracking the ﬁve ‘quarter marks’ through the transformations to plot y = f (x), we can use ﬁve other pieces of information: the phase shift, vertical shift, amplitude, period and basic shape of the cosine curve. Turning our attention now to the function g in Example 10.5.1, we ﬁrst need to use the odd property of the sine function to write it in the form required by Theorem 10.23 ω = − −π/2 g(x) = 1 2 sin(π − 2x) + 3 2 = 1 2 sin(−(2x − π)) + 3 2 = − 1 2 sin(2x − π) + 3 2 = − 1 2 sin(2x + (−π)) + 3 2 = 1 2 , the phase shift is − −π 2 . The period is then 2π Instead of the graph starting at x = π 2 , ω = 2, φ = −π and B = 3 2 = π We ﬁnd A = − 1 2 = π, the amplitude is − 1 2 (indicating a shift right π 2 units) and the vertical shift is up 2 3 2 . Note that, in this case, all of the data match our graph of y = g(x) with the exception of the phase shift. 2 , it ends there. Remember, however, that the graph presented in Example 10.5.1 is only one portion of the graph of y = g(x). Indeed, another complete cycle begins at x = π 2 , and this is the cycle Theorem 10.23 is detecting. The reason for the discrepancy is that, in order to apply Theorem 10.23, we had to rewrite the formula for g(x) using the odd property of the sine function. Note that whether we graph y = g(x) using the ‘quarter marks’ approach or using the Theorem 10.23, we get one complete cycle of the graph, which means we have completely determined the sinusoid. Example 10.5.2. Below is the graph of one complete cycle of a sinusoid y = f (x). −1, 5 2 y 3 2 1 51 1 2 3 4 5 x −1 −2 2, − 3 2 One cycle of y = f (x). 1. Find a cosine function whose graph matches the graph of y = f (x). 10.5 Graphs of the Trigonometric Functions 797 2. Find a sine function whose graph matches the graph of y = f (x). Solution. ω , so that ω = π 1. We ﬁt the data to a function of the form C(x) = A cos(ωx + φ) + B. Since one cycle is graphed over the interval [−1, 5], its period is 5 − (−1) = 6. According to Theorem 10.23, 3 . Next, we see that the phase shift is −1, so we have − φ 6 = 2π ω = −1, or . As a result the amplitude A = 1 2 (4) = 2. Finally, to determine the vertical shift, we 2 average the endpoints of the range to ﬁnd B = 1 2 . Our ﬁnal answer is 2 + 1 C(x) = 2 cos π 2 . 3 . To ﬁnd the amplitude, note that the range of the sinusoid is − 3 2 + − 3 5 2 (1. Most of the work to ﬁt the data to a function of the form S(x) = A sin(ωx + φ) + B is done. The period, amplitude and vertical shift are the same as before with ω = π 3 , A = 2 and B = 1 2 . The trickier part is ﬁnding the phase shift. To that end, we imagine extending the graph of the given sinusoid as in the ﬁgure below so that we can identify a cycle beginning at 7 = − 7π 2 , we get − φ 6 . + 1 Hence, our answer is S(x) = 2 sin π 2 . . Taking the phase shift to be 7 3 x − 7π 2 , or , 5 2 − 10 x 7 2 , 1 2 13 2 , 1 2 19 2 , 5 2 −1 −2 8, − 3 2 Extending the graph of y = f (x). Note that each of the answers given in Example 10.5.2 is one choice out of many possible answers. taking For example, when ﬁtting a sine function to the data, we could have chosen to start at 1 + 1 6 for an answer of S(x) = −2 sin π A = −2. In this case, the phase shift is 1 2 . Alternatively, we could have extended the graph of y = f (x) to the left and considered a sine , and so on. Each of these formulas determine the same sinusoid curve function starting at − 5 2 and their formulas are all equivalent using identities. Speaking of identities, if we use the sum identity for cosine, we can expand the formula to yield 2 , 1 3 x − π 2 so (x) = A cos(ωx + φ) + B = A cos(ωx) cos(φ) − A sin(ωx) sin(φ) + B. 798 Foundations of Trigonometry Similarly, using the sum identity for sine, we get S(x) = A sin(ωx + φ) + B = A sin(ωx) cos(φ) + A cos(ωx) sin(φ) + B. Making these observations allows us to recognize (and graph) functions as sinusoids which, at ﬁrst glance, don’t appear to ﬁt the forms of either C(x) or S(x). Example 10.5.3. Consider the function f (x) = cos(2x) − √ 3 sin(2x). Find a formula for f (x): 1. in the form C(x) = A cos(ωx + φ) + B for ω > 0 2. in the form S(x) = A sin(ωx + φ) + B for ω > 0 Check your answers analytically using identities and graphically using a calculator. Solution. 1. The key to this problem is to use the expanded forms of the sinusoid formulas and match up 3 sin(2x) with the expanded form of corresponding coeﬃcients. Equating f (x) = cos(2x) − C(x) = A cos(ωx + φ) + B, we get √ √ cos(2x) − 3 sin(2x) = A cos(ωx) cos(φ) − A sin(ωx) sin(φ) + B It should be clear that we can take ω = 2 and B = 0 to get √ cos(2x) − 3 sin(2x) = A cos(2x) cos(φ) − A sin(2x) sin(φ) To determine A and φ, a bit more work is involved. We get started by equating the coeﬃcients of the trigonometric functions on either side of the equation. On the left hand side, the coeﬃcient of cos(2x) is 1, while on the right hand side, it is A cos(φ). Since this equation is to hold for all real numbers, we must have8 that A cos(φ) = 1. Similarly, we ﬁnd by equating the coeﬃcients of sin(2x) that A sin(φ) = 3. What we have here is a system of nonlinear equations! We can temporarily eliminate the dependence on φ by using the Pythagorean Identity. We know cos2(φ) + sin2(φ) = 1, so multiplying this by A2 gives A2 cos2(φ)+A2 sin2(φ) = A2. Since A cos(φ) = 1 and A sin(φ) = 3)2 = 4 or A = ±2. Choosing A = 2, we have 2 cos(φ) = 1 and 2 sin(φ) = 3 or, after some √ 3 rearrangement, cos(φ) = 1 2 . One such angle φ which satisﬁes this criteria is . We can easily φ = π check our answer using the sum formula for cosine 3 . Hence, one way to write f (x) as a sinusoid is f (x) = 2 cos 2x + π 3, we get A2 = 12+( √ 2 and sin(φ) = √ √ √ 3 f (x) = 2 cos 2x + π 3 = 2 cos(2x) cos π 3 cos(2x) 1 − sin(2x) 2 √ 3 sin(2x) = 2 = cos(2x) − − sin(2x) sin π 3 √ 3 2 8This should remind you of equation coeﬃcients of like powers of x in Section 8.6. 10.5 Graphs of the Trigonometric Functions 799 2. Proceeding as before, we equate f (x) = cos(2x) − S(x) = A sin(ωx + φ) + B to get √ 3 sin(2x) with the expanded form of √ cos(2x) − 3 sin(2x) = A sin(ωx) cos(φ) + A cos(ωx) sin(φ) + B Once again, we may take ω = 2 and B = 0 so that √ cos(2x) − 3 sin(2x) = A sin(2x) cos(φ) + A cos(2x) sin(φ) We equate9 the coeﬃcients of cos(2x) on either side and get A sin(φ) = 1 and A cos(φ) = − 3. Using A2 cos2(φ) + A2 sin2(φ) = A2 as before, we get A = ±2, and again we choose A = 2. √ 3 This means 2 sin(φ) = 1, or sin(φ) = 1 2 . . One such angle which meets these criteria is φ = 5π 6 Checking our work analytically, we have 3, which means cos(φ) = − 6 . Hence, we have f (x) = 2 sin 2x + 5π 2 , and 2 cos(φ) = − √ √ f (x) = 2 sin 2x + 5π 6 + cos(2x) sin 5π = 2 sin(2x) cos 5π 6 6 √ + cos(2x) 1 3 2 2 − = 2 sin(2x) = cos(2x) − √ 3 sin(2x) Graphing the three formulas for f (x) result in the identical curve, verifying our analytic work. √ It is important to note that in order for the technique presented in Example 10.5.3 to ﬁt a function into one of the forms in Theorem 10.23, the arguments of the cosine and sine function much match. 3 sin(3x) is not.10 It That is, while f (x) = cos(2x) − is also worth mentioning that, had we chosen A = −2 instead of A = 2 as we worked through Example 10.5.3, our ﬁnal answers would have looked diﬀerent. The reader is encouraged to rework Example 10.5.3 using A = −2 to see what these diﬀerences are, and then for a challenging exercise, use identities to show that the formulas are all equivalent. The general equations to ﬁt a function of the form f (x) = a cos(ωx) + b sin(ωx) + B into one of the forms in Theorem 10.23 are explored in Exercise 35. 3 sin(2x) is a sinusoid, g(x) = cos(2x) − √ 9Be careful here! 10This graph does, however, exhibit sinusoid-like characteristics! Check it out! 800 Foundations of Trigonometry 10.5.2 Graphs of the Secant and Cosecant Functions We now turn our attention to graphing y = sec(x). Since sec(x) = 1 cos(x) , we can use our table of values for the graph of y = cos(x) and take reciprocals. We know from Section 10.3.1 that the domain of F (x) = sec(x) excludes all odd multiples of π 2 , and sure enough, we run into trouble at 2 and x = 3π x = π 2 since cos(x) = 0 at these values. Using the notation introduced in Section 4.2, −, cos(x) → 0+, so sec(x) → ∞. (See Section 10.3.1 for a more detailed we have that as x → π 2 analysis.) Similarly, we ﬁnd that as x → π , sec(x) → −∞; and as 2 x → 3π , sec(x) → ∞. This means we have a pair of vertical asymptotes to the graph of y = sec(x), 2 x = π 2 . Since cos(x) is periodic with period 2π, it follows that sec(x) is also.11 Below we graph a fundamental cycle of y = sec(x) along with a more complete graph obtained by the usual ‘copying and pasting.’12 +, sec(x) → −∞; as x → 3π 2 2 and x = 3π − + y x 0 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π √ 2 − − √ √ sec(x) 1 cos(x) 1 √ 2 2 0 undeﬁned 2 2 −1 √ 2 2 0 undeﬁned 2 2 1 −1 √ √ − − 1 2 √ (x, sec(x)) (0, 1) √ 2 π 4 , 2 3π 2 5π √ 2 4 , − (π, −1) √ 2 4 , − √ 2 7π 4 , (2π, 1) 3 2 1 −1 −2 −3 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π x The ‘fundamental cycle’ of y = sec(x). y x The graph of y = sec(x). 11Provided sec(α) and sec(β) are deﬁned, sec(α) = sec(β) if and only if cos(α) = cos(β). He
nce, sec(x) inherits its period from cos(x). 12In Section 10.3.1, we argued the range of F (x) = sec(x) is (−∞, −1] ∪ [1, ∞). We can now see this graphically. 10.5 Graphs of the Trigonometric Functions 801 As one would expect, to graph y = csc(x) we begin with y = sin(x) and take reciprocals of the corresponding y-values. Here, we encounter issues at x = 0, x = π and x = 2π. Proceeding with the usual analysis, we graph the fundamental cycle of y = csc(x) below along with the dotted graph of y = sin(x) for reference. Since y = sin(x) and y = cos(x) are merely phase shifts of each other, so too are y = csc(x) and y = sec(x). y x 0 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π sin(x) √ 1 2 √ √ csc(x) 0 undeﬁned 2 2 1 √ 2 2 0 undeﬁned 2 2 −1 √ 2 2 0 undeﬁned −x, csc(x)) √ 3π 4 , 2 5π 2 7π √ 2 4 , − 2 , −1 3π √ 2 4 , − 3 2 1 −1 −2 −3 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π x Once again, our domain and range work in Section 10.3.1 is veriﬁed geometrically in the graph of y = G(x) = csc(x). The ‘fundamental cycle’ of y = csc(x). y x The graph of y = csc(x). Note that, on the intervals between the vertical asymptotes, both F (x) = sec(x) and G(x) = csc(x) are continuous and smooth. In other words, they are continuous and smooth on their domains.13 The following theorem summarizes the properties of the secant and cosecant functions. Note that 13Just like the rational functions in Chapter 4 are continuous and smooth on their domains because polynomials are continuous and smooth everywhere, the secant and cosecant functions are continuous and smooth on their domains since the cosine and sine functions are continuous and smooth everywhere. 802 Foundations of Trigonometry all of these properties are direct results of them being reciprocals of the cosine and sine functions, respectively. Theorem 10.24. Properties of the Secant and Cosecant Functions The function F (x) = sec(x) – has domain x : x = π 2 + πk, k is an integer = ∞ k=−∞ (2k + 1)π 2 , (2k + 3)π 2 – has range {y : \|y\| ≥ 1} = (−∞, −1] ∪ [1, ∞) – is continuous and smooth on its domain – is even – has period 2π The function G(x) = csc(x) ∞ k=−∞ (kπ, (k + 1)π) – has domain {x : x = πk, k is an integer} = – has range {y : \|y\| ≥ 1} = (−∞, −1] ∪ [1, ∞) – is continuous and smooth on its domain – is odd – has period 2π In the next example, we discuss graphing more general secant and cosecant curves. Example 10.5.4. Graph one cycle of the following functions. State the period of each. 1. f (x) = 1 − 2 sec(2x) Solution. 2. g(x) = csc(π − πx) − 5 3 1. To graph y = 1 − 2 sec(2x), we follow the same procedure as in Example 10.5.1. First, we set 2 and 2π and solve for x. the argument of secant, 2x, equal to the ‘quarter marks’ 0, π 2 , π, 3π a 2x = a 0 2x = 0 π 2x = π 2 2 π 2x = π 3π 2x = 3π 2 2 2π 2x = 2π x 0 π 4 π 2 3π 4 π 10.5 Graphs of the Trigonometric Functions 803 Next, we substitute these x values into f (x). If f (x) exists, we have a point on the graph; otherwise, we have found a vertical asymptote. In addition to these points and asymptotes, we have graphed the associated cosine curve – in this case y = 1 − 2 cos(2x) – dotted in the picture below. Since one cycle is graphed over the interval [0, π], the period is π − 0 = π. y 3 2 1 −1 π 4 π 2 3π 4 π x x 0 π 4 π 2 3π 4 π f (x) −1 (x, f (x)) (0, −1) undeﬁned 3 π 2 , 3 undeﬁned −1 (π, −1) 2. Proceeding as before, we set the argument of cosecant in g(x) = csc(π−πx)−5 3 equal to the quarter marks and solve for x. One cycle of y = 1 − 2 sec(2x). a π − πx = a x π − πx = 0 1 0 π − πx = π 1 π 2 2 2 π − πx = π 0 π 2 − 1 π − πx = 3π 3π 2 2 2π π − πx = 2π −1 Substituting these x-values into g(x), we generate the graph below and ﬁnd the period to be 1 − (−1) = 2. The associated sine curve, y = sin(π−πx)−5 , is dotted in as a reference1 g(x) undeﬁned − 4 3 undeﬁned (x, g(x)) 1 2 , − 4 3 −2 − 1 2 , −2 undeﬁned y −1 − 1 2 1 1 2 x −1 −2 One cycle of y = csc(π−πx)−5 3 . 804 Foundations of Trigonometry Before moving on, we note that it is possible to speak of the period, phase shift and vertical shift of secant and cosecant graphs and use even/odd identities to put them in a form similar to the sinusoid forms mentioned in Theorem 10.23. Since these quantities match those of the corresponding cosine and sine curves, we do not spell this out explicitly. Finally, since the ranges of secant and cosecant are unbounded, there is no amplitude associated with these curves. 10.5.3 Graphs of the Tangent and Cotangent Functions 2 and x = 3π Finally, we turn our attention to the graphs of the tangent and cotangent functions. When constructing a table of values for the tangent function, we see that J(x) = tan(x) is undeﬁned at −, sin(x) → 1− x = π 2 , in accordance with our ﬁndings in Section 10.3.1. As x → π and cos(x) → 0+, so that tan(x) = sin(x) 2 . Using a + similar analysis, we get that as x → π , 2 tan(x) → −∞. Plotting this information and performing the usual ‘copy and paste’ produces: cos(x) → ∞ producing a vertical asymptote at x = π +, tan(x) → −∞; as x → 3π 2 , tan(x) → ∞; and as x → 3π 3π 4 π 5π 4 3π 2 7π 4 2π tan(x) 0 1 (x, tan(x)) (0, 0) 4 , 1 π undeﬁned −1 0 1 undeﬁned −1 0 4 , −1 3π (π, 0) 4 , 1 5π 4 , −1 7π (2π, 0) 1 −1 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π x The graph of y = tan(x) over [0, 2π]. y x The graph of y = tan(x). 10.5 Graphs of the Trigonometric Functions 805 From the graph, it appears as if the tangent function is periodic with period π. To prove that this is the case, we appeal to the sum formula for tangents. We have: tan(x + π) = tan(x) + tan(π) 1 − tan(x) tan(π) = tan(x) + 0 1 − (tan(x))(0) = tan(x), which tells us the period of tan(x) is at most π. To show that it is exactly π, suppose p is a positive real number so that tan(x + p) = tan(x) for all real numbers x. For x = 0, we have tan(p) = tan(0 + p) = tan(0) = 0, which means p is a multiple of π. The smallest positive multiple of π is π itself, so we have established the result. We take as our fundamental cycle for y = tan(x) the interval − π 2 , − π 2 . From the graph, we see conﬁrmation of our domain and range work in Section 10.3.1. , and use as our ‘quarter marks’ x = − π 4 and π 4 , 0, π 2 , π 2 It should be no surprise that K(x) = cot(x) behaves similarly to J(x) = tan(x). Plotting cot(x) over the interval [0, 2π] results in the graph below. y 1 −1 x 0 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π cot(x) undeﬁned (x, cot(x)) 1 0 −1 undeﬁned 1 0 −1 undeﬁned 1 3π 4 , 1 5π 3π 2 , 0 4 , −1 7π π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π x The graph of y = cot(x) over [0, 2π]. From these data, it clearly appears as if the period of cot(x) is π, and we leave it to the reader to prove this.14 We take as one fundamental cycle the interval (0, π) with quarter marks: x = 0, π 4 , π 4 and π. A more complete graph of y = cot(x) is below, along with the fundamental cycle highlighted as usual. Once again, we see the domain and range of K(x) = cot(x) as read from the graph matches with what we found analytically in Section 10.3.1. 2 , 3π 14Certainly, mimicking the proof that the period of tan(x) is an option; for another approach, consider transforming tan(x) to cot(x) using identities. 806 Foundations of Trigonometry y x The graph of y = cot(x). The properties of the tangent and cotangent functions are summarized below. As with Theorem 10.24, each of the results below can be traced back to properties of the cosine and sine functions and the deﬁnition of the tangent and cotangent functions as quotients thereof. Theorem 10.25. Properties of the Tangent and Cotangent Functions The function J(x) = tan(x) – has domain x : x = π 2 + πk, k is an integer = ∞ k=−∞ (2k + 1)π 2 , (2k + 3)π 2 – has range (−∞, ∞) – is continuous and smooth on its domain – is odd – has period π The function K(x) = cot(x) – has domain {x : x = πk, k is an integer} = – has range (−∞, ∞) – is continuous and smooth on its domain – is odd – has period π ∞ k=−∞ (kπ, (k + 1)π) 10.5 Graphs of the Trigonometric Functions 807 Example 10.5.5. Graph one cycle of the following functions. Find the period. 1. f (x) = 1 − tan x 2 . 2. g(x) = 2 cot π 2 x + π + 1. Solution. 1. We proceed as we have in all of the previous graphing examples by setting the argument of 4 , 0, π 2 , equal to each of the ‘quarter marks’ − π , namely x 2 , − π 4 tangent in f (x) = 1 − tan x 2 and π 2 , and solving for x Substituting these x-values into f (x), we ﬁnd points on the graph and the vertical asymptotes. x −x) undeﬁned (x, f (x)) 2 1 0 − π 2 , 2 (0, 1) 2 , 0 π undeﬁned y 2 1 −1 −2 −π − π 2 π 2 π x We see that the period is π − (−π) = 2π. One cycle of y = 1 − tan x 2 . 2. The ‘quarter marks’ for the fundamental cycle of the cotangent curve are 0, π To graph g(x) = 2 cot π and solving for x. 2 x + π + 1, we begin by setting π 4 and π. 2 x + π equal to each quarter mark 4 , π 2 , 3π 808 Foundations of Trigonometry a 0 π 4 π 2 3π 1 4 − 1 2 x + π = 3π We now use these x-values to generate our graph. x −2 − 3 2 −1 − 1 2 0 g(x) undeﬁned (x, g(x)) 3 1 2 , 3 − 3 (−1, 1) 2 , −1 −1 − 1 undeﬁned y 3 2 1 −2 −1 x −1 One cycle of y = 2 cot π 2 x + π + 1. We ﬁnd the period to be 0 − (−2) = 2. As with the secant and cosecant functions, it is possible to extend the notion of period, phase shift and vertical shift to the tangent and cotangent functions as we did for the cosine and sine functions in Theorem 10.23. Since the number of classical applications involving sinusoids far outnumber those involving tangent and cotangent functions, we omit this. The ambitious reader is invited to formulate such a theorem, however. 10.5 Graphs of the Trigonometric Functions 809 10.5.4 Exercises In Exercises 1 - 12, graph one cycle of the given function. State the period, amplitude, phase shift and vertical shift of the function. 1. y = 3 sin(x) 2. y = sin(3x) 3. y = −2 cos(x) 4. y = cos x − 7. y = − 1 3 cos π 2 1 2 x + π 3 5. y = − sin x + π 3 6. y = sin(2x − π) 8. y = cos(3x − 2π) + 4 9. y = sin −x − π 4 − 2 10. y = 2 3 cos π 2 − 4x + 1 11. y = − cos 2x + 3 2 π 3 − 1 2 12. y = 4 sin(−2πx + π) In Exercises 13
- 24, graph one cycle of the given function. State the period of the function. 13. y = tan 16. y = sec x − x − π 3 π 2 14. y = 2 tan 17. y = − csc x − 3 1 4 x + π 3 15. y = 1 3 18. y = − 19. y = csc(2x − π) 20. y = sec(3x − 2π) + 4 21. y = csc −x − tan(−2x − π) + 1 1 3 sec 22. y = cot x + π 6 23. y = −11 cot x 1 5 24. y = cot 2x + 3π 2 + 1 1 3 In Exercises 25 - 34, use Example 10.5.3 as a guide to show that the function is a sinusoid by rewriting it in the forms C(x) = A cos(ωx + φ) + B and S(x) = A sin(ωx + φ) + B for ω > 0 and 0 ≤ φ < 2π. 25. f (x) = √ 2 sin(x) + √ 2 cos(x) + 1 26. f (x) = 3 √ 3 sin(3x) − 3 cos(3x) 27. f (x) = − sin(x) + cos(x) − 2 28. f (x) = − 1 2 sin(2x) − √ 3 2 cos(2x) 29. f (x) = 2 √ 3 cos(x) − 2 sin(x) 31. f (x) = − 1 2 cos(5x) − √ 3 2 sin(5x) 30. f (x) = 3 2 cos(2x) − √ 3 3 2 sin(2x) + 6 32. f (x) = −6 √ 3 cos(3x) − 6 sin(3x) − 3 810 Foundations of Trigonometry 33. f (x) = √ 5 2 2 sin(x) − √ 5 2 2 cos(x) 34. f (x) = 3 sin x 6 √ − 3 3 cos x 6 35. In Exercises 25 - 34, you should have noticed a relationship between the phases φ for the S(x) and C(x). Show that if f (x) = A sin(ωx + α) + B, then f (x) = A cos(ωx + β) + B where β = α − . π 2 36. Let φ be an angle measured in radians and let P (a, b) be a point on the terminal side of φ when it is drawn in standard position. Use Theorem 10.3 and the sum identity for sine in Theorem 10.15 to show that f (x) = a sin(ωx) + b cos(ωx) + B (with ω > 0) can be rewritten as f (x) = a2 + b2 sin(ωx + φ) + B. √ 37. With the help of your classmates, express the domains of the functions in Examples 10.5.4 and 10.5.5 using extended interval notation. (We will revisit this in Section 10.7.) In Exercises 38 - 43, verify the identity by graphing the right and left hand sides on a calculator. 38. sin2(x) + cos2(x) = 1 39. sec2(x) − tan2(x) = 1 40. cos(x) = sin 41. tan(x + π) = tan(x) 42. sin(2x) = 2 sin(x) cos(x) 43. tan x 2 = − x π 2 sin(x) 1 + cos(x) In Exercises 44 - 50, graph the function with the help of your calculator and discuss the given questions with your classmates. 44. f (x) = cos(3x) + sin(x). Is this function periodic? If so, what is the period? 45. f (x) = sin(x) x . What appears to be the horizontal asymptote of the graph? 46. f (x) = x sin(x). Graph y = ±x on the same set of axes and describe the behavior of f . 47. f (x) = sin 1 x . What’s happening as x → 0? 48. f (x) = x − tan(x). Graph y = x on the same set of axes and describe the behavior of f . 49. f (x) = e−0.1x (cos(2x) + sin(2x)). Graph y = ±e−0.1x on the same set of axes and describe the behavior of f . 50. f (x) = e−0.1x (cos(2x) + 2 sin(x)). Graph y = ±e−0.1x on the same set of axes and describe the behavior of f . 51. Show that a constant function f is periodic by showing that f (x + 117) = f (x) for all real numbers x. Then show that f has no period by showing that you cannot ﬁnd a smallest number p such that f (x + p) = f (x) for all real numbers x. Said another way, show that f (x + p) = f (x) for all real numbers x for ALL values of p > 0, so no smallest value exists to satisfy the deﬁnition of ‘period’. 10.5 Graphs of the Trigonometric Functions 811 10.5.5 Answers 1. y = 3 sin(x) Period: 2π Amplitude: 3 Phase Shift: 0 Vertical Shift: 0 2. y = sin(3x) 2π 3 Period: Amplitude: 1 Phase Shift: 0 Vertical Shift: 0 3. y = −2 cos(x) Period: 2π Amplitude: 2 Phase Shift: 0 Vertical Shift: 0 4. y = cos x − π 2 Period: 2π Amplitude: 1 Phase Shift: Vertical Shift: 0 π 2 y 3 −3 y 1 −1 y 2 −2 y 1 −1 π 2 π 3π 2 x 2π π 6 π 3 π 2 2π 3 x π 2 π 3π 2 2π x π 2 π 3π 2 2π 5π 2 x 812 5. y = − sin x + Period: 2π Amplitude: 1 π Phase Shift: − 3 Vertical Shift: 0 π 3 Foundations of Trigonometry y 1 − π 3 π 6 2π 3 7π 6 5π 3 x 6. y = sin(2x − π) Period: π Amplitude: 1 Phase Shift: Vertical Shift: 0 π 2 7. y = − 1 3 cos 1 2 x + π 3 Period: 4π Amplitude: 1 3 2π Phase Shift: − 3 Vertical Shift: 0 8. y = cos(3x − 2π) + 4 2π 3 Period: Amplitude: 1 2π Phase Shift: 3 Vertical Shift: 4 −1 y 1 −1 y 1 3 π 2 3π 4 π 5π 4 3π 2 x − 2π 3 π 3 4π 3 7π 3 x 10π 3 − 1 3 y 5 4 3 2π 3 5π 6 π 7π 6 4π 3 x 10.5 Graphs of the Trigonometric Functions 813 9. y = sin −x − π 4 − 2 Period: 2π Amplitude: 1 Phase Shift: − π 4 π x + 4 Vertical Shift: −2 y = − sin (You need to use − 2 to ﬁnd this.)15 − 4x + 1 10. y = 2 3 Period: π 2 cos π 2 Amplitude: 2 3 π 8 4x − cos y = Vertical Shift: 1 Phase Shift: 2 3 (You need to use π 2 + 1 to ﬁnd this.)16 − 9π 4 − 7π 4 − 5π 4 − 3π 4 y π 4 3π 4 5π 4 x 7π 4 − π 4 −1 −2 − 3π 8 π 2 5π 8 x − 3π 8 − π 4 − π 8 11. y = − cos 2x + 3 2 π 3 − 1 2 y 1 Period: π Amplitude: 3 2 Phase Shift: − π 6 Vertical Shift: − 1 2 12. y = 4 sin(−2πx + π) Period: 1 Amplitude: 4 1 2 (You need to use Phase Shift: y = −4 sin(2πx − π) to ﬁnd this.)17 Vertical Shift: 0 − π 6 − 1 2 π 12 π 3 7π 12 x 5π 6 −4 15Two cycles of the graph are shown to illustrate the discrepancy discussed on page 796. 16Again, we graph two cycles to illustrate the discrepancy discussed on page 796. 17This will be the last time we graph two cycles to illustrate the discrepancy discussed on page 796. 814 Foundations of Trigonometry 13. y = tan x − Period: π π 3 14. y = 2 tan x − 3 1 4 Period: 4π y 1 − π 6 −1 π 12 π 3 7π 12 5π 6 x −2π −π y −1 −3 −5 π 2π x 15. y = tan(−2x − π) + 1 1 3 is equivalent to 1 3 tan(2x + π) + 1 y = − via the Even / Odd identity for tangent. Period − 3π 8 − π 4 x − 3π 4 − 5π 8 10.5 Graphs of the Trigonometric Functions 815 16. y = sec x − π 2 Start with y = cos x − π 2 Period: 2π 17. y = − csc x + π 3 Start with y = − sin Period: 2π x + π 3 y 1 −1 y 1 π 2 π 3π 2 2π x 5π 2 − π 3 −1 π 6 2π 3 7π 6 x 5π 3 18. y = − 1 3 sec 1 2 x + Start with y = − Period: 4π 1 3 π 3 cos 1 2 x + π 3 y 1 3 − 2π 3 − 1 3 π 3 4π 3 7π 3 x 10π 3 816 Foundations of Trigonometry 19. y = csc(2x − π) Start with y = sin(2x − π) Period: π 20. y = sec(3x − 2π) + 4 Start with y = cos(3x − 2π) + 4 Period: 2π 3 21. y = csc −x − π 4 Start with y = sin Period: 2π − 2 −x − π 4 − 2 y 1 −1 −2 −3 π 2 3π 4 π 5π 4 x 3π 2 2π 3 5π 6 π 7π 6 4π 3 x π 4 3π 4 5π 4 x 7π 4 10.5 Graphs of the Trigonometric Functions 817 22. y = cot x + Period: π π 6 23. y = −11 cot Period: 5π 1 5 x y 1 − π 6 π 12 −1 π 3 7π 12 x 5π 6 y 11 −11 5π 4 5π 2 15π 4 5π x 24. y = 1 3 Period: cot π 2 2x + 3π − 3π 8 − π 4 x − 3π 4 − 5π 8 818 Foundations of Trigonometry 25. f (x) = √ 2 sin(x) + √ 2 cos(x) + 1 = 2 sin x + π 4 + 1 = 2 cos x + 7π 4 + 1 26. f (x) = 3 √ 3 sin(3x) − 3 cos(3x) = 6 sin 3x + 11π 6 = 6 cos 3x + 4π 3 27. f (x) = − sin(x) + cos(x) − 2 = √ 2 sin x + 3π 4 − 2 = √ x + 2 cos π 4 − 2 28. f (x) = − 1 2 sin(2x) − √ 3 2 cos(2x) = sin 2x + 4π 3 = cos 2x + 5π 6 29. f (x) = 2 √ 3 cos(x) − 2 sin(x) = 4 sin x + 30. f (x) = 3 2 cos(2x) − 31. f (x) = − 1 2 cos(5x) − 3 2 √ 3 3 2 √ 2π 3 = 4 cos x + π 6 5π 6 + 6 = 3 cos = cos 5x + 2π 3 sin(2x) + 6 = 3 sin 2x + sin(5x) = sin 5x + 7π 6 2x + π 3 + 6 32. f (x) = −6 √ 3 cos(3x) − 6 sin(3x) − 3 = 12 sin 3x + 4π 3 − 3 = 12 cos 3x + 5π 6 − 3 33. f (x) = √ 5 2 2 sin(x) − √ 5 2 2 cos(x) = 5 sin x + 34. f (x) = 3 sin x 6 √ − 3 3 cos x 6 = 6 sin x 6 + 7π 4 5π 3 = 5 cos x + 5π 4 = 6 cos x 6 + 7π 6 10.6 The Inverse Trigonometric Functions 819 10.6 The Inverse Trigonometric Functions As the title indicates, in this section we concern ourselves with ﬁnding inverses of the (circular) trigonometric functions. Our immediate problem is that, owing to their periodic nature, none of the six circular functions is one-to-one. To remedy this, we restrict the domains of the circular functions in the same way we restricted the domain of the quadratic function in Example 5.2.3 in Section 5.2 to obtain a one-to-one function. We ﬁrst consider f (x) = cos(x). Choosing the interval [0, π] allows us to keep the range as [−1, 1] as well as the properties of being smooth and continuous. y x Restricting the domain of f (x) = cos(x) to [0, π]. 1 Recall from Section 5.2 that the inverse of a function f is typically denoted f −1. For this reason, some textbooks use the notation f −1(x) = cos−1(x) for the inverse of f (x) = cos(x). The obvious pitfall here is our convention of writing (cos(x))2 as cos2(x), (cos(x))3 as cos3(x) and so on. It is far too easy to confuse cos−1(x) with cos(x) = sec(x) so we will not use this notation in our text.1 Instead, we use the notation f −1(x) = arccos(x), read ‘arc-cosine of x’. To understand the ‘arc’ in ‘arccosine’, recall that an inverse function, by deﬁnition, reverses the process of the original function. The function f (t) = cos(t) takes a real number input t, associates it with the angle θ = t radians, and returns the value cos(θ). Digging deeper,2 we have that cos(θ) = cos(t) is the x-coordinate of the terminal point on the Unit Circle of an oriented arc of length \|t\| whose initial point is (1, 0). Hence, we may view the inputs to f (t) = cos(t) as oriented arcs and the outputs as x-coordinates on the Unit Circle. The function f −1, then, would take x-coordinates on the Unit Circle and return oriented arcs, hence the ‘arc’ in arccosine. Below are the graphs of f (x) = cos(x) and f −1(x) = arccos(x), where we obtain the latter from the former by reﬂecting it across the line y = x, in accordance with Theorem 5.3. y 1 −x) = cos(x), 0 ≤ x ≤ π reﬂect across y = x −−−−−−−−−−−−→ switch x and y coordinates −1 1 x f −1(x) = arccos(x). 1But be aware that many books do! As always, be sure to check the context! 2See page 704 if you need a review of how we associate real numbers with angles in radian measure. 820 Foundations of Trigonometry We restrict g(x) = sin(x) in a similar manner, although the interval of choice is − π 2 , π 2 y . x Restricting the domain of f (x) = sin(x) to − π 2 , π 2 . It should be no surprise that we call g−1(x) = arcsin(x), which is read ‘arc-sine of x’. y 1 − π 2 x π 2 −1 g(x) = sin(x), − 1 1 x reﬂect across y = x −−−−−−−−−−−−→ switch x and y coordinates − π 2 g−1(x) = arcsin(x). We list some important facts about the arccosine and arcsine functions in the following theorem. Theorem 10.26. Properties of the Arccosine and Arcsine Functi
ons Properties of F (x) = arccos(x) – Domain: [−1, 1] – Range: [0, π] – arccos(x) = t if and only if 0 ≤ t ≤ π and cos(t) = x – cos(arccos(x)) = x provided −1 ≤ x ≤ 1 – arccos(cos(x)) = x provided 0 ≤ x ≤ π Properties of G(x) = arcsin(x) – Domain: [−1, 1] – Range – arcsin(x) = t if and only if − π – sin(arcsin(x)) = x provided − – arcsin(sin(x)) = x provided − π – additionally, arcsine is odd 2 2 and sin(t) = x 10.6 The Inverse Trigonometric Functions 821 Everything in Theorem 10.26 is a direct consequence of the facts that f (x) = cos(x) for 0 ≤ x ≤ π and F (x) = arccos(x) are inverses of each other as are g(x) = sin(x) for − π 2 and G(x) = arcsin(x). It’s about time for an example. 2 ≤ x ≤ π Example 10.6.1. 1. Find the exact values of the following. (a) arccos 1 2 (c) arccos − √ 2 2 (e) arccos cos π 6 (g) cos arccos − 3 5 (b) arcsin √ 2 2 (d) arcsin − 1 2 (f) arccos cos 11π 6 (h) sin arccos − 3 5 2. Rewrite the following as algebraic expressions of x and state the domain on which the equiv- alence is valid. (a) tan (arccos (x)) (b) cos (2 arcsin(x)) Solution. 1. (a) To ﬁnd arccos 1 2 , we need to ﬁnd the real number t (or, equivalently, an angle measuring 3 meets these 2 . We know t = π 2 and π 2 with sin(t) = √ 2 2 . The (b) The value of arcsin t radians) which lies between 0 and π with cos(t) = 1 = π criteria, so arccos 1 3 . 2 √ 2 is a real number t between − π 2 √ 2 number we seek is t = π 4 . Hence, arcsin 2 = π 4 . √ (c) The number t = arccos √ − 2 2 lies in the interval [0, π] with cos(t) = − √ 2 2 . Our answer 2 2 − is arccos (d) To ﬁnd arcsin − 1 2 answer is t = − π 6 so that arcsin − 1 = 3π 4 . , we seek the number t in the interval − , we could simply invoke Theorem 10.26 to get arccos cos π 6 . However, in order to make sure we understand why this is the case, we choose to work the example through using the deﬁnition of arccosine. Working from the inside out, arccos cos π is the real number t with 0 ≤ t ≤ π 6 with sin(t) = − 1 (e) Since . The 2 , π 2 6 2 = arccos √ 3 2 . We ﬁnd t = π . Now, arccos 6 , so that arccos cos π 6 = π 6 . and cos(t) = 822 Foundations of Trigonometry (f) Since 11π 6 does not fall between 0 and π, Theorem 10.26 does not apply. We are forced to . From work through from the inside out starting with arccos cos 11π 6 = arccos 6 . Hence, arccos cos 11π is to use Theorem 10.26 directly. Since − the previous problem, we know arccos (g) One way to simplify cos arccos − 3 5 between −1 and 1, we have that cos arccos − 3 5 before, to really understand why this cancellation occurs, we let t = arccos − 3 5 by deﬁnition, cos(t) = − 3 in (nearly) the same amount of time. 5 is 5 and we are done. However, as . Then, 5 , and we are ﬁnished 5 . Hence, cos arccos − 3 = cos(t) = − 3 = − 3 5 2. (h) As in the previous example, we let t = arccos − 3 5 5 for some t where 0 ≤ t ≤ π. Since cos(t) < 0, we can narrow this down a bit and conclude that π 2 < t < π, so that t corresponds to an angle in Quadrant II. In terms of t, then, we need to ﬁnd = sin(t). Using the Pythagorean Identity cos2(t) + sin2(t) = 1, we get sin arccos − 3 5 − 3 5 . Since t corresponds to a Quadrants II angle, we 5 choose sin(t) = 4 + sin2(t) = 1 or sin(t) = ± 4 5 . Hence, sin arccos − 3 so that cos(ta) We begin this problem in the same manner we began the previous two problems. To help us see the forest for the trees, we let t = arccos(x), so our goal is to ﬁnd a way to express tan (arccos (x)) = tan(t) in terms of x. Since t = arccos(x), we know cos(t) = x where 0 ≤ t ≤ π, but since we are after an expression for tan(t), we know we need to throw out t = π 2 < t ≤ π so that, geometrically, t corresponds to an angle in Quadrant I or Quadrant II. One approach3 to ﬁnding tan(t) is to use the quotient identity tan(t) = sin(t) cos(t) . Substituting cos(t) = x into the Pythagorean Identity cos2(t) + sin2(t) = 1 gives x2 + sin2(t) = 1, from which we 1 − x2. Since t corresponds to angles in Quadrants I and II, sin(t) ≥ 0, get sin(t) = ± so we choose sin(t) = 2 from consideration. Hence, either 0 ≤ t < π 1 − x2. Thus, 2 or π √ √ tan(t) = sin(t) cos(t) = √ 1 − x2 x To determine the values of x for which this equivalence is valid, we consider our substitution t = arccos(x). Since the domain of arccos(x) is [−1, 1], we know we must restrict −1 ≤ x ≤ 1. Additionally, since we had to discard t = π 2 , we need to discard x = cos π 2 = 0. Hence, tan (arccos (x)) = is valid for x in [−1, 0) ∪ (0, 1]. 1−x2 x √ 2 , π (b) We proceed as in the previous problem by writing t = arcsin(x) so that t lies in the with sin(t) = x. We aim to express cos (2 arcsin(x)) = cos(2t) in terms interval − π of x. Since cos(2t) is deﬁned everywhere, we get no additional restrictions on t as we did in the previous problem. We have three choices for rewriting cos(2t): cos2(t) − sin2(t), 2 cos2(t) − 1 and 1 − 2 sin2(t). Since we know x = sin(t), it is easiest to use the last form: 2 cos (2 arcsin(x)) = cos(2t) = 1 − 2 sin2(t) = 1 − 2x2 3Alternatively, we could use the identity: 1 + tan2(t) = sec2(t). Since x = cos(t), sec(t) = 1 cos(t) = 1 x . The reader is invited to work through this approach to see what, if any, diﬃculties arise. 10.6 The Inverse Trigonometric Functions 823 To ﬁnd the restrictions on x, we once again appeal to our substitution t = arcsin(x). Since arcsin(x) is deﬁned only for −1 ≤ x ≤ 1, the equivalence cos (2 arcsin(x)) = 1−2x2 is valid only on [−1, 1]. A few remarks about Example 10.6.1 are in order. Most of the common errors encountered in dealing with the inverse circular functions come from the need to restrict the domains of the original functions so that they are one-to-one. One instance of this phenomenon is the fact that arccos cos 11π 6 as opposed to 11π 6 . This is the exact same phenomenon discussed in Section 6 (−2)2 = 2 as opposed to −2. Additionally, even though the expression we 5.2 when we saw arrived at in part 2b above, namely 1 − 2x2, is deﬁned for all real numbers, the equivalence cos (2 arcsin(x)) = 1 − 2x2 is valid for only −1 ≤ x ≤ 1. This is akin to the fact that while the x)2 = x is valid only for x ≥ 0. For expression x is deﬁned for all real numbers, the equivalence ( this reason, it pays to be careful when we determine the intervals where such equivalences are valid. = π √ The next pair of functions we wish to discuss are the inverses of tangent and cotangent, which are named arctangent and arccotangent, respectively. First, we restrict f (x) = tan(x) to its fundamental cycle on − π to obtain f −1(x) = arctan(x). Among other things, note that the 2 , π 2 vertical asymptotes x = − π 2 and x = π 2 of the graph of f (x) = tan(x) become the horizontal 2 and y = π asymptotes y = − π 2 of the graph of f −1(x) = arctan(x). y 1 −x) = tan(x), − π 2 < x < π 2 . reﬂect across y = x −−−−−−−−−−−−→ switch x and y coordinates y π 2 π 4 −1(x) = arctan(x). Next, we restrict g(x) = cot(x) to its fundamental cycle on (0, π) to obtain g−1(x) = arccot(x). Once again, the vertical asymptotes x = 0 and x = π of the graph of g(x) = cot(x) become the horizontal asymptotes y = 0 and y = π of the graph of g−1(x) = arccot(x). We show these graphs on the next page and list some of the basic properties of the arctangent and arccotangent functions. 824 Foundations of Trigonometry y 1 −1 π 4 π 2 3π 4 π x y π 3π 4 π 2 π 4 g(x) = cot(x), 0 < x < π. reﬂect across y = x −−−−−−−−−−−−→ switch x and y coordinates g−1(x) = arccot(x). −1 1 x Theorem 10.27. Properties of the Arctangent and Arccotangent Functions Properties of F (x) = arctan(x) – Domain: (−∞, ∞) – Range: − π 2 , π 2 – as x → −∞, arctan(x) → − π 2 – arctan(x) = t if and only if − π for x > 0 – arctan(x) = arccot 1 x – tan (arctan(x)) = x for all real numbers x 2 < x < π – arctan(tan(x)) = x provided − π – additionally, arctangent is odd 2 +; as x → ∞, arctan(x and tan(t) = x − Properties of G(x) = arccot(x) – Domain: (−∞, ∞) – Range: (0, π) – as x → −∞, arccot(x) → π−; as x → ∞, arccot(x) → 0+ – arccot(x) = t if and only if 0 < t < π and cot(t) = x – arccot(x) = arctan 1 x – cot (arccot(x)) = x for all real numbers x for x > 0 – arccot(cot(x)) = x provided 0 < x < π 10.6 The Inverse Trigonometric Functions 825 Example 10.6.2. 1. Find the exact values of the following. √ (a) arctan( 3) (c) cot(arccot(−5)) √ 3) (b) arccot(− (d) sin arctan − 3 4 2. Rewrite the following as algebraic expressions of x and state the domain on which the equiv- alence is valid. (a) tan(2 arctan(x)) (b) cos(arccot(2x)) Solution. 1. 2. (a) We know arctan( 3 , so arctan( t = π 3) is the real number t between − π 3) = π 3 . √ √ √ (b) The real number t = arccot(− 3) = 5π 6 . arccot(− √ 2 and π 2 with tan(t) = √ 3. We ﬁnd 3) lies in the interval (0, π) with cot(t) = − 3. We get √ (c) We can apply Theorem 10.27 directly and obtain cot(arccot(−5)) = −5. However, working it through provides us with yet another opportunity to understand why this is the case. Letting t = arccot(−5), we have that t belongs to the interval (0, π) and cot(t) = −5. Hence, cot(arccot(−5)) = cot(t) = −5. (a) If we let t = arctan(x), then − π (d) We start simplifying sin arctan − 3 4 . Then tan(t) = − 3 by letting t = arctan − , we choose csc(t) = − 5 2 3 , so sin(t) = − 3 3 . Substituting, we get 1 + − 4 2 . Since tan(t) < 0, we know, in fact, − π 4 for some − π 2 < t < 0. One way to proceed is to use The Pythagorean Identity, 1+cot2(t) = csc2(t), since this relates the reciprocals of tan(t) and sin(t) and is valid for all t under consideration.4 From tan(t) = − 3 4 , we get cot(t) = − 4 3 . Since = − 3 − π 5 . = csc2(t) so that csc(t) = ± 5 5 . Hence, sin arctan − 3 2 and tan(t) = x. We look for a way to express tan(2 arctan(x)) = tan(2t) in terms of x. Before we get started using identities, we note that tan(2t) is undeﬁned when 2t = π 2 + πk for integers k. Dividing both sides of this equation by 2 tells us we need to exclude values of t where t = π 4 + π 2 k, where k is are t = ± π an integer. The only members of this family which lie in − π 2 , π
4 , which 2 . Returning ∪ − π means the values of t under consideration are − to arctan(2t), we note the double angle identity tan(2t) = 2 tan(t) 1−tan2(t) , is valid for all the values of t under consideration, hence we get tan(2 arctan(x)) = tan(2t) = 2 tan(t) 1 − tan2(t) = 2x 1 − x2 4It’s always a good idea to make sure the identities used in these situations are valid for all values t under consideration. Check our work back in Example 10.6.1. Were the identities we used there valid for all t under consideration? A pedantic point, to be sure, but what else do you expect from this book? 826 Foundations of Trigonometry To ﬁnd where this equivalence is valid we check back with our substitution t = arctan(x). Since the domain of arctan(x) is all real numbers, the only exclusions come from the values of t we discarded earlier, t = ± π 4 . Since x = tan(t), this means we exclude x = tan ± π 1−x2 holds for all x in 4 (−∞, −1) ∪ (−1, 1) ∪ (1, ∞). = ±1. Hence, the equivalence tan(2 arctan(x)) = 2x (b) To get started, we let t = arccot(2x) so that cot(t) = 2x where 0 < t < π. In terms of t, cos(arccot(2x)) = cos(t), and our goal is to express the latter in terms of x. Since cos(t) is always deﬁned, there are no additional restrictions on t, so we can begin using identities to relate cot(t) to cos(t). The identity cot(t) = cos(t) sin(t) is valid for t in (0, π), so our strategy is to obtain sin(t) in terms of x, then write cos(t) = cot(t) sin(t). The identity 1 + cot2(t) = csc2(t) holds for all t in (0, π) and relates cot(t) and csc(t) = 1 sin(t) . Substituting cot(t) = 2x, we get 1 + (2x)2 = csc2(t), or csc(t) = ± 4x2 + 1. Since t is . Hence, between 0 and π, csc(t) > 0, so csc(t) = 4x2 + 1 which gives sin(t) = 1√ √ √ 4x2+1 cos(arccot(2x)) = cos(t) = cot(t) sin(t) = √ 2x 4x2 + 1 Since arccot(2x) is deﬁned for all real numbers x and we encountered no additional restrictions on t, we have cos (arccot(2x)) = 2x√ for all real numbers x. 4x2+1 The last two functions to invert are secant and cosecant. A portion of each of their graphs, which were ﬁrst discussed in Subsection 10.5.2, are given below with the fundamental cycles highlighted. y y x x The graph of y = sec(x). The graph of y = csc(x). It is clear from the graph of secant that we cannot ﬁnd one single continuous piece of its graph which covers its entire range of (−∞, −1] ∪ [1, ∞) and restricts the domain of the function so that it is one-to-one. The same is true for cosecant. Thus in order to deﬁne the arcsecant and arccosecant functions, we must settle for a piecewise approach wherein we choose one piece to cover the top of the range, namely [1, ∞), and another piece to cover the bottom, namely (−∞, −1]. There are two generally accepted ways make these choices which restrict the domains of these functions so that they are one-to-one. One approach simpliﬁes the Trigonometry associated with the inverse functions, but complicates the Calculus; the other makes the Calculus easier, but the Trigonometry less so. We present both points of view. 10.6 The Inverse Trigonometric Functions 827 10.6.1 Inverses of Secant and Cosecant: Trigonometry Friendly Approach In this subsection, we restrict the secant and cosecant functions to coincide with the restrictions on cosine and sine, respectively. For f (x) = sec(x), we restrict the domain to 0x) = sec(x) on 0, π 2 ∪ π 2 , π reﬂect across y = x −−−−−−−−−−−−→ switch x and y coordinates −1 1 x f −1(x) = arcsec(x) and we restrict g(x) = csc(x) to − π 2 , 0 ∪ 0, π 2 . y 1 −1 − π 2 x π 2 g(x) = csc(x) on − π 2 , 0 ∪ 0, π 2 y π 2 −1 1 x reﬂect across y = x −−−−−−−−−−−−→ switch x and y coordinates − π 2 g−1(x) = arccsc(x) Note that for both arcsecant and arccosecant, the domain is (−∞, −1] ∪ [1, ∞). Taking a page from Section 2.2, we can rewrite this as {x : \|x\| ≥ 1}. This is often done in Calculus textbooks, so we include it here for completeness. Using these deﬁnitions, we get the following properties of the arcsecant and arccosecant functions. 828 Foundations of Trigonometry Theorem 10.28. Properties of the Arcsecant and Arccosecant Functionsa Properties of F (x) = arcsec(x) ∪ π – Domain: {x : \|x\| ≥ 1} = (−∞, −1] ∪ [1, ∞) – Range: 0, π 2 , π 2 – as x → −∞, arcsec(x) → π 2 – arcsec(x) = t if and only if 0 ≤ t < π – arcsec(x) = arccos 1 x – sec (arcsec(x)) = x provided \|x\| ≥ 1 – arcsec(sec(x)) = x provided 0 ≤ x < π provided \|x\| ≥ 1 2 or π 2 or π 2 < x ≤ π +; as x → ∞, arcsec(x and sec(t) = x Properties of G(x) = arccsc(x) 2 2 , 0 ∪ 0, π – Domain: {x : \|x\| ≥ 1} = (−∞, −1] ∪ [1, ∞) – Range: − π – as x → −∞, arccsc(x) → 0−; as x → ∞, arccsc(x) → 0+ – arccsc(x) = t if and only if − π – arccsc(x) = arcsin 1 x – csc (arccsc(x)) = x provided \|x\| ≥ 1 – arccsc(csc(x)) = x provided − π – additionally, arccosecant is odd 2 ≤ x < 0 or or 0 < t ≤ π provided \|x\| ≥ 1 2 2 and csc(t) = x a. . . assuming the “Trigonometry Friendly” ranges are used. Example 10.6.3. 1. Find the exact values of the following. (a) arcsec(2) (b) arccsc(−2) (c) arcsec sec 5π 4 (d) cot (arccsc (−3)) 2. Rewrite the following as algebraic expressions of x and state the domain on which the equiv- alence is valid. (a) tan(arcsec(x)) (b) cos(arccsc(4x)) 10.6 The Inverse Trigonometric Functions 829 Solution. 1. 2. (a) Using Theorem 10.28, we have arcsec(2) = arccos 1 2 (b) Once again, Theorem 10.28 comes to our aid giving arccsc(−2) = arcsin − 1 2 (c) Since 5π 4 doesn’t fall between 0 and π 2 and π, we cannot use the inverse property stated in Theorem 10.28. We can, nevertheless, begin by working ‘inside out’ which yields arcsec sec 5π 4 = arcsec(− = − π 6 . 2) = arccos = π 3 . 2 or π = 3π 4 . 2 2 √ − √ (d) One way to begin to simplify cot (arccsc (−3)) is to let t = arccsc(−3). Then, csc(t) = −3 and, since this is negative, we have that t lies in the interval − π 2 , 0. We are after cot (arccsc (−3)) = cot(t), so we use the Pythagorean Identity 1 + cot2(t) = csc2(t). Substituting, we have 1 + cot2(t) = (−3)2, or cot(t) = ± 2 ≤ t < 0, cot(t) < 0, so we get cot (arccsc (−3)) = −2 2. Since − π 8 = ±2 √ √ √ 2. ∪ π (a) We begin simplifying tan(arcsec(x)) by letting t = arcsec(x). Then, sec(t) = x for t in 0, π 2 , π, and we seek a formula for tan(t). Since tan(t) is deﬁned for all t values 2 under consideration, we have no additional restrictions on t. To relate sec(t) to tan(t), we use the identity 1 + tan2(t) = sec2(t). This is valid for all values of t under consideration, √ and when we substitute sec(t) = x, we get 1 + tan2(t) = x2. Hence, tan(t) = ± x2 − 1. 2 , π then then tan(t) ≥ 0; if, on the the other hand, t belongs to π If t belongs to 0, π 2 tan(t) ≤ 0. As a result, we get a piecewise deﬁned function for tan(t) tan(t) = √ √ − x2 − 1, x2 − 1, if 0 ≤ t < π 2 if π 2 < t ≤ π Now we need to determine what these conditions on t mean for x. Since x = sec(t), when , x ≤ −1. Since we encountered no further restrictions on t, the equivalence below holds for all x in (−∞, −1] ∪ [1, ∞). 2 , x ≥ 1, and when π tan(arcsec(x)) = √ √ − x2 − 1, if x ≥ 1 x2 − 1, if x ≤ −1 2 , 0∪0, π (b) To simplify cos(arccsc(4x)), we start by letting t = arccsc(4x). Then csc(t) = 4x for t in , and we now set about ﬁnding an expression for cos(arccsc(4x)) = cos(t). − π Since cos(t) is deﬁned for all t, we do not encounter any additional restrictions on t. From csc(t) = 4x, we get sin(t) = 1 4x , so to ﬁnd cos(t), we can make use if the identity cos2(t) + sin2(t) = 1. Substituting sin(t) = 1 = 1. Solving, we get 2 2 cos(t) = ± 4x gives cos2(t) + 1 √ 4x 16x2 − 1 16x2 = ± 16x2 − 1 4\|x\| Since t belongs to − π . (The absolute values here are necessary, since x could be negative.) To ﬁnd the values for , we know cos(t) ≥ 0, so we choose cos(t) = 2 , 0 ∪ 0, π 16−x2 4\|x\| 2 √ 830 Foundations of Trigonometry which this equivalence is valid, we look back at our original substution, t = arccsc(4x). Since the domain of arccsc(x) requires its argument x to satisfy \|x\| ≥ 1, the domain of arccsc(4x) requires \|4x\| ≥ 1. Using Theorem 2.4, we rewrite this inequality and solve to get x ≤ − 1 4 . Since we had no additional restrictions on t, the equivalence ∪ 1 cos(arccsc(4x)) = holds for all x in −∞, − 1 4 4 or x ≥ 1 16x2−1 4\|x\| 4 , ∞. √ Inverses of Secant and Cosecant: Calculus Friendly Approach 10.6.2 In this subsection, we restrict f (x) = sec(x) to 0, π 2 y 1 −1 π 2 π 3π 2 x ∪ π, 3π 2 y 3π 2 π π 2 f (x) = sec(x) on 0, π 2 ∪ π, 3π 2 reﬂect across y = x −−−−−−−−−−−−→ switch x and y coordinates −1 1 x f −1(x) = arcsec(x) and we restrict g(x) = csc(x) to 0, π 2 ∪ π, 3π 2 . y 1 −1 π 2 π 3π 2 x y 3π 2 π π 2 g(x) = csc(x) on 0, π 2 ∪ π, 3π 2 reﬂect across y = x −−−−−−−−−−−−→ switch x and y coordinates −1 1 x g−1(x) = arccsc(x) Using these deﬁnitions, we get the following result. 10.6 The Inverse Trigonometric Functions 831 Theorem 10.29. Properties of the Arcsecant and Arccosecant Functionsa Properties of F (x) = arcsec(x) ∪ π, 3π 2 – Domain: {x : \|x\| ≥ 1} = (−∞, −1] ∪ [1, ∞) – Range: 0, π 2 – as x → −∞, arcsec(x) → 3π 2 – arcsec(x) = t if and only if 0 ≤ t < π – arcsec(x) = arccos 1 for x ≥ 1 onlyb x – sec (arcsec(x)) = x provided \|x\| ≥ 1 – arcsec(sec(x)) = x provided 0 ≤ x < π − 2 or π ≤ x < 3π 2 ; as x → ∞, arcsec(x) → π 2 2 or π ≤ t < 3π − 2 and sec(t) = x Properties of G(x) = arccsc(x) ∪ π, 3π 2 – Domain: {x : \|x\| ≥ 1} = (−∞, −1] ∪ [1, ∞) – Range: 0, π 2 – as x → −∞, arccsc(x) → π+; as x → ∞, arccsc(x) → 0+ – arccsc(x) = t if and only if 0 < t ≤ π for x ≥ 1 onlyc – arccsc(x) = arcsin 1 x – csc (arccsc(x)) = x provided \|x\| ≥ 1 – arccsc(csc(x)) = x provided 0 < x ≤ π 2 or π < t ≤ 3π 2 or π < x ≤ 3π 2 2 and csc(t) = x a. . . assuming the “Calculus Friendly” ranges are used. bCompare this with the similar result in Theorem 10.28. cCompare this with the similar result in Theorem 10.28. Our next example is a duplicate of Example 10.6.3. The interested reader is invited to compare and contrast the solution to each. Example 10.6.4. 1. Find the exact values of the following. (a) arcsec(2) (b) arccsc(−2) (c) arcsec sec 5π 4 (d) c
ot (arccsc (−3)) 2. Rewrite the following as algebraic expressions of x and state the domain on which the equiv- alence is valid. (a) tan(arcsec(x)) (b) cos(arccsc(4x)) 832 Solution. Foundations of Trigonometry 1. 2. (a) Since 2 ≥ 1, we may invoke Theorem 10.29 to get arcsec(2) = arccos 1 2 (b) Unfortunately, −2 is not greater to or equal to 1, so we cannot apply Theorem 10.29 to arccsc(−2) and convert this into an arcsine problem. Instead, we appeal to the deﬁnition. and satisﬁes csc(t) = −2. The t The real number t = arccsc(−2) lies in 0, π 2 we’re after is t = 7π ∪ π, 3π 2 6 , so arccsc(−2) = 7π 6 . = π 3 . lies between π and 3π (c) Since 5π 4 arcsec sec 5π 4 tion as we have done in the previous examples to see how it goes. 2 , we may apply Theorem 10.29 directly to simplify 4 . We encourage the reader to work this through using the deﬁni- = 5π (d) To simplify cot (arccsc (−3)) we let t = arccsc (−3) so that cot (arccsc (−3)) = cot(t). . Using the identity 2. Since √ 2. We know csc(t) = −3, and since this is negative, t lies in π, 3π 2 1 + cot2(t) = csc2(t), we ﬁnd 1 + cot2(t) = (−3)2 so that cot(t) = ± t is in the interval π, 3π 2 , we know cot(t) > 0. Our answer is cot (arccsc (−3)) = 2 8 = ±2 √ √ ∪ π, 3π 2 (a) We begin simplifying tan(arcsec(x)) by letting t = arcsec(x). Then, sec(t) = x for t in 0, π , and we seek a formula for tan(t). Since tan(t) is deﬁned for all t values 2 under consideration, we have no additional restrictions on t. To relate sec(t) to tan(t), we use the identity 1 + tan2(t) = sec2(t). This is valid for all values of t under consideration, and when we substitute sec(t) = x, we get 1 + tan2(t) = x2. Hence, tan(t) = ± x2 − 1. Since t lies in 0, π , tan(t) ≥ 0, so we choose tan(t) = x2 − 1. Since we found √ 2 x2 − 1 holds for all x no additional restrictions on t, the equivalence tan(arcsec(x)) = in the domain of t = arcsec(x), namely (−∞, −1] ∪ [1, ∞). ∪ π, 3π 2 √ √ ∪ π, 3π 2 (b) To simplify cos(arccsc(4x)), we start by letting t = arccsc(4x). Then csc(t) = 4x for t in , and we now set about ﬁnding an expression for cos(arccsc(4x)) = cos(t). 0, π 2 Since cos(t) is deﬁned for all t, we do not encounter any additional restrictions on t. From csc(t) = 4x, we get sin(t) = 1 4x , so to ﬁnd cos(t), we can make use if the identity cos2(t) + sin2(t) = 1. Substituting sin(t) = 1 = 1. Solving, we get 2 cos(t) = ± 4x gives cos2(t) + 1 √ 4x 16x2 − 1 16x2 = ± 16x2 − 1 4\|x\| , then cos(t) ≥ 0, and we choose cos(t) = in which case cos(t) ≤ 0, so, we choose cos(t) = − If t lies in 0, π 2 to π, 3π 2 (momentarily) piecewise deﬁned function for cos(t) √ 16x2−1 4\|x\| √ . Otherwise, t belongs 16x2−1 4\|x\| This leads us to a cos(t) = √ √ −    16x2 − 1 4\|x\| 16x2 − 1 4\|x\| , , if 0 ≤ t ≤ π 2 if π < t ≤ 3π 2 10.6 The Inverse Trigonometric Functions 833 We now see what these restrictions mean in terms of x. Since 4x = csc(t), we get that for 0 ≤ t ≤ π 2 , 4x ≥ 1, or x ≥ 1 4 . In this case, we can simplify \|x\| = x so √ √ cos(t) = 16x2 − 1 4\|x\| = 16x2 − 1 4x Similarly, for π < t ≤ 3π get 2 , we get 4x ≤ −1, or x ≤ − 1 4 . In this case, \|x\| = −x, so we also cos(t) = − √ 16x2 − 1 4\|x\| = − √ 16x2 − 1 4(−x) = √ 16x2 − 1 4x 16x2−1 Hence, in all cases, cos(arccsc(4x)) = 4x the domain of t = arccsc(4x), namely −∞, − 1 4 , and this equivalence is valid for all x in ∪ 1 4 , ∞ √ 10.6.3 Calculators and the Inverse Circular Functions. In the sections to come, we will have need to approximate the values of the inverse circular functions. On most calculators, only the arcsine, arccosine and arctangent functions are available and they are usually labeled as sin−1, cos−1 and tan−1, respectively. If we are asked to approximate these values, it is a simple matter to punch up the appropriate decimal on the calculator. If we are asked for an arccotangent, arcsecant or arccosecant, however, we often need to employ some ingenuity, as our next example illustrates. Example 10.6.5. 1. Use a calculator to approximate the following values to four decimal places. (a) arccot(2) (b) arcsec(5) (c) arccot(−2) (d) arccsc − 3 2 2. Find the domain and range of the following functions. Check your answers using a calculator. (a) f (x) = π 2 − arccos x 5 (b) f (x) = 3 arctan (4x). (c) f (x) = arccot x 2 + π Solution. 1. (a) Since 2 > 0, we can use the property listed in Theorem 10.27 to rewrite arccot(2) as arccot(2) = arctan 1 2 . In ‘radian’ mode, we ﬁnd arccot(2) = arctan 1 2 ≈ 0.4636. (b) Since 5 ≥ 1, we can use the property from either Theorem 10.28 or Theorem 10.29 to write arcsec(5) = arccos 1 5 ≈ 1.3694. 834 Foundations of Trigonometry (c) Since the argument −2 is negative, we cannot directly apply Theorem 10.27 to help us ﬁnd arccot(−2). Let t = arccot(−2). Then t is a real number such that 0 < t < π and cot(t) = −2. Moreover, since cot(t) < 0, we know π 2 < t < π. Geometrically, this means t corresponds to a Quadrant II angle θ = t radians. This allows us to proceed using a ‘reference angle’ approach. Consider α, the reference angle for θ, as pictured below. By deﬁnition, α is an acute angle so 0 < α < π 2 , and the Reference Angle Theorem, Theorem 10.2, tells us that cot(α) = 2. This means α = arccot(2) radians. Since the argument of arccotangent is now a positive 2, we can use Theorem 10.27 to get α = arccot(2) = arctan 1 ≈ 2.6779 radians, 2 we get arccot(−2) ≈ 2.6779. radians. Since θ = π − α = π − arctan 1 2 y 1 θ = arccot(−2) radians α 1 x . By deﬁnition, the real number Another way to attack the problem is to use arctan − 1 2 t = arctan − 1 satisﬁes tan(t) = − 1 2 . Since tan(t) < 0, we know 2 more speciﬁcally that − π 2 < t < 0, so t corresponds to an angle β in Quadrant IV. To ﬁnd the value of arccot(−2), we once again visualize the angle θ = arccot(−2) radians and note that it is a Quadrant II angle with tan(θ) = − 1 2 . This means it is exactly π units away from β, and we get θ = π + β = π + arctan − 1 ≈ 2.6779 radians. Hence, 2 as before, arccot(−2) ≈ 2.6779. 2 < t < π 2 with − π 10.6 The Inverse Trigonometric Functions 835 y 1 θ = arccot(−2) radians , π (d) If the range of arccosecant is taken to be − π = arcsin − 2 3 ∪ π, 3π 2 , we can use Theorem 10.28 to ≈ −0.7297. If, on the other hand, the range of arccosecant get arccsc − 3 2 is taken to be 0, π , then we proceed as in the previous problem by letting 2 . Then t is a real number with csc(t) = − 3 t = arccsc − 3 2 . Since csc(t) < 0, we have 2 that π < θ ≤ 3π 2 , so t corresponds to a Quadrant III angle, θ. As above, we let α be 2 , which means α = arccsc 3 the reference angle for θ. Then 0 < α < π 2 radians. Since the argument of arccosecant is now positive, we may use Theorem 10.29 to get α = arccsc 3 ≈ 3.8713 = arcsin 2 2 3 ≈ 3.8713. radians, arccsc − 3 2 radians. Since θ = π + α = π + arcsin 2 3 2 and csc(α) = 3 y 1 θ = arccsc − 3 2 radians α 1 x 836 2. Foundations of Trigonometry 5 by setting the argument of the arccosine, in this case x (a) Since the domain of F (x) = arccos(x) is −1 ≤ x ≤ 1, we can ﬁnd the domain of 2 − arccos x f (x) = π 5 , between −1 and 1. Solving −1 ≤ x 5 ≤ 1 gives −5 ≤ x ≤ 5, so the domain is [−5, 5]. To determine the range of f , we take a cue from Section 1.7. Three ‘key’ points on the graph of and (1, 0) . Following the procedure outlined in F (x) = arccos(x) are (−1, π), 0, π 2 Theorem 1.7, we track these points to −5, − π . Plotting these values 2 tells us that the range5 of f is − π , (0, 0) and 5, π 2 . Our graph conﬁrms our results. 2 , π 2 (b) To ﬁnd the domain and range of f (x) = 3 arctan (4x), we note that since the domain of F (x) = arctan(x) is all real numbers, the only restrictions, if any, on the domain of f (x) = 3 arctan (4x) come from the argument of the arctangent, in this case, 4x. Since 4x is deﬁned for all real numbers, we have established that the domain of f is all real numbers. To determine the range of f , we can, once again, appeal to Theorem 1.7. Choosing our ‘key’ point to be (0, 0) and tracking the horizontal asymptotes y = − π 2 and y = π 2 , we ﬁnd that the graph of y = f (x) = 3 arctan (4x) diﬀers from the graph of y = F (x) = arctan(x) by a horizontal compression by a factor of 4 and a vertical stretch by a factor of 3. It is the latter which aﬀects the range, producing a range of − 3π . We conﬁrm our ﬁndings on the calculator below. 2 , 3π 2 y = f (x) = − arccos y = f (x) = 3 arctan (4x) π 2 x 5 (c) To ﬁnd the domain of g(x) = arccot x + π, we proceed as above. Since the domain of 2 G(x) = arccot(x) is (−∞, ∞), and x 2 is deﬁned for all x, we get that the domain of g is (−∞, ∞) as well. As for the range, we note that the range of G(x) = arccot(x), like that of F (x) = arctan(x), is limited by a pair of horizontal asymptotes, in this case y = 0 and y = π. Following Theorem 1.7, we graph y = g(x) = arccot x + π starting with 2 y = G(x) = arccot(x) and ﬁrst performing a horizontal expansion by a factor of 2 and following that with a vertical shift upwards by π. This latter transformation is the one which aﬀects the range, making it now (π, 2π). To check this graphically, we encounter a bit of a problem, since on many calculators, there is no shortcut button corresponding to the arccotangent function. Taking a cue from number 1c, we attempt to rewrite +π in terms of the arctangent function. Using Theorem 10.27, we have g(x) = arccot x 2 when x = arctan 2 that arccot x 2 > 0, or, in this case, when x > 0. Hence, for x > 0, x 2 + π. When x we have g(x) = arctan 2 2 < 0, we can use the same argument in number x . 1c that gave us arccot(−2) = π + arctan − 1 2 to give us arccot x 2 = π + arctan 2 x 5It also conﬁrms our domain! 10.6 The Inverse Trigonometric Functions 837 Hence, for x < 0, g(x) = π + arctan 2 + 2π. What about x = 0? We x know g(0) = arccot(0) + π = π, and neither of the formulas for g involving arctangent will produce this result.6 Hence, in order to graph y = g(x) on our calculators, we need to write it as a piecewise deﬁned function: + π = arctan 2 x g(x) = arccot x 2 + π =    We show the
input and the result below. arctan 2 x arctan 2 x + 2π, π, + π, if x < 0 if x = 0 if x > 0 y = g(x) in terms of arctangent y = g(x) = arccot x 2 + π The inverse trigonometric functions are typically found in applications whenever the measure of an angle is required. One such scenario is presented in the following example. Example 10.6.6. 7 The roof on the house below has a ‘6/12 pitch’. This means that when viewed from the side, the roof line has a rise of 6 feet over a run of 12 feet. Find the angle of inclination from the bottom of the roof to the top of the roof. Express your answer in decimal degrees, rounded to the nearest hundredth of a degree. Front View Side View Solution. If we divide the side view of the house down the middle, we ﬁnd that the roof line forms the hypotenuse of a right triangle with legs of length 6 feet and 12 feet. Using Theorem 10.10, we 6Without Calculus, of course . . . 7The authors would like to thank Dan Stitz for this problem and associated graphics. 838 Foundations of Trigonometry ﬁnd the angle of inclination, labeled θ below, satisﬁes tan(θ) = 6 we can use the arctangent function and we ﬁnd θ = arctan 1 2 12 = 1 radians ≈ 26.56◦. 2 . Since θ is an acute angle, 6 feet θ 12 feet 10.6.4 Solving Equations Using the Inverse Trigonometric Functions. In Sections 10.2 and 10.3, we learned how to solve equations like sin(θ) = 1 2 for angles θ and tan(t) = −1 for real numbers t. In each case, we ultimately appealed to the Unit Circle and relied on the fact that the answers corresponded to a set of ‘common angles’ listed on page 724. If, on the other hand, we had been asked to ﬁnd all angles with sin(θ) = 1 3 or solve tan(t) = −2 for real numbers t, we would have been hard-pressed to do so. With the introduction of the inverse trigonometric functions, however, we are now in a position to solve these equations. A good parallel to keep in mind is how the square root function can be used to solve certain quadratic equations. The equation x2 = 4 is a lot like sin(θ) = 1 2 in that it has friendly, ‘common value’ answers x = ±2. The equation x2 = 7, on the other hand, is a lot like sin(θ) = 1 3 . We know8 there are answers, but we can’t express them using ‘friendly’ numbers.9 To solve x2 = 7, we make use of the square root function and write x = ± 7. We can certainly approximate these answers using a calculator, but as far as exact answers go, we leave them as x = ± 7. In the same way, we will use the arcsine function to solve sin(θ) = 1 3 , as seen in the following example. √ √ Example 10.6.7. Solve the following equations. 1. Find all angles θ for which sin(θ) = 1 3 . 2. Find all real numbers t for which tan(t) = −2 3. Solve sec(x) = − 5 3 for x. Solution. 1. If sin(θ) = 1 3 , then the terminal side of θ, when plotted in standard position, intersects the Unit Circle at y = 1 3 . Geometrically, we see that this happens at two places: in Quadrant I and Quadrant II. If we let α denote the acute solution to the equation, then all the solutions 8How do we know this again? 9This is all, of course, a matter of opinion. For the record, the authors ﬁnd ± √ 7 just as ‘nice’ as ±2. 10.6 The Inverse Trigonometric Functions 839 to this equation in Quadrant I are coterminal with α, and α serves as the reference angle for all of the solutions to this equation in Quadrant II = arcsin 1 3 radians α 1 x Since 1 3 isn’t the sine of any of the ‘common angles’ discussed earlier, we use the arcsine functions to express our answers. The real number t = arcsin 1 is deﬁned so it satisﬁes 3 radians. Since the solutions in Quadrant I 0 < t < π + 2πk are all coterminal with α, we get part of our solution to be θ = α + 2πk = arcsin 1 3 for integers k. Turning our attention to Quadrant II, we get one solution to be π − α. Hence, the Quadrant II solutions are θ = π − α + 2πk = π − arcsin 1 3 3 . Hence, α = arcsin 1 + 2πk, for integers k. 2 with sin(t) = 1 3 2. We may visualize the solutions to tan(t) = −2 as angles θ with tan(θ) = −2. Since tangent is negative only in Quadrants II and IV, we focus our eﬀorts there = arctan(−2) radians π β Since −2 isn’t the tangent of any of the ‘common angles’, we need to use the arctangent function to express our answers. The real number t = arctan(−2) satisﬁes tan(t) = −2 and − π 2 < t < 0. If we let β = arctan(−2) radians, we see that all of the Quadrant IV solutions 840 Foundations of Trigonometry to tan(θ) = −2 are coterminal with β. Moreover, the solutions from Quadrant II diﬀer by exactly π units from the solutions in Quadrant IV, so all the solutions to tan(θ) = −2 are of the form θ = β + πk = arctan(−2) + πk for some integer k. Switching back to the variable t, we record our ﬁnal answer to tan(t) = −2 as t = arctan(−2) + πk for integers k. 3. The last equation we are asked to solve, sec(x) = − 5 3 , poses two immediate problems. First, we are not told whether or not x represents an angle or a real number. We assume the latter, but note that we will use angles and the Unit Circle to solve the equation regardless. Second, as we have mentioned, there is no universally accepted range of the arcsecant function. For that reason, we adopt the advice given in Section 10.3 and convert this to the cosine problem cos(x) = − 3 5 . Adopting an angle approach, we consider the equation cos(θ) = − 3 5 and note that our solutions lie in Quadrants II and III. Since − 3 5 isn’t the cosine of any of the ‘common angles’, we’ll need to express our solutions in terms of the arccosine function. The real number 5 . If we let β = arccos − 3 t = arccos − 3 5 radians, we see that β is a Quadrant II angle. To obtain a Quadrant III angle solution, we may simply use −β = − arccos − 3 . Since all angle solutions are coterminal with β 5 + 2πk or 5 to be θ = β + 2πk = arccos − 3 or −β, we get our solutions to cos(θ) = − 3 + 2πk for integers k. Switching back to the variable x, we θ = −β + 2πk = − arccos − 3 5 + 2πk record our ﬁnal answer to sec(x) = − 5 for integers k. + 2πk or x = − arccos − 3 5 2 < t < π with cos(t) = − 3 3 as x = arccos − 3 is deﬁned so that π 5 5 5 y 1 y 1 β = arccos − 3 5 radians β = arccos − 3 5 radians 1 x 1 −β = − arccos − 3 5 x radians The reader is encouraged to check the answers found in Example 10.6.7 - both analytically and with the calculator (see Section 10.6.3). With practice, the inverse trigonometric functions will become as familiar to you as the square root function. Speaking of practice . . . 10.6 The Inverse Trigonometric Functions 841 10.6.5 Exercises In Exercises 1 - 40, ﬁnd the exact value. 1. arcsin (−1) 2. arcsin − √ 3 2 5. arcsin (0) 6. arcsin 1 2 3. arcsin − √ 2 2 7. arcsin √ 2 2 4. arcsin − 1 2 8. arcsin √ 3 2 9. arcsin (1) 10. arccos (−1) 11. arccos − √ 3 2 12. arccos − √ 2 2 13. arccos − 1 2 17. arccos √ 3 2 21. arctan − √ 3 3 14. arccos (0) 15. arccos 1 2 16. arccos √ 2 2 18. arccos (1) 19. arctan − √ 3 20. arctan (−1) 22. arctan (0) 23. arctan 25. arctan √ 3 26. arccot − √ 3 29. arccot (0) 30. arccot √ 3 3 33. arcsec (2) 34. arccsc (2) 37. arcsec √ 2 3 3 38. arccsc √ 2 3 3 √ 3 3 24. arctan (1) 27. arccot (−1) 28. arccot − √ 3 3 31. arccot (1) 32. arccot √ 3 35. arcsec √ 2 36. arccsc √ 2 39. arcsec (1) 40. arccsc (1) In Exercises 41 - 48, assume that the range of arcsecant is 0, π 2 arccosecant is 0, π 2 when ﬁnding the exact value. ∪ π, 3π 2 ∪ π, 3π 2 and that the range of 41. arcsec (−2) 45. arccsc (−2) 42. arcsec − √ 46. arccsc − √ 2 2 43. arcsec − 47. arccsc − √ 2 3 3 √ 2 3 3 44. arcsec (−1) 48. arccsc (−1) 842 Foundations of Trigonometry In Exercises 49 - 56, assume that the range of arcsecant is 0, π 2 arccosecant is − π when ﬁnding the exact value. 2 , 0 ∪ 0, π 2 ∪ π 2 , π and that the range of 49. arcsec (−2) 53. arccsc (−2) 50. arcsec − √ 54. arccsc − √ 2 2 51. arcsec − 55. arccsc − √ 2 3 3 √ 2 3 3 52. arcsec (−1) 56. arccsc (−1) In Exercises 57 - 86, ﬁnd the exact value or state that it is undeﬁned. 57. sin arcsin 1 2 58. sin arcsin − √ 2 2 60. sin (arcsin (−0.42)) 61. sin arcsin 5 4 63. cos arccos − 1 2 64. cos arccos 5 13 66. cos (arccos (π)) 67. tan (arctan (−1)) 59. sin arcsin 3 5 √ 62. cos arccos 2 2 65. cos (arccos (−0.998)) 68. tan arctan √ 3 70. tan (arctan (0.965)) 71. tan (arctan (3π)) 69. tan arctan 5 12 72. cot (arccot (1)) 73. cot arccot − √ 3 75. cot (arccot (−0.001)) 76. cot arccot 78. sec (arcsec (−1)) 79. sec arcsec 17π 4 1 2 82. csc arccsc √ 2 81. sec (arcsec (117π)) 84. csc arccsc √ 2 2 74. cot arccot − 7 24 77. sec (arcsec (2)) 80. sec (arcsec (0.75)) 83. csc arccsc − √ 2 3 3 85. csc (arccsc (1.0001)) 86. csc arccsc π 4 In Exercises 87 - 106, ﬁnd the exact value or state that it is undeﬁned. 87. arcsin sin π 6 88. arcsin − sin π 3 89. arcsin sin 3π 4 10.6 The Inverse Trigonometric Functions 843 90. arcsin sin 93. arccos 96. arccos cos cos 11π 6 2π 3 5π 4 91. arcsin sin 94. arccos 97. arctan 4π 3 3π 2 cos 92. arccos 95. arccos cos π 4 cos − π 6 tan tan π 3 π 2 98. arctan tan 101. arctan tan − π 4 2π 3 99. arctan (tan (π)) 100. arctan 102. arccot 105. arccot cot cot π 3 π 2 103. arccot cot 106. arccot cot − π 4 2π 3 104. arccot (cot (π)) In Exercises 107 - 118, assume that the range of arcsecant is 0, π 2 arccosecant is 0, π 2 when ﬁnding the exact value. ∪ π, 3π 2 117. arcsec sec 118. arccsc csc In Exercises 119 - 130, assume that the range of arcsecant is 0, π 2 arccosecant is − π when ﬁnding the exact value. 2 , 0 ∪ 0, π 2 sec π 4 107. arcsec 110. arcsec sec 113. arccsc csc 116. arccsc csc − π 2 5π 4 11π 6 sec π 4 119. arcsec 122. arcsec sec 125. arccsc csc 128. arccsc csc − π 2 5π 4 11π 6 108. arcsec sec 111. arcsec sec 114. arccsc csc 120. arcsec sec 123. arcsec sec 126. arccsc csc 4π 3 5π 3 2π 3 11π 12 4π 3 5π 3 2π 3 11π 12 ∪ π, 3π 2 and that the range of 109. arcsec sec 5π 6 csc π 6 112. arccsc 115. arccsc csc − π 2 9π 8 ∪ π 2 , π and that the range of 121. arcsec sec 5π 6 csc π 6 124. arccsc 127. arccsc csc − π 2 9π 8 129. arcsec sec 130. arccsc csc 844 Foundations of Trigonometry In Exercises 131 - 154, ﬁnd the exact value o
r state that it is undeﬁned. 131. sin arccos − 1 2 132. sin arccos 3 5 134. sin arccot √ 5 137. cos arctan √ 7 135. sin (arccsc (−3)) 140. tan arcsin − √ 2 5 5 141. tan arccos − 1 2 143. tan (arccot (12)) 144. cot arcsin 12 13 138. cos (arccot (3)) 139. cos (arcsec (5)) 133. sin (arctan (−2)) 136. cos arcsin − 5 13 142. tan arcsec 145. cot arccos 5 3 √ 3 2 √ 3 2 146. cot arccsc √ 5 147. cot (arctan (0.25)) 148. sec arccos 149. sec arcsin − 12 13 150. sec (arctan (10)) 151. sec arccot − √ 10 10 152. csc (arccot (9)) 153. csc arcsin 3 5 154. csc arctan − 2 3 In Exercises 155 - 164, ﬁnd the exact value or state that it is undeﬁned. 155. sin arcsin 5 13 + π 4 156. cos (arcsec(3) + arctan(2)) 157. tan arctan(3) + arccos − 3 5 158. sin 2 arcsin − 4 5 159. sin 2arccsc 13 5 161. cos 2 arcsin 3 5 163. cos 2arccot − √ 5 160. sin (2 arctan (2)) 162. cos 2arcsec 25 7 164. sin arctan(2) 2 10.6 The Inverse Trigonometric Functions 845 In Exercises 165 - 184, rewrite the quantity as algebraic expressions of x and state the domain on which the equivalence is valid. 165. sin (arccos (x)) 166. cos (arctan (x)) 167. tan (arcsin (x)) 168. sec (arctan (x)) 169. csc (arccos (x)) 170. sin (2 arctan (x)) 171. sin (2 arccos (x)) 172. cos (2 arctan (x)) 173. sin(arccos(2x)) 174. sin arccos x 5 175. cos arcsin x 2 176. cos (arctan (3x)) 177. sin(2 arcsin(7x)) 178. sin 2 arcsin √ x 3 3 179. cos(2 arcsin(4x)) 180. sec(arctan(2x)) tan(arctan(2x)) 181. sin (arcsin(x) + arccos(x)) 182. cos (arcsin(x) + arctan(x)) 183. tan (2 arcsin(x)) 184. sin arctan(x) 1 2 185. If sin(θ) = 186. If tan(θ) = 187. If sec(θ) = x 2 x 7 x 4 for − π 2 for − π 2 < θ < π 2 , ﬁnd an expression for θ + sin(2θ) in terms of x. < θ < π 2 , ﬁnd an expression for 1 2 θ − 1 2 sin(2θ) in terms of x. for 0 < θ < π 2 , ﬁnd an expression for 4 tan(θ) − 4θ in terms of x. In Exercises 188 - 207, solve the equation using the techniques discussed in Example 10.6.7 then approximate the solutions which lie in the interval [0, 2π) to four decimal places. 188. sin(x) = 7 11 189. cos(x) = − 2 9 190. sin(x) = −0.569 191. cos(x) = 0.117 192. sin(x) = 0.008 193. cos(x) = 194. tan(x) = 117 195. cot(x) = −12 196. sec(x) = 197. csc(x) = − 200. cos(x) = − 90 17 7 16 198. tan(x) = − √ 10 199. sin(x) = 201. tan(x) = 0.03 202. sin(x) = 0.3502 359 360 3 2 3 8 846 Foundations of Trigonometry 203. sin(x) = −0.721 204. cos(x) = 0.9824 205. cos(x) = −0.5637 206. cot(x) = 1 117 207. tan(x) = −0.6109 In Exercises 208 - 210, ﬁnd the two acute angles in the right triangle whose sides have the given lengths. Express your answers using degree measure rounded to two decimal places. 208. 3, 4 and 5 209. 5, 12 and 13 210. 336, 527 and 625 211. A guy wire 1000 feet long is attached to the top of a tower. When pulled taut it touches level ground 360 feet from the base of the tower. What angle does the wire make with the ground? Express your answer using degree measure rounded to one decimal place. 212. At Cliﬀs of Insanity Point, The Great Sasquatch Canyon is 7117 feet deep. From that point, a ﬁre is seen at a location known to be 10 miles away from the base of the sheer canyon wall. What angle of depression is made by the line of sight from the canyon edge to the ﬁre? Express your answer using degree measure rounded to one decimal place. 213. Shelving is being built at the Utility Muﬃn Research Library which is to be 14 inches deep. An 18-inch rod will be attached to the wall and the underside of the shelf at its edge away from the wall, forming a right triangle under the shelf to support it. What angle, to the nearest degree, will the rod make with the wall? 214. A parasailor is being pulled by a boat on Lake Ippizuti. The cable is 300 feet long and the parasailor is 100 feet above the surface of the water. What is the angle of elevation from the boat to the parasailor? Express your answer using degree measure rounded to one decimal place. 215. A tag-and-release program to study the Sasquatch population of the eponymous Sasquatch National Park is begun. From a 200 foot tall tower, a ranger spots a Sasquatch lumbering through the wilderness directly towards the tower. Let θ denote the angle of depression from the top of the tower to a point on the ground. If the range of the riﬂe with a tranquilizer dart is 300 feet, ﬁnd the smallest value of θ for which the corresponding point on the ground is in range of the riﬂe. Round your answer to the nearest hundreth of a degree. In Exercises 216 - 221, rewrite the given function as a sinusoid of the form S(x) = A sin(ωx + φ) using Exercises 35 and 36 in Section 10.5 for reference. Approximate the value of φ (which is in radians, of course) to four decimal places. 216. f (x) = 5 sin(3x) + 12 cos(3x) 217. f (x) = 3 cos(2x) + 4 sin(2x) 218. f (x) = cos(x) − 3 sin(x) 219. f (x) = 7 sin(10x) − 24 cos(10x) 10.6 The Inverse Trigonometric Functions 847 220. f (x) = − cos(x) − 2 √ 2 sin(x) 221. f (x) = 2 sin(x) − cos(x) In Exercises 222 - 233, ﬁnd the domain of the given function. Write your answers in interval notation. 222. f (x) = arcsin(5x) 223. f (x) = arccos 3x − 1 2 224. f (x) = arcsin 2x2 225. f (x) = arccos 1 x2 − 4 226. f (x) = arctan(4x) 227. f (x) = arccot 2x x2 − 9 228. f (x) = arctan(ln(2x − 1)) 229. f (x) = arccot( √ 2x − 1) 230. f (x) = arcsec(12x) 231. f (x) = arccsc(x + 5) 232. f (x) = arcsec x3 8 233. f (x) = arccsc e2x 234. Show that arcsec(x) = arccos of f (x) = arcsec(x). 1 x for \|x\| ≥ 1 as long as we use 0, ∪ π 2 , π π 2 as the range 235. Show that arccsc(x) = arcsin of f (x) = arccsc(x). 1 x for \|x\| ≥ 1 as long as we use − π 2 , 0 0, ∪ π 2 as the range 236. Show that arcsin(x) + arccos(x) = π 2 for −1 ≤ x ≤ 1. 237. Discuss with your classmates why arcsin 1 2 = 30◦. 238. Use the following picture and the series of exercises on the next page to show that arctan(1) + arctan(2) + arctan(3) = π y D(2, 3) A(0, 1) α β γ B(1, 0) x C(2, 0) O(0, 0) 848 Foundations of Trigonometry (a) Clearly AOB and BCD are right triangles because the line through O and A and the line through C and D are perpendicular to the x-axis. Use the distance formula to show that BAD is also a right triangle (with ∠BAD being the right angle) by showing that the sides of the triangle satisfy the Pythagorean Theorem. (b) Use AOB to show that α = arctan(1) (c) Use BAD to show that β = arctan(2) (d) Use BCD to show that γ = arctan(3) (e) Use the fact that O, B and C all lie on the x-axis to conclude that α + β + γ = π. Thus arctan(1) + arctan(2) + arctan(3) = π. 10.6 The Inverse Trigonometric Functions 849 10.6.6 Answers 1. arcsin (−1) = − π 2 2. arcsin − √ 3 2 = − π 3 3. arcsin − √ 2 2 = − π 4 4. arcsin 7. arcsin − . arcsin (0) = 0 8. arcsin √ 3 2 10. arccos (−1) = π 11. arccos − √ 3 2 = π 3 = 5π 6 13. arccos 16. arccos = 2π 14. arccos (0) = 17. arccos √ 3 2 π 2 = π 6 19. arctan − √ 3 = − π 3 20. arctan (−1) = − π 4 22. arctan (0) = 0 23. arctan √ 3 3 = π 6 25. arctan √ 3 = √ π 3 28. arccot − = 3 3 26. arccot − √ 3 = 5π 6 2π 3 29. arccot (0) = π 2 31. arccot (1) = 34. arccsc (2) = 37. arcsec √ 2 3 3 π 4 π 6 32. arccot √ 35. arcsec √ 38. arccsc . arcsin 1 2 = π 6 9. arcsin (1) = π 2 12. arccos − √ 2 2 = 3π 4 15. arccos 1 2 = π 3 18. arccos (1) = 0 21. arctan − √ 3 3 = − π 6 24. arctan (1) = π 4 27. arccot (−1) = 3π 4 30. arccot √ 3 3 = π 3 33. arcsec (2) = 36. arccsc √ π 3 2 = π 4 39. arcsec (1) = 0 40. arccsc (1) = π 2 41. arcsec (−2) = 4π 3 42. arcsec − √ 2 = 5π 4 850 Foundations of Trigonometry 44. arcsec (−1) = π 45. arccsc (−2) = 47. arccsc − √ 3 2 3 = 4π 3 48. arccsc (−1) = 7π 6 3π 2 50. arcsec − 2 = √ 51. arcsec 2 − √ 3 3 = 5π 6 54. arccsc − √ 2 = − π 4 3π 4 π 6 π 2 52. arcsec (−1) = π 53. arccsc (−2) = − 55. arccsc − √ 3 2 = − π 3 56. arccsc (−1) = − 43. arcsec − √ 3 2 3 = 7π 6 46. arccsc − √ 2 = 5π 4 49. arcsec (−2) = 2π 3 57. sin arcsin 59. sin arcsin 61. sin arcsin 58. sin arcsin − √ 2 2 √ 2 2 = − 60. sin (arcsin (−0.42)) = −0.42 √ 2 2 √ 2 2 = is undeﬁned. 62. cos arccos 63. cos arccos − 1 2 = − 1 2 65. cos (arccos (−0.998)) = −0.998 67. tan (arctan (−1)) = −1 69. tan arctan 5 12 = 5 12 64. cos arccos 5 13 = 5 13 66. cos (arccos (π)) is undeﬁned. 68. tan arctan √ 3 = √ 3 70. tan (arctan (0.965)) = 0.965 71. tan (arctan (3π)) = 3π 72. cot (arccot (1)) = 1 73. cot arccot − √ √ 3 = − 3 75. cot (arccot (−0.001)) = −0.001 74. cot arccot 76. cot arccot − 7 24 17π 4 = − 7 24 = 17π 4 77. sec (arcsec (2)) = 2 78. sec (arcsec (−1)) = −1 10.6 The Inverse Trigonometric Functions 851 79. sec arcsec 1 2 is undeﬁned. 81. sec (arcsec (117π)) = 117π 83. csc arccsc 2 − √ 3 3 √ 3 2 3 = − 85. csc (arccsc (1.0001)) = 1.0001 87. arcsin sin 89. arcsin sin 91. arcsin sin 93. arccos cos π 6 3π 4 4π 3 2π 3 95. arccos − cos π 6 = π 4 = − π 3 = 2π 3 = π 6 = π 6 97. arctan tan π 3 = π 3 99. arctan (tan (π)) = 0 101. arctan tan 2π 3 = − π 3 80. sec (arcsec (0.75)) is undeﬁned. 82. csc arccsc √ √ 2 = √ 2 84. csc arccsc is undeﬁned. 2 2 86. csc arccsc π 4 is undeﬁned. 88. arcsin sin 90. arcsin sin − π 3 11π 6 = − π 3 = − π 6 92. arccos cos 94. arccos cos 96. arccos cos π 4 3π 2 5π 4 = − tan 98. arctan 100. arctan tan π 4 π 4 = = π 2 3π 4 = − π 4 102. arccot cot π 2 π 3 is undeﬁned = π 3 103. arccot cot 105. arccot cot − π 4 3π 2 = 107. arcsec sec 109. arcsec sec 111. arcsec sec π 4 5π 6 5π 3 = 3π 4 π 2 7π 6 π 3 104. arccot (cot (π)) is undeﬁned 106. arccot cot 108. arcsec sec 2π 3 4π 3 = = 2π 3 4π 3 110. arcsec 112. arccsc sec − π 2 is undeﬁned. csc π 6 = π 6 = π 4 = = 852 Foundations of Trigonometry 113. arccsc csc 5π 4 = 5π 4 115. arccsc csc 117. arcsec sec 119. arcsec sec 121. arcsec sec 123. arcsec sec 125. arccsc csc 127. arccsc csc 129. arcsec sec − π 2 11π 12 = 3π 2 = 13π 12 π 4 = = 5π 5π 6 5π 3 5π 4 − π 2 11π 12 = 11π 12 131. sin arccos − = 1 2 133. sin (arctan (−2)) = − 135. sin (arccsc (−3)) = − 137. cos arctan √ 7 = 139. cos (arcsec (5)) = 1 5 114. arccsc csc 116. arccsc csc 118. arccsc csc 120. arcsec sec 2π 3 11π 6 9π 8 4π 3 = π 3 = 7π 6 = = 9π 8 2π 3 122. arcsec sec − π 2 is undeﬁned csc csc csc 128.
arccsc 124. arccsc 126. arccsc π 6 2π 3 11π 6 9π 8 3 5 134. sin arccot √ 130. arccsc 132. sin arccos csc 136. cos arcsin − 5 = = 4 5 √ 6 6 5 13 = 12 13 √ 3 10 138. cos (arccot (3)) = 140. tan arcsin − √ 2 5 = −2 10 5 = 4 3 141. tan arccos − √ 3 = − 1 2 143. tan (arccot (12)) = 1 12 142. tan arcsec 144. cot arcsin 5 3 12 13 = 5 12 10.6 The Inverse Trigonometric Functions 853 145. cot arccos √ 3 2 √ 3 = 147. cot (arctan (0.25)) = 4 = 13 5 149. sec arcsin − 12 13 √ 151. sec arccot − 10 10 153. csc arcsin 155. sin arcsin 3 5 = 5 3 5 13 + π 4 = 157. tan arctan(3) + arccos − 146. cot arccsc √ 5 = 2 148. sec arccos √ 3 2 √ 3 2 3 = 150. sec (arctan (10)) = √ 101 √ = − 11 152. csc (arccot (9)) = √ 82 154. csc arctan − 2 3 = − √ 13 2 156. cos (arcsec(3) + arctan(2)) = √ 10 √ 5 − 4 15 √ 17 2 26 3 5 = 1 3 158. sin 2 arcsin − 4 5 = − 24 25 159. sin 2arccsc 161. cos 2 arcsin 13 5 3 5 = 120 169 = 7 25 160. sin (2 arctan (2)) = 4 5 162. cos 2arcsec 25 7 = − 163. cos 2arccot − √ 5 = 2 3 165. sin (arccos (x)) = √ 1 − x2 for −1 ≤ x ≤ 1 164. sin arctan(2) 2 = 5 − 10 527 625 √ 5 166. cos (arctan (x)) = √ 167. tan (arcsin (x)) = √ 1 1 + x2 x 1 − x2 for all x for −1 < x < 1 168. sec (arctan (x)) = √ 1 + x2 for all x 169. csc (arccos (x)) = √ 1 1 − x2 for −1 < x < 1 170. sin (2 arctan (x)) = for all x 2x x2 + 1 √ 171. sin (2 arccos (x)) = 2x 1 − x2 for −1 ≤ x ≤ 1 854 Foundations of Trigonometry 172. cos (2 arctan (x)) = 173. sin(arccos(2x)) = 174. sin arccos 175. cos arcsin x 5 x 2 = = √ 1 − x2 1 + x2 for all x √ 1 − 4x2 for − 1 √ 25 − x2 5 2 ≤ x ≤ 1 2 for −5 ≤ x ≤ 5 4 − x2 2 for −2 ≤ x ≤ 2 176. cos (arctan (3x)) = √ for all x 1 1 + 9x2 √ 177. sin(2 arcsin(7x)) = 14x 1 − 49x2 for − 1 7 ≤ x ≤ 1 7 178. sin 2 arcsin √ x 3 3 √ 2x = 3 − x2 3 √ for − 3 ≤ x ≤ √ 3 179. cos(2 arcsin(4x)) = 1 − 32x2 for − 1 4 1 4 ≤ x ≤ √ 180. sec(arctan(2x)) tan(arctan(2x)) = 2x 1 + 4x2 for all x 181. sin (arcsin(x) + arccos(x)) = 1 for −1 ≤ x ≤ 1 182. cos (arcsin(x) + arctan(x)) = √ 1 − x2 − x2 √ 1 + x2 for −1 ≤ x ≤ 1 183. 10 tan (2 arcsin(x)) = √ 2x 1 − x2 1 − 2x2 for x in −1 184. sin arctan(x) = 1 2    √ x2 + 1 − 1 √ x2 + 1 2 √ x2 + 1 − 1 √ x2 + 1 2 − for x ≥ 0 for x < 0 185. If sin(θ) = 186. If tan(θ) = x 2 x 7 for − π 2 for − π 2 < θ < π 2 , then θ + sin(2θ) = arcsin x 2 x + √ 4 − x2 2 < θ < π 2 , then 1 2 θ − 1 2 sin(2θ) = 1 2 arctan x 7 − 7x x2 + 49 10The equivalence for x = ±1 can be veriﬁed independently of the derivation of the formula, but Calculus is required to fully understand what is happening at those x values. You’ll see what we mean when you work through the details of the identity for tan(2t). For now, we exclude x = ±1 from our answer. 10.6 The Inverse Trigonometric Functions 855 187. If sec(θ) = 188. x = arcsin x 4 7 11 for 0 < θ < π 2 , then 4 tan(θ) − 4θ = √ x2 − 16 − 4arcsec x 4 + 2πk or x = π − arcsin 7 11 + 2πk, in [0, 2π), x ≈ 0.6898, 2.4518 189. x = arccos − 2 9 + 2πk or x = − arccos − 2 9 + 2πk, in [0, 2π), x ≈ 1.7949, 4.4883 190. x = π + arcsin(0.569) + 2πk or x = 2π − arcsin(0.569) + 2πk, in [0, 2π), x ≈ 3.7469, 5.6779 191. x = arccos(0.117) + 2πk or x = 2π − arccos(0.117) + 2πk, in [0, 2π), x ≈ 1.4535, 4.8297 192. x = arcsin(0.008) + 2πk or x = π − arcsin(0.008) + 2πk, in [0, 2π), x ≈ 0.0080, 3.1336 193. x = arccos 359 360 + 2πk or x = 2π − arccos 359 360 + 2πk, in [0, 2π), x ≈ 0.0746, 6.2086 194. x = arctan(117) + πk, in [0, 2π), x ≈ 1.56225, 4.70384 195. x = arctan − 1 12 + πk, in [0, 2π), x ≈ 3.0585, 6.2000 196. x = arccos 2 3 + 2πk or x = 2π − arccos 2 3 + 2πk, in [0, 2π), x ≈ 0.8411, 5.4422 197. x = π + arcsin 198. x = arctan − 3 8 199. x = arcsin 200. x = arccos − 7 16 + 2πk or x = 2π − arcsin 17 90 10 + πk, in [0, 2π), x ≈ 1.8771, 5.0187 17 90 √ + 2πk or x = π − arcsin 3 8 + 2πk, in [0, 2π), x ≈ 0.3844, 2.7572 7 16 + 2πk, in [0, 2π), x ≈ 2.0236, 4.2596 + 2πk or x = − arccos − + 2πk, in [0, 2π), x ≈ 3.3316, 6.0932 201. x = arctan(0.03) + πk, in [0, 2π), x ≈ 0.0300, 3.1716 202. x = arcsin(0.3502) + 2πk or x = π − arcsin(0.3502) + 2πk, in [0, 2π), x ≈ 0.3578, 2.784 203. x = π + arcsin(0.721) + 2πk or x = 2π − arcsin(0.721) + 2πk, in [0, 2π), x ≈ 3.9468, 5.4780 204. x = arccos(0.9824) + 2πk or x = 2π − arccos(0.9824) + 2πk, in [0, 2π), x ≈ 0.1879, 6.0953 205. x = arccos(−0.5637) + 2πk or x = − arccos(−0.5637) + 2πk, in [0, 2π), x ≈ 2.1697, 4.1135 206. x = arctan(117) + πk, in [0, 2π), x ≈ 1.5622, 4.7038 207. x = arctan(−0.6109) + πk, in [0, 2π), x ≈ 2.5932, 5.7348 856 Foundations of Trigonometry 208. 36.87◦ and 53.13◦ 209. 22.62◦ and 67.38◦ 210. 32.52◦ and 57.48◦ 211. 68.9◦ 212. 7.7◦ 213. 51◦ 216. f (x) = 5 sin(3x) + 12 cos(3x) = 13 sin 3x + arcsin 214. 19.5◦ 12 13 ≈ 13 sin(3x + 1.1760) 215. 41.81◦ 217. f (x) = 3 cos(2x) + 4 sin(2x) = 5 sin 2x + arcsin 3 5 ≈ 5 sin(2x + 0.6435) 218. f (x) = cos(x) − 3 sin(x) = √ 10 sin x + arccos − 10 219. f (x) = 7 sin(10x) − 24 cos(10x) = 25 sin 10x + arcsin − √ 3 10 √ ≈ 10 sin(x + 2.8198) 24 25 ≈ 25 sin(10x − 1.2870) 220. f (x) = − cos(x) − 2 √ 2 sin(x) = 3 sin x + π + arcsin 221. f (x) = 2 sin(x) − cos(x) = √ 5 sin x + arcsin − 1 3 ≈ 3 sin(x + 3.4814) √ ≈ 5 sin(x − 0.4636) √ 5 5 223. − , 1 1 3 1 5 222. − 224. − , 1 5 √ √ 2 2 , 225. (−∞, − √ √ 5] ∪ [− √ 3, √ 3] ∪ [ 5, ∞) 2 2 226. (−∞, ∞) 227. (−∞, −3) ∪ (−3, 3) ∪ (3, ∞) 228. , ∞ 1 2 230. −∞, − 1 12 ∪ 1 12 , ∞ 229. , ∞ 1 2 231. (−∞, −6] ∪ [−4, ∞) 232. (−∞, −2] ∪ [2, ∞) 233. [0, ∞) 10.7 Trigonometric Equations and Inequalities 857 10.7 Trigonometric Equations and Inequalities In Sections 10.2, 10.3 and most recently 10.6, we solved some basic equations involving the trigonometric functions. Below we summarize the techniques we’ve employed thus far. Note that we use the neutral letter ‘u’ as the argument1 of each circular function for generality. Strategies for Solving Basic Equations Involving Trigonometric Functions To solve cos(u) = c or sin(u) = c for −1 ≤ c ≤ 1, ﬁrst solve for u in the interval [0, 2π) and add integer multiples of the period 2π. If c < −1 or of c > 1, there are no real solutions. To solve sec(u) = c or csc(u) = c for c ≤ −1 or c ≥ 1, convert to cosine or sine, respectively, and solve as above. If −1 < c < 1, there are no real solutions. To solve tan(u) = c for any real number c, ﬁrst solve for u in the interval − π 2 , π 2 and add integer multiples of the period π. To solve cot(u) = c for c = 0, convert to tangent and solve as above. If c = 0, the solution to cot(u) = 0 is u = π 2 + πk for integers k. 2 and ﬁnd the solution x = π 2 , we know the solutions take the form u = π 6 + 2πk for integers k. How do we solve something like sin(3x) = 1 Using the above guidelines, we can comfortably solve sin(x) = 1 or x = 5π has the form sin(u) = 1 integers k. Since the argument of sine here is 3x, we have 3x = π integers k. To solve for x, we divide both sides2 of these equations by 3, and obtain x = π or x = 5π 6 + 2πk 2 ? Since this equation 6 + 2πk for 6 + 2πk for 18 + 2π 3 k 3 k for integers k. This is the technique employed in the example below. Example 10.7.1. Solve the following equations and check your answers analytically. List the solutions which lie in the interval [0, 2π) and verify them using a graphing utility. 2. csc 1 6 + 2πk or u = 5π 6 + 2πk or 3x = 5π 3. cot (3x) = 0 18 + 2π 1. cos(2x3 5. tan x 2 6. sin(2x) = 0.87 4. sec2(x) = 4 Solution. 1. The solutions to cos(u) = − √ the argument of cosine here is 2x, this means 2x = 5π k. Solving for x gives x = 5π analytically, we substitute them into the original equation. For any integer k we have 6 + 2πk for integers k. Since 6 + 2πk for integers 12 + πk for integers k. To check these answers 6 + 2πk or 2x = 7π 12 + πk or x = 7π 6 + 2πk or u = 7π 2 are u = 5π 3 cos 2 5π 12 + πk = cos 5π = cos 5π 6 √ = − 3 2 6 + 2πk (the period of cosine is 2π) 1See the comments at the beginning of Section 10.5 for a review of this concept. 2Don’t forget to divide the 2πk by 3 as well! 858 Foundations of Trigonometry 12 + πk = cos 7π √ Similarly, we ﬁnd cos 2 7π 3 2 . To determine which of our solutions lie in [0, 2π), we substitute integer values for k. The solutions we keep come from the values of k = 0 and k = 1 and are x = 5π 12 , 17π 12 . Using a √ 3 calculator, we graph y = cos(2x) and y = − 2 over [0, 2π) and examine where these two graphs intersect. We see that the x-coordinates of the intersection points correspond to the decimal representations of our exact answers. 6 + 2πk = cos 7π 12 and 19π = − 12 , 7π 6 2. Since this equation has the form csc(u) = 2, we rewrite this as sin(u) = u = π 4 + 2πk or u = 3π 4 + 2πk for integers k. Since the argument of cosecant here is 1 √ 2 2 and ﬁnd 3 x − π + 2πk or 1 3 x − π = 3π 4 + 2πk To solve 1 3 x − π = π 4 + 2πk, we ﬁrst add π to both sides 1 3 x = π 4 + 2πk + π A common error is to treat the ‘2πk’ and ‘π’ terms as ‘like’ terms and try to combine them when they are not.3 We can, however, combine the ‘π’ and ‘ π 4 ’ terms to get We now ﬁnish by multiplying both sides by 3 to get 1 3 x = 5π 4 + 2πk x = 3 5π 4 + 2πk = 15π 4 + 6πk Solving the other equation, 1 check the ﬁrst family of answers, we substitute, combine line terms, and simplify. 4 + 2πk produces x = 21π 3 x − π = 3π 4 + 6πk for integers k. To csc 1 3 15π 4 + 6πk − π = csc 5π = csc π = csc π √ 4 2 = 4 + 2πk − π 4 + 2πk (the period of cosecant is 2π) The family x = 21π 4 + 6πk checks similarly. Despite having inﬁnitely many solutions, we ﬁnd that none of them lie in [0, 2π). To verify this graphically, we use a reciprocal identity to 2 do not intersect at rewrite the cosecant as a sine and we ﬁnd that y = and y = √ 1 sin( 1 3 x−π) all over the interval [0, 2π). 3Do you see why? 10.7 Trigonometric Equations and Inequalities 859 y = cos(2x) and y = − √ 3 2 y = 1 sin( 1 3 x−π) and y = √ 2 3. Since cot(3x) = 0 has the form cot(u) = 0, we know u = π 2 + πk, so, in this case, 3x = π 2 + πk for integers k. Solving for x yields x = π 3 k. Checking our answers, we get 6 + π 2 + πk cot 3 π 6 + π 3 k = cot π = cot π 2 = 0 (the period of cotangent is π) As k runs through the integer
s, we obtain six answers, corresponding to k = 0 through k = 5, which lie in [0, 2π): x = π 2 and 11π 6 . To conﬁrm these graphically, we must be careful. On many calculators, there is no function button for cotangent. We choose4 to use sin(3x) . Graphing y = cos(3x) the quotient identity cot(3x) = cos(3x) sin(3x) and y = 0 (the x-axis), we see that the x-coordinates of the intersection points approximately match our solutions. 6 , 3π 6 , 7π 2 , 5π 6 , π 4. The complication in solving an equation like sec2(x) = 4 comes not from the argument of secant, which is just x, but rather, the fact the secant is being squared. To get this equation to look like one of the forms listed on page 857, we extract square roots to get sec(x) = ±2. Converting to cosines, we have cos(x) = ± 1 3 + 2πk or x = 5π 3 + 2πk for integers k. For cos(x) = − 1 3 + 2πk for integers k. If we take a step back and think of these families of solutions geometrically, we see we are ﬁnding the measures of all angles with a reference angle of π 3 . As a result, these solutions can be combined and we may write our solutions as x = π 3 + πk for integers k. To check the ﬁrst family of solutions, we note that, depending on the integer . However, it is true that for all integers k, k, sec π sec π 3 = ±2. (Can you show this?) As a result, 3 + πk doesn’t always equal sec π 2 . For cos(x) = 1 2 , we get x = 2π 2 , we get x = π 3 + 2πk or x = 4π 3 + πk and x = 2π 3 + πk = ± sec π 3 sec2 π 3 + πk = ± sec π 3 2 = (±2)2 = 4 The same holds for the family x = 2π the values k = 0 and k = 1, namely x = π 3 + πk. The solutions which lie in [0, 2π) come from 3 . To conﬁrm graphically, we use 3 and 5π 3 , 4π 3 , 2π 4The reader is encouraged to see what happens if we had chosen the reciprocal identity cot(3x) = 1 tan(3x) instead. The graph on the calculator appears identical, but what happens when you try to ﬁnd the intersection points? 860 Foundations of Trigonometry a reciprocal identity to rewrite the secant as cosine. The x-coordinates of the intersection points of y = (cos(x))2 and y = 4 verify our answers. 1 y = cos(3x) sin(3x) and y = 0 y = 1 cos2(x) and y = 4 5. The equation tan x 2 Hence, x = −3 has the form tan(u) = −3, whose solution is u = arctan(−3)+πk. 2 = arctan(−3) + πk, so x = 2 arctan(−3) + 2πk for integers k. To check, we note tan 2 arctan(−3)+2πk 2 = tan (arctan(−3) + πk) = tan (arctan(−3)) = −3 (the period of tangent is π) (See Theorem 10.27) To determine which of our answers lie in the interval [0, 2π), we ﬁrst need to get an idea of the value of 2 arctan(−3). While we could easily ﬁnd an approximation using a calculator,5 we proceed analytically. Since −3 < 0, it follows that − π 2 < arctan(−3) < 0. Multiplying through by 2 gives −π < 2 arctan(−3) < 0. We are now in a position to argue which of the solutions x = 2 arctan(−3) + 2πk lie in [0, 2π). For k = 0, we get x = 2 arctan(−3) < 0, so we discard this answer and all answers x = 2 arctan(−3) + 2πk where k < 0. Next, we turn our attention to k = 1 and get x = 2 arctan(−3) + 2π. Starting with the inequality −π < 2 arctan(−3) < 0, we add 2π and get π < 2 arctan(−3) + 2π < 2π. This means x = 2 arctan(−3) + 2π lies in [0, 2π). Advancing k to 2 produces x = 2 arctan(−3) + 4π. Once again, we get from −π < 2 arctan(−3) < 0 that 3π < 2 arctan(−3) + 4π < 4π. Since this is outside the interval [0, 2π), we discard x = 2 arctan(−3) + 4π and all solutions of the form x = 2 arctan(−3) + 2πk for k > 2. Graphically, we see y = tan x and y = −3 intersect only 2 once on [0, 2π) at x = 2 arctan(−3) + 2π ≈ 3.7851. 6. To solve sin(2x) = 0.87, we ﬁrst note that it has the form sin(u) = 0.87, which has the family of solutions u = arcsin(0.87) + 2πk or u = π − arcsin(0.87) + 2πk for integers k. Since the argument of sine here is 2x, we get 2x = arcsin(0.87) + 2πk or 2x = π − arcsin(0.87) + 2πk 2 − 1 which gives x = 1 2 arcsin(0.87) + πk for integers k. To check, 2 arcsin(0.87) + πk or x = π 5Your instructor will let you know if you should abandon the analytic route at this point and use your calculator. But seriously, what fun would that be? 10.7 Trigonometric Equations and Inequalities 861 sin 2 1 2 arcsin(0.87) + πk = sin (arcsin(0.87) + 2πk) = sin (arcsin(0.87)) (the period of sine is 2π) = 0.87 (See Theorem 10.26) For the family x = π 2 − 1 2 arcsin(0.87) + πk , we get sin 2 π 2 − 1 2 arcsin(0.87) + πk = sin (π − arcsin(0.87) + 2πk) = sin (π − arcsin(0.87)) = sin (arcsin(0.87)) = 0.87 (the period of sine is 2π) (sin(π − t) = sin(t)) (See Theorem 10.26) 2 arcsin(0.87) < π 2 so that multiplying through by 1 4 . Starting with the family of solutions x = 1 To determine which of these solutions lie in [0, 2π), we ﬁrst need to get an idea of the value of x = 1 2 arcsin(0.87). Once again, we could use the calculator, but we adopt an analytic route here. By deﬁnition, 0 < arcsin(0.87) < π 2 gives us 0 < 1 2 arcsin(0.87) + πk, we use the same kind of arguments as in our solution to number 5 above and ﬁnd only the solutions corresponding to k = 0 and k = 1 lie in [0, 2π): x = 1 2 arcsin(0.87) + π. Next, we move to the family x = π 2 arcsin(0.87) + πk for integers k. Here, we need to get a better estimate of π 2 arcsin(0.87) < π 4 , we ﬁrst multiply through by −1 and then add π 4 , or 4 < π π 2 . Proceeding with the usual arguments, we ﬁnd the only solutions which lie in [0, 2π) correspond to k = 0 and k = 1, namely x = π 2 arcsin(0.87) and x = 3π 2 arcsin(0.87). All told, we have found four solutions to sin(2x) = 0.87 in [0, 2π): x = 1 2 arcsin(0.87). By graphing y = sin(2x) and y = 0.87, we conﬁrm our results. 2 arcsin(0.87). From the inequality 0 < 1 2 − 1 2 arcsin(0.87), x = 1 2 arcsin(0.87) and x = 3π 2 arcsin(0.87) + π, x = π 2 arcsin(0.87) and x = 1 2 arcsin(0.87) > π 2 arcsin(0.87) < π 2 to get = tan x 2 and y = −3 y = sin(2x) and y = 0.87 862 Foundations of Trigonometry Each of the problems in Example 10.7.1 featured one trigonometric function. If an equation involves two diﬀerent trigonometric functions or if the equation contains the same trigonometric function but with diﬀerent arguments, we will need to use identities and Algebra to reduce the equation to the same form as those given on page 857. Example 10.7.2. Solve the following equations and list the solutions which lie in the interval [0, 2π). Verify your solutions on [0, 2π) graphically. 1. 3 sin3(x) = sin2(x) 3. cos(2x) = 3 cos(x) − 2 5. cos(3x) = cos(5x) 7. sin(x) cos x 2 + cos(x) sin x 2 = 1 Solution. 2. sec2(x) = tan(x) + 3 4. cos(3x) = 2 − cos(x) 6. sin(2x) = √ 3 cos(x) 8. cos(x) − √ 3 sin(x) = 2 1. We resist the temptation to divide both sides of 3 sin3(x) = sin2(x) by sin2(x) (What goes wrong if you do?) and instead gather all of the terms to one side of the equation and factor. 3 sin3(x) = sin2(x) 3 sin3(x) − sin2(x) = 0 sin2(x)(3 sin(x) − 1) = 0 Factor out sin2(x) from both terms. We get sin2(x) = 0 or 3 sin(x) − 1 = 0. Solving for sin(x), we ﬁnd sin(x) = 0 or sin(x) = 1 3 . The solution to the ﬁrst equation is x = πk, with x = 0 and x = π being the two solutions +2πk which lie in [0, 2π). To solve sin(x) = 1 + 2πk for integers k. We ﬁnd the two solutions here which lie in [0, 2π) or x = π − arcsin 1 3 . To check graphically, we plot y = 3(sin(x))3 and and x = π − arcsin 1 to be x = arcsin 1 3 3 y = (sin(x))2 and ﬁnd the x-coordinates of the intersection points of these two curves. Some extra zooming is required near x = 0 and x = π to verify that these two curves do in fact intersect four times.6 3 , we use the arcsine function to get x = arcsin 1 3 2. Analysis of sec2(x) = tan(x) + 3 reveals two diﬀerent trigonometric functions, so an identity is in order. Since sec2(x) = 1 + tan2(x), we get sec2(x) = tan(x) + 3 1 + tan2(x) = tan(x) + 3 (Since sec2(x) = 1 + tan2(x).) tan2(x) − tan(x) − 2 = 0 u2 − u − 2 = 0 (u + 1)(u − 2) = 0 Let u = tan(x). 6Note that we are not counting the point (2π, 0) in our solution set since x = 2π is not in the interval [0, 2π). In the forthcoming solutions, remember that while x = 2π may be a solution to the equation, it isn’t counted among the solutions in [0, 2π). 10.7 Trigonometric Equations and Inequalities 863 This gives u = −1 or u = 2. Since u = tan(x), we have tan(x) = −1 or tan(x) = 2. From tan(x) = −1, we get x = − π 4 + πk for integers k. To solve tan(x) = 2, we employ the arctangent function and get x = arctan(2) + πk for integers k. From the ﬁrst set of solutions, we get x = 3π 4 as our answers which lie in [0, 2π). Using the same sort of argument we saw in Example 10.7.1, we get x = arctan(2) and x = π + arctan(2) as answers from our second set of solutions which lie in [0, 2π). Using a reciprocal identity, we rewrite the secant as a cosine and graph y = (cos(x))2 and y = tan(x) + 3 to ﬁnd the x-values of the points where they intersect. 4 and x = 7π 1 y = 3(sin(x))3 and y = (sin(x))2 y = 1 (cos(x))2 and y = tan(x) + 3 3. In the equation cos(2x) = 3 cos(x) − 2, we have the same circular function, namely cosine, on both sides but the arguments diﬀer. Using the identity cos(2x) = 2 cos2(x) − 1, we obtain a ‘quadratic in disguise’ and proceed as we have done in the past. cos(2x) = 3 cos(x) − 2 2 cos2(x) − 1 = 3 cos(x) − 2 (Since cos(2x) = 2 cos2(x) − 1.) 2 cos2(x) − 3 cos(x) + 1 = 0 2u2 − 3u + 1 = 0 (2u − 1)(u − 1) = 0 Let u = cos(x). 2 , we get x = π 2 or u = 1. Since u = cos(x), we get cos(x) = 1 This gives u = 1 2 or cos(x) = 1. Solving cos(x) = 1 3 + 2πk or x = 5π 3 + 2πk for integers k. From cos(x) = 1, we get x = 2πk for integers k. The answers which lie in [0, 2π) are x = 0, π 3 . Graphing y = cos(2x) and y = 3 cos(x) − 2, we ﬁnd, after a little extra eﬀort, that the curves intersect in three places on [0, 2π), and the x-coordinates of these points conﬁrm our results. 3 , and 5π 4. To solve cos(3x) = 2 − cos(x), we use the same technique as in the previous problem. From Example 10.4.3, number 4, we know that cos(3x) = 4 cos3(x) − 3 cos(x). This transforms the equation into a polynomial in terms of cos(
x). cos(3x) = 2 − cos(x) 4 cos3(x) − 3 cos(x) = 2 − cos(x) 2 cos3(x) − 2 cos(x) − 2 = 0 4u3 − 2u − 2 = 0 Let u = cos(x). 864 Foundations of Trigonometry To solve 4u3 − 2u − 2 = 0, we need the techniques in Chapter 3 to factor 4u3 − 2u − 2 into (u − 1) 4u2 + 4u + 2. We get either u − 1 = 0 or 4u2 + 2u + 2 = 0, and since the discriminant of the latter is negative, the only real solution to 4u3 − 2u − 2 = 0 is u = 1. Since u = cos(x), we get cos(x) = 1, so x = 2πk for integers k. The only solution which lies in [0, 2π) is x = 0. Graphing y = cos(3x) and y = 2 − cos(x) on the same set of axes over [0, 2π) shows that the graphs intersect at what appears to be (0, 1), as required. y = cos(2x) and y = 3 cos(x) − 2 y = cos(3x) and y = 2 − cos(x) 5. While we could approach cos(3x) = cos(5x) in the same manner as we did the previous two problems, we choose instead to showcase the utility of the Sum to Product Identities. From cos(3x) = cos(5x), we get cos(5x) − cos(3x) = 0, and it is the presence of 0 on the right hand side that indicates a switch to a product would be a good move.7 Using Theorem 10.21, we have that cos(5x) − cos(3x) = −2 sin 5x+3x = −2 sin(4x) sin(x). Hence, 2 the equation cos(5x) = cos(3x) is equivalent to −2 sin(4x) sin(x) = 0. From this, we get sin(4x) = 0 or sin(x) = 0. Solving sin(4x) = 0 gives x = π 4 k for integers k, and the solution to sin(x) = 0 is x = πk for integers k. The second set of solutions is contained in the ﬁrst set of solutions,8 so our ﬁnal solution to cos(5x) = cos(3x) is x = π 4 k for integers k. There are eight of these answers which lie in [0, 2π): x = 0, π 4 , 3π 4 , π, 5π 4 . Our plot of the graphs of y = cos(3x) and y = cos(5x) below (after some careful zooming) bears this out. sin 5x−3x 2 and 7π 2 , 3π 4 , π 2 √ 6. In examining the equation sin(2x) = 3 cos(x), not only do we have diﬀerent circular functions involved, namely sine and cosine, we also have diﬀerent arguments to contend with, namely 2x and x. Using the identity sin(2x) = 2 sin(x) cos(x) makes all of the arguments the same and we proceed as we would solving any nonlinear equation – gather all of the nonzero terms on one side of the equation and factor. sin(2x) = 2 sin(x) cos(x) = √ 2 sin(x) cos(x) − cos(x)(2 sin(x) − 3 cos(x) = 0 3) = 0 √ √ √ 3 cos(x) 3 cos(x) (Since sin(2x) = 2 sin(x) cos(x).) from which we get cos(x) = 0 or sin(x) = 2 , we get x = π integers k. From sin(x) = √ 3 3 2 . From cos(x) = 0, we obtain x = π 2 + πk for 3 +2πk for integers k. The answers 3 +2πk or x = 2π √ 7As always, experience is the greatest teacher here! 8As always, when in doubt, write it out! 10.7 Trigonometric Equations and Inequalities which lie in [0, 2π) are x = π after some careful zooming, verify our answers. 3 and 2π 2 , 3π 2 , π 3 . We graph y = sin(2x) and y = 865 √ 3 cos(x) and, y = cos(3x) and y = cos(5x) y = sin(2x) and y = √ 3 cos(x) 7. Unlike the previous problem, there seems to be no quick way to get the circular functions or their arguments to match in the equation sin(x) cos x = 1. If we stare at 2 it long enough, however, we realize that the left hand side is the expanded form of the sum formula for sin x + x 2 x = 1. Solving, 2 we ﬁnd x = π 3 and x = 5π 3 . Graphing y = sin(x) cos x 2 . Hence, our original equation is equivalent to sin 3 3 k for integers k. Two of these solutions lie in [0, 2π): x = π and y = 1 validates our solutions. + cos(x) sin x 2 + cos(x) sin x 2 3 + 4π 8. With the absence of double angles or squares, there doesn’t seem to be much we can do. However, since the arguments of the cosine and sine are the same, we can rewrite the left hand side of this equation as a sinusoid.9 To ﬁt f (x) = cos(x) − 3 sin(x) to the form A sin(ωt + φ) + B, we use what we learned in Example 10.5.3 and ﬁnd A = 2, B = 0, ω = 1 = 2, 3 sin(x) = 2 as 2 sin x + 5π and φ = 5π 6 or sin x + 5π 3 + 2πk for integers k. Only one of 6 these solutions, x = 5π 3 , which corresponds to k = 1, lies in [0, 2π). Geometrically, we see that y = cos(x) − 6 . Hence, we can rewrite the equation cos(x) − = 1. Solving the latter, we get x = − π 3 sin(x) and y = 2 intersect just once, supporting our answer. √ √ √ y = sin(x) cos x 2 + cos(x) sin x 2 and y = 1 y = cos(x) − √ 3 sin(x) and y = 2 We repeat here the advice given when solving systems of nonlinear equations in section 8.7 – when it comes to solving equations involving the trigonometric functions, it helps to just try something. 9We are essentially ‘undoing’ the sum / diﬀerence formula for cosine or sine, depending on which form we use, so this problem is actually closely related to the previous one! 866 Foundations of Trigonometry Next, we focus on solving inequalities involving the trigonometric functions. Since these functions are continuous on their domains, we may use the sign diagram technique we’ve used in the past to solve the inequalities.10 Example 10.7.3. Solve the following inequalities on [0, 2π). Express your answers using interval notation and verify your answers graphically. 1. 2 sin(x) ≤ 1 2. sin(2x) > cos(x) 3. tan(x) ≥ 3 Solution. 1. We begin solving 2 sin(x) ≤ 1 by collecting all of the terms on one side of the equation and zero on the other to get 2 sin(x) − 1 ≤ 0. Next, we let f (x) = 2 sin(x) − 1 and note that our original inequality is equivalent to solving f (x) ≤ 0. We now look to see where, if ever, f is undeﬁned and where f (x) = 0. Since the domain of f is all real numbers, we can immediately set about ﬁnding the zeros of f . Solving f (x) = 0, we have 2 sin(x) − 1 = 0 or sin(x) = 1 2 . The solutions here are x = π 6 + 2πk for integers k. Since we are restricting our attention to [0, 2π), only x = π 6 are of concern to us. Next, we choose test values in [0, 2π) other than the zeros and determine if f is positive or negative there. For = 1 and for x = π we get f (π) = −1. 2 we get f π x = 0 we have f (0) = −1, for x = π Since our original inequality is equivalent to f (x) ≤ 0, we are looking for where the function is negative (−) or 0, and we get the intervals 0, π 6 , 2π. We can conﬁrm our answer 6 graphically by seeing where the graph of y = 2 sin(x) crosses or is below the graph of y = 1. 6 + 2πk and x = 5π 6 and x = 5π ∪ 5π 2 (−) 0 (+) π 6 0 (−) 5π 6 2π 0 y = 2 sin(x) and y = 1 2. We ﬁrst rewrite sin(2x) > cos(x) as sin(2x) − cos(x) > 0 and let f (x) = sin(2x) − cos(x). Our original inequality is thus equivalent to f (x) > 0. The domain of f is all real numbers, so we can advance to ﬁnding the zeros of f . Setting f (x) = 0 yields sin(2x) − cos(x) = 0, which, by way of the double angle identity for sine, becomes 2 sin(x) cos(x) − cos(x) = 0 or 2 +πk for integers k of which only x = π cos(x)(2 sin(x)−1) = 0. From cos(x) = 0, we get x = π 2 and x = 3π 2 which gives x = π 6 + 2πk or x = 5π 6 lie in [0, 2π). Next, we choose 2 lie in [0, 2π). For 2 sin(x) − 1 = 0, we get sin(x) = 1 6 + 2πk for integers k. Of those, only x = π 6 and x = 5π 10See page 214, Example 3.1.5, page 321, page 399, Example 6.3.2 and Example 6.4.2 for discussion of this technique. 10.7 Trigonometric Equations and Inequalities 867 4 we get f 3π our test values. For x = 0 we ﬁnd f (0) = −1; when x = π = −1 + for x = 3π ; when x = π we have f (π) = 1, and lastly, for √ , so this is 2 x = 7π 6 , π our answer. We can use the calculator to check that the graph of y = sin(2x) is indeed above the graph of y = cos(x) on those intervals. 2−2 2 . We see f (x) > 0 on π 2 √ 2 2 = 2 = −2− 4 we get f 7π = −1 − 2 = 2− ∪ 5π 6 , 3π 4 we get ; √ √ (−) 0 (+) π 6 0 (−) π 2 0 (+) 5π 6 0 (−) 3π 2 2π 0 y = sin(2x) and y = cos(x) 2 and 3π 3. Proceeding as in the last two problems, we rewrite tan(x) ≥ 3 as tan(x) − 3 ≥ 0 and let f (x) = tan(x) − 3. We note that on [0, 2π), f is undeﬁned at x = π 2 , so those values will need the usual disclaimer on the sign diagram.11 Moving along to zeros, solving f (x) = tan(x) − 3 = 0 requires the arctangent function. We ﬁnd x = arctan(3) + πk for integers k and of these, only x = arctan(3) and x = arctan(3) + π lie in [0, 2π). Since 3 > 0, we know 0 < arctan(3) < π 2 which allows us to position these zeros correctly on the sign diagram. To choose test values, we begin with x = 0 and ﬁnd f (0) = −3. Finding a is a bit more challenging. Keep in mind convenient test value in the interval arctan(3), π 2 that the arctangent function is increasing and is bounded above by π 2 . This means that the number x = arctan(117) is guaranteed12 to lie between arctan(3) and π 2 . We see that f (arctan(117)) = tan(arctan(117)) − 3 = 114. For our next test value, we take x = π and ﬁnd f (π) = −3. To ﬁnd our next test value, we note that since arctan(3) < arctan(117) < π 2 , it follows13 that arctan(3) + π < arctan(117) + π < 3π 2 . Evaluating f at x = arctan(117) + π yields f (arctan(117) + π) = tan(arctan(117) + π) − 3 = tan(arctan(117)) − 3 = 114. We = −4. Since we want f (x) ≥ 0, we 4 and ﬁnd f 7π choose our last test value to be x = 7π . Using the graphs of y = tan(x) ∪ arctan(3) + π, 3π see that our answer is arctan(3), π 2 2 and y = 3, we see when the graph of the former is above (or meets) the graph of the latter. 4 11See page 321 for a discussion of the non-standard character known as the interrobang. 12We could have chosen any value arctan(t) where t > 3. 13. . . by adding π through the inequality . . . 868 Foundations of Trigonometry (−) 0 (+) 0 arctan(3) (−) π 2 (arctan(3) + π) 0 (+) (−) 3π 2 2π y = tan(x) and y = 3 Our next example puts solving equations and inequalities to good use – ﬁnding domains of functions. Example 10.7.4. Express the domain of the following functions using extended interval notation.14 1. f (x) = csc 2x + π 3 2. f (x) = sin(x) 2 cos(x) − 1 3. f (x) = 1 − cot(x) Solution. 1. To ﬁnd the domain of f (x) = csc 2x + π 3 , we rewrite f in terms of sine as f (x) = 1 sin(2x+ π 3 ) Since the sine function is deﬁned everywhere, our only concern comes from zeros in the denominator. Solving sin 2x + π 2 k for integers k. In set-builder notation, 3 our domain is
x : x = − π 2 k for integers k. To help visualize the domain, we follow the 6 , . . ., old mantra ‘When in doubt, write it out!’ We get x : x = − π where we have kept the denominators 6 throughout to help see the pattern. Graphing the situation on a numberline, we have = 0, we get x = − π 6 + π 6 , − 7π 6 , − 4π 6 + π 6 , 5π 6 , 8π 6 , 2π . − 7π 6 − 4π 6 − π 6 2π 6 5π 6 8π 6 Proceeding as we did on page 756 in Section 10.3.1, we let xk denote the kth number excluded from the domain and we have xk = − π for integers k. The intervals which comprise the domain are of the form (xk, xk + 1) = integers. Using extended interval notation, we have that the domain is as k runs through the 2 k = (3k−1)π 6 (3k−1)π 6 , (3k+2)π 6 6 + π ∞ k=−∞ (3k − 1)π 6 , (3k + 2)π 6 We can check our answer by substituting in values of k to see that it matches our diagram. 14See page 756 for details about this notation. 10.7 Trigonometric Equations and Inequalities 869 2. Since the domains of sin(x) and cos(x) are all real numbers, the only concern when ﬁnding 2 cos(x)−1 is division by zero so we set the denominator equal to zero and 3 +2πk for integers 3 + 2πk for integers k, the domain of f (x) = sin(x) solve. From 2 cos(x)−1 = 0 we get cos(x) = 1 k. Using set-builder notation, the domain is x : x = π or x : x = ± π 3 , ± 11π 3 , . . ., so we have 3 + 2πk and x = 5π 3 +2πk or x = 5π 2 so that x = π 3 , ± 7π 3 , ± 5π − 11π 3 − 7π 3 − 5π 3 − π 3 π 3 5π 3 7π 3 11π 3 Unlike the previous example, we have two diﬀerent families of points to consider, and we present two ways of dealing with this kind of situation. One way is to generalize what we did in the previous example and use the formulas we found in our domain work to describe the intervals. To that end, we let ak = π for integers k. The goal now is to write the domain in terms of the a’s an b’s. We ﬁnd a0 = π 3 , a1 = 7π 3 and b−2 = − 7π 3 . Hence, in terms of the a’s and b’s, our domain is 3 + 2πk = (6k+5)π 3 + 2πk = (6k+1)π 3 , a−2 = − 11π 3 , a−1 = − 5π 3 , b−1 = − π 3 , a2 = 13π 3 , b1 = 11π 3 , b0 = 5π 3 , b2 = 17π and bk = 5π 3 3 . . . (a−2, b−2) ∪ (b−2, a−1) ∪ (a−1, b−1) ∪ (b−1, a0) ∪ (a0, b0) ∪ (b0, a1) ∪ (a1, b1) ∪ . . . If we group these intervals in pairs, (a−2, b−2)∪(b−2, a−1), (a−1, b−1)∪(b−1, a0), (a0, b0)∪(b0, a1) and so forth, we see a pattern emerge of the form (ak, bk) ∪ (bk, ak + 1) for integers k so that our domain can be written as ∞ k=−∞ (ak, bk) ∪ (bk, ak + 1) = ∞ k=−∞ (6k + 1)π 3 , (6k + 5)π 3 ∪ (6k + 5)π 3 , (6k + 7)π 3 A second approach to the problem exploits the periodic nature of f . Since cos(x) and sin(x) have period 2π, it’s not too diﬃcult to show the function f repeats itself every 2π units.15 This means if we can ﬁnd a formula for the domain on an interval of length 2π, we can express the entire domain by translating our answer left and right on the x-axis by adding integer multiples of 2π. One such interval that arises from our domain work is π . The portion . Adding integer multiples of 2π, we get the family of of the domain here is π 3 , 7π 3 + 2πk for integers k. We leave it to the reader intervals π 3 + 2πk, 7π to show that getting common denominators leads to our previous answer. ∪ 5π 3 + 2πk ∪ 5π 3 + 2πk, 5π 3 , 5π 3 , 7π 3 3 3 15This doesn’t necessarily mean the period of f is 2π. The tangent function is comprised of cos(x) and sin(x), but its period is half theirs. The reader is invited to investigate the period of f . 870 Foundations of Trigonometry 3. To ﬁnd the domain of f (x) = 1 − cot(x), we ﬁrst note that, due to the presence of the cot(x) term, x = πk for integers k. Next, we recall that for the square root to be deﬁned, we need 1−cot(x) ≥ 0. Unlike the inequalities we solved in Example 10.7.3, we are not restricted here to a given interval. Our strategy is to solve this inequality over (0, π) (the same interval which generates a fundamental cycle of cotangent) and then add integer multiples of the period, in this case, π. We let g(x) = 1 − cot(x) and set about making a sign diagram for g over the interval (0, π) to ﬁnd where g(x) ≥ 0. We note that g is undeﬁned for x = πk for integers k, in particular, at the endpoints of our interval x = 0 and x = π. Next, we look for the zeros of g. Solving g(x) = 0, we get cot(x) = 1 or x = π 4 + πk for integers k and 6 and x = π only one of these, x = π 2 , we get √ 3, and g π g π 2 6 4 , lies in (0, π). Choosing the test values x = π = 1. = 1 − (−) 0 0 (+) π 4 π We ﬁnd g(x) ≥ 0 on π the intervals π express our ﬁnal answer as 4 + πk, π + πk = 4 , π. Adding multiples of the period we get our solution to consist of . Using extended interval notation, we , (k + 1)π (4k+1)π 4 ∞ k=−∞ (4k + 1)π 4 , (k + 1)π We close this section with an example which demonstrates how to solve equations and inequalities involving the inverse trigonometric functions. Example 10.7.5. Solve the following equations and inequalities analytically. Check your answers using a graphing utility. 1. arcsin(2x) = π 3 2. 4 arccos(x) − 3π = 0 3. 3 arcsec(2x − 1) + π = 2π 4. 4 arctan2(x) − 3π arctan(x) − π2 = 0 5. π2 − 4 arccos2(x) < 0 6. 4 arccot(3x) > π Solution. 1. To solve arcsin(2x) = π 3 is in the range of the arcsine function (so a solution exists!) Next, we exploit the inverse property of sine and arcsine from Theorem 10.26 3 , we ﬁrst note that π 10.7 Trigonometric Equations and Inequalities 871 arcsin(2x) = π 3 sin (arcsin(2x)) = sin π 3 2x = x = √ 3 2 √ 3 4 Since sin(arcsin(u)) = u Graphing y = arcsin(2x) and the horizontal line y = π √ 3 4 ≈ 0.4430. 2. Our ﬁrst step in solving 4 arccos(x) − 3π = 0 is to isolate the arccosine. Doing so, we get 3 , we see they intersect at arccos(x) = 3π 4 . Since 3π 4 is in the range of arccosine, we may apply Theorem 10.26 arccos(x) = 3π 4 cos (arccos(x)) = cos 3π 4 x = − √ 2 2 Since cos(arccos(u)) = u The calculator conﬁrms y = 4 arccos(x) − 3π crosses y = 0 (the x-axis) at − √ 2 2 ≈ −0.7071. y = arcsin(2x) and y = π 3 y = 4 arccos(x) − 3π 3. From 3 arcsec(2x − 1) + π = 2π, we get arcsec(2x − 1) = π 3 . As we saw in Section 10.6, there are two possible ranges for the arcsecant function. Fortunately, both ranges contain π 3 . Applying Theorem 10.28 / 10.29, we get arcsec(2x − 1) = π 3 sec(arcsec(2x − 1)) = sec π 3 2x − 1 = 2 x = 3 2 Since sec(arcsec(u)) = u To check using our calculator, we need to graph y = 3 arcsec(2x − 1) + π. To do so, we make from Theorems 10.28 and 10.29.16 We see the graph use of the identity arcsec(u) = arccos 1 u of y = 3 arccos + π and the horizontal line y = 2π intersect at 3 1 2 = 1.5. 2x−1 16Since we are checking for solutions where arcsecant is positive, we know u = 2x − 1 ≥ 1, and so the identity applies in both cases. 872 Foundations of Trigonometry 4. With the presence of both arctan2(x) ( = (arctan(x))2) and arctan(x), we substitute u = arctan(x). The equation 4 arctan2(x) − 3π arctan(x) − π2 = 0 becomes 4u2 − 3πu − π2 = 0. Factoring,17 we get (4u + π)(u − π) = 0, so u = arctan(x) = − π 4 or u = arctan(x) = π. Since − π 4 is in the range of arctangent, but π is not, we only get solutions from the ﬁrst equation. Using Theorem 10.27, we get arctan(x) = − π 4 tan(arctan(x)) = tan − π 4 x = −1 Since tan(arctan(u)) = u. The calculator veriﬁes our result. y = 3 arcsec(2x − 1) + π and y = 2π y = 4 arctan2(x) − 3π arctan(x) − π2 5. Since the inverse trigonometric functions are continuous on their domains, we can solve inequalities featuring these functions using sign diagrams. Since all of the nonzero terms of π2 − 4 arccos2(x) < 0 are on one side of the inequality, we let f (x) = π2 − 4 arccos2(x) and note the domain of f is limited by the arccos(x) to [−1, 1]. Next, we ﬁnd the zeros of f by setting f (x) = π2 − 4 arccos2(x) = 0. We get arccos(x) = ± π 2 , and since the range of arccosine is = 0 [0, π], we focus our attention on arccos(x) = π as our only zero. Hence, we have two test intervals, [−1, 0) and (0, 1]. Choosing test values x = ±1, we get f (−1) = −3π2 < 0 and f (1) = π2 > 0. Since we are looking for where f (x) = π2 − 4 arccos2(x) < 0, our answer is [−1, 0). The calculator conﬁrms that for these values of x, the graph of y = π2 − 4 arccos2(x) is below y = 0 (the x-axis.) 2 . Using Theorem 10.26, we get x = cos π 2 17It’s not as bad as it looks... don’t let the π throw you! 10.7 Trigonometric Equations and Inequalities 873 (−) (+) 0 0 1 −1 y = π2 − 4 arccos2(x) 6. To begin, we rewrite 4 arccot(3x) > π as 4 arccot(3x) − π > 0. We let f (x) = 4 arccot(3x) − π, and note the domain of f is all real numbers, (−∞, ∞). To ﬁnd the zeros of f , we set f (x) = 4 arccot(3x) − π = 0 and solve. We get arccot(3x) = π 4 is in the range of arccotangent, we may apply Theorem 10.27 and solve 4 , and since π arccot(3x) = π 4 cot(arccot(3x)) = cot π 4 3x = 1 x = 1 3 Since cot(arccot(u)) = u. 3 , we have two test intervals, −∞, 1 Next, we make a sign diagram for f . Since the domain of f is all real numbers, and there is 3 , ∞. Ideally, we wish only one zero of f , x = 1 to ﬁnd test values x in these intervals so that arccot(4x) corresponds to one of our oft-used ‘common’ angles. After a bit of computation,18 we choose x = 0 for x < 1 3 , we √ 3 = − π choose x = 3 < 0. Since we are looking for where 3 . We ﬁnd f (0) = π > 0 and f . To check graphically, we use the f (x) = 4 arccot(3x) − π > 0, we get our answer −∞, 1 3 technique in number 2c of Example 10.6.5 in Section 10.6 to graph y = 4 arccot(3x) and we see it is above the horizontal line y = π on −∞, 1 3 3 and for x > 1 = −∞, 0.3. and 1 √ 3 3 3 (−) (+) 0 1 3 y = 4 arccot(3x) and y = π 18Set 3x equal to the cotangents of the ‘common angles’ and choose accordingly. 874 Foundations of Trigonometry 10.7.1 Exercises In Exercises 1 - 18, ﬁnd all of the exact solutions of the equation and then list those solutions which are in the interval [0, 2π). 1. sin (5x) = 0 2. cos (3x) = 1 2 4. tan (6x) = 1 5. csc (4x) = −1 √ 3 3 = 0 7. cot (2x) = − 10. cos x + 5π 6 13. csc(x) = 0 16. sec2 (x) = 4 3 8. cos (9x) = 9 11. sin 2x − π 3 = −
1 2 14. tan (2x − π) = 1 17. cos2 (x. sin (−2x) = 6. sec (3x) = 9. sin = x 3 12. 2 cos x + √ 3 = 7π 4 15. tan2 (x) = 3 18. sin2 (x) = 3 4 In Exercises 19 - 42, solve the equation, giving the exact solutions which lie in [0, 2π) 19. sin (x) = cos (x) 21. sin (2x) = cos (x) 23. cos (2x) = cos (x) 20. sin (2x) = sin (x) 22. cos (2x) = sin (x) 24. cos(2x) = 2 − 5 cos(x) 25. 3 cos(2x) + cos(x) + 2 = 0 26. cos(2x) = 5 sin(x) − 2 27. 3 cos(2x) = sin(x) + 2 29. tan2(x) = 1 − sec(x) 31. sec(x) = 2 csc(x) 33. sin(2x) = tan(x) 35. cos(2x) + csc2(x) = 0 37. tan2 (x) = 3 2 sec (x) 39. tan(2x) − 2 cos(x) = 0 28. 2 sec2(x) = 3 − tan(x) 30. cot2(x) = 3 csc(x) − 3 32. cos(x) csc(x) cot(x) = 6 − cot2(x) 34. cot4(x) = 4 csc2(x) − 7 36. tan3 (x) = 3 tan (x) 38. cos3 (x) = − cos (x) 40. csc3(x) + csc2(x) = 4 csc(x) + 4 41. 2 tan(x) = 1 − tan2(x) 42. tan (x) = sec (x) 10.7 Trigonometric Equations and Inequalities 875 In Exercises 43 - 58, solve the equation, giving the exact solutions which lie in [0, 2π) 43. sin(6x) cos(x) = − cos(6x) sin(x) 44. sin(3x) cos(x) = cos(3x) sin(x) 45. cos(2x) cos(x) + sin(2x) sin(x) = 1 46. cos(5x) cos(3x) − sin(5x) sin(3x) = √ 3 2 47. sin(x) + cos(x) = 1 √ 49. 2 cos(x) − √ 2 sin(x) = 1 51. cos(2x) − √ 3 sin(2x) = √ 2 48. sin(x) + √ 3 cos(x) = 1 √ 50. 3 sin(2x) + cos(2x) = 1 √ 52. 3 3 sin(3x) − 3 cos(3x) = 3 √ 3 53. cos(3x) = cos(5x) 54. cos(4x) = cos(2x) 55. sin(5x) = sin(3x) 56. cos(5x) = − cos(2x) 57. sin(6x) + sin(x) = 0 58. tan(x) = cos(x) In Exercises 59 - 68, solve the equation. 59. arccos(2x) = π 60. π − 2 arcsin(x) = 2π 61. 4 arctan(3x − 1) − π = 0 62. 6 arccot(2x) − 5π = 0 63. 4 arcsec x 2 = π 64. 12 arccsc x 3 = 2π 65. 9 arcsin2(x) − π2 = 0 66. 9 arccos2(x) − π2 = 0 67. 8 arccot2(x) + 3π2 = 10π arccot(x) 68. 6 arctan(x)2 = π arctan(x) + π2 In Exercises 69 - 80, solve the inequality. Express the exact answer in interval notation, restricting your attention to 0 ≤ x ≤ 2π. 69. sin (x) ≤ 0 72. cos2 (x) > 75. cot2 (x) ≥ 1 2 1 3 78. cos(3x) ≤ 1 70. tan (x) ≥ √ 3 73. cos (2x) ≤ 0 76. 2 cos(x) ≥ 1 79. sec(x) ≤ √ 2 71. sec2 (x) ≤ 4 74. sin x + π 3 > 1 2 77. sin(5x) ≥ 5 80. cot(x) ≤ 4 876 Foundations of Trigonometry In Exercises 81 - 86, solve the inequality. Express the exact answer in interval notation, restricting your attention to −π ≤ x ≤ π. 81. cos (x) > 84. sin2 (x) < √ 3 2 3 4 82. sin(x) > 1 3 83. sec (x) ≤ 2 85. cot (x) ≥ −1 86. cos(x) ≥ sin(x) In Exercises 87 - 92, solve the inequality. Express the exact answer in interval notation, restricting your attention to −2π ≤ x ≤ 2π. 87. csc (x) > 1 90. tan2 (x) ≥ 1 88. cos(x) ≤ 5 3 89. cot(x) ≥ 5 91. sin(2x) ≥ sin(x) 92. cos(2x) ≤ sin(x) In Exercises 93 - 98, solve the given inequality. 93. arcsin(2x) > 0 94. 3 arccos(x) ≤ π 95. 6 arccot(7x) ≥ π 96. π > 2 arctan(x) 97. 2 arcsin(x)2 > π arcsin(x) 98. 12 arccos(x)2 + 2π2 > 11π arccos(x) In Exercises 99 - 107, express the domain of the function using the extended interval notation. (See page 756 in Section 10.3.1 for details.) 99. f (x) = 1 cos(x) − 1 100. f (x) = cos(x) sin(x) + 1 101. f (x) = tan2(x) − 1 102. f (x) = 2 − sec(x) 103. f (x) = csc(2x) 104. f (x) = sin(x) 2 + cos(x) 105. f (x) = 3 csc(x) + 4 sec(x) 106. f (x) = ln (\| cos(x)\|) 107. f (x) = arcsin(tan(x)) 108. With the help of your classmates, determine the number of solutions to sin(x) = 1 2 and sin(4x) = 1 2 in [0, 2π). Then ﬁnd the number of solutions to sin(2x) = 1 2 in [0, 2π). A pattern should emerge. Explain how this pattern would help you solve equations like sin(11x) = 1 2 . What do you ﬁnd? 2 , sin(3x) = 1 2 and sin 5x 2 , sin 3x = 1 = 1 2 2 = 1 2 . Now consider sin x with −1 and repeat the whole exploration. 2 Replace 1 2 10.7 Trigonometric Equations and Inequalities 877 10.7.2 Answers 1. x = πk 5 ; x = 0, π 5 , 2π 5 , 3π 5 , 4π 5 , π, 6π 5 , 7π 5 , 8π 5 , 9π 5 2. x = π 9 + 2πk 3 or x = + πk or x = 5π 9 5π 6 + 2πk 3 ; x = π 9 , 5π 9 , 7π 9 , 11π 9 , 13π 9 , 17π 9 + πk; x = 2π 3 , 5π 6 , 5π 3 , 11π 6 + + πk 6 πk 2 ; x = π 24 , 5π 24 , 3π 8 , 13π 24 , 17π 24 , 7π 8 , 25π 24 , 29π 24 , 11π 8 , 37π 24 , 41π 24 , 15π 8 ; x = 3π 8 , 7π 8 , 11π 8 , 15π 8 + 2πk 3 or x = 7π 12 + 2πk 3 ; x = π 12 , 7π 12 , 3π 4 , 5π 4 , 17π 12 , 23π 12 3. x = 4. x = 5. x = 6. x = 2π 3 π 24 3π 8 π 12 7. x = π 3 + πk 2 ; x = π 3 , 5π 6 , 4π 3 , 11π 6 8. No solution 9. x = 3π 4 10. x = − π 3 11. x = 3π 4 + 6πk or x = 9π 4 + 6πk; x = 3π 4 + πk; x = 2π 3 , 5π 3 + πk or x = 13π 12 + πk; x = π 12 , 3π 4 , 13π 12 , 7π 4 12. x = − 19π 12 + 2πk or x = π 12 + 2πk; x = π 12 , 5π 12 13. No solution 14. x = 5π 8 + πk 2 ; x = π 8 , 5π 8 , 9π 8 , 13π 8 15. x = 16. x = 17. x = 18 + πk or x = + πk or x = 2π 3 5π 6 + πk; x = + πk; x = π 3 π 6 , , 2π 3 5π 6 , , 4π 3 7π 6 , , 5π 3 11π 6 + πk 2 ; x = π 4 , 3π 4 , 5π 4 , 7π 4 + πk or x = 2π 3 + πk; x = π 3 , 2π 3 , 4π 3 , 5π 3 878 19. x = 21. x = π 4 π 6 23. x = 0, , , 5π 4 π 2 2π 3 , , 3π 2 , 5π 6 4π 3 25. x = 27. x = 2π 3 7π 6 , , 4π 3 , arccos 11π 6 , arcsin 1 3 1 3 , 2π − arccos , π − arcsin 1 3 1 3 29. x = 0, 2π 3 , 4π 3 31. x = arctan(2), π + arctan(2) Foundations of Trigonometry 20. x = 0, 22. x = 24. x = 26, 5π 3 3π 2 , π 3 5π 6 5π 3 5π 6 , , , 28. x = 3π 4 , 7π 4 , arctan 1 2 , π + arctan 1 2 30. x = 32. x = 34. x = π 6 π 6 π 6 36. x = 0, 38. x = 40. x = π 2 π 6 , , , , , , 5π 6 7π 6 π 4 π 3 3π 2 5π 6 , , , π 2 5π 6 3π 4 2π 3 , , 11π 6 5π 6 4π 3 , , 7π 6 5π 3 , , π, , 7π 6 , 3π 2 , 11π 6 5π 4 , 7π 4 , 11π 6 33. x = 0, π, , 3π 4 , 5π 4 , 7π 4 π 4 3π 2 5π 3 π 2 5π 8 , 35. x = 37. x = 39. x = 41 , , , , , 5π 6 9π 8 , 3π 2 13π 8 4π 7 , , , 3π 7 3π 2 13π 48 , 43. x = 0, 44. x = 0, π 7 π 2 , 2π 7 , π, 46. x = π 48 , 11π 48 47. x = 0, π 2 49. x = 51. x = , π 12 17π 24 17π 12 41π 24 , , 23π 24 , 47π 24 42. No solution , 5π 7 , 6π 7 , π, 8π 7 , 9π 7 , 10π 7 , 11π 7 , 12π 7 , 13π 7 45. x = 0 , 23π 48 , 25π 48 , 35π 48 , 37π 48 , 47π 48 , 49π 48 , 59π 48 , 48. x = , 71π 48 , 73π 48 , 83π 48 , 85π 48 , 95π 48 61π 48 π 2 , 11π 6 π 3 5π 18 , , 50. x = 0, π, 52. x = π 6 , 4π 3 5π 6 , 17π 18 , 3π 2 , 29π 18 10.7 Trigonometric Equations and Inequalities 879 53. x = 0, 55. x = 0 3π 8 3π 4 5π 8 , , π, , 5π 4 7π 8 , 3π 2 9π 8 7π 4 11π 8 , , 13π 8 , 15π 8 54. x = 0, π 3 , 2π 3 , π, 4π 3 , 5π 3 , , π, 56. x = π 7 , π 3 , 3π 7 , 5π 7 , π, 9π 7 , 11π 7 , 5π 3 , 13π 7 57. x = 0, 2π 7 , 58. x = arcsin 59. x = − 1 2 61. x = 2 3 63. x = 2 √ 2 65. x = ± √ 3 2 67. x = −1, 0 , 6π 7 , 4π 7 −1 + 2 8π 7 √ 5 , 10π 7 , 12π 7 , π 5 , 3π 5 , π, , 7π 5 ≈ 0.6662, π − arcsin 9π 5 −1 + 2 √ 5 ≈ 2.4754 60. x = −1 62. x = − √ 3 2 64. x = 6 66. x = 1 2 √ 68. x = − 3 69. [π, 2π] 71. 73. 0, π 3 ∪ π 4 , 3π 4 , 2π 3 5π 4 ∪ , 7π 4 4π 3 ∪ 5π 3 , 2π 0, 75. π 3 ∪ 2π 3 , π ∪ π, 4π 3 ∪ 5π 3 , 2π 70. 72. 74. 76. π 3 , π 2 ∪ 0, 0, 0 , , 4π 3 3π 4 11π 6 5π 3 , 2π 3π 2 5π 4 ∪ 7π 4 , 2π , 2π 77. No solution 78. [0, 2π] 0, 79. π 4 ∪ π 2 , 3π 2 ∪ 7π 4 , 2π 81. − π 6 , π 6 −π, − 83 80. [arccot(4), π) ∪ [π + arccot(4), 2π) 82. arcsin 1 3 , π − arcsin 1 3 84. − 2π 3 , − π 3 ∪ π 3 , 2π 3 880 Foundations of Trigonometry −π, − 85. π 4 ∪ 0, 3π 4 −2π, − 87. 3π 2 ∪ − 3π 2 , −π ∪ 0, π 2 ∪ π 2 , π 86. − 3π 4 , π 4 88. [−2π, 2π] 89. (−2π, arccot(5) − 2π] ∪ (−π, arccot(5) − π] ∪ (0, arccot(5)] ∪ (π, π + arccot(5)] − 3π 2 , − 5π 4 − ∪ 3π 4 , − −π, − 0 , 3π 4 ∪ 5π 4 , 3π 2 ∪ 3π 2 , 7π 4 ∪ π, 5π 3 90. − 7π 4 , − 3π 2 −2π, − 91. 5π 3 ∪ 92. − 93. 0, 1 2 , − 7π 6 11π 6 ∪ ∪ π 6 , 5π 6 ∪, , − π 3π 2 2 2 , 1 94. 1 96. (−∞, ∞) 97. [−1, 0) 99. ∞ k=−∞ (2kπ, (2k + 2)π) 100. ∞ k=−∞ (4k − 1)π 2 , (4k + 3)π 2 −∞, √ 3 7 95. 98. −14k + 1)π k=−∞ 4 ∞ (6k − 1)π 3 (2k + 1)π 2 ∪ (2k + 1)π 2 , (4k + 3)π 4 (6k + 1)π 3 ∪ (4k + 1)π 2 , (4k + 3)π 2 , , kπ 2 kπ 2 , , (k + 1)π 2 (k + 1)π 2 (4k − 1)π 4 , (4k + 1)π 4 104. (−∞, ∞) 106. ∞ k=−∞ (2k − 1)π 2 , (2k + 1)π 2 101. 102. 103. 105. 107. k=−∞ ∞ k=−∞ ∞ k=−∞ ∞ k=−∞ Chapter 11 Applications of Trigonometry 11.1 Applications of Sinusoids In the same way exponential functions can be used to model a wide variety of phenomena in nature,1 the cosine and sine functions can be used to model their fair share of natural behaviors. In section 10.5, we introduced the concept of a sinusoid as a function which can be written either in the form C(x) = A cos(ωx+φ)+B for ω > 0 or equivalently, in the form S(x) = A sin(ωx+φ)+B for ω > 0. At the time, we remained undecided as to which form we preferred, but the time for such indecision is over. For clarity of exposition we focus on the sine function2 in this section and switch to the independent variable t, since the applications in this section are time-dependent. We reintroduce and summarize all of the important facts and deﬁnitions about this form of the sinusoid below. Properties of the Sinusoid S(t) = A sin(ωt + φ) + B The amplitude is \|A\| The angular frequency is ω and the ordinary frequency is f = ω 2π The period is T = 1 f = 2π ω The phase is φ and the phase shift is − φ ω The vertical shift or baseline is B Along with knowing these formulas, it is helpful to remember what these quantities mean in context. The amplitude measures the maximum displacement of the sine wave from its baseline (determined by the vertical shift), the period is the length of time it takes to complete one cycle of the sinusoid, the angular frequency tells how many cycles are completed over an interval of length 2π, and the ordinary frequency measures how many cycles occur per unit of time. The phase indicates what 1See Section 6.5. 2Sine haters can use the co-function identity cos π 2 − θ = sin(θ) to turn all of the sines into cosines. 882 Applications of Trigonometry angle φ corresponds to t = 0, and the phase shift represents how much of a ‘head start’ the sinusoid has over the un-shifted sine function. The ﬁgure below is repeated from Section 10.5. amplitude baseline period In Section 10.1.1, we introduced the concept of circular motion and in Section 10.2.1, we developed formulas for circular motion. Our ﬁrst foray into sinusoidal motion puts these notions to good use. Example 11.1.1. Recall from Exercise 55 in Section 10.1 that The Giant Wheel at Ceda
r Point is a circle with diameter 128 feet which sits on an 8 foot tall platform making its overall height 136 feet. It completes two revolutions in 2 minutes and 7 seconds. Assuming that the riders are at the edge of the circle, ﬁnd a sinusoid which describes the height of the passengers above the ground t seconds after they pass the point on the wheel closest to the ground. Solution. We sketch the problem situation below and assume a counter-clockwise rotation.3 θ Q h P O 3Otherwise, we could just observe the motion of the wheel from the other side. 11.1 Applications of Sinusoids 883 We know from the equations given on page 732 in Section 10.2.1 that the y-coordinate for counterclockwise motion on a circle of radius r centered at the origin with constant angular velocity (frequency) ω is given by y = r sin(ωt). Here, t = 0 corresponds to the point (r, 0) so that θ, the angle measuring the amount of rotation, is in standard position. In our case, the diameter of the wheel is 128 feet, so the radius is r = 64 feet. Since the wheel completes two revolutions in 2 minutes and 7 seconds (which is 127 seconds) the period T = 1 seconds. Hence, the angular frequency is ω = 2π T = 4π 127 radians per second. Putting these two pieces of information together, we have that y = 64 sin 4π 127 t describes the y-coordinate on the Giant Wheel after t seconds, assuming it is centered at (0, 0) with t = 0 corresponding to the point Q. In order to ﬁnd an expression for h, we take the point O in the ﬁgure as the origin. Since the base of the Giant Wheel ride is 8 feet above the ground and the Giant Wheel itself has a radius of 64 feet, its center is 72 feet above the ground. To account for this vertical shift upward,4 we add 72 to our formula for y to obtain the new formula h = y + 72 = 64 sin 4π 127 t + 72. Next, we need to adjust things so that t = 0 corresponds to the point P instead of the point Q. This is where the phase comes into play. Geometrically, we need to shift the angle θ in the ﬁgure back π 2 radians. From Section 10.2.1, we know θ = ωt = 4π 127 t, so we (temporarily) write the height in terms of θ as h = 64 sin (θ) + 72. + 72. We 127 t − π Subtracting π 2 . can check the reasonableness of our answer by graphing y = h(t) over the interval 0, 127 2 2 from θ gives the ﬁnal answer h(t) = 64 sin θ − π + 72 = 64 sin 4π 2 (127) = 127 2 2 y 136 72 8 t 127 2 A few remarks about Example 11.1.1 are in order. First, note that the amplitude of 64 in our answer corresponds to the radius of the Giant Wheel. This means that passengers on the Giant Wheel never stray more than 64 feet vertically from the center of the Wheel, which makes sense. 8 = 15.875. This represents the Second, the phase shift of our answer works out to be ‘time delay’ (in seconds) we introduce by starting the motion at the point P as opposed to the point Q. Said diﬀerently, passengers which ‘start’ at P take 15.875 seconds to ‘catch up’ to the point Q. 4π/127 = 127 π/2 Our next example revisits the daylight data ﬁrst introduced in Section 2.5, Exercise 6b. 4We are readjusting our ‘baseline’ from y = 0 to y = 72. 884 Applications of Trigonometry Example 11.1.2. According to the U.S. Naval Observatory website, the number of hours H of daylight that Fairbanks, Alaska received on the 21st day of the nth month of 2009 is given below. Here t = 1 represents January 21, 2009, t = 2 represents February 21, 2009, and so on. Month Number Hours of Daylight 1 2 3 4 5 6 7 8 9 10 11 12 5.8 9.3 12.4 15.9 19.4 21.8 19.4 15.6 12.4 9.1 5.6 3.3 1. Find a sinusoid which models these data and use a graphing utility to graph your answer along with the data. 2. Compare your answer to part 1 to one obtained using the regression feature of a calculator. Solution. 1. To get a feel for the data, we plot it below. H 22 20 18 16 14 12 10 10 11 12 6 The data certainly appear sinusoidal,5 but when it comes down to it, ﬁtting a sinusoid to data manually is not an exact science. We do our best to ﬁnd the constants A, ω, φ and B so that the function H(t) = A sin(ωt + φ) + B closely matches the data. We ﬁrst go after the vertical shift B whose value determines the baseline. In a typical sinusoid, the value of B is the average of the maximum and minimum values. So here we take B = 3.3+21.8 = 12.55. Next is the amplitude A which is the displacement from the baseline to the maximum (and minimum) values. We ﬁnd A = 21.8 − 12.55 = 12.55 − 3.3 = 9.25. At this point, we have H(t) = 9.25 sin(ωt + φ) + 12.55. Next, we go after the angular frequency ω. Since the data collected is over the span of a year (12 months), we take the period T = 12 months.6 This 2 5Okay, it appears to be the ‘∧’ shape we saw in some of the graphs in Section 2.2. Just humor us. 6Even though the data collected lies in the interval [1, 12], which has a length of 11, we need to think of the data point at t = 1 as a representative sample of the amount of daylight for every day in January. That is, it represents H(t) over the interval [0, 1]. Similarly, t = 2 is a sample of H(t) over [1, 2], and so forth. 11.1 Applications of Sinusoids 885 12 = π T = 2π means ω = 2π 6 . The last quantity to ﬁnd is the phase φ. Unlike the previous example, it is easier in this case to ﬁnd the phase shift − φ ω . Since we picked A > 0, the phase shift corresponds to the ﬁrst value of t with H(t) = 12.55 (the baseline value).7 Here, we choose t = 3, since its corresponding H value of 12.4 is closer to 12.55 than the next value, 15.9, which corresponds to t = 4. Hence, − φ 2 . We have H(t) = 9.25 sin π + 12.55. Below is a graph of our data with the curve y = H(t). ω = 3, so φ = −3ω = −. Using the ‘SinReg’ command, we graph the calculator’s regression below. While both models seem to be reasonable ﬁts to the data, the calculator model is possibly the better ﬁt. The calculator does not give us an r2 value like it did for linear regressions in Section 2.5, nor does it give us an R2 value like it did for quadratic, cubic and quartic regressions as in Section 3.1. The reason for this, much like the reason for the absence of R2 for the logistic model in Section 6.5, is beyond the scope of this course. We’ll just have to use our own good judgment when choosing the best sinusoid model. 11.1.1 Harmonic Motion One of the major applications of sinusoids in Science and Engineering is the study of harmonic motion. The equations for harmonic motion can be used to describe a wide range of phenomena, from the motion of an object on a spring, to the response of an electronic circuit. In this subsection, we restrict our attention to modeling a simple spring system. Before we jump into the Mathematics, there are some Physics terms and concepts we need to discuss. In Physics, ‘mass’ is deﬁned as a measure of an object’s resistance to straight-line motion whereas ‘weight’ is the amount of force (pull) gravity exerts on an object. An object’s mass cannot change,8 while its weight could change. 7See the ﬁgure on page 882. 8Well, assuming the object isn’t subjected to relativistic speeds . . . 886 Applications of Trigonometry An object which weighs 6 pounds on the surface of the Earth would weigh 1 pound on the surface of the Moon, but its mass is the same in both places. In the English system of units, ‘pounds’ (lbs.) is a measure of force (weight), and the corresponding unit of mass is the ‘slug’. In the SI system, the unit of force is ‘Newtons’ (N) and the associated unit of mass is the ‘kilogram’ (kg). We convert between mass and weight using the formula9 w = mg. Here, w is the weight of the object, m is the mass and g is the acceleration due to gravity. In the English system, g = 32 feet second2 , and in the SI system, g = 9.8 meters second2 . Hence, on Earth a mass of 1 slug weighs 32 lbs. and a mass of 1 kg weighs 9.8 N.10 Suppose we attach an object with mass m to a spring as depicted below. The weight of the object will stretch the spring. The system is said to be in ‘equilibrium’ when the weight of the object is perfectly balanced with the restorative force of the spring. How far the spring stretches to reach equilibrium depends on the spring’s ‘spring constant’. Usually denoted by the letter k, the spring constant relates the force F applied to the spring to the amount d the spring stretches in accordance with Hooke’s Law11 F = kd. If the object is released above or below the equilibrium position, or if the object is released with an upward or downward velocity, the object will bounce up and down on the end of the spring until some external force stops it. If we let x(t) denote the object’s displacement from the equilibrium position at time t, then x(t) = 0 means the object is at the equilibrium position, x(t) < 0 means the object is above the equilibrium position, and x(t) > 0 means the object is below the equilibrium position. The function x(t) is called the ‘equation of motion’ of the object.12 x(t) = 0 at the equilibrium position x(t) < 0 above the equilibrium position x(t) > 0 below the equilibrium position If we ignore all other inﬂuences on the system except gravity and the spring force, then Physics tells us that gravity and the spring force will battle each other forever and the object will oscillate indeﬁnitely. In this case, we describe the motion as ‘free’ (meaning there is no external force causing the motion) and ‘undamped’ (meaning we ignore friction caused by surrounding medium, which in our case is air). The following theorem, which comes from Diﬀerential Equations, gives x(t) as a function of the mass m of the object, the spring constant k, the initial displacement x0 of the 9This is a consequence of Newton’s Second Law of Motion F = ma where F is force, m is mass and a is acceleration. In our present setting, the force involved is weight which is caused by the acceleration due to gravity. 10Note that 1 pound = 1 slug foot 11Look familiar? We saw Hooke’s Law in Section 4.3.1. 12To keep units compatible, if we are using the English system, we use fee
t (ft.) to measure displacement. If we second2 and 1 Newton = 1 kg meter second2 . are in the SI system, we measure displacement in meters (m). Time is always measured in seconds (s). 11.1 Applications of Sinusoids 887 object and initial velocity v0 of the object. As with x(t), x0 = 0 means the object is released from the equilibrium position, x0 < 0 means the object is released above the equilibrium position and x0 > 0 means the object is released below the equilibrium position. As far as the initial velocity v0 is concerned, v0 = 0 means the object is released ‘from rest,’ v0 < 0 means the object is heading upwards and v0 > 0 means the object is heading downwards.13 Theorem 11.1. Equation for Free Undamped Harmonic Motion: Suppose an object of mass m is suspended from a spring with spring constant k. If the initial displacement from the equilibrium position is x0 and the initial velocity of the object is v0, then the displacement x from the equilibrium position at time t is given by x(t) = A sin(ωt + φ) where ω = k m and A = x2 0 + 2 v0 ω A sin(φ) = x0 and Aω cos(φ) = v0. It is a great exercise in ‘dimensional analysis’ to verify that the formulas given in Theorem 11.1 work out so that ω has units 1 s and A has units ft. or m, depending on which system we choose. Example 11.1.3. Suppose an object weighing 64 pounds stretches a spring 8 feet. 1. If the object is attached to the spring and released 3 feet below the equilibrium position from rest, ﬁnd the equation of motion of the object, x(t). When does the object ﬁrst pass through the equilibrium position? Is the object heading upwards or downwards at this instant? 2. If the object is attached to the spring and released 3 feet below the equilibrium position with an upward velocity of 8 feet per second, ﬁnd the equation of motion of the object, x(t). What is the longest distance the object travels above the equilibrium position? When does this ﬁrst happen? Conﬁrm your result using a graphing utility. Solution. In order to use the formulas in Theorem 11.1, we ﬁrst need to determine the spring constant k and the mass of the object m. To ﬁnd k, we use Hooke’s Law F = kd. We know the object weighs 64 lbs. and stretches the spring 8 ft.. Using F = 64 and d = 8, we get 64 = k · 8, or k = 8 lbs. s2 . We get m = 2 slugs. We can now proceed to apply Theorem 11.1. ft. . To ﬁnd m, we use w = mg with w = 64 lbs. and g = 32 ft. k 8 1. With k = 8 and m = 2, we get ω = 2 = 2. We are told that the object is released 3 feet below the equilibrium position ‘from rest.’ This means x0 = 3 and v0 = 0. Therefore, 32 + 02 = 3. To determine the phase φ, we have A sin(φ) = x0, 2 and angles coterminal to it A = which in this case gives 3 sin(φ) = 3 so sin(φ) = 1. Only φ = π 0 + v0 x2 m = 2 √ = ω 13The sign conventions here are carried over from Physics. If not for the spring, the object would fall towards the ground, which is the ‘natural’ or ‘positive’ direction. Since the spring force acts in direct opposition to gravity, any movement upwards is considered ‘negative’. 888 Applications of Trigonometry = 0. Going through the usual analysis we ﬁnd t = − π satisfy this condition, so we pick14 the phase to be φ = π 2 . Hence, the equation of motion is x(t) = 3 sin 2t + π . To ﬁnd when the object passes through the equilibrium position we 2 solve x(t) = 3 sin 2t + π 2 k for 2 integers k. Since we are interested in the ﬁrst time the object passes through the equilibrium position, we look for the smallest positive t value which in this case is t = π 4 ≈ 0.78 seconds after the start of the motion. Common sense suggests that if we release the object below the equilibrium position, the object should be traveling upwards when it ﬁrst passes through it. To check this answer, we graph one cycle of x(t). Since our applied domain in this situation is t ≥ 0, and the period of x(t) is T = 2π 2 = π, we graph x(t) over the interval [0, π]. Remembering that x(t) > 0 means the object is below the equilibrium position and x(t) < 0 means the object is above the equilibrium position, the fact our graph is crossing through the t-axis from positive x to negative x at t = π ω = 2π 4 + π 4 conﬁrms our answer. ω = 2 0 + v0 x2 5 . From Aω cos(φ) = v0, we get 10 cos(φ) = −8, or cos(φ) = − 4 2. The only diﬀerence between this problem and the previous problem is that we now release the object with an upward velocity of 8 ft s . We still have ω = 2 and x0 = 3, but now we have v0 = −8, the negative indicating the velocity is directed upwards. Here, we get 32 + (−4)2 = 5. From A sin(φ) = x0, we get 5 sin(φ) = 3 which gives A = sin(φ) = 3 5 . This means that φ is a Quadrant II angle which we can describe in terms of either arcsine or arccosine. Since x(t) is expressed in terms of sine, we choose to express φ = π − arcsin 3 . Hence, 5 . Since the amplitude of x(t) is 5, the object will travel x(t) = 5 sin 2t + π − arcsin 3 5 at most 5 feet above the equilibrium position. To ﬁnd when this happens, we solve the equation x(t) = 5 sin 2t + π − arcsin 3 = −5, the negative once again signifying that 5 the object is above the equilibrium position. Going through the usual machinations, we get 2 arcsin 3 t = 1 4 + πk for integers k. The smallest of these values occurs when k = 0, 5 that is, t = 1 4 ≈ 1.107 seconds after the start of the motion. To check our answer using the calculator, we graph y = 5 sin 2x + π − arcsin 3 on a graphing utility 5 and conﬁrm the coordinates of the ﬁrst relative minimum to be approximately (1.107, −5). + π 2 arcsin 3 + π 5 x π 4 π 2 3π 4 π t 3 2 1 −1 −2 −3 x(t) = 3 sin 2t + π 2 y = 5 sin 2x + π − arcsin 3 5 It is possible, though beyond the scope of this course, to model the eﬀects of friction and other external forces acting on the system.15 While we may not have the Physics and Calculus background 14For conﬁrmation, we note that Aω cos(φ) = v0, which in this case reduces to 6 cos(φ) = 0. 15Take a good Diﬀerential Equations class to see this! 11.1 Applications of Sinusoids 889 to derive equations of motion for these scenarios, we can certainly analyze them. We examine three cases in the following example. Example 11.1.4. 1. Write x(t) = 5e−t/5 cos(t) + 5e−t/5 using a graphing utility. √ 3 sin(t) in the form x(t) = A(t) sin(ωt + φ). Graph x(t) 2. Write x(t) = (t + 3) √ 2 cos(2t) + (t + 3) √ x(t) using a graphing utility. 2 sin(2t) in the form x(t) = A(t) sin(ωt + φ). Graph 3. Find the period of x(t) = 5 sin(6t) − 5 sin (8t). Graph x(t) using a graphing utility. Solution. √ √ √ 1. We start rewriting x(t) = 5e−t/5 cos(t) + 5e−t/5 3 sin(t) by factoring out 5e−t/5 from both terms to get x(t) = 5e−t/5 cos(t) + 3 sin(t). We convert what’s left in parentheses to the required form using the formulas introduced in Exercise 36 from Section 10.5. We ﬁnd so that x(t) = 10e−t/5 sin t + π cos(t) + . Graphing this on the 3 sin(t) = 2 sin t + π 3 3 calculator as y = 10e−x/5 sin x + π reveals some interesting behavior. The sinusoidal nature 3 continues indeﬁnitely, but it is being attenuated. In the sinusoid A sin(ωx + φ), the coeﬃcient , we can think A of the sine function is the amplitude. In the case of y = 10e−x/5 sin x + π 3 of the function A(x) = 10e−x/5 as the amplitude. As x → ∞, 10e−x/5 → 0 which means the amplitude continues to shrink towards zero. Indeed, if we graph y = ±10e−x/5 along with y = 10e−x/5 sin x + π , we see this attenuation taking place. This equation corresponds to 3 the motion of an object on a spring where there is a slight force which acts to ‘damp’, or slow the motion. An example of this kind of force would be the friction of the object against the air. In this model, the object oscillates forever, but with smaller and smaller amplitude. y = 10e−x/5 sin x + π 3 y = 10e−x/5 sin x + π 3 , y = ±10e−x/5 √ 2. Proceeding as in the ﬁrst example, we factor out (t + 3) 2 from each term in the function 2(cos(2t) + sin(2t)). We ﬁnd x(t) = (t + 3) . Graphing this on the (cos(2t) + sin(2t)) = calculator as y = 2(x + 3) sin 2x + π , we ﬁnd the sinusoid’s amplitude growing. Since our 4 amplitude function here is A(x) = 2(x + 3) = 2x + 6, which continues to grow without bound , so x(t) = 2(t + 3) sin 2t + π 4 √ 2 sin 2t + π 4 2 sin(2t) to get x(t) = (t + 3) 2 cos(2t) + (t + 3) √ √ √ 890 Applications of Trigonometry as x → ∞, this is hardly surprising. The phenomenon illustrated here is ‘forced’ motion. That is, we imagine that the entire apparatus on which the spring is attached is oscillating as well. In this case, we are witnessing a ‘resonance’ eﬀect – the frequency of the external oscillation matches the frequency of the motion of the object on the spring.16 y = 2(x + 3) sin 2x + π 4 y = 2(x + 3) sin 2x + π 4 y = ±2(x + 3) 8 = 3 3. Last, but not least, we come to x(t) = 5 sin(6t) − 5 sin(8t). To ﬁnd the period of this function, we need to determine the length of the smallest interval on which both f (t) = 5 sin(6t) and g(t) = 5 sin(8t) complete a whole number of cycles. To do this, we take the ratio of their frequencies and reduce to lowest terms: 6 4 . This tells us that for every 3 cycles f makes, In other words, the period of x(t) is three times the period of f (t) (which is g makes 4. four times the period of g(t)), or π. We graph y = 5 sin(6x) − 5 sin(8x) over [0, π] on the calculator to check this. This equation of motion also results from ‘forced’ motion, but here the frequency of the external oscillation is diﬀerent than that of the object on the spring. Since the sinusoids here have diﬀerent frequencies, they are ‘out of sync’ and do not amplify each other as in the previous example. Taking things a step further, we can use a sum to product identity to rewrite x(t) = 5 sin(6t) − 5 sin(8t) as x(t) = −10 sin(t) cos(7t). The lower frequency factor in this expression, −10 sin(t), plays an interesting role in the graph of x(t). Below we graph y = 5 sin(6x) − 5 sin(8x) and y = ±10 sin(x) over [0, 2π]. This is an example of the ‘beat’ phenomena, and the curious reader is invited to explore this co
ncept as well.17 y = 5 sin(6x) − 5 sin(8x) over [0, π] y = 5 sin(6x) − 5 sin(8x) and y = ±10 sin(x) over [0, 2π] 16The reader is invited to investigate the destructive implications of resonance. 17A good place to start is this article on beats. 11.1 Applications of Sinusoids 891 11.1.2 Exercises 1. The sounds we hear are made up of mechanical waves. The note ‘A’ above the note ‘middle second . Find a sinusoid which C’ is a sound wave with ordinary frequency f = 440 Hertz = 440 cycles models this note, assuming that the amplitude is 1 and the phase shift is 0. 2. The voltage V in an alternating current source has amplitude 220 √ 2 and ordinary frequency f = 60 Hertz. Find a sinusoid which models this voltage. Assume that the phase is 0. 3. The London Eye is a popular tourist attraction in London, England and is one of the largest Ferris Wheels in the world. It has a diameter of 135 meters and makes one revolution (counterclockwise) every 30 minutes. It is constructed so that the lowest part of the Eye reaches ground level, enabling passengers to simply walk on to, and oﬀ of, the ride. Find a sinsuoid which models the height h of the passenger above the ground in meters t minutes after they board the Eye at ground level. 4. On page 732 in Section 10.2.1, we found the x-coordinate of counter-clockwise motion on a circle of radius r with angular frequency ω to be x = r cos(ωt), where t = 0 corresponds to the point (r, 0). Suppose we are in the situation of Exercise 3 above. Find a sinsusoid which models the horizontal displacement x of the passenger from the center of the Eye in meters t minutes after they board the Eye. Here we take x(t) > 0 to mean the passenger is to the right of the center, while x(t) < 0 means the passenger is to the left of the center. 5. In Exercise 52 in Section 10.1, we introduced the yo-yo trick ‘Around the World’ in which a yo-yo is thrown so it sweeps out a vertical circle. As in that exercise, suppose the yo-yo string is 28 inches and it completes one revolution in 3 seconds. If the closest the yo-yo ever gets to the ground is 2 inches, ﬁnd a sinsuoid which models the height h of the yo-yo above the ground in inches t seconds after it leaves its lowest point. 6. Suppose an object weighing 10 pounds is suspended from the ceiling by a spring which stretches 2 feet to its equilibrium position when the object is attached. (a) Find the spring constant k in lbs. (b) Find the equation of motion of the object if it is released from 1 foot below the equilibrium position from rest. When is the ﬁrst time the object passes through the equilibrium position? In which direction is it heading? ft. and the mass of the object in slugs. (c) Find the equation of motion of the object if it is released from 6 inches above the equilibrium position with a downward velocity of 2 feet per second. Find when the object passes through the equilibrium position heading downwards for the third time. 892 Applications of Trigonometry 7. Consider the pendulum below. Ignoring air resistance, the angular displacement of the pen- dulum from the vertical position, θ, can be modeled as a sinusoid.18 θ The amplitude of the sinusoid is the same as the initial angular displacement, θ0, of the pendulum and the period of the motion is given by T = 2π l g where l is the length of the pendulum and g is the acceleration due to gravity. (a) Find a sinusoid which gives the angular displacement θ as a function of time, t. Arrange things so θ(0) = θ0. (b) In Exercise 40 section 5.3, you found the length of the pendulum needed in Jeﬀ’s antique Seth-Thomas clock to ensure the period of the pendulum is 1 2 of a second. Assuming the initial displacement of the pendulum is 15◦, ﬁnd a sinusoid which models the displacement of the pendulum θ as a function of time, t, in seconds. 8. The table below lists the average temperature of Lake Erie as measured in Cleveland, Ohio on the ﬁrst of the month for each month during the years 1971 – 2000.19 For example, t = 3 represents the average of the temperatures recorded for Lake Erie on every March 1 for the years 1971 through 2000. Month Number, t Temperature (◦ F), 10 11 12 36 33 34 38 47 57 67 74 73 67 56 46 (a) Using the techniques discussed in Example 11.1.2, ﬁt a sinusoid to these data. (b) Using a graphing utility, graph your model along with the data set to judge the reason- ableness of the ﬁt. 18Provided θ is kept ‘small.’ Carl remembers the ‘Rule of Thumb’ as being 20◦ or less. Check with your friendly neighborhood physicist to make sure. 19See this website: http://www.erh.noaa.gov/cle/climate/cle/normals/laketempcle.html. 11.1 Applications of Sinusoids 893 (c) Use the model you found in part 8a to predict the average temperature recorded for Lake Erie on April 15th and September 15th during the years 1971–2000.20 (d) Compare your results to those obtained using a graphing utility. 9. The fraction of the moon illuminated at midnight Eastern Standard Time on the tth day of June, 2009 is given in the table below.21 Day of June, t Fraction Illuminated, F 3 6 9 12 15 18 21 24 27 30 0.81 0.98 0.98 0.83 0.57 0.27 0.04 0.03 0.26 0.58 (a) Using the techniques discussed in Example 11.1.2, ﬁt a sinusoid to these data.22 (b) Using a graphing utility, graph your model along with the data set to judge the reason- ableness of the ﬁt. (c) Use the model you found in part 9a to predict the fraction of the moon illuminated on June 1, 2009. 23 (d) Compare your results to those obtained using a graphing utility. 10. With the help of your classmates, research the phenomena mentioned in Example 11.1.4, namely resonance and beats. 11. With the help of your classmates, research Amplitude Modulation and Frequency Modulation. 12. What other things in the world might be roughly sinusoidal? Look to see what models you can ﬁnd for them and share your results with your class. 20The computed average is 41◦F for April 15th and 71◦F for September 15th. 21See this website: http://www.usno.navy.mil/USNO/astronomical-applications/data-services/frac-moon-ill. 22You may want to plot the data before you ﬁnd the phase shift. 23The listed fraction is 0.62. 894 Applications of Trigonometry 11.1.3 Answers 1. S(t) = sin (880πt) 2. V (t) = 220 √ 2 sin (120πt) 3. h(t) = 67.5 sin π 15 t − π 2 + 67.5 4. x(t) = 67.5 cos π 15 t − π 2 = 67.5 sin π 15 t 5. h(t) = 28 sin 2π + 30 2 3 t − π ft. and m = 5 16 slugs 6. (a) k = 5 lbs. (b) x(t) = sin 4t + π 2 . The object ﬁrst passes through the equilibrium point when t = π 8 ≈ 0.39 seconds after the motion starts. At this time, the object is heading upwards. (c) x(t) = √ 2 2 sin 4t + 7π 4 . The object passes through the equilibrium point heading down- wards for the third time when t = 17π 16 ≈ 3.34 seconds. 7. (a) θ(t) = θ0 sin g l t + π 2 (b) θ(t) = π 12 sin 4πt + π 2 8. (a) T (t) = 20.5 sin π (b) Our function and the data set are graphed below. The sinusoid seems to be shifted to 6 t − π + 53.5 the right of our data. (c) The average temperature on April 15th is approximately T (4.5) ≈ 39.00◦F and the average temperature on September 15th is approximately T (9.5) ≈ 73.38◦F. (d) Using a graphing calculator, we get the following This model predicts the average temperature for April 15th to be approximately 42.43◦F and the average temperature on September 15th to be approximately 70.05◦F. This model appears to be more accurate. 11.1 Applications of Sinusoids 895 9. (a) Based on the shape of the data, we either choose A < 0 or we ﬁnd the second value of t which closely approximates the ‘baseline’ value, F = 0.505. We choose the latter to obtain F (t) = 0.475 sin π 15 t − 2π + 0.505 = 0.475 sin π 15 t + 0.505 (b) Our function and the data set are graphed below. It’s a pretty good ﬁt. (c) The fraction of the moon illuminated on June 1st, 2009 is approximately F (1) ≈ 0.60 (d) Using a graphing calculator, we get the following. This model predicts that the fraction of the moon illuminated on June 1st, 2009 is approximately 0.59. This appears to be a better ﬁt to the data than our ﬁrst model. 896 Applications of Trigonometry 11.2 The Law of Sines Trigonometry literally means ‘measuring triangles’ and with Chapter 10 under our belts, we are more than prepared to do just that. The main goal of this section and the next is to develop theorems which allow us to ‘solve’ triangles – that is, ﬁnd the length of each side of a triangle and the measure of each of its angles. In Sections 10.2, 10.3 and 10.6, we’ve had some experience solving right triangles. The following example reviews what we know. Example 11.2.1. Given a right triangle with a hypotenuse of length 7 units and one leg of length 4 units, ﬁnd the length of the remaining side and the measures of the remaining angles. Express the angles in decimal degrees, rounded to the nearest hundreth of a degree. Solution. For deﬁnitiveness, we label the triangle below √ To ﬁnd the length of the missing side a, we use the Pythagorean Theorem to get a2 + 42 = 72 which then yields a = 33 units. Now that all three sides of the triangle are known, there are several ways we can ﬁnd α using the inverse trigonometric functions. To decrease the chances of propagating error, however, we stick to using the data given to us in the problem. In this case, the lengths 4 and 7 were given, so we want to relate these to α. According to Theorem 10.4, cos(α) = 4 7 . Since α is an acute angle, α = arccos 4 radians. Converting to degrees, we ﬁnd α ≈ 55.15◦. Now 7 that we have the measure of angle α, we could ﬁnd the measure of angle β using the fact that α and β are complements so α + β = 90◦. Once again, we opt to use the data given to us in the radians and we problem. According to Theorem 10.4, we have that sin(β) = 4 have β ≈ 34.85◦. 7 so β = arcsin 4 7 A few remarks about Example 11.2.1 are in order. First, we adhere to the convention that a lower case Greek letter denotes an angle1 and the corresponding lowercase English letter represents the side2 opposite that angle. Thus, a is t
he side opposite α, b is the side opposite β and c is the side opposite γ. Taken together, the pairs (α, a), (β, b) and (γ, c) are called angle-side opposite pairs. Second, as mentioned earlier, we will strive to solve for quantities using the original data given in the problem whenever possible. While this is not always the easiest or fastest way to proceed, it 1as well as the measure of said angle 2as well as the length of said side 11.2 The Law of Sines 897 minimizes the chances of propagated error.3 Third, since many of the applications which require solving triangles ‘in the wild’ rely on degree measure, we shall adopt this convention for the time being.4 The Pythagorean Theorem along with Theorems 10.4 and 10.10 allow us to easily handle any given right triangle problem, but what if the triangle isn’t a right triangle? In certain cases, we can use the Law of Sines to help. Theorem 11.2. The Law of Sines: Given a triangle with angle-side opposite pairs (α, a), (β, b) and (γ, c), the following ratios hold or, equivalently, sin(α) a = sin(β) b = sin(γ) c a sin(α) = b sin(β) = c sin(γ) The proof of the Law of Sines can be broken into three cases. For our ﬁrst case, consider the triangle ABC below, all of whose angles are acute, with angle-side opposite pairs (α, a), (β, b) and (γ, c). If we drop an altitude from vertex B, we divide the triangle into two right triangles: ABQ and BCQ. If we call the length of the altitude h (for height), we get from Theorem 10.4 that sin(α) = h a so that h = c sin(α) = a sin(γ). After some rearrangement of the last equation, we get sin(α) . If we drop an altitude from vertex A, we can proceed as above using the triangles ABQ and ACQ to get sin(β) B , completing the proof for this case. c and sin(γ) = h a = sin(γ) b = sin(γ For our next case consider the triangle ABC below with obtuse angle α. Extending an altitude from vertex A gives two right triangles, as in the previous case: ABQ and ACQ. Proceeding as before, we get h = b sin(γ) and h = c sin(β) so that sin(β) . b = sin(γ 3Your Science teachers should thank us for this. 4Don’t worry! Radians will be back before you know it! 898 Applications of Trigonometry Dropping an altitude from vertex B also generates two right triangles, ABQ and BCQ. We know that sin(α) = h c so that h = c sin(α). Since α = 180◦ − α, sin(α) = sin(α), so in fact, we have h = c sin(α). Proceeding to BCQ, we get sin(γ) = h a so h = a sin(γ). Putting this together with the previous equation, we get sin(γ) , and we are ﬁnished with this case. c = sin(α The remaining case is when ABC is a right triangle. In this case, the Law of Sines reduces to the formulas given in Theorem 10.4 and is left to the reader. In order to use the Law of Sines to solve a triangle, we need at least one angle-side opposite pair. The next example showcases some of the power, and the pitfalls, of the Law of Sines. Example 11.2.2. Solve the following triangles. Give exact answers and decimal approximations (rounded to hundredths) and sketch the triangle. 1. α = 120◦, a = 7 units, β = 45◦ 2. α = 85◦, β = 30◦, c = 5.25 units 3. α = 30◦, a = 1 units, c = 4 units 4. α = 30◦, a = 2 units, c = 4 units 5. α = 30◦, a = 3 units, c = 4 units 6. α = 30◦, a = 4 units, c = 4 units Solution. b sin(45◦) = 1. Knowing an angle-side opposite pair, namely α and a, we may proceed in using the Law of √ 6 Sines. Since β = 45◦, we use 3 ≈ 5.72 units. Now that we have two angle-side pairs, it is time to ﬁnd the third. To ﬁnd γ, we use the fact that the sum of the measures of the angles in a triangle is 180◦. Hence, γ = 180◦ − 120◦ − 45◦ = 15◦. To ﬁnd c, we have no choice but to used the derived value γ = 15◦, yet we can minimize the propagation of error here by using the given angle-side opposite pair (α, a). The Law of Sines gives us sin(120◦) so b = 7 sin(45◦) sin(120◦) so that c = 7 sin(15◦) sin(120◦) ≈ 2.09 units.5 sin(120◦) = 7 c sin(15◦) = 7 7 2. In this example, we are not immediately given an angle-side opposite pair, but as we have the measures of α and β, we can solve for γ since γ = 180◦ − 85◦ − 30◦ = 65◦. As in the previous example, we are forced to use a derived value in our computations since the only 5The exact value of sin(15◦) could be found using the diﬀerence identity for sine or a half-angle formula, but that becomes unnecessarily messy for the discussion at hand. Thus “exact” here means 7 sin(15◦) sin(120◦) . 11.2 The Law of Sines 899 angle-side pair available is (γ, c). The Law of Sines gives rearrangement, we get a = 5.25 sin(85◦) which yields sin(65◦) . After the usual sin(65◦) ≈ 5.77 units. To ﬁnd b we use the angle-side pair (γ, c) sin(65◦) hence b = 5.25 sin(30◦) sin(65◦) ≈ 2.90 units. sin(30◦) = 5.25 b a sin(85◦) = 5.25 β = 45◦ a = 7 c ≈ 2.09 α = 120◦ γ = 15◦ b ≈ 5.72 Triangle for number 1 β = 30◦ c = 5.25 a ≈ 5.77 α = 85◦ γ = 65◦ b ≈ 2.90 Triangle for number 2 1 4 = sin(30◦) 3. Since we are given (α, a) and c, we use the Law of Sines to ﬁnd the measure of γ. We start with sin(γ) and get sin(γ) = 4 sin (30◦) = 2. Since the range of the sine function is [−1, 1], there is no real number with sin(γ) = 2. Geometrically, we see that side a is just too short to make a triangle. The next three examples keep the same values for the measure of α and the length of c while varying the length of a. We will discuss this case in more detail after we see what happens in those examples. 2 4 = sin(30◦) 4. In this case, we have the measure of α = 30◦, a = 2 and c = 4. Using the Law of Sines, we get sin(γ) so sin(γ) = 2 sin (30◦) = 1. Now γ is an angle in a triangle which also contains α = 30◦. This means that γ must measure between 0◦ and 150◦ in order to ﬁt inside the triangle with α. The only angle that satisﬁes this requirement and has sin(γ) = 1 is γ = 90◦. In other words, we have a right triangle. We ﬁnd the measure of β to be β = 180◦ − 30◦ − 90◦ = 60◦ and then determine b using the Law of Sines. We ﬁnd b = 2 sin(60◦) 3 ≈ 3.46 units. In this case, the side a is precisely long enough to form a sin(30◦) = 2 unique right triangle. √ c = 4 a = 1 α = 30◦ c = 4 β = 60◦ a = 2 α = 30◦ b ≈ 3.46 Diagram for number 3 Triangle for number 4 5. Proceeding as we have in the previous two examples, we use the Law of Sines to ﬁnd γ. In this 3 . Since γ lies in a triangle with α = 30◦, or sin(γ) = 4 sin(30◦) case, we have sin(γ) 4 = sin(30◦) = 2 3 3 900 Applications of Trigonometry 3 3 : γ = arcsin 2 we must have that 0◦ < γ < 150◦. There are two angles γ that fall in this range and have radians ≈ 138.19◦. At radians ≈ 41.81◦ and γ = π − arcsin 2 sin(γ) = 2 3 this point, we pause to see if it makes sense that we actually have two viable cases to consider. As we have discussed, both candidates for γ are ‘compatible’ with the given angle-side pair (α, a) = (30◦, 3) in that both choices for γ can ﬁt in a triangle with α and both have a sine of 2 3 . The only other given piece of information is that c = 4 units. Since c > a, it must be true that γ, which is opposite c, has greater measure than α which is opposite a. In both cases, γ > α, so both candidates for γ are compatible with this last piece of given information as radians ≈ 41.81◦, we well. Thus have two triangles on our hands. In the case γ = arcsin 2 3 ﬁnd6 β ≈ 180◦ − 30◦ − 41.81◦ = 108.19◦. Using the Law of Sines with the angle-side opposite radians pair (α, a) and β, we ﬁnd b ≈ 3 sin(108.19◦) ≈ 138.19◦, we repeat the exact same steps and ﬁnd β ≈ 11.81◦ and b ≈ 1.23 units.7 Both triangles are drawn below. sin(30◦) ≈ 5.70 units. In the case γ = π − arcsin 2 3 β ≈ 11.81◦ c = 4 β ≈ 108.19◦ a = 3 α = 30◦ γ ≈ 41.81◦ α = 30◦ c = 4 a = 3 γ ≈ 138.19◦ b ≈ 5.70 b ≈ 1.23 6. For this last problem, we repeat the usual Law of Sines routine to ﬁnd that sin(γ) so that sin(γ) = 1 2 . Since γ must inhabit a triangle with α = 30◦, we must have 0◦ < γ < 150◦. Since the measure of γ must be strictly less than 150◦, there is just one angle which satisﬁes both required conditions, namely γ = 30◦. So β = 180◦ − 30◦ − 30◦ = 120◦ and, using the √ Law of Sines one last time, b = 4 sin(120◦) 3 ≈ 6.93 units. 4 = sin(30◦) 4 sin(30◦) = 4 c = 4 β = 120◦ a = 4 α = 30◦ γ = 30◦ b ≈ 6.93 Some remarks about Example 11.2.2 are in order. We ﬁrst note that if we are given the measures of two of the angles in a triangle, say α and β, the measure of the third angle γ is uniquely 6 radians, γ = arcsin 2 6To ﬁnd an exact expression for β, we convert everything back to radians: α = 30◦ = π 3 radians and 180◦ = π radians. Hence, β = π − π 7An exact answer for β in this case is β = arcsin 2 6 − arcsin 2 − π 3 6 − arcsin 2 = 5π 6 radians ≈ 11.81◦. 3 3 radians ≈ 108.19◦. 11.2 The Law of Sines 901 determined using the equation γ = 180◦ − α − β. Knowing the measures of all three angles of a triangle completely determines its shape. If in addition we are given the length of one of the sides of the triangle, we can then use the Law of Sines to ﬁnd the lengths of the remaining two sides to determine the size of the triangle. Such is the case in numbers 1 and 2 above. In number 1, the given side is adjacent to just one of the angles – this is called the ‘Angle-Angle-Side’ (AAS) case.8 In number 2, the given side is adjacent to both angles which means we are in the so-called ‘Angle-Side-Angle’ (ASA) case. If, on the other hand, we are given the measure of just one of the angles in the triangle along with the length of two sides, only one of which is adjacent to the given angle, we are in the ‘Angle-Side-Side’ (ASS) case.9 In number 3, the length of the one given side a was too short to even form a triangle; in number 4, the length of a was just long enough to form a right triangle; in 5, a was long enough, but not too long, so that two triangles were possible; and in number 6, side a was long enough to form a triangle but too long to swing back and form two. These four cases exemplify all of the possibilities in the Angle-Side-Side case which are summarized in the following theorem. Theorem 11.3. Suppose (α,
a) and (γ, c) are intended to be angle-side pairs in a triangle where α, a and c are given. Let h = c sin(α) If a < h, then no triangle exists which satisﬁes the given criteria. If a = h, then γ = 90◦ so exactly one (right) triangle exists which satisﬁes the criteria. If h < a < c, then two distinct triangles exist which satisfy the given criteria. If a ≥ c, then γ is acute and exactly one triangle exists which satisﬁes the given criteria Theorem 11.3 is proved on a case-by-case basis. If a < h, then a < c sin(α). If a triangle were c = sin(α) to exist, the Law of Sines would have sin(γ) a = 1, which is impossible. In the ﬁgure below, we see geometrically why this is the case. so that sin(γ) = c sin(α) > a a a a h = c sin(α sin(α) a < h, no triangle a = h, γ = 90◦ Simply put, if a < h the side a is too short to connect to form a triangle. This means if a ≥ h, we are always guaranteed to have at least one triangle, and the remaining parts of the theorem 8If this sounds familiar, it should. From high school Geometry, we know there are four congruence conditions for triangles: Angle-Angle-Side (AAS), Angle-Side-Angle (ASA), Side-Angle-Side (SAS) and Side-Side-Side (SSS). If we are given information about a triangle that meets one of these four criteria, then we are guaranteed that exactly one triangle exists which satisﬁes the given criteria. 9In more reputable books, this is called the ‘Side-Side-Angle’ or SSA case. 902 Applications of Trigonometry c a a = a a = sin(γ) < 1 which means there are two solutions to sin(γ) = c sin(α) tell us what kind and how many triangles to expect in each case. If a = h, then a = c sin(α) and so that sin(γ) = c sin(α) the Law of Sines gives sin(α) a = 1. Here, γ = 90◦ as required. Moving along, now suppose h < a < c. As before, the Law of Sines10 gives sin(γ) = c sin(α) . Since h < a, c sin(α) < a or c sin(α) : an acute angle which we’ll call γ0, and its supplement, 180◦ − γ0. We need to argue that each of these angles ‘ﬁt’ into a triangle with α. Since (α, a) and (γ0, c) are angle-side opposite pairs, the assumption c > a in this case gives us γ0 > α. Since γ0 is acute, we must have that α is acute as well. This means one triangle can contain both α and γ0, giving us one of the triangles promised in the theorem. If we manipulate the inequality γ0 > α a bit, we have 180◦ −γ0 < 180◦ −α which gives (180◦ − γ0) + α < 180◦. This proves a triangle can contain both of the angles α and (180◦ − γ0), giving us the second triangle predicted in the theorem. To prove the last case in the theorem, we assume a ≥ c. Then α ≥ γ, which forces γ to be an acute angle. Hence, we get only one triangle in this case, completing the proof. a a c a a h α γ0 γ0 h < a < c, two triangles c α h a γ a ≥ c, one triangle One last comment before we use the Law of Sines to solve an application problem. In the AngleSide-Side case, if you are given an obtuse angle to begin with then it is impossible to have the two triangle case. Think about this before reading further. Example 11.2.3. Sasquatch Island lies oﬀ the coast of Ippizuti Lake. Two sightings, taken 5 miles apart, are made to the island. The angle between the shore and the island at the ﬁrst observation point is 30◦ and at the second point the angle is 45◦. Assuming a straight coastline, ﬁnd the distance from the second observation point to the island. What point on the shore is closest to the island? How far is the island from this point? Solution. We sketch the problem below with the ﬁrst observation point labeled as P and the second as Q. In order to use the Law of Sines to ﬁnd the distance d from Q to the island, we ﬁrst need to ﬁnd the measure of β which is the angle opposite the side of length 5 miles. To that end, we note that the angles γ and 45◦ are supplemental, so that γ = 180◦ − 45◦ = 135◦. We can now 5 ﬁnd β = 180◦ − 30◦ − γ = 180◦ − 30◦ − 135◦ = 15◦. By the Law of Sines, we have sin(15◦) which gives d = 5 sin(30◦) sin(15◦) ≈ 9.66 miles. Next, to ﬁnd the point on the coast closest to the island, which we’ve labeled as C, we need to ﬁnd the perpendicular distance from the island to the coast.11 d sin(30◦) = 10Remember, we have already argued that a triangle exists in this case! 11Do you see why C must lie to the right of Q? 11.2 The Law of Sines 903 Let x denote the distance from the second observation point Q to the point C and let y denote the distance from C to the island. Using Theorem 10.4, we get sin (45◦) = y d . After some rearranging, we ﬁnd y = d sin (45◦) ≈ 9.66 ≈ 6.83 miles. Hence, the island is approximately 6.83 miles from the coast. To ﬁnd the distance from Q to C, we note that β = 180◦ − 90◦ − 45◦ = 45◦ so by symmetry,12 we get x = y ≈ 6.83 miles. Hence, the point on the shore closest to the island is approximately 6.83 miles down the coast from the second observation point. √ 2 2 Sasquatch Island Sasquatch Island β β d ≈ 9.66 miles d ≈ 9.66 miles y miles γ Q 30◦ 5 miles P 45◦ Shoreline 45◦ Q C x miles We close this section with a new formula to compute the area enclosed by a triangle. Its proof uses the same cases and diagrams as the proof of the Law of Sines and is left as an exercise. Theorem 11.4. Suppose (α, a), (β, b) and (γ, c) are the angle-side opposite pairs of a triangle. Then the area A enclosed by the triangle is given by A = 1 2 bc sin(α) = 1 2 ac sin(β) = 1 2 ab sin(γ) Example 11.2.4. Find the area of the triangle in Example 11.2.2 number 1. Solution. From our work in Example 11.2.2 number 1, we have all three angles and all three sides to work with. However, to minimize propagated error, we choose A = 1 2 ac sin(β) from Theorem 11.4 because it uses the most pieces of given information. We are given a = 7 and β = 45◦, and we calculated c = 7 sin(15◦) sin (45◦) =≈ 5.18 square units. The reader is encouraged to check this answer against the results obtained using the other formulas in Theorem 11.4. sin(120◦) . Using these values, we ﬁnd A = 1 7 sin(15◦) sin(120◦) 2 (7) 12Or by Theorem 10.4 again . . . 904 Applications of Trigonometry 11.2.1 Exercises In Exercises 1 - 20, solve for the remaining side(s) and angle(s) if possible. As in the text, (α, a), (β, b) and (γ, c) are angle-side opposite pairs. 1. α = 13◦, β = 17◦, a = 5 2. α = 73.2◦, β = 54.1◦, a = 117 3. α = 95◦, β = 85◦, a = 33.33 4. α = 95◦, β = 62◦, a = 33.33 5. α = 117◦, a = 35, b = 42 6. α = 117◦, a = 45, b = 42 7. α = 68.7◦, a = 88, b = 92 8. α = 42◦, a = 17, b = 23.5 9. α = 68.7◦, a = 70, b = 90 10. α = 30◦, a = 7, b = 14 11. α = 42◦, a = 39, b = 23.5 12. γ = 53◦, α = 53◦, c = 28.01 13. α = 6◦, a = 57, b = 100 14. γ = 74.6◦, c = 3, a = 3.05 15. β = 102◦, b = 16.75, c = 13 16. β = 102◦, b = 16.75, c = 18 17. β = 102◦, γ = 35◦, b = 16.75 18. β = 29.13◦, γ = 83.95◦, b = 314.15 19. γ = 120◦, β = 61◦, c = 4 20. α = 50◦, a = 25, b = 12.5 21. Find the area of the triangles given in Exercises 1, 12 and 20 above. (Another Classic Application: Grade of a Road) The grade of a road is much like the pitch of a roof (See Example 10.6.6) in that it expresses the ratio of rise/run. In the case of a road, this ratio is always positive because it is measured going uphill and it is usually given as a percentage. For example, a road which rises 7 feet for every 100 feet of (horizontal) forward progress is said to have a 7% grade. However, if we want to apply any Trigonometry to a story problem involving roads going uphill or downhill, we need to view the grade as an angle with respect to the horizontal. In Exercises 22 - 24, we ﬁrst have you change road grades into angles and then use the Law of Sines in an application. 22. Using a right triangle with a horizontal leg of length 100 and vertical leg with length 7, show that a 7% grade means that the road (hypotenuse) makes about a 4◦ angle with the horizontal. (It will not be exactly 4◦, but it’s pretty close.) 23. What grade is given by a 9.65◦ angle made by the road and the horizontal?13 13I have friends who live in Paciﬁca, CA and their road is actually this steep. It’s not a nice road to drive. 11.2 The Law of Sines 905 24. Along a long, straight stretch of mountain road with a 7% grade, you see a tall tree standing perfectly plumb alongside the road.14 From a point 500 feet downhill from the tree, the angle of inclination from the road to the top of the tree is 6◦. Use the Law of Sines to ﬁnd the height of the tree. (Hint: First show that the tree makes a 94◦ angle with the road.) (Another Classic Application: Bearings) In the next several exercises we introduce and work with the navigation tool known as bearings. Simply put, a bearing is the direction you are heading according to a compass. The classic nomenclature for bearings, however, is not given as an angle in standard position, so we must ﬁrst understand the notation. A bearing is given as an acute angle of rotation (to the east or to the west) away from the north-south (up and down) line of a compass rose. For example, N40◦E (read “40◦ east of north”) is a bearing which is rotated clockwise 40◦ from due north. If we imagine standing at the origin in the Cartesian Plane, this bearing would have us heading into Quadrant I along the terminal side of θ = 50◦. Similarly, S50◦W would point into Quadrant III along the terminal side of θ = 220◦ because we started out pointing due south (along θ = 270◦) and rotated clockwise 50◦ back to 220◦. Counter-clockwise rotations would be found in the bearings N60◦W (which is on the terminal side of θ = 150◦) and S27◦E (which lies along the terminal side of θ = 297◦). These four bearings are drawn in the plane below. N N40◦E N60◦W 60◦ 40◦ W E 50◦ 27◦ S50◦W S27◦E S The cardinal directions north, south, east and west are usually not given as bearings in the fashion described above, but rather, one just refers to them as ‘due north’, ‘due south’, ‘due east’ and ‘due west’, respectively, and it is assumed that you know which quadrantal angle goes with each cardinal direction. (Hint: Look at the diagram above.) 25. Find the angle θ
in standard position with 0◦ ≤ θ < 360◦ which corresponds to each of the bearings given below. (a) due west (b) S83◦E (c) N5.5◦E (d) due south 14The word ‘plumb’ here means that the tree is perpendicular to the horizontal. 906 Applications of Trigonometry (e) N31.25◦W (f) S72◦4112W15 (g) N45◦E (h) S45◦W 26. The Colonel spots a campﬁre at a of bearing N42◦E from his current position. Sarge, who is positioned 3000 feet due east of the Colonel, reckons the bearing to the ﬁre to be N20◦W from his current position. Determine the distance from the campﬁre to each man, rounded to the nearest foot. 27. A hiker starts walking due west from Sasquatch Point and gets to the Chupacabra Trailhead before she realizes that she hasn’t reset her pedometer. From the Chupacabra Trailhead she hikes for 5 miles along a bearing of N53◦W which brings her to the Muﬃn Ridge Observatory. From there, she knows a bearing of S65◦E will take her straight back to Sasquatch Point. How far will she have to walk to get from the Muﬃn Ridge Observatory to Sasquach Point? What is the distance between Sasquatch Point and the Chupacabra Trailhead? 28. The captain of the SS Bigfoot sees a signal ﬂare at a bearing of N15◦E from her current location. From his position, the captain of the HMS Sasquatch ﬁnds the signal ﬂare to be at a bearing of N75◦W. If the SS Bigfoot is 5 miles from the HMS Sasquatch and the bearing from the SS Bigfoot to the HMS Sasquatch is N50◦E, ﬁnd the distances from the ﬂare to each vessel, rounded to the nearest tenth of a mile. 29. Carl spies a potential Sasquatch nest at a bearing of N10◦E and radios Jeﬀ, who is at a bearing of N50◦E from Carl’s position. From Jeﬀ’s position, the nest is at a bearing of S70◦W. If Jeﬀ and Carl are 500 feet apart, how far is Jeﬀ from the Sasquatch nest? Round your answer to the nearest foot. 30. A hiker determines the bearing to a lodge from her current position is S40◦W. She proceeds to hike 2 miles at a bearing of S20◦E at which point she determines the bearing to the lodge is S75◦W. How far is she from the lodge at this point? Round your answer to the nearest hundredth of a mile. 31. A watchtower spots a ship oﬀ shore at a bearing of N70◦E. A second tower, which is 50 miles from the ﬁrst at a bearing of S80◦E from the ﬁrst tower, determines the bearing to the ship to be N25◦W. How far is the boat from the second tower? Round your answer to the nearest tenth of a mile. 32. Skippy and Sally decide to hunt UFOs. One night, they position themselves 2 miles apart on an abandoned stretch of desert runway. An hour into their investigation, Skippy spies a UFO hovering over a spot on the runway directly between him and Sally. He records the angle of inclination from the ground to the craft to be 75◦ and radios Sally immediately to ﬁnd the angle of inclination from her position to the craft is 50◦. How high oﬀ the ground is the UFO at this point? Round your answer to the nearest foot. (Recall: 1 mile is 5280 feet.) 15See Example 10.1.1 in Section 10.1 for a review of the DMS system. 11.2 The Law of Sines 907 33. The angle of depression from an observer in an apartment complex to a gargoyle on the building next door is 55◦. From a point ﬁve stories below the original observer, the angle of inclination to the gargoyle is 20◦. Find the distance from each observer to the gargoyle and the distance from the gargoyle to the apartment complex. Round your answers to the nearest foot. (Use the rule of thumb that one story of a building is 9 feet.) 34. Prove that the Law of Sines holds when ABC is a right triangle. 35. Discuss with your classmates why knowing only the three angles of a triangle is not enough to determine any of the sides. 36. Discuss with your classmates why the Law of Sines cannot be used to ﬁnd the angles in the triangle when only the three sides are given. Also discuss what happens if only two sides and the angle between them are given. (Said another way, explain why the Law of Sines cannot be used in the SSS and SAS cases.) 37. Given α = 30◦ and b = 10, choose four diﬀerent values for a so that (a) the information yields no triangle (b) the information yields exactly one right triangle (c) the information yields two distinct triangles (d) the information yields exactly one obtuse triangle Explain why you cannot choose a in such a way as to have α = 30◦, b = 10 and your choice of a yield only one triangle where that unique triangle has three acute angles. 38. Use the cases and diagrams in the proof of the Law of Sines (Theorem 11.2) to prove the area formulas given in Theorem 11.4. Why do those formulas yield square units when four quantities are being multiplied together? 908 Applications of Trigonometry 11.2.2 Answers 1. 3. 5. 7. α = 13◦ β = 17◦ a = 5 γ = 150◦ b ≈ 6.50 c ≈ 11.11 Information does not produce a triangle Information does not produce a triangle α = 68.7◦ β ≈ 76.9◦ γ ≈ 34.4◦ c ≈ 53.36 a = 88 α = 68.7◦ β ≈ 103.1◦ γ ≈ 8.2◦ c ≈ 13.47 a = 88 b = 92 b = 92 9. Information does not produce a triangle 11. 13. 15. 17. 19. α = 42◦ β ≈ 23.78◦ γ ≈ 114.22◦ a = 39 c ≈ 53.15 b = 23.5 α = 6◦ β ≈ 169.43◦ γ ≈ 4.57◦ c ≈ 43.45 a = 57 b = 100 α = 6◦ β ≈ 10.57◦ γ ≈ 163.43◦ c ≈ 155.51 a = 57 b = 100 α ≈ 28.61◦ β = 102◦ a ≈ 8.20 b = 16.75 c = 13 γ ≈ 49.39◦ α = 43◦ γ = 35◦ β = 102◦ a ≈ 11.68 b = 16.75 c ≈ 9.82 Information does not produce a triangle 2. 4. 6. 8. 10. 12. 14. 16. 18. 20. α = 73.2◦ β = 54.1◦ γ = 52.7◦ c ≈ 97.22 b ≈ 99.00 a = 117 β = 62◦ α = 95◦ a = 33.33 b ≈ 29.54 c ≈ 13.07 γ = 23◦ α = 117◦ β ≈ 56.3◦ γ ≈ 6.7◦ c ≈ 5.89 b = 42 a = 45 b = 23.5 α = 42◦ β ≈ 67.66◦ γ ≈ 70.34◦ c ≈ 23.93 a = 17 α = 42◦ β ≈ 112.34◦ γ ≈ 25.66◦ c ≈ 11.00 a = 17 b = 23.5 α = 30◦ β = 90◦ γ = 60◦ √ 3 b = 14 a = 7 c = 7 α = 53◦ β = 74◦ a = 28.01 b ≈ 33.71 c = 28.01 γ = 53◦ α ≈ 78.59◦ β ≈ 26.81◦ γ = 74.6◦ b ≈ 1.40 a = 3.05 α ≈ 101.41◦ β ≈ 3.99◦ γ = 74.6◦ b ≈ 0.217 a = 3.05 c = 3 c = 3 Information does not produce a triangle α = 66.92◦ β = 29.13◦ γ = 83.95◦ c ≈ 641.75 a ≈ 593.69 b = 314.15 α = 50◦ β ≈ 22.52◦ γ ≈ 107.48◦ a = 25 c ≈ 31.13 b = 12.5 21. The area of the triangle from Exercise 1 is about 8.1 square units. The area of the triangle from Exercise 12 is about 377.1 square units. The area of the triangle from Exercise 20 is about 149 square units. ≈ 0.699 radians, which is equivalent to 4.004◦ 22. arctan 7 100 23. About 17% 24. About 53 feet 11.2 The Law of Sines 909 25. (a) θ = 180◦ (b) θ = 353◦ (c) θ = 84.5◦ (d) θ = 270◦ (e) θ = 121.25◦ (f) θ = 197◦1848 (g) θ = 45◦ (h) θ = 225◦ 26. The Colonel is about 3193 feet from the campﬁre. Sarge is about 2525 feet to the campﬁre. 27. The distance from the Muﬃn Ridge Observatory to Sasquach Point is about 7.12 miles. The distance from Sasquatch Point to the Chupacabra Trailhead is about 2.46 miles. 28. The SS Bigfoot is about 4.1 miles from the ﬂare. The HMS Sasquatch is about 2.9 miles from the ﬂare. 29. Jeﬀ is about 371 feet from the nest. 30. She is about 3.02 miles from the lodge 31. The boat is about 25.1 miles from the second tower. 32. The UFO is hovering about 9539 feet above the ground. 33. The gargoyle is about 44 feet from the observer on the upper ﬂoor. The gargoyle is about 27 feet from the observer on the lower ﬂoor. The gargoyle is about 25 feet from the other building. 910 Applications of Trigonometry 11.3 The Law of Cosines In Section 11.2, we developed the Law of Sines (Theorem 11.2) to enable us to solve triangles in the ‘Angle-Angle-Side’ (AAS), the ‘Angle-Side-Angle’ (ASA) and the ambiguous ‘Angle-Side-Side’ (ASS) cases. In this section, we develop the Law of Cosines which handles solving triangles in the ‘Side-Angle-Side’ (SAS) and ‘Side-Side-Side’ (SSS) cases.1 We state and prove the theorem below. Theorem 11.5. Law of Cosines: Given a triangle with angle-side opposite pairs (α, a), (β, b) and (γ, c), the following equations hold a2 = b2 + c2 − 2bc cos(α) b2 = a2 + c2 − 2ac cos(β) c2 = a2 + b2 − 2ab cos(γ) or, solving for the cosine in each equation, we have cos(α) = b2 + c2 − a2 2bc cos(β) = a2 + c2 − b2 2ac cos(γ) = a2 + b2 − c2 2ab To prove the theorem, we consider a generic triangle with the vertex of angle α at the origin with side b positioned along the positive x-axis. B = (c cos(α), c sin(α)) c α a A = (0, 0) b C = (b, 0) From this set-up, we immediately ﬁnd that the coordinates of A and C are A(0, 0) and C(b, 0). From Theorem 10.3, we know that since the point B(x, y) lies on a circle of radius c, the coordinates 1Here, ‘Side-Angle-Side’ means that we are given two sides and the ‘included’ angle - that is, the given angle is adjacent to both of the given sides. 11.3 The Law of Cosines 911 of B are B(x, y) = B(c cos(α), c sin(α)). (This would be true even if α were an obtuse or right angle so although we have drawn the case when α is acute, the following computations hold for any angle α drawn in standard position where 0 < α < 180◦.) We note that the distance between the points B and C is none other than the length of side a. Using the distance formula, Equation 1.1, we get a = (c cos(α) − b)2 + (c sin(α) − 0)2 2 (c cos(α) − b)2 + c2 sin2(α) a2 = a2 = (c cos(α) − b)2 + c2 sin2(α) a2 = c2 cos2(α) − 2bc cos(α) + b2 + c2 sin2(α) a2 = c2 cos2(α) + sin2(α) + b2 − 2bc cos(α) a2 = c2(1) + b2 − 2bc cos(α) a2 = c2 + b2 − 2bc cos(α) Since cos2(α) + sin2(α) = 1 The remaining formulas given in Theorem 11.5 can be shown by simply reorienting the triangle to place a diﬀerent vertex at the origin. We leave these details to the reader. What’s important about a and α in the above proof is that (α, a) is an angle-side opposite pair and b and c are the sides adjacent to α – the same can be said of any other angle-side opposite pair in the triangle. Notice that the proof of the Law of Cosines relies on the distance formula which has its roots in the Pythagorean Theorem. That being said, the Law of Cosines can be thought of as a generalization of the Pythagorean Theorem. If we have a triangle in which γ = 90◦, then cos(γ) = cos (90◦) = 0 so we get the familiar rela
tionship c2 = a2 + b2. What this means is that in the larger mathematical sense, the Law of Cosines and the Pythagorean Theorem amount to pretty much the same thing.2 Example 11.3.1. Solve the following triangles. Give exact answers and decimal approximations (rounded to hundredths) and sketch the triangle. 1. β = 50◦, a = 7 units, c = 2 units 2. a = 4 units, b = 7 units, c = 5 units Solution. 1. We are given the lengths of two sides, a = 7 and c = 2, and the measure of the included angle, β = 50◦. With no angle-side opposite pair to use, we apply the Law of Cosines. We get b2 = 72 + 22 − 2(7)(2) cos (50◦) which yields b = In order to determine the measures of the remaining angles α and γ, we are forced to used the derived value for b. There are two ways to proceed at this point. We could use the Law of Cosines again, or, since we have the angle-side opposite pair (β, b) we could use the Law of Sines. The advantage to using the Law of Cosines over the Law of Sines in cases like this is that unlike the sine function, the cosine function distinguishes between acute and obtuse angles. The cosine of an acute is positive, whereas the cosine of an obtuse angle is negative. Since the sine of both acute and obtuse angles are positive, the sine of an angle alone is not 53 − 28 cos (50◦) ≈ 5.92 units. 2This shouldn’t come as too much of a shock. All of the theorems in Trigonometry can ultimately be traced back to the deﬁnition of the circular functions along with the distance formula and hence, the Pythagorean Theorem. 912 Applications of Trigonometry enough to determine if the angle in question is acute or obtuse. Since both authors of the textbook prefer the Law of Cosines, we proceed with this method ﬁrst. When using the Law of Cosines, it’s always best to ﬁnd the measure of the largest unknown angle ﬁrst, since this will give us the obtuse angle of the triangle if there is one. Since the largest angle is opposite the longest side, we choose to ﬁnd α ﬁrst. To that end, we use the formula cos(α) = b2+c2−a2 53 − 28 cos (50◦) and c = 2. We get3 and substitute a = 7, b = 2bc cos(α) = 2 − 7 cos (50◦) 53 − 28 cos (50◦) Since α is an angle in a triangle, we know the radian measure of α must lie between 0 and π radians. This matches the range of the arccosine function, so we have α = arccos 2 − 7 cos (50◦) 53 − 28 cos (50◦) radians ≈ 114.99◦ At this point, we could ﬁnd γ using γ = 180◦ − α − β ≈ 180◦ − 114.99◦ − 50◦ = 15.01◦, that is if we trust our approximation for α. To minimize propagation of error, however, we could use the Law of Cosines again,4 in this case using cos(γ) = a2+b2−c2 . Plugging in a = 7, b = 53 − 28 cos (50◦) and c = 2, we get γ = arccos sketch the triangle below. √ 7−2 cos(50◦) 53−28 cos(50◦) 2ab radians ≈ 15.01◦. We β = 50◦ a = 7 c = 2 α ≈ 114.99◦ γ ≈ 15.01◦ b ≈ 5.92 As we mentioned earlier, once we’ve determined b it is possible to use the Law of Sines to ﬁnd the remaining angles. Here, however, we must proceed with caution as we are in the ambiguous (ASS) case. It is advisable to ﬁrst ﬁnd the smallest of the unknown angles, since we are guaranteed it will be acute.5 In this case, we would ﬁnd γ since the side opposite γ is smaller than the side opposite the other unknown angle, α. Using the angle-side opposite pair (β, b), we get sin(γ) . The usual calculations produces γ ≈ 15.01◦ and α = 180◦ − β − γ ≈ 180◦ − 50◦ − 15.01◦ = 114.99◦. sin(50◦) 53−28 cos(50◦) 2 = √ 2. Since all three sides and no angles are given, we are forced to use the Law of Cosines. Following our discussion in the previous problem, we ﬁnd β ﬁrst, since it is opposite the longest side, radians ≈ 101.54◦. As in b. We get cos(β) = a2+c2−b2 5 , so we get β = arccos − 1 = − 1 2ac 5 3after simplifying . . . 4Your instructor will let you know which procedure to use. It all boils down to how much you trust your calculator. 5There can only be one obtuse angle in the triangle, and if there is one, it must be the largest. 11.3 The Law of Cosines 913 the previous problem, now that we have obtained an angle-side opposite pair (β, b), we could proceed using the Law of Sines. The Law of Cosines, however, oﬀers us a rare opportunity to ﬁnd the remaining angles using only the data given to us in the statement of the problem. Using this, we get γ = arccos 5 7 radians ≈ 44.42◦ and α = arccos 29 35 radians ≈ 34.05◦. β ≈ 101.54◦ c = 5 a = 4 α ≈ 34.05◦ γ ≈ 44.42◦ b = 7 We note that, depending on how many decimal places are carried through successive calculations, and depending on which approach is used to solve the problem, the approximate answers you obtain may diﬀer slightly from those the authors obtain in the Examples and the Exercises. A great example of this is number 2 in Example 11.3.1, where the approximate values we record for the measures of the angles sum to 180.01◦, which is geometrically impossible. Next, we have an application of the Law of Cosines. Example 11.3.2. A researcher wishes to determine the width of a vernal pond as drawn below. From a point P , he ﬁnds the distance to the eastern-most point of the pond to be 950 feet, while the distance to the western-most point of the pond from P is 1000 feet. If the angle between the two lines of sight is 60◦, ﬁnd the width of the pond. 1000 feet 950 feet 60◦ P Solution. We are given the lengths of two sides and the measure of an included angle, so we may apply the Law of Cosines to ﬁnd the length of the missing side opposite the given angle. Calling this length w (for width), we get w2 = 9502 + 10002 − 2(950)(1000) cos (60◦) = 952500 from which we get w = 952500 ≈ 976 feet. √ 914 Applications of Trigonometry In Section 11.2, we used the proof of the Law of Sines to develop Theorem 11.4 as an alternate formula for the area enclosed by a triangle. In this section, we use the Law of Cosines to derive another such formula - Heron’s Formula. Theorem 11.6. Heron’s Formula: Suppose a, b and c denote the lengths of the three sides of a triangle. Let s be the semiperimeter of the triangle, that is, let s = 1 2 (a + b + c). Then the area A enclosed by the triangle is given by A = s(s − a)(s − b)(s − c) We prove Theorem 11.6 using Theorem 11.4. Using the convention that the angle γ is opposite the side c, we have A = 1 2 ab sin(γ) from Theorem 11.4. In order to simplify computations, we start by manipulating the expression for A2. A2 = 2 ab sin(γ) 1 2 a2b2 sin2(γ) = = 1 4 a2b2 4 1 − cos2(γ) since sin2(γ) = 1 − cos2(γ). The Law of Cosines tells us cos(γ) = a2+b2−c2 2ab , so substituting this into our equation for A2 gives A2 = = = = = = = = 1 − 1 − 2 1 − cos2(γ) a2 + b2 − c2 2ab a2 + b2 − c22 4a2b2 4a2b2 − a2 + b2 − c22 4a2b2 a2b2 4 a2b2 4 a2b2 4 a2b2 4 4a2b2 − a2 + b2 − c22 16 (2ab)2 − a2 + b2 − c22 16 2ab − a2 + b2 − c2 2ab + a2 + b2 − c2 16 c2 − a2 + 2ab − b2 a2 + 2ab + b2 − c2 16 diﬀerence of squares. 11.3 The Law of Cosines 915 A2 = = = = = c2 − a2 − 2ab + b2 a2 + 2ab + b2 − c2 16 c2 − (a − b)2 (a + b)2 − c2 16 (c − (a − b))(c + (a − b))((a + b) − c)((a + b) + c) 16 perfect square trinomials. diﬀerence of squares. (b + c − a)(a + c − b)(a + b − c)(a + b + c) 16 (a + c − b) 2 (b + c − a) 2 (a + b − c) 2 · · · (a + b + c) 2 At this stage, we recognize the last factor as the semiperimeter, s = 1 complete the proof, we note that 2 (a + b + c) = a+b+c 2 . To (s − a − 2a 2 = b + c − a 2 Similarly, we ﬁnd (s − b) = a+c−b 2 and (s − c) = a+b−c 2 . Hence, we get A2 = (b + c − a) 2 · (a + c − b) 2 · (a + b − c) 2 · (a + b + c) 2 = (s − a)(s − b)(s − c)s so that A = s(s − a)(s − b)(s − c) as required. We close with an example of Heron’s Formula. Example 11.3.3. Find the area enclosed of the triangle in Example 11.3.1 number 2. Solution. We are given a = 4, b = 7 and c = 5. Using these values, we ﬁnd s = 1 2 (4 + 7 + 5) = 8, (s − a) = 8 − 4 = 4, (s − b) = 8 − 7 = 1 and (s − c) = 8 − 5 = 3. Using Heron’s Formula, we get √ A = s(s − a)(s − b)(s − c) = 6 ≈ 9.80 square units. (8)(4)(1)(3) = 96 = 4 √ 916 Applications of Trigonometry 11.3.1 Exercises In Exercises 1 - 10, use the Law of Cosines to ﬁnd the remaining side(s) and angle(s) if possible. 1. a = 7, b = 12, γ = 59.3◦ 2. α = 104◦, b = 25, c = 37 3. a = 153, β = 8.2◦, c = 153 4. a = 3, b = 4, γ = 90◦ 5. α = 120◦, b = 3, c = 4 6. a = 7, b = 10, c = 13 7. a = 1, b = 2, c = 5 8. a = 300, b = 302, c = 48 9. a = 5, b = 5, c = 5 10. a = 5, b = 12, ; c = 13 In Exercises 11 - 16, solve for the remaining side(s) and angle(s), if possible, using any appropriate technique. 11. a = 18, α = 63◦, b = 20 12. a = 37, b = 45, c = 26 13. a = 16, α = 63◦, b = 20 14. a = 22, α = 63◦, b = 20 15. α = 42◦, b = 117, c = 88 16. β = 7◦, γ = 170◦, c = 98.6 17. Find the area of the triangles given in Exercises 6, 8 and 10 above. 18. The hour hand on my antique Seth Thomas schoolhouse clock in 4 inches long and the minute hand is 5.5 inches long. Find the distance between the ends of the hands when the clock reads four o’clock. Round your answer to the nearest hundredth of an inch. 19. A geologist wants to measure the diameter of a crater. From her camp, it is 4 miles to the northern-most point of the crater and 2 miles to the southern-most point. If the angle between the two lines of sight is 117◦, what is the diameter of the crater? Round your answer to the nearest hundredth of a mile. 20. From the Pedimaxus International Airport a tour helicopter can ﬂy to Cliﬀs of Insanity Point by following a bearing of N8.2◦E for 192 miles and it can ﬂy to Bigfoot Falls by following a bearing of S68.5◦E for 207 miles.6 Find the distance between Cliﬀs of Insanity Point and Bigfoot Falls. Round your answer to the nearest mile. 21. Cliﬀs of Insanity Point and Bigfoot Falls from Exericse 20 above both lie on a straight stretch of the Great Sasquatch Canyon. What bearing would the tour helicopter need to follow to go directly from Bigfoot Falls to Cliﬀs of Insanity Point? Round your angle to the nearest tenth of a degree. 6Please refer to Page 905 in Section 11.2 for an introduction to bearings. 11.3
The Law of Cosines 917 22. A naturalist sets oﬀ on a hike from a lodge on a bearing of S80◦W. After 1.5 miles, she changes her bearing to S17◦W and continues hiking for 3 miles. Find her distance from the lodge at this point. Round your answer to the nearest hundredth of a mile. What bearing should she follow to return to the lodge? Round your angle to the nearest degree. 23. The HMS Sasquatch leaves port on a bearing of N23◦E and travels for 5 miles. It then changes course and follows a heading of S41◦E for 2 miles. How far is it from port? Round your answer to the nearest hundredth of a mile. What is its bearing to port? Round your angle to the nearest degree. 24. The SS Bigfoot leaves a harbor bound for Nessie Island which is 300 miles away at a bearing of N32◦E. A storm moves in and after 100 miles, the captain of the Bigfoot ﬁnds he has drifted oﬀ course. If his bearing to the harbor is now S70◦W, how far is the SS Bigfoot from Nessie Island? Round your answer to the nearest hundredth of a mile. What course should the captain set to head to the island? Round your angle to the nearest tenth of a degree. 25. From a point 300 feet above level ground in a ﬁretower, a ranger spots two ﬁres in the Yeti National Forest. The angle of depression7 made by the line of sight from the ranger to the ﬁrst ﬁre is 2.5◦ and the angle of depression made by line of sight from the ranger to the second ﬁre is 1.3◦. The angle formed by the two lines of sight is 117◦. Find the distance between the two ﬁres. Round your answer to the nearest foot. (Hint: In order to use the 117◦ angle between the lines of sight, you will ﬁrst need to use right angle Trigonometry to ﬁnd the lengths of the lines of sight. This will give you a Side-Angle-Side case in which to apply the Law of Cosines.) ﬁre 117◦ ﬁre ﬁretower 26. If you apply the Law of Cosines to the ambiguous Angle-Side-Side (ASS) case, the result is a quadratic equation whose variable is that of the missing side. If the equation has no positive real zeros then the information given does not yield a triangle. If the equation has only one positive real zero then exactly one triangle is formed and if the equation has two distinct positive real zeros then two distinct triangles are formed. Apply the Law of Cosines to Exercises 11, 13 and 14 above in order to demonstrate this result. 27. Discuss with your classmates why Heron’s Formula yields an area in square units even though four lengths are being multiplied together. 7See Exercise 78 in Section 10.3 for the deﬁnition of this angle. 918 Applications of Trigonometry 11.3.2 Answers 1. 3. 5. 7. 9. 11. 13. 15. α ≈ 35.54◦ β ≈ 85.16◦ γ = 59.3◦ c ≈ 10.36 a = 7 b = 12 α ≈ 85.90◦ β = 8.2◦ a = 153 b ≈ 21.88 c = 153 γ ≈ 85.90◦ α = 120◦ β ≈ 25.28◦ γ ≈ 34.72◦ a = c = 4 b = 3 37 √ Information does not produce a triangle α = 60◦ β = 60◦ γ = 60 = 20 α = 63◦ β ≈ 98.11◦ γ ≈ 18.89◦ a = 18 α = 63◦ β ≈ 81.89◦ γ ≈ 35.11◦ c ≈ 11.62 a = 18 c ≈ 6.54 b = 20 Information does not produce a triangle α = 42◦ a ≈ 78.30 b = 117 β ≈ 89.23◦ γ ≈ 48.77◦ c = 88 α = 104◦ a ≈ 49.41 b = 25 β ≈ 29.40◦ γ ≈ 46.60◦ c = 37 α ≈ 36.87◦ β ≈ 53.13◦ γ = 90 ≈ 32.31◦ β ≈ 49.58◦ γ ≈ 98.21◦ a = 7 c = 13 b = 10 α ≈ 83.05◦ β ≈ 87.81◦ γ ≈ 9.14◦ b = 302 a = 300 c = 48 α ≈ 22.62◦ β ≈ 67.38◦ γ = 90◦ c = 13 b = 12 a = 5 α ≈ 55.30◦ β ≈ 89.40◦ γ ≈ 35.30◦ a = 37 c = 26 b = 45 α = 63◦ β ≈ 54.1◦ γ ≈ 62.9◦ c ≈ 21.98 a = 22 b = 20 α ≈ 3◦ γ = 170◦ β = 7◦ a ≈ 29.72 b ≈ 69.2 c = 98.6 √ 2. 4. 6. 8. 10. 12. 14. 16. √ √ 17. The area of the triangle given in Exercise 6 is The area of the triangle given in Exercise 8 is The area of the triangle given in Exercise 10 is exactly 30 square units. 1200 = 20 51764375 ≈ 7194.75 square units. 3 ≈ 34.64 square units. 18. The distance between the ends of the hands at four o’clock is about 8.26 inches. 19. The diameter of the crater is about 5.22 miles. 20. About 313 miles 21. N31.8◦W 22. She is about 3.92 miles from the lodge and her bearing to the lodge is N37◦E. 23. It is about 4.50 miles from port and its heading to port is S47◦W. 24. It is about 229.61 miles from the island and the captain should set a course of N16.4◦E to reach the island. 25. The ﬁres are about 17456 feet apart. (Try to avoid rounding errors.) 11.4 Polar Coordinates 919 11.4 Polar Coordinates In Section 1.1, we introduced the Cartesian coordinates of a point in the plane as a means of assigning ordered pairs of numbers to points in the plane. We deﬁned the Cartesian coordinate plane using two number lines – one horizontal and one vertical – which intersect at right angles at a point we called the ‘origin’. To plot a point, say P (−3, 4), we start at the origin, travel horizontally to the left 3 units, then up 4 units. Alternatively, we could start at the origin, travel up 4 units, then to the left 3 units and arrive at the same location. For the most part, the ‘motions’ of the Cartesian system (over and up) describe a rectangle, and most points can be thought of as the corner diagonally across the rectangle from the origin.1 For this reason, the Cartesian coordinates of a point are often called ‘rectangular’ coordinates. In this section, we introduce a new system for assigning coordinates to points in the plane – polar coordinates. We start with an origin point, called the pole, and a ray called the polar axis. We then locate a point P using two coordinates, (r, θ), where r represents a directed distance from the pole2 and θ is a measure of rotation from the polar axis. Roughly speaking, the polar coordinates (r, θ) of a point measure ‘how far out’ the point is from the pole (that’s r), and ‘how far to rotate’ from the polar axis, (that’s θ). y P (−3, 4) P (r, θ) 3 2 1 r θ −4 −3 −2 −1 −1 1 2 3 4 x Pole r Polar Axis −2 −3 −4 For example, if we wished to plot the point P with polar coordinates 4, 5π 6 move out along the polar axis 4 units, then rotate 5π , we’d start at the pole, 6 radians counter-clockwise. P 4, 5π 6 r = 4 Pole θ = 5π 6 Pole Pole We may also visualize this process by thinking of the rotation ﬁrst.3 To plot P 4, 5π 6 we rotate 5π this way, 6 counter-clockwise from the polar axis, then move outwards from the pole 4 units. 1Excluding, of course, the points in which one or both coordinates are 0. 2We will explain more about this momentarily. 3As with anything in Mathematics, the more ways you have to look at something, the better. The authors encourage the reader to take time to think about both approaches to plotting points given in polar coordinates. 920 Applications of Trigonometry Essentially we are locating a point on the terminal side of 5π 6 which is 4 units away from the pole. θ = 5π 6 Pole θ = 5π 6 Pole P 4, 5π 6 Pole If r < 0, we begin by moving in the opposite direction on the polar axis from the pole. For example, to plot Q −3.5, π 4 we have r = −3.5 Pole θ = π 4 Pole Pole Q −3.5, π 4 If we interpret the angle ﬁrst, we rotate π Here we are locating a point 3.5 units away from the pole on the terminal side of 5π 4 radians, then move back through the pole 3.5 units. 4 , not π 4 . θ = π 4 Pole θ = π 4 Pole Pole Q −3.5, π 4 As you may have guessed, θ < 0 means the rotation away from the polar axis is clockwise instead of counter-clockwise. Hence, to plot R 3.5, − 3π 4 we have the following. r = 3.5 Pole Pole θ = − 3π 4 Pole R 3.5, − 3π 4 From an ‘angles ﬁrst’ approach, we rotate − 3π R is the point on the terminal side of θ = − 3π 4 then move out 3.5 units from the pole. We see that 4 which is 3.5 units from the pole. Pole θ = − 3π 4 Pole θ = − 3π 4 Pole R 3.5, − 3π 4 11.4 Polar Coordinates 921 The points Q and R above are, in fact, the same point despite the fact that their polar coordinate representations are diﬀerent. Unlike Cartesian coordinates where (a, b) and (c, d) represent the same point if and only if a = c and b = d, a point can be represented by inﬁnitely many polar coordinate pairs. We explore this notion more in the following example. Example 11.4.1. For each point in polar coordinates given below plot the point and then give two additional expressions for the point, one of which has r > 0 and the other with r < 0. 4. P −3, − π 4 3. P 117, − 5π 2 2. P −4, 7π 6 1. P (2, 240◦) Solution. 1. Whether we move 2 units along the polar axis and then rotate 240◦ or rotate 240◦ then move out 2 units from the pole, we plot P (2, 240◦) below. θ = 240◦ Pole Pole P (2, 240◦) We now set about ﬁnding alternate descriptions (r, θ) for the point P . Since P is 2 units from the pole, r = ±2. Next, we choose angles θ for each of the r values. The given representation for P is (2, 240◦) so the angle θ we choose for the r = 2 case must be coterminal with 240◦. (Can you see why?) One such angle is θ = −120◦ so one answer for this case is (2, −120◦). For the case r = −2, we visualize our rotation starting 2 units to the left of the pole. From this position, we need only to rotate θ = 60◦ to arrive at location coterminal with 240◦. Hence, our answer here is (−2, 60◦). We check our answers by plotting them. Pole θ = −120◦ θ = 60◦ Pole P (2, −120◦) P (−2, 60◦) 2. We plot −4, 7π 6 6 radians. Since r = −4 < 0, we ﬁnd our point lies 4 units from the pole on the terminal side of π 6 . by ﬁrst moving 4 units to the left of the pole and then rotating 7π P −4, 7π 6 Pole Pole θ = 7π 6 922 Applications of Trigonometry To ﬁnd alternate descriptions for P , we note that the distance from P to the pole is 4 units, so any representation (r, θ) for P must have r = ±4. As we noted above, P lies on the terminal as one of our answers. To ﬁnd a diﬀerent side of π representation for P with r = −4, we may choose any angle coterminal with the angle in the as our second answer. original representation of P −4, 7π 6 6 , so this, coupled with r = 4, gives us 4, π . We pick − 5π 6 and get −4, − 5π 6 6 P 4, π 6 θ = π 6 Pole θ = − 5π 6 Pole P −4, − 5π 6 , we move along the polar axis 117 units from the pole and rotate 3. To plot P 117, − 5π 2 clockwise 5π 2 radians as illustrated below. Pole θ = − 5π 2 Pole P 117,
− 5π 2 Since P is 117 units from the pole, any representation (r, θ) for P satisﬁes r = ±117. For the r = 117 case, we can take θ to be any angle coterminal with − 5π 2 . In this case, we choose as one answer. For the r = −117 case, we visualize moving left 117 θ = 3π units from the pole and then rotating through an angle θ to reach P . We ﬁnd that θ = π 2 satisﬁes this requirement, so our second answer is −117, π 2 2 , and get 117, 3π . 2 Pole θ = 3π 2 Pole θ = π 2 P 117, 3π 2 P −117, π 2 11.4 Polar Coordinates 923 4. We move three units to the left of the pole and follow up with a clockwise rotation of π 4 radians to plot P −3, − π 4 . We see that P lies on the terminal side of 3π 4 . P −3, − π 4 θ = − π 4 Pole Pole . To Since P lies on the terminal side of 3π ﬁnd a diﬀerent representation for P with r = −3, we may choose any angle coterminal with − π 4 , one alternative representation for P is 3, 3π 4 for our ﬁnal answer −3, 7π 4 . We choose θ = 7π . 4 4 P 3, 3π 4 P −3, 7π 4 θ = 3π 4 Pole θ = 7π 4 Pole Now that we have had some practice with plotting points in polar coordinates, it should come as no surprise that any given point expressed in polar coordinates has inﬁnitely many other representations in polar coordinates. The following result characterizes when two sets of polar coordinates determine the same point in the plane. It could be considered as a deﬁnition or a theorem, depending on your point of view. We state it as a property of the polar coordinate system. Equivalent Representations of Points in Polar Coordinates Suppose (r, θ) and (r, θ) are polar coordinates where r = 0, r = 0 and the angles are measured in radians. Then (r, θ) and (r, θ) determine the same point P if and only if one of the following is true: r = r and θ = θ + 2πk for some integer k r = −r and θ = θ + (2k + 1)π for some integer k All polar coordinates of the form (0, θ) represent the pole regardless of the value of θ. The key to understanding this result, and indeed the whole polar coordinate system, is to keep in mind that (r, θ) means (directed distance from pole, angle of rotation). If r = 0, then no matter how much rotation is performed, the point never leaves the pole. Thus (0, θ) is the pole for all 924 Applications of Trigonometry values of θ. Now let’s assume that neither r nor r is zero. If (r, θ) and (r, θ) determine the same point P then the (non-zero) distance from P to the pole in each case must be the same. Since this distance is controlled by the ﬁrst coordinate, we have that either r = r or r = −r. If r = r, then when plotting (r, θ) and (r, θ), the angles θ and θ have the same initial side. Hence, if (r, θ) and (r, θ) determine the same point, we must have that θ is coterminal with θ. We know that this means θ = θ + 2πk for some integer k, as required. If, on the other hand, r = −r, then when plotting (r, θ) and (r, θ), the initial side of θ is rotated π radians away from the initial side of θ. In this case, θ must be coterminal with π + θ. Hence, θ = π + θ + 2πk which we rewrite as θ = θ + (2k + 1)π for some integer k. Conversely, if r = r and θ = θ + 2πk for some integer k, then the points P (r, θ) and P (r, θ) lie the same (directed) distance from the pole on the terminal sides of coterminal angles, and hence are the same point. Now suppose r = −r and θ = θ + (2k + 1)π for some integer k. To plot P , we ﬁrst move a directed distance r from the pole; to plot P , our ﬁrst step is to move the same distance from the pole as P , but in the opposite direction. At this intermediate stage, we have two points equidistant from the pole rotated exactly π radians apart. Since θ = θ + (2k + 1)π = (θ + π) + 2πk for some integer k, we see that θ is coterminal to (θ + π) and it is this extra π radians of rotation which aligns the points P and P . Next, we marry the polar coordinate system with the Cartesian (rectangular) coordinate system. To do so, we identify the pole and polar axis in the polar system to the origin and positive x-axis, respectively, in the rectangular system. We get the following result. Theorem 11.7. Conversion Between Rectangular and Polar Coordinates: Suppose P is represented in rectangular coordinates as (x, y) and in polar coordinates as (r, θ). Then x = r cos(θ) and y = r sin(θ) x2 + y2 = r2 and tan(θ) = y x (provided x = 0) In the case r > 0, Theorem 11.7 is an immediate consequence of Theorem 10.3 along with the quotient identity tan(θ) = sin(θ) If r < 0, then we know an alternate representation for (r, θ) cos(θ) . is (−r, θ + π). Since cos(θ + π) = − cos(θ) and sin(θ + π) = − sin(θ), applying the theorem to (−r, θ + π) gives x = (−r) cos(θ + π) = (−r)(− cos(θ)) = r cos(θ) and y = (−r) sin(θ + π) = (−r)(− sin(θ)) = r sin(θ). Moreover, x2 + y2 = (−r)2 = r2, and y x = tan(θ + π) = tan(θ), so the theorem is true in this case, too. The remaining case is r = 0, in which case (r, θ) = (0, θ) is the pole. Since the pole is identiﬁed with the origin (0, 0) in rectangular coordinates, the theorem in this case amounts to checking ‘0 = 0.’ The following example puts Theorem 11.7 to good use. Example 11.4.2. Convert each point in rectangular coordinates given below into polar coordinates with r ≥ 0 and 0 ≤ θ < 2π. Use exact values if possible and round any approximate values to two decimal places. Check your answer by converting them back to rectangular coordinates. 1. P 2, −2 √ 3 2. Q(−3, −3) 3. R(0, −3) 4. S(−3, 4) 11.4 Polar Coordinates 925 Solution. 1. Even though we are not explicitly told to do so, we can avoid many common mistakes by taking 3 shows that the time to plot the points before we do any calculations. Plotting P 2, −2 √ it lies in Quadrant IV. With x = 2 and y = −2 = 4 + 12 = 16 so r = ±4. Since we are asked for r ≥ 0, we choose r = 4. To ﬁnd θ, we have that tan(θ) = y 3 , and since P lies in Quadrant IV, we know θ is a Quadrant IV angle. We are asked to have 0 ≤ θ < 2π, . To check, we convert (r, θ) = 4, 5π 3 . Hence, our answer is 4, 5π so we choose θ = 5π 3 back to rectangular coordinates and we ﬁnd x = r cos(θ) = 4 cos 5π = 2 and 3 y = r sin(θ) = 4 sin 5π 3 3. This tells us θ has a reference angle of π 3, we get r2 = x2 + y2 = (2)2 + −, as required. x = −2 = 4 = −2 32 √ √ 18 = ±3 2. The point Q(−3, −3) lies in Quadrant III. Using x = y = −3, we get r2 = (−3)2 + (−3)2 = 18 2. We ﬁnd 4 . Since Q lies in Quadrant III, 4 , which satisﬁes the requirement that 0 ≤ θ < 2π. Our ﬁnal answer is . To check, we ﬁnd x = r cos(θ) = (3 = −3 so r = ± tan(θ) = −3 we choose θ = 5π (r, θ) = 3 −3 = 1, which means θ has a reference angle of π 2. Since we are asked for r ≥ 0, we choose r = 3 = (3 2) √ √ √ √ − √ 2) cos 5π 4 2, 5π 4 2 2 and y = r sin(θ) = (3 √ 2) sin 5π 4 √ = (3 − √ 2 2 2) = −3, so we are done. y y θ = 5π 3 x θ = 5π 4 x P Q P has rectangular coordinates (2, −2 P has polar coordinates 4, 5π 3 √ 3) Q has rectangular coordinates (−3, −3) Q has polar coordinates 3 √ 2, 5π 4 3. The point R(0, −3) lies along the negative y-axis. While we could go through the usual computations4 to ﬁnd the polar form of R, in this case we can ﬁnd the polar coordinates of R using the deﬁnition. Since the pole is identiﬁed with the origin, we can easily tell the point R is 3 units from the pole, which means in the polar representation (r, θ) of R we know r = ±3. Since we require r ≥ 0, we choose r = 3. Concerning θ, the angle θ = 3π 2 satisﬁes 0 ≤ θ < 2π 4Since x = 0, we would have to determine θ geometrically. 926 Applications of Trigonometry with its terminal side along the negative y-axis, so our answer is 3, 3π 2 x = r cos(θ) = 3 cos 3π 2 = (3)(0) = 0 and y = r sin(θ) = 3 sin 3π 2 = 3(−1) = −3. . To check, we note −3 = − 4 4. The point S(−3, 4) lies in Quadrant II. With x = −3 and y = 4, we get r2 = (−3)2 +(4)2 = 25 so r = ±5. As usual, we choose r = 5 ≥ 0 and proceed to determine θ. We have tan(θ) = y x = 4 3 , and since this isn’t the tangent of one the common angles, we resort to using the arctangent function. Since θ lies in Quadrant II and must satisfy 0 ≤ θ < 2π, we choose θ = π − arctan 4 ≈ (5, 2.21). To radians. Hence, our answer is (r, θ) = 5, π − arctan 4 3 3 check our answers requires a bit of tenacity since we need to simplify expressions of the form: cos π − arctan 4 . These are good review exercises and are hence 3 left to the reader. We ﬁnd cos π − arctan 4 5 , so that 3 x = r cos(θ) = (5) − 3 5 = − 3 = −3 and y = r sin(θ) = (5) 4 5 = 4 which conﬁrms our answer. and sin π − arctan 4 3 5 and sin π − arctan 4 = 4 3 y y S θ = 3π 2 x R θ = π − arctan 4 3 x R has rectangular coordinates (0, −3) R has polar coordinates 3, 3π 2 S has rectangular coordinates (−3, 4) S has polar coordinates 5, π − arctan 4 3 Now that we’ve had practice converting representations of points between the rectangular and polar coordinate systems, we now set about converting equations from one system to another. Just as we’ve used equations in x and y to represent relations in rectangular coordinates, equations in the variables r and θ represent relations in polar coordinates. We convert equations between the two systems using Theorem 11.7 as the next example illustrates. Example 11.4.3. 1. Convert each equation in rectangular coordinates into an equation in polar coordinates. (a) (x − 3)2 + y2 = 9 (b) y = −x (c) y = x2 2. Convert each equation in polar coordinates into an equation in rectangular coordinates. (a) r = −3 (b) θ = 4π 3 (c) r = 1 − cos(θ) 11.4 Polar Coordinates 927 Solution. 1. One strategy to convert an equation from rectangular to polar coordinates is to replace every occurrence of x with r cos(θ) and every occurrence of y with r sin(θ) and use identities to simplify. This is the technique we employ below. (a) We start by substituting x = r cos(θ) and y = sin(θ) into (x−3)2+y2 = 9 and simplifying. With no real direction in which to proceed, we follow our mathematical instincts and see where they take us.5 (r cos(θ) − 3)2 + (r sin(θ))2 = 9 r2 cos2(θ) − 6r cos(θ) + 9 + r2 sin2(θ) = 9 r2 cos2(θ) + sin2(θ) − 6r cos(θ) =
0 Subtract 9 from both sides. Since cos2(θ) + sin2(θ) = 1 Factor. r2 − 6r cos(θ) = 0 r(r − 6 cos(θ)) = 0 We get r = 0 or r = 6 cos(θ). From Section 7.2 we know the equation (x − 3)2 + y2 = 9 describes a circle, and since r = 0 describes just a point (namely the pole/origin), we choose r = 6 cos(θ) for our ﬁnal answer.6 (b) Substituting x = r cos(θ) and y = r sin(θ) into y = −x gives r sin(θ) = −r cos(θ). Rearranging, we get r cos(θ) + r sin(θ) = 0 or r(cos(θ) + sin(θ)) = 0. This gives r = 0 or cos(θ) + sin(θ) = 0. Solving the latter equation for θ, we get θ = − π 4 + πk for integers k. As we did in the previous example, we take a step back and think geometrically. We know y = −x describes a line through the origin. As before, r = 0 describes the origin, but nothing else. Consider the equation θ = − π 4 . In this equation, the variable r is free,7 meaning it can assume any and all values including r = 0. If we imagine plotting points (r, − π 4 ) for all conceivable values of r (positive, negative and zero), we are essentially drawing the line containing the terminal side of θ = − π 4 which is none other than y = −x. Hence, we can take as our ﬁnal answer θ = − π 4 here.8 (c) We substitute x = r cos(θ) and y = r sin(θ) into y = x2 and get r sin(θ) = (r cos(θ))2, or r2 cos2(θ) − r sin(θ) = 0. Factoring, we get r(r cos2(θ) − sin(θ)) = 0 so that either r = 0 or r cos2(θ) = sin(θ). We can solve the latter equation for r by dividing both sides of the equation by cos2(θ), but as a general rule, we never divide through by a quantity that may be 0. In this particular case, we are safe since if cos2(θ) = 0, then cos(θ) = 0, and for the equation r cos2(θ) = sin(θ) to hold, then sin(θ) would also have to be 0. Since there are no angles with both cos(θ) = 0 and sin(θ) = 0, we are not losing any 5Experience is the mother of all instinct, and necessity is the mother of invention. Study this example and see what techniques are employed, then try your best to get your answers in the homework to match Jeﬀ’s. 2 into r = 6 cos(θ), we recover the point r = 0, so we aren’t losing anything 6Note that when we substitute θ = π by disregarding r = 0. 7See Section 8.1. 8We could take it to be any of θ = − π 4 + πk for integers k. 928 Applications of Trigonometry information by dividing both sides of r cos2(θ) = sin(θ) by cos2(θ). Doing so, we get r = sin(θ) cos2(θ) , or r = sec(θ) tan(θ). As before, the r = 0 case is recovered in the solution r = sec(θ) tan(θ) (let θ = 0), so we state the latter as our ﬁnal answer. 2. As a general rule, converting equations from polar to rectangular coordinates isn’t as straight forward as the reverse process. We could solve r2 = x2 + y2 for r to get r = ± x2 + y2 + πk for and solving tan(θ) = y integers k. Neither of these expressions for r and θ are especially user-friendly, so we opt for a second strategy – rearrange the given polar equation so that the expressions r2 = x2 + y2, r cos(θ) = x, r sin(θ) = y and/or tan(θ) = y x requires the arctangent function to get θ = arctan y x present themselves. x (a) Starting with r = −3, we can square both sides to get r2 = (−3)2 or r2 = 9. We may now substitute r2 = x2 + y2 to get the equation x2 + y2 = 9. As we have seen,9 squaring an equation does not, in general, produce an equivalent equation. The concern here is that the equation r2 = 9 might be satisﬁed by more points than r = −3. On the surface, this appears to be the case since r2 = 9 is equivalent to r = ±3, not just r = −3. However, any point with polar coordinates (3, θ) can be represented as (−3, θ + π), which means any point (r, θ) whose polar coordinates satisfy the relation r = ±3 has an equivalent10 representation which satisﬁes r = −3. (b) We take the tangent of both sides the equation θ = 4π Since tan(θ) = y if, geometrically, the equations θ = 4π 3 and y = x The same argument presented in number 1b applies equally well here so we are done. 3 to get tan(θ) = tan 4π 3. 3. Of course, we pause a moment to wonder 3 generate the same set of points.11 x , we get y 3 or c) Once again, we need to manipulate r = 1 − cos(θ) a bit before using the conversion formulas given in Theorem 11.7. We could square both sides of this equation like we did in part 2a above to obtain an r2 on the left hand side, but that does nothing helpful for the right hand side. Instead, we multiply both sides by r to obtain r2 = r − r cos(θ). We now have an r2 and an r cos(θ) in the equation, which we can easily handle, but we also have another r to deal with. Rewriting the equation as r = r2 + r cos(θ) and squaring both sides yields r2 = r2 + r cos(θ)2 . Substituting r2 = x2 + y2 and r cos(θ) = x gives x2 + y2 = x2 + y2 + x2 . Once again, we have performed some 9Exercise 5.3.1 in Section 5.3, for instance . . . 10Here, ‘equivalent’ means they represent the same point in the plane. As ordered pairs, (3, 0) and (−3, π) are diﬀerent, but when interpreted as polar coordinates, they correspond to the same point in the plane. Mathematically speaking, relations are sets of ordered pairs, so the equations r2 = 9 and r = −3 represent diﬀerent relations since they correspond to diﬀerent sets of ordered pairs. Since polar coordinates were deﬁned geometrically to describe the location of points in the plane, however, we concern ourselves only with ensuring that the sets of points in the plane generated by two equations are the same. This was not an issue, by the way, when we ﬁrst deﬁned relations as sets of points in the plane in Section 1.2. Back then, a point in the plane was identiﬁed with a unique ordered pair given by its Cartesian coordinates. 11In addition to taking the tangent of both sides of an equation (There are inﬁnitely many solutions to tan(θ) = √ √ 3 is only one of them!), we also went from y x = 3, in which x cannot be 0, to y = x √ 3, 3 in which we assume and θ = 4π x can be 0. 11.4 Polar Coordinates 929 algebraic maneuvers which may have altered the set of points described by the original equation. First, we multiplied both sides by r. This means that now r = 0 is a viable solution to the equation. In the original equation, r = 1 − cos(θ), we see that θ = 0 gives r = 0, so the multiplication by r doesn’t introduce any new points. The squaring of both sides of this equation is also a reason to pause. Are there points with coordinates (r, θ) which satisfy r2 = r2 + r cos(θ)2 but do not satisfy r = r2 + r cos(θ)? Suppose (r, θ) satisﬁes r2 = r2 + r cos(θ)2 . Then r = ± (r)2 + r cos(θ). If we have that r = (r)2+r cos(θ), we are done. What if r = − (r)2 + r cos(θ) = −(r)2−r cos(θ)? We claim that the coordinates (−r, θ + π), which determine the same point as (r, θ), satisfy r = r2 + r cos(θ). We substitute r = −r and θ = θ + π into r = r2 + r cos(θ) to see if we get a true statement. −r − −(r)2 − r cos(θ) (r)2 + r cos(θ) (r)2 + r cos(θ) ? = (−r)2 + (−r cos(θ + π)) ? = (r)2 − r cos(θ + π) ? = (r)2 − r(− cos(θ)) = (r)2 + r cos(θ) Since r = −(r)2 − r cos(θ) Since cos(θ + π) = − cos(θ) Since both sides worked out to be equal, (−r, θ + π) satisﬁes r = r2 + r cos(θ) which means that any point (r, θ) which satisﬁes r2 = r2 + r cos(θ)2 has a representation which satisﬁes r = r2 + r cos(θ), and we are done. In practice, much of the pedantic veriﬁcation of the equivalence of equations in Example 11.4.3 is left unsaid. Indeed, in most textbooks, squaring equations like r = −3 to arrive at r2 = 9 happens without a second thought. Your instructor will ultimately decide how much, if any, justiﬁcation is warranted. If you take anything away from Example 11.4.3, it should be that relatively nice things in rectangular coordinates, such as y = x2, can turn ugly in polar coordinates, and vice-versa. In the next section, we devote our attention to graphing equations like the ones given in Example 11.4.3 number 2 on the Cartesian coordinate plane without converting back to rectangular coordinates. If nothing else, number 2c above shows the price we pay if we insist on always converting to back to the more familiar rectangular coordinate system. 930 Applications of Trigonometry 11.4.1 Exercises In Exercises 1 - 16, plot the point given in polar coordinates and then give three diﬀerent expressions (c) r > 0 and θ ≥ 2π for the point such that (a) r < 0 and 0 ≤ θ ≤ 2π, (b) r > 0 and θ ≤ 0 2, 1. π 3 5. 12, − 7π 6 9. (−20, 3π) 2. 5, 7π 4 6. 3, − 10. −4, 5π 4 5π 4 3. 1 3 , 3π 2 √ 7. 2 2, −π 11. −1, 2π 3 13. −3, − 11π 6 14. −2.5, − π 4 15. √ − 5, − 4π 3 4. 8. 5 2 7 2 , 5π 6 , − 13π 6 −3, 12. π 2 16. (−π, −π) In Exercises 17 - 36, convert the point from polar coordinates into rectangular coordinates. 18. 2, π 3 22. −4, 5π 6 7π 6 19. 11, − 23. 9, 7π 2 20. (−20, 3π) 24. −5, − 9π 4 26. (−117, 117π) 27. (6, arctan(2)) 28. (10, arctan(3)) 17. 5, 7π 4 21. 3 5 , π 2 25. 42, 13π 6 29. −3, arctan 31. 33. 2, π − arctan −1, π + arctan 4 3 1 2 3 4 34. 2 3 35. (π, arctan(π)) 36. 13, arctan 30. 5, arctan − 4 3 32. − 1 2 , π − arctan (5) , π + arctan 2 √ 2 12 5 In Exercises 37 - 56, convert the point from rectangular coordinates into polar coordinates with r ≥ 0 and 0 ≤ θ < 2π. 37. (0, 5) 41. (−3, 0) 38. (3, √ 3) 39. (7, −7) √ 42. − √ 2 2, 43. −4, −4 √ 3 √ 3) 40. (−3, − √ 44. 3 4 , − 1 4 11.4 Polar Coordinates 931 45. − √ 3 3 10 3 10 , − √ 46. − √ 5, − 5 47. (6, 8) √ √ 5) 5, 2 48. ( 49. (−8, 1) √ 50. (−2 √ 10, 6 10) 53. (24, −7) 54. (12, −9) 51. (−5, −12) √ 2 4 √ 6 4 , 55. 52. − √ 5 15 √ 5 2 15 , − 56. − √ 65 5 2 , √ 65 5 In Exercises 57 - 76, convert the equation from rectangular coordinates into polar coordinates. Solve for r in all but #60 through #63. In Exercises 60 - 63, you need to solve for θ 57. x = 6 61. y = −x 58. x = −3 √ 3 62. y = x 59. y = 7 60. y = 0 63. y = 2x 64. x2 + y2 = 25 65. x2 + y2 = 117 66. y = 4x − 19 67. x = 3y + 1 68. y = −3x2 69. 4x = y2 70. x2 + y2 − 2y = 0 71. x2 − 4x + y2 = 0 72. x2 + y2 = x 73. y2 = 7y − x2 75. x2 + (y − 3)2 = 9 74. (x + 2)2 + y2 = 4 76. 4x2 + 4 y − 2 1 2 = 1 In Exercises 77 - 96, convert the equation from polar coordinate
s into rectangular coordinates. 77. r = 7 81. θ = 2π 3 78. r = −3 82. θ = π 79. r = 83. θ = √ 2 3π 2 80. θ = π 4 84. r = 4 cos(θ) 85. 5r = cos(θ) 86. r = 3 sin(θ) 87. r = −2 sin(θ) 88. r = 7 sec(θ) 89. 12r = csc(θ) 90. r = −2 sec(θ) √ 91. r = − 5 csc(θ) 92. r = 2 sec(θ) tan(θ) 93. r = − csc(θ) cot(θ) 94. r2 = sin(2θ) 95. r = 1 − 2 cos(θ) 96. r = 1 + sin(θ) 97. Convert the origin (0, 0) into polar coordinates in four diﬀerent ways. 98. With the help of your classmates, use the Law of Cosines to develop a formula for the distance between two points in polar coordinates. 932 Applications of Trigonometry 11.4.2 Answers 1. 2, , π 3 2, − 5π 3 −2, , 4π 3 7π 3 2, 2. 5, 7π 4 π 4 , , −5, 5, 3π 4 15π 4 5, − 3. 1 3 1 3 4. 5 2 5 2 , , , 3π , , 5π 6 7π 7π 2 , , 11π 6 17π 6 y y 2 1 1 −1 1 2 x −1 1 2 3 x −1 −2 −3 −2 −1 −1 1 2 3 x −2 −3 11.4 Polar Coordinates 933 5. 12, − 12, − , −12, , 12, 7π 6 19π 6 11π 6 17π 6 6. 3, − 3, − , −3, , 3, 5π 4 13π 4 7π 4 11π 4 √ √ 7. 2 2 √ 2, −π , −2 √ 2, −3π , 2 2, 0 2, 3π 8. 7 2 7 2 , − , − 13π 23π 6 5π 6 y 6 3 −12 −9 −6 −3 x y 3 2 1 −3 −2 −1 1 2 3 x −1 −2 −3 3 2 1 y −3 −2 −1 1 2 3 x −1 −2 −3 4 3 2 1 y −4 −3 −2 −1 1 2 3 4 x −1 −2 −3 −4 934 Applications of Trigonometry 9. (−20, 3π), (−20, π) (20, −2π), (20, 4π) 10. −4, 4, − 5π 4 7π 4 11. −1, 1, − 2π 3 π 3 13π 4 , , −4, 4, 9π 4 8π 3 , , 1, −1, 11π 3 12. −3, 3, − , , π 2 π 2 5π 2 −3, 3, 7π 2 10 x 20 −20 −10 y y 1 −1 4 3 2 1 −4 −3 −2 −1 −1 1 2 3 4 x −2 −3 −4 2 1 y −2 −1 1 2 x −1 −2 y 3 2 1 −3 −2 −1 −1 1 2 3 x −2 −3 11.4 Polar Coordinates 935 13. −3, − 11π 6 , 3, − 5π 6 π 6 −3, 19π 6 , 3, , , 14. −2.5, − 2.5, − π 4 5π 4 −2.5, 2.5, 7π 4 11π 4 √ 15. − 5, − √ 5, − π 3 4π 3 , , √ √ − 2π 3 5, 5, 11π 3 16. (−π, −π) , (−π, π) (π, −2π) , (π, 2π3 −2 −1 −1 −2 −3 2 1 −2 −1 −1 −2 2 1 −2 −1 −1 −2 3 2 1 −3 −2 −1 1 2 3 x −1 −2 −3 936 Applications of Trigonometry √ 5 2 2 , − √ 2 5 2 17. 18. 1, √ 3 19. − √ 11 2 3 , 11 2 22. 2 √ 3, −2 23. (0, −9) 20. (20, 0) 24 28. √ 10, 3 √ 10 √ 26 52 √ 5 26 52 , − 32. 36. (5, 12) √ 2 3, 40. 44. 1 2 , 11π 6 7π 6 4 3 48. (5, arctan (2)) 26. (117, 0) 30. (3, −4 34. √ 2 3, 38. 42. 2, √ 46. 3π 4 10, π 6 5π 4 √ 5 6 5 5 √ 12 5 , 27. 31 35. π√ 1+π2 , √ π2 1+π2 7π 4 √ 7 2, 39. 43. 8, 4π 3 47. 10, arctan 21. 0, 3 5 √ 25. 21 3, 21 29. − 9 5 , − 12 5 33. 4 5 , 3 5 37. 5, π 2 41. (3, π) 4π 3 , 3 5 √ 51. 13, π + arctan 25, 2π − arctan 1 8 12 5 7 24 45. 49. 53. 55. 65, π − arctan 50. (20, π − arctan(3)) , π + arctan (2) 52. 1 3 54. 15, 2π − arctan 3 4 √ 2 2 , π 3 56. √ 13, π − arctan(2) 57. r = 6 sec(θ) 58. r = −3 sec(θ) 59. r = 7 csc(θ) 60. θ = 0 61. θ = 3π 4 √ 65. r = 117 62. θ = π 3 63. θ = arctan(2) 64. r = 5 66. r = 19 4 cos(θ)−sin(θ) 67. x = 1 cos(θ)−3 sin(θ) 68. r = − sec(θ) tan(θ) 3 69. r = 4 csc(θ) cot(θ) 70. r = 2 sin(θ) 71. r = 4 cos(θ) 72. r = cos(θ) 11.4 Polar Coordinates 937 73. r = 7 sin(θ) 74. r = −4 cos(θ) 75. r = 6 sin(θ) 76. r = sin(θ) 77. x2 + y2 = 49 78. x2 + y2 = 9 79. x2 + y2 = 2 80. y = x √ 81. y = − 3x 83. x = 0 82. y = 0 84. x2 + y2 = 4x or (x − 2)2 + y2 = 4 85. 5x2 + 5y2 = x or x − 2 1 10 + y2 = 1 100 86. x2 + y2 = 3y or x2 + y − 2 3 2 = 9 4 87. x2 + y2 = −2y or x2 + (y + 1)2 = 1 88. x = 7 89. y = 1 12 √ 91. y = − 5 93. y2 = −x 90. x = −2 92. x2 = 2y 94. x2 + y22 = 2xy 95. x2 + 2x + y22 = x2 + y2 96. x2 + y2 + y2 97. Any point of the form (0, θ) will work, e.g. (0, π), (0, −117), 0, = x2 + y2 23π 4 and (0, 0). 938 Applications of Trigonometry 11.5 Graphs of Polar Equations In this section, we discuss how to graph equations in polar coordinates on the rectangular coordinate plane. Since any given point in the plane has inﬁnitely many diﬀerent representations in polar coordinates, our ‘Fundamental Graphing Principle’ in this section is not as clean as it was for graphs of rectangular equations on page 23. We state it below for completeness. The Fundamental Graphing Principle for Polar Equations The graph of an equation in polar coordinates is the set of points which satisfy the equation. That is, a point P (r, θ) is on the graph of an equation if and only if there is a representation of P , say (r, θ), such that r and θ satisfy the equation. Our ﬁrst example focuses on some of the more structurally simple polar equations. Example 11.5.1. Graph the following polar equations. 1. r = 4 2. r = −3 √ 2 3. θ = 5π 4 4. θ = − 3π 2 Solution. In each of these equations, only one of the variables r and θ is present making the other variable free.1 This makes these graphs easier to visualize than others. 1. In the equation r = 4, θ is free. The graph of this equation is, therefore, all points which have a polar coordinate representation (4, θ), for any choice of θ. Graphically this translates into tracing out all of the points 4 units away from the origin. This is exactly the deﬁnition of circle, centered at the origin, with a radius of 44 x 4 In r = 4, θ is free The graph of r = 4 −4 2. Once again we have θ being free in the equation r = −3 2, θ) gives us a circle of radius 3 form (−3 √ √ 2 centered at the origin. 2. Plotting all of the points of the √ 1See the discussion in Example 11.4.3 number 2a. 11.5 Graphs of Polar Equations 939 4 x 4 In r = −3 √ 2, θ is free −4 The graph of r = −3 √ 2 3. In the equation θ = 5π . 4 , r is free, so we plot all of the points with polar representation r, 5π What we ﬁnd is that we are tracing out the line which contains the terminal side of θ = 5π 4 when plotted in standard position. 4 y r < 0 θ = 5π 4 y 4 r = 0 x −4 x 4 r > 0 In θ = 5π 4 , r is free The graph of θ = 5π 4 −4 4. As in the previous example, the variable r is free in the equation θ = − 3π 2 . Plotting r, − 3π 2 for various values of r shows us that we are tracing out the y-axis. 940 Applications of Trigonometry y r > 0 r = 0 θ = − 3π 2 y 4 x −4 x 4 r < 0 −4 In θ = − 3π 2 , r is free The graph of θ = − 3π 2 Hopefully, our experience in Example 11.5.1 makes the following result clear. Theorem 11.8. Graphs of Constant r and θ: Suppose a and α are constants, a = 0. The graph of the polar equation r = a on the Cartesian plane is a circle centered at the origin of radius \|a\|. The graph of the polar equation θ = α on the Cartesian plane is the line containing the terminal side of α when plotted in standard position. Suppose we wish to graph r = 6 cos(θ). A reasonable way to start is to treat θ as the independent variable, r as the dependent variable, evaluate r = f (θ) at some ‘friendly’ values of θ and plot the resulting points.2 We generate the table below. θ 0 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π r = 6 cos(θ) 6 √ 3 2 3 √ √ (r, θ) (6, 0) 2, π 4 0, π 2 2, 3π 4 (−6, π) √ 2, 5π 4 0, 3π 2 √ 2, 7π 4 (6, 2π) 0 2 −3 √ −3 −6 √ −3 √ 3 2 6 2 −3 0 3 y 3 −3 3 x 6 2For a review of these concepts and this process, see Sections 1.4 and 1.6. 11.5 Graphs of Polar Equations 941 Despite having nine ordered pairs, we get only four distinct points on the graph. For this reason, we employ a slightly diﬀerent strategy. We graph one cycle of r = 6 cos(θ) on the θr-plane3 and use it to help graph the equation on the xy-plane. We see that as θ ranges from 0 to π 2 , r ranges from 6 to 0. In the xy-plane, this means that the curve starts 6 units from the origin on the positive x-axis (θ = 0) and gradually returns to the origin by the time the curve reaches the y-axis (θ = π 2 ). The arrows drawn in the ﬁgure below are meant to help you visualize this process. In the θr-plane, the arrows are drawn from the θ-axis to the curve r = 6 cos(θ). In the xy-plane, each of these arrows starts at the origin and is rotated through the corresponding angle θ, in accordance with how we plot polar coordinates. It is a less-precise way to generate the graph than computing the actual function values, but it is markedly faster. r 6 3 −3 −6 π 2 π 3π 2 2π θ y θ runs from 0 to π 2 x Next, we repeat the process as θ ranges from π 2 to π. Here, the r values are all negative. This means that in the xy-plane, instead of graphing in Quadrant II, we graph in Quadrant IV, with all of the angle rotations starting from the negative x-axis. r 6 3 −3 −6 y θ runs from π 2 to π π 2 π 3π 2 2π θ x r < 0 so we plot here As θ ranges from π to 3π 2 , the r values are still negative, which means the graph is traced out in Quadrant I instead of Quadrant III. Since the \|r\| for these values of θ match the r values for θ in 3The graph looks exactly like y = 6 cos(x) in the xy-plane, and for good reason. At this stage, we are just graphing the relationship between r and θ before we interpret them as polar coordinates (r, θ) on the xy-plane. 942 Applications of Trigonometry , we have that the curve begins to retrace itself at this point. Proceeding further, we ﬁnd 2 ≤ θ ≤ 2π, we retrace the portion of the curve in Quadrant IV that we ﬁrst traced 2 ≤ θ ≤ π. The reader is invited to verify that plotting any range of θ outside the interval 0, π 2 that when 3π out as π [0, π] results in retracting some portion of the curve.4 We present the ﬁnal graph below. r 6 3 −3 −6 π 2 π θ y 3 −3 3 x 6 r = 6 cos(θ) in the θr-plane r = 6 cos(θ) in the xy-plane Example 11.5.2. Graph the following polar equations. 1. r = 4 − 2 sin(θ) 2. r = 2 + 4 cos(θ) 3. r = 5 sin(2θ) 4. r2 = 16 cos(2θ) Solution. 1. We ﬁrst plot the fundamental cycle of r = 4 − 2 sin(θ) on the θr-axes. To help us visualize what is going on graphically, we divide up [0, 2π] into the usual four subintervals 0, π 2 , π, 2 π, 3π 2 , r decreases from 2 4 to 2. This means that the curve in the xy-plane starts 4 units from the origin on the positive x-axis and gradually pulls in towards the origin as it moves towards the positive y-axis. 2 , 2π, and proceed as we did above. As θ ranges from 0 to π and 3π , π r 6 4 2 π 2 π 3π 2 2π θ y θ runs from 0 to π 2 x 4The graph of r = 6 cos(θ) looks suspiciously like a circle, for good reason. See number 1a in Example 11.4.3. 11.5 Graphs of Polar Equations 943 Next, as θ runs from π oﬀ, we gradually pull the graph away from the origin until we reach the negative x-axis. 2 to π, we see that r increases from 2 to 4. Picking up where w
e left θ runs from π 2 to 3π 2 2π θ , we see that r increases from 4 to 6. On the xy-plane, the curve Over the interval π, 3π 2 sweeps out away from the origin as it travels from the negative x-axis to the negative y-axis 3π 2 2π θ θ runs from π to 3π 2 Finally, as θ takes on values from 3π xy-plane pulls in from the negative y-axis to ﬁnish where we started. 2 to 2π, r decreases from 6 back to 4. The graph on the r 6 4 2 y x π 2 π 3π 2 2π θ θ runs from 3π 2 to 2π We leave it to the reader to verify that plotting points corresponding to values of θ outside the interval [0, 2π] results in retracing portions of the curve, so we are ﬁnished. 944 Applications of Trigonometry sin(θ) in the θr-plane 3π 2 2π y 2 −4 x 4 −6 r = 4 − 2 sin(θ) in the xy-plane. 2. The ﬁrst thing to note when graphing r = 2 + 4 cos(θ) on the θr-plane over the interval [0, 2π] is that the graph crosses through the θ-axis. This corresponds to the graph of the curve passing through the origin in the xy-plane, and our ﬁrst task is to determine when this happens. Setting r = 0 we get 2 + 4 cos(θ) = 0, or cos(θ) = − 1 2 . Solving for θ in [0, 2π] gives θ = 2π 3 . Since these values of θ are important geometrically, we break the interval [0, 2π] into six subintervals: 0, π 2 , 2π. As 3 , π, π, 4π 2 θ ranges from 0 to π 2 , r decreases from 6 to 2. Plotting this on the xy-plane, we start 6 units out from the origin on the positive x-axis and slowly pull in towards the positive y-axis. 3 and θ = 4π and 3π , 2π , 4π 3 , 3π , π 2 , 2π 3 3 2 r 6 4 2 −2 2π 3 4π 3 π 2 π 3π 2 2π θ y θ runs from 0 to π 2 x On the interval π will eventually cross through) the origin. Not only do we reach the origin when θ = 2π theorem from Calculus5 states that the curve hugs the line θ = 2π , r decreases from 2 to 0, which means the graph is heading into (and 3 , a 3 as it approaches the origin. 2 , 2π 3 5The ‘tangents at the pole’ theorem from second semester Calculus. 11.5 Graphs of Polar Equations 945 r 6 4 2 −2 y θ = 2π 3 2π 3 4π 3 π 2 π 3π 2 2π θ x 3 , π, r ranges from 0 to −2. Since r ≤ 0, the curve passes through the On the interval 2π origin in the xy-plane, following the line θ = 2π 3 and continues upwards through Quadrant IV towards the positive x-axis.6 Since \|r\| is increasing from 0 to 2, the curve pulls away from the origin to ﬁnish at a point on the positive x-axis. r 6 4 2 −2 y θ = 2π 3 2π 3 4π 3 π 2 π 3π 2 2π θ x Next, as θ progresses from π to 4π graph in the ﬁrst quadrant, heading into the origin along the line θ = 4π 3 . 3 , r ranges from −2 to 0. Since r ≤ 0, we continue our 6Recall that one way to visualize plotting polar coordinates (r, θ) with r < 0 is to start the rotation from the left 3 and π radians from the negative x-axis in side of the pole - in this case, the negative x-axis. Rotating between 2π this case determines the region between the line θ = 2π 3 and the x-axis in Quadrant IV. 946 Applications of Trigonometry r 6 4 2 −2 2π 3 4π 3 π 2 π 3π 2 2π θ y θ = 4π 3 x On the interval 4π line θ = 4π 3 , 3π 2 , r returns to positive values and increases from 0 to 2. We hug the 3 as we move through the origin and head towards the negative y-axis. r 6 4 2 −2 2π 3 4π 3 π 2 π 3π 2 2π θ y θ = 4π 3 x As we round out the interval, we ﬁnd that as θ runs through 3π to 6, and we end up back where we started, 6 units from the origin on the positive x-axis. 2 to 2π, r increases from 2 out y r 6 4 2 −2 2π 3 4π 3 π 2 π 3π 2 2π θ x θ runs from 3π 2 to 2π 11.5 Graphs of Polar Equations 947 Again, we invite the reader to show that plotting the curve for values of θ outside [0, 2π] results in retracing a portion of the curve already traced. Our ﬁnal graph is below. r 6 4 2 −2 2π 3 4π 3 π 2 π 3π 2 2π θ y θ = 2π 3 2 θ = 4π 3 −2 2 6 x r = 2 + 4 cos(θ) in the θr-plane r = 2 + 4 cos(θ) in the xy-plane 3. As usual, we start by graphing a fundamental cycle of r = 5 sin(2θ) in the θr-plane, which in this case, occurs as θ ranges from 0 to π. We partition our interval into subintervals to help us with the graphing, namely 0, π , π 4 , r 4 increases from 0 to 5. This means that the graph of r = 5 sin(2θ) in the xy-plane starts at the origin and gradually sweeps out so it is 5 units away from the origin on the line θ = π 4 . 4 , π. As θ ranges from 0 to π and 3π , π 2 , 3π 3π 4 π θ x r 5 −5 Next, we see that r decreases from 5 to 0 as θ runs through π heading negative as θ crosses π as the curve heads to the origin. 2 . Hence, we draw the curve hugging the line θ = π , and furthermore, r is 2 (the y-axis) 4 , π 2 948 Applications of Trigonometry y π 4 π 2 3π 4 π θ x r 5 −5 As θ runs from π pulls away from the negative y-axis into Quadrant IV. 2 to 3π 4 , r becomes negative and ranges from 0 to −5. Since r ≤ 0, the curve y π 4 π 2 3π 4 π θ x r 5 −5 For 3π 4 ≤ θ ≤ π, r increases from −5 to 0, so the curve pulls back to the origin. y π 4 π 2 3π 4 π θ x r 5 −5 11.5 Graphs of Polar Equations 949 Even though we have ﬁnished with one complete cycle of r = 5 sin(2θ), if we continue plotting beyond θ = π, we ﬁnd that the curve continues into the third quadrant! Below we present a graph of a second cycle of r = 5 sin(2θ) which continues on from the ﬁrst. The boxed labels on the θ-axis correspond to the portions with matching labels on the curve in the xy-plane. r 5 π 1 5π 4 2 3π 2 3 7π 4 4 2π θ 4 1 −5 We have the ﬁnal graph below 3π 4 π 5π 4 3π 2 7π 4 2π θ −5 x 5 −5 −5 r = 5 sin(2θ) in the θr-plane r = 5 sin(2θ) in the xy-plane 4. Graphing r2 = 16 cos(2θ) is complicated by the r2, so we solve to get r = ± 16 cos(2θ) = ±4 cos(2θ). How do we sketch such a curve? First oﬀ, we sketch a fundamental period cos(2θ) is of r = cos(2θ) which we have dotted in the ﬁgure below. When cos(2θ) < 0, . On the intervals which remain, undeﬁned, so we don’t have any values on the interval π cos(2θ) ranges from 0 to 1 as well.7 From cos(2θ) ranges from 0 to 1, inclusive. Hence, cos(2θ) ranges continuously from 0 to ±4, respectively. Below we this, we know r = ±4 cos(2θ) on the θr plane and use them to sketch the graph both r = 4 corresponding pieces of the curve r2 = 16 cos(2θ) in the xy-plane. As we have seen in earlier cos(2θ) and r = −4 4 , 3π 4 7Owing to the relationship between y = x and y = √ x over [0, 1], we also know cos(2θ) ≥ cos(2θ) wherever the former is deﬁned. 950 Applications of Trigonometry examples, the lines θ = π serve as guides for us to draw the curve as is passes through the origin. 4 , which are the zeros of the functions r = ±4 4 and θ = 3π cos(2θ), θ = 3π 4 π 4 π 2 3π 4 4 3 π θ r = 4 cos(2θ) and r = −4cos(2θ) As we plot points corresponding to values of θ outside of the interval [0, π], we ﬁnd ourselves retracing parts of the curve,8 so our ﬁnal answer is below. r 4 θ = 3π 3π 4 π θ −4 x 4 −4 r = ±4 cos(2θ) in the θr-plane −4 r2 = 16 cos(2θ) in the xy-plane A few remarks are in order. First, there is no relation, in general, between the period of the function f (θ) and the length of the interval required to sketch the complete graph of r = f (θ) in the xyplane. As we saw on page 941, despite the fact that the period of f (θ) = 6 cos(θ) is 2π, we sketched the complete graph of r = 6 cos(θ) in the xy-plane just using the values of θ as θ ranged from 0 to π. In Example 11.5.2, number 3, the period of f (θ) = 5 sin(2θ) is π, but in order to obtain the complete graph of r = 5 sin(2θ), we needed to run θ from 0 to 2π. While many of the ‘common’ polar graphs can be grouped into families,9 the authors truly feel that taking the time to work through each graph in the manner presented here is the best way to not only understand the polar 8In this case, we could have generated the entire graph by using just the plot r = 4cos(2θ), but graphed over the interval [0, 2π] in the θr-plane. We leave the details to the reader. 9Numbers 1 and 2 in Example 11.5.2 are examples of ‘lima¸cons,’ number 3 is an example of a ‘polar rose,’ and number 4 is the famous ‘Lemniscate of Bernoulli.’ 11.5 Graphs of Polar Equations 951 coordinate system, but also prepare you for what is needed in Calculus. Second, the symmetry seen in the examples is also a common occurrence when graphing polar equations. In addition to the usual kinds of symmetry discussed up to this point in the text (symmetry about each axis and the origin), it is possible to talk about rotational symmetry. We leave the discussion of symmetry to the Exercises. In our next example, we are given the task of ﬁnding the intersection points of polar curves. According to the Fundamental Graphing Principle for Polar Equations on page 938, in order for a point P to be on the graph of a polar equation, it must have a representation P (r, θ) which satisﬁes the equation. What complicates matters in polar coordinates is that any given point has inﬁnitely many representations. As a result, if a point P is on the graph of two diﬀerent polar equations, it is entirely possible that the representation P (r, θ) which satisﬁes one of the equations does not satisfy the other equation. Here, more than ever, we need to rely on the Geometry as much as the Algebra to ﬁnd our solutions. Example 11.5.3. Find the points of intersection of the graphs of the following polar equations. 1. r = 2 sin(θ) and r = 2 − 2 sin(θ) 2. r = 2 and r = 3 cos(θ) 3. r = 3 and r = 6 cos(2θ) Solution. 4. r = 3 sin θ 2 and r = 3 cos θ 2 1. Following the procedure in Example 11.5.2, we graph r = 2 sin(θ) and ﬁnd it to be a circle centered at the point with rectangular coordinates (0, 1) with a radius of 1. The graph of r = 2 − 2 sin(θ) is a special kind of lima¸con called a ‘cardioid.’10 y 2 −2 2 x −4 r = 2 − 2 sin(θ) and r = 2 sin(θ) It appears as if there are three intersection points: one in the ﬁrst quadrant, one in the second quadrant, and the origin. Our next task is to ﬁnd polar representations of these points. In 10Presumably, the name is derived from its resemblance to a stylized human heart. 952 Applications of Trigonometry 2 . From this, we get θ = π 6 into r = 2 sin(θ), we get r = 2 sin
π order for a point P to be on the graph of r = 2 sin(θ), it must have a representation P (r, θ) which satisﬁes r = 2 sin(θ). If P is also on the graph of r = 2−2 sin(θ), then P has a (possibly diﬀerent) representation P (r, θ) which satisﬁes r = 2 sin(θ). We ﬁrst try to see if we can ﬁnd any points which have a single representation P (r, θ) that satisﬁes both r = 2 sin(θ) and r = 2 − 2 sin(θ). Assuming such a pair (r, θ) exists, then equating11 the expressions for r gives 2 sin(θ) = 2 − 2 sin(θ) or sin(θ) = 1 6 + 2πk = 1, which for integers k. Plugging θ = π is also the value we obtain when we substitute it into r = 2 − 2 sin(θ). Hence, 1, π is one 6 representation for the point of intersection in the ﬁrst quadrant. For the point of intersection , so this is in the second quadrant, we try θ = 5π our answer here. What about the origin? We know from Section 11.4 that the pole may be represented as (0, θ) for any angle θ. On the graph of r = 2 sin(θ), we start at the origin when θ = 0 and return to it at θ = π, and as the reader can verify, we are at the origin exactly when θ = πk for integers k. On the curve r = 2 − 2 sin(θ), however, we reach the origin when θ = π 2 + 2πk for integers k. There is no integer value of k for which πk = π 2 + 2πk which means while the origin is on both graphs, the point is never reached simultaneously. In any case, we have determined the three points of intersection to be 1, π 6 6 . Both equations give us the point 1, 5π 2 , and more generally, when θ = π 6 + 2πk or θ = 5π = 2 1 2 and the origin. , 1, 5π 6 6 6 2. As before, we make a quick sketch of r = 2 and r = 3 cos(θ) to get feel for the number and location of the intersection points. The graph of r = 2 is a circle, centered at the origin, with a radius of 2. The graph of r = 3 cos(θ) is also a circle - but this one is centered at the point with rectangular coordinates 3 2 , 0 and has a radius of 3 2 . y 2 −2 2 3 x −2 r = 2 and r = 3 cos(θ) We have two intersection points to ﬁnd, one in Quadrant I and one in Quadrant IV. Proceeding as above, we ﬁrst determine if any of the intersection points P have a representation (r, θ) which satisﬁes both r = 2 and r = 3 cos(θ). Equating these two expressions for r, we get cos(θ) = 2 3 . To solve this equation, we need the arccosine function. We get 11We are really using the technique of substitution to solve the system of equations r = 2 sin(θ) r = 2 − 2 sin(θ) 11.5 Graphs of Polar Equations 953 θ = arccos 2 + 2πk or θ = 2π − arccos 2 3 3 2, arccos 2 as one representation for our answer in Quadrant I, and 2, 2π − arccos 2 3 3 as one representation for our answer in Quadrant IV. The reader is encouraged to check these results algebraically and geometrically. + 2πk for integers k. From these solutions, we get 3. Proceeding as above, we ﬁrst graph r = 3 and r = 6 cos(2θ) to get an idea of how many intersection points to expect and where they lie. The graph of r = 3 is a circle centered at the origin with a radius of 3 and the graph of r = 6 cos(2θ) is another four-leafed rose.12 y 6 3 −6 −3 3 x 6 −3 −6 r = 3 and r = 6 cos(2θ) , 3, 5π 6 , 3, 7π 6 and 3, 11π 6 6 + πk or θ = 5π 2 . Solving, we get θ = π It appears as if there are eight points of intersection - two in each quadrant. We ﬁrst look to see if there any points P (r, θ) with a representation that satisﬁes both r = 3 and r = 6 cos(2θ). For these points, 6 cos(2θ) = 3 or cos(2θ) = 1 6 + πk for integers k. Out of all of these solutions, we obtain just four distinct points represented by 3, π . To determine the coordinates of the remaining four 6 points, we have to consider how the representations of the points of intersection can diﬀer. We know from Section 11.4 that if (r, θ) and (r, θ) represent the same point and r = 0, then either r = r or r = −r. If r = r, then θ = θ+2πk, so one possibility is that an intersection point P has a representation (r, θ) which satisﬁes r = 3 and another representation (r, θ+2πk) for some integer, k which satisﬁes r = 6 cos(2θ). At this point,13 if we replace every occurrence of θ in the equation r = 6 cos(2θ) with (θ+2πk) and then see if, by equating the resulting expressions for r, we get any more solutions for θ. Since cos(2(θ + 2πk)) = cos(2θ + 4πk) = cos(2θ) for every integer k, however, the equation r = 6 cos(2(θ + 2πk)) reduces to the same equation we had before, r = 6 cos(2θ), which means we get no additional solutions. Moving on to the case where r = −r, we have that θ = θ + (2k + 1)π for integers k. We look to see if we can ﬁnd points P which have a representation (r, θ) that satisﬁes r = 3 and another, 12See Example 11.5.2 number 3. 13The authors have chosen to replace θ with θ + 2πk in the equation r = 6 cos(2θ) for illustration purposes only. We could have just as easily chosen to do this substitution in the equation r = 3. Since there is no θ in r = 3, however, this case would reduce to the previous case instantly. The reader is encouraged to follow this latter procedure in the interests of eﬃciency. 954 Applications of Trigonometry (−r, θ + (2k + 1)π), that satisﬁes r = 6 cos(2θ). To do this, we substitute14 (−r) for r and (θ + (2k + 1)π) for θ in the equation r = 6 cos(2θ) and get −r = 6 cos(2(θ + (2k + 1)π)). Since cos(2(θ + (2k + 1)π)) = cos(2θ + (2k + 1)(2π)) = cos(2θ) for all integers k, the equation −r = 6 cos(2(θ + (2k + 1)π)) reduces to −r = 6 cos(2θ), or r = −6 cos(2θ). Coupling this equation with r = 3 gives −6 cos(2θ) = 3 or cos(2θ) = − 1 3 + πk. From these solutions, we obtain15 the remaining four intersection points with representations −3, π 3 , which we can readily check graphically. 3 + πk or θ = 2π 2 . We get θ = π and −3, 5π 3 , −3, 2π 3 , −3, 4π 3 . Using the techniques 4. As usual, we begin by graphing r = 3 sin θ 2 presented in Example 11.5.2, we ﬁnd that we need to plot both functions as θ ranges from 0 to 4π to obtain the complete graph. To our surprise and/or delight, it appears as if these two equations describe the same curve! and r = 3 cos θ 2 y 3 −3 3 x −3 r = 3 sin θ 2 and r = 3 cos θ 2 appear to determine the same curve in the xy-plane and r = 3 cos θ 2 To verify this incredible claim,16 we need to show that, in fact, the graphs of these two equations intersect at all points on the plane. Suppose P has a representation (r, θ) which satisﬁes both r = 3 sin θ . Equating these two expressions for r gives 2 . While normally we discourage dividing by a variable = 3 cos θ the equation 3 sin θ 2 2 expression (in case it could be 0), we note here that if 3 cos θ = 0, then for our equation 2 = 0 as well. Since no angles have both cosine and sine equal to zero, to hold, 3 sin θ 2 to get we are safe to divide both sides of the equation 3 sin θ 2 tan θ 2 + 2πk for integers k. From these solutions, however, we 2 = 1 which gives θ = π by 3 cos θ 2 = 3 cos θ 2 14Again, we could have easily chosen to substitute these into r = 3 which would give −r = 3, or r = −3. 15We obtain these representations by substituting the values for θ into r = 6 cos(2θ), once again, for illustration purposes. Again, in the interests of eﬃciency, we could ‘plug’ these values for θ into r = 3 (where there is no θ) and represents the same point get the list of points: 3, π as −3, π , we still get the same set of solutions. . While it is not true that 3, π and 3, 5π , 3, 4π , 3, 2π 3 3 3 3 3 16A quick sketch of r = 3 sin θ and r = 3 cos θ in the θr-plane will convince you that, viewed as functions of r, 3 2 2 these are two diﬀerent animals. 11.5 Graphs of Polar Equations 955 √ 3 2 2 , π 2 2 + πk = 3 cos θ 2 [θ + 2πk] = 3 cos θ get only one intersection point which can be represented by . We now investigate other representations for the intersection points. Suppose P is an intersection point with a representation (r, θ) which satisﬁes r = 3 sin θ and the same point P has a diﬀerent 2 . Substituting representation (r, θ + 2πk) for some integer k which satisﬁes r = 3 cos θ 2 into the latter, we get r = 3 cos 1 2 + πk. Using the sum formula for , since sin (πk) = ±3 cos θ cosine, we expand 3 cos θ 2 sin(πk) = 0 for all integers k, and cos (πk) = ±1 for all integers k. If k is an even integer, we get the same equation r = 3 cos θ as before. If k is odd, we get r = −3 cos θ . This 2 2 = −1. Solving, latter expression for r leads to the equation 3 sin θ 2 √ we get θ = − π 2 , − π . Next, we assume P has a representation (r, θ) which satisﬁes r = 3 sin θ and a representation 2 for some integer k. Substituting (−r) for (−r, θ + (2k + 1)π) which satisﬁes r = 3 cos θ 2 r and (θ + (2k + 1)π) in for θ into r = 3 cos θ 2 [θ + (2k + 1)π]. Once gives −r = 3 cos 1 2 again, we use the sum formula for cosine to get 2 + 2πk for integers k, which gives the intersection point cos(πk) − 3 sin θ 2 , or tan θ 2 3 = −3 cos θ 2 2 2 2 cos 1 2 [θ + (2k + 1)π] = cos 2 θ 2 + (2k+1)π cos (2k+1)π 2 = cos θ 2 = ± sin θ 2 − sin θ 2 sin (2k+1)π 2 = 0 and sin (2k+1)π (2k+1)π where the last equality is true since cos 2 2 2 [θ + (2k + 1)π] can be rewritten as r = ±3 sin θ Hence, −r = 3 cos 1 (2k+1)π = 1, and the equation −r = 3 cos 1 = sin π then sin 2 2 reduces to −r = −3 sin θ , or r = 3 sin θ 2 2 What this means is that if a polar representation (r, θ) for the point P satisﬁes r = 3 sin( θ then the representation (−r, θ + π) for P automatically satisﬁes r = 3 cos θ 2 equations r = 3 sin( θ = ±1 for integers k. . If we choose k = 0, 2 [θ + (2k + 1)π] in this case which is the other equation under consideration! 2 ), . Hence the 2 ) determine the same set of points in the plane. 2 ) and r = 3 cos( θ 2 Our work in Example 11.5.3 justiﬁes the following. Guidelines for Finding Points of Intersection of Graphs of Polar Equations To ﬁnd the points of intersection of the graphs of two polar equations E1 and E2: Sketch the graphs of E1 and E2. Check to see if the curves intersect at the origin (pole). Solve for pairs (r, θ) which satisfy both E1 and E2. Substitute (θ + 2πk) for θ in either one of E1 or E2 (but not both) and solve for pairs (r,
θ) which satisfy both equations. Keep in mind that k is an integer. Substitute (−r) for r and (θ + (2k + 1)π) for θ in either one of E1 or E2 (but not both) and solve for pairs (r, θ) which satisfy both equations. Keep in mind that k is an integer. 956 Applications of Trigonometry Our last example ties together graphing and points of intersection to describe regions in the plane. Example 11.5.4. Sketch the region in the xy-plane described by the following sets. 1. (r, θ) \| 0 ≤ r ≤ 5 sin(2θ), 0 ≤ θ ≤ π 2 2. (r, θ) \| 3 ≤ r ≤ 6 cos(2θ), 0 ≤ θ ≤ π 6 3. (r, θ) \| 2 + 4 cos(θ) ≤ r ≤ 0, 2π 4. (r, θ) \| 0 ≤ r ≤ 2 sin(θ), 0 ≤ θ ≤ π 6 3 ≤ θ ≤ 4π 3 ∪ (r, θ) \| 0 ≤ r ≤ 2 − 2 sin(θ), π 6 ≤ θ ≤ π 2 Solution. Our ﬁrst step in these problems is to sketch the graphs of the polar equations involved to get a sense of the geometric situation. Since all of the equations in this example are found in either Example 11.5.2 or Example 11.5.3, most of the work is done for us. 1. We know from Example 11.5.2 number 3 that the graph of r = 5 sin(2θ) is a rose. Moreover, we know from our work there that as 0 ≤ θ ≤ π 2 , we are tracing out the ‘leaf’ of the rose which lies in the ﬁrst quadrant. The inequality 0 ≤ r ≤ 5 sin(2θ) means we want to capture all of the points between the origin (r = 0) and the curve r = 5 sin(2θ) as θ runs through 0, π 2 . Hence, the region we seek is the leaf itself. r 5 −5 π 4 π 2 3π 4 π θ y x (r, θ) \| 0 ≤ r ≤ 5 sin(2θ), 0 ≤ θ ≤ π 2 2. We know from Example 11.5.3 number 3 that r = 3 and r = 6 cos(2θ) intersect at θ = π 6 , so the region that is being described here is the set of points whose directed distance r from the origin is at least 3 but no more than 6 cos(2θ) as θ runs from 0 to π 6 . In other words, we are looking at the points outside or on the circle (since r ≥ 3) but inside or on the rose (since r ≤ 6 cos(2θ)). We shade the region below and r = 6 cos(2θ) (r, θ) \| 3 ≤ r ≤ 6 cos(2θ), 0 ≤ θ ≤ π 6 11.5 Graphs of Polar Equations 957 3. From Example 11.5.2 number 2, we know that the graph of r = 2 + 4 cos(θ) is a lima¸con whose ‘inner loop’ is traced out as θ runs through the given values 2π 3 . Since the values r takes on in this interval are non-positive, the inequality 2 + 4 cos(θ) ≤ r ≤ 0 makes sense, and we are looking for all of the points between the pole r = 0 and the lima¸con as θ ranges over the interval 2π r . In other words, we shade in the inner loop of the lima¸con. 3 to 4π 3 , 4π 3 6 4 2 −2 2π 3 4π 3 π 2 π 3π 2 2π θ y θ = 2π 3 θ = 4π 3 x (r, θ) \| 2 + 4 cos(θ) ≤ r ≤ 0, 2π 3 ≤ θ ≤ 4π 3 4. We have two regions described here connected with the union symbol ‘∪.’ We shade each in turn and ﬁnd our ﬁnal answer by combining the two. In Example 11.5.3, number 1, we found that the curves r = 2 sin(θ) and r = 2 − 2 sin(θ) intersect when θ = π 6 . Hence, for the , we are shading the region between the origin ﬁrst region, (r, θ) \| 0 ≤ r ≤ 2 sin(θ), 0 ≤ θ ≤ π 6 (r = 0) out to the circle (r = 2 sin(θ)) as θ ranges from 0 to π 6 , which is the angle of intersection of the two curves. For the second region, (r, θ) \| 0 ≤ r ≤ 2 − 2 sin(θ), π , θ picks up 6 ≤ θ ≤ π where it left oﬀ at π 2 . In this case, however, we are shading from the origin (r = 0) out to the cardioid r = 2 − 2 sin(θ) which pulls into the origin at θ = π 2 . Putting these two regions together gives us our ﬁnal answer. 6 and continues to sin(θ) and r = 2 sin(θ) (r, θ) \| 0 ≤ r ≤ 2 sin(θ), 0 ≤ θ ≤ π 6 (r, θ) \| 0 ≤ r ≤ 2 − 2 sin(θ), π ∪ 6 ≤ θ ≤ π 2 958 Applications of Trigonometry 11.5.1 Exercises In Exercises 1 - 20, plot the graph of the polar equation by hand. Carefully label your graphs. 1. Circle: r = 6 sin(θ) 2. Circle: r = 2 cos(θ) 3. Rose: r = 2 sin(2θ) 4. Rose: r = 4 cos(2θ) 5. Rose: r = 5 sin(3θ) 6. Rose: r = cos(5θ) 7. Rose: r = sin(4θ) 8. Rose: r = 3 cos(4θ) 9. Cardioid: r = 3 − 3 cos(θ) 10. Cardioid: r = 5 + 5 sin(θ) 11. Cardioid: r = 2 + 2 cos(θ) 12. Cardioid: r = 1 − sin(θ) 13. Lima¸con: r = 1 − 2 cos(θ) 14. Lima¸con: r = 1 − 2 sin(θ) 15. Lima¸con: r = 2 √ 3 + 4 cos(θ) 16. Lima¸con: r = 3 − 5 cos(θ) 17. Lima¸con: r = 3 − 5 sin(θ) 18. Lima¸con: r = 2 + 7 sin(θ) 19. Lemniscate: r2 = sin(2θ) 20. Lemniscate: r2 = 4 cos(2θ) In Exercises 21 - 30, ﬁnd the exact polar coordinates of the points of intersection of graphs of the polar equations. Remember to check for intersection at the pole (origin). 21. r = 3 cos(θ) and r = 1 + cos(θ) 22. r = 1 + sin(θ) and r = 1 − cos(θ) 23. r = 1 − 2 sin(θ) and r = 2 24. r = 1 − 2 cos(θ) and r = 1 25. r = 2 cos(θ) and r = 2 27. r2 = 4 cos(2θ) and r = 2 √ 3 sin(θ) √ 26. r = 3 cos(θ) and r = sin(θ) 28. r2 = 2 sin(2θ) and r = 1 29. r = 4 cos(2θ) and r = 2 30. r = 2 sin(2θ) and r = 1 In Exercises 31 - 40, sketch the region in the xy-plane described by the given set. 31. {(r, θ) \| 0 ≤ r ≤ 3, 0 ≤ θ ≤ 2π} 32. {(r, θ) \| 0 ≤ r ≤ 4 sin(θ), 0 ≤ θ ≤ π} 33. (r, θ) \| 0 ≤ r ≤ 3 cos(θ), − π 2 ≤ θ ≤ π 2 34. (r, θ) \| 0 ≤ r ≤ 2 sin(2θ), 0 ≤ θ ≤ π 2 11.5 Graphs of Polar Equations 959 35. (r, θ) \| 0 ≤ r ≤ 4 cos(2θ), − π 4 ≤ θ ≤ π 37. (r, θ) \| 1 + cos(θ) ≤ r ≤ 3 cos(θ), − π 4 3 ≤ θ ≤ π 3 36. (r, θ) \| 1 ≤ r ≤ 1 − 2 cos(θ), π 2 ≤ θ ≤ 3π 2 38. (r, θ) \| 1 ≤ r ≤ 2 sin(2θ), 13π √ 12 ≤ θ ≤ 17π 12 39. (r, θ) \| 0 ≤ r ≤ 2 ∪ (r, θ) \| 0 ≤ r ≤ 2 cos(θ), π 3 sin(θ), 0 ≤ θ ≤ π 6 ∪ (r, θ) \| 0 ≤ r ≤ 1, π 40. (r, θ) \| 0 ≤ r ≤ 2 sin(2θ), 0 ≤ θ ≤ π 12 12 ≤ In Exercises 41 - 50, use set-builder notation to describe the polar region. Assume that the region contains its bounding curves. 41. The region inside the circle r = 5. 42. The region inside the circle r = 5 which lies in Quadrant III. 43. The region inside the left half of the circle r = 6 sin(θ). 44. The region inside the circle r = 4 cos(θ) which lies in Quadrant IV. 45. The region inside the top half of the cardioid r = 3 − 3 cos(θ) 46. The region inside the cardioid r = 2 − 2 sin(θ) which lies in Quadrants I and IV. 47. The inside of the petal of the rose r = 3 cos(4θ) which lies on the positive x-axis 48. The region inside the circle r = 5 but outside the circle r = 3. 49. The region which lies inside of the circle r = 3 cos(θ) but outside of the circle r = sin(θ) 50. The region in Quadrant I which lies inside both the circle r = 3 as well as the rose r = 6 sin(2θ) While the authors truly believe that graphing polar curves by hand is fundamental to your understanding of the polar coordinate system, we would be derelict in our duties if we totally ignored the graphing calculator. Indeed, there are some important polar curves which are simply too difﬁcult to graph by hand and that makes the calculator an important tool for your further studies in Mathematics, Science and Engineering. We now give a brief demonstration of how to use the graphing calculator to plot polar curves. The ﬁrst thing you must do is switch the MODE of your calculator to POL, which stands for “polar”. 960 Applications of Trigonometry This changes the “Y=” menu as seen above in the middle. Let’s plot the polar rose given by r = 3 cos(4θ) from Exercise 8 above. We type the function into the “r=” menu as seen above on the right. We need to set the viewing window so that the curve displays properly, but when we look at the WINDOW menu, we ﬁnd three extra lines. In order for the calculator to be able to plot r = 3 cos(4θ) in the xy-plane, we need to tell it not only the dimensions which x and y will assume, but we also what values of θ to use. From our previous work, we know that we need 0 ≤ θ ≤ 2π, so we enter the data you see above. (I’ll say more about the θ-step in just a moment.) Hitting GRAPH yields the curve below on the left which doesn’t look quite right. The issue here is that the calculator screen is 96 pixels wide but only 64 pixels tall. To get a true geometric perspective, we need to hit ZOOM SQUARE (seen below in the middle) to produce a more accurate graph which we present below on the right. In function mode, the calculator automatically divided the interval [Xmin, Xmax] into 96 equal subintervals. In polar mode, however, we must specify how to split up the interval [θmin, θmax] using the θstep. For most graphs, a θstep of 0.1 is ﬁne. If you make it too small then the calculator takes a long time to graph. It you make it too big, you get chunky garbage like this. You will need to experiment with the settings in order to get a nice graph. Exercises 51 - 60 give you some curves to graph using your calculator. Notice that some of them have explicit bounds on θ and others do not. 11.5 Graphs of Polar Equations 961 51. r = θ, 0 ≤ θ ≤ 12π 52. r = ln(θ), 1 ≤ θ ≤ 12π 53. r = e.1θ, 0 ≤ θ ≤ 12π 54. r = θ3 − θ, −1.2 ≤ θ ≤ 1.2 55. r = sin(5θ) − 3 cos(θ) 57. r = arctan(θ), −π ≤ θ ≤ π 59. r = 1 2 − cos(θ) 56. r = sin3 θ 2 + cos2 θ 3 58. r = 1 1 − cos(θ) 60. r = 1 2 − 3 cos(θ) 61. How many petals does the polar rose r = sin(2θ) have? What about r = sin(3θ), r = sin(4θ) and r = sin(5θ)? With the help of your classmates, make a conjecture as to how many petals the polar rose r = sin(nθ) has for any natural number n. Replace sine with cosine and repeat the investigation. How many petals does r = cos(nθ) have for each natural number n? Looking back through the graphs in the section, it’s clear that many polar curves enjoy various forms of symmetry. However, classifying symmetry for polar curves is not as straight-forward as it was for equations back on page 26. In Exercises 62 - 64, we have you and your classmates explore some of the more basic forms of symmetry seen in common polar curves. 62. Show that if f is even17 then the graph of r = f (θ) is symmetric about the x-axis. (a) Show that f (θ) = 2 + 4 cos(θ) is even and verify that the graph of r = 2 + 4 cos(θ) is indeed symmetric about the x-axis. (See Example 11.5.2 number 2.) (b) Show that f (θ) = 3 sin θ 2 is not even, yet the graph of r = 3 sin θ 2 is symmetric about the x-axis. (See Example 11.5.3 number 4.) 63. Show that if f is odd18 then the graph of r = f (θ) is symmetric about the origin. (a) Show that f (θ) = 5 sin(2θ) is odd and verify that the graph of r = 5 s
in(2θ) is indeed symmetric about the origin. (See Example 11.5.2 number 3.) (b) Show that f (θ) = 3 cos θ 2 is not odd, yet the graph of r = 3 cos θ 2 is symmetric about the origin. (See Example 11.5.3 number 4.) 64. Show that if f (π − θ) = f (θ) for all θ in the domain of f then the graph of r = f (θ) is symmetric about the y-axis. (a) For f (θ) = 4 − 2 sin(θ), show that f (π − θ) = f (θ) and the graph of r = 4 − 2 sin(θ) is symmetric about the y-axis, as required. (See Example 11.5.2 number 1.) 17Recall that this means f (−θ) = f (θ) for θ in the domain of f . 18Recall that this means f (−θ) = −f (θ) for θ in the domain of f . 962 Applications of Trigonometry (b) For f (θ) = 5 sin(2θ), show that f π − π 4 = f π 4 , yet the graph of r = 5 sin(2θ) is symmetric about the y-axis. (See Example 11.5.2 number 3.) In Section 1.7, we discussed transformations of graphs. classmates explore transformations of polar graphs. In Exercise 65 we have you and your 65. For Exercises 65a and 65b below, let f (θ) = cos(θ) and g(θ) = 2 − sin(θ). (a) Using your graphing calculator, compare the graph of r = f (θ) to each of the graphs of . Repeat this process for g(θ). In general, how do you think the graph of r = f (θ + α) compares with the graph of r = f (θ)? and r = f θ − 3π 4 , r = f θ + 3π 4 (b) Using your graphing calculator, compare the graph of r = f (θ) to each of the graphs of r = 2f (θ), r = 1 2 f (θ), r = −f (θ) and r = −3f (θ). Repeat this process for g(θ). In general, how do you think the graph of r = k · f (θ) compares with the graph of r = f (θ)? (Does it matter if k > 0 or k < 0?) 66. In light of Exercises 62 - 64, how would the graph of r = f (−θ) compare with the graph of r = f (θ) for a generic function f ? What about the graphs of r = −f (θ) and r = f (θ)? What about r = f (θ) and r = f (π − θ)? Test out your conjectures using a variety of polar functions found in this section with the help of a graphing utility. 67. With the help of your classmates, research cardioid microphones. 68. Back in Section 1.2, in the paragraph before Exercise 53, we gave you this link to a fascinating list of curves. Some of these curves have polar representations which we invite you and your classmates to research. 11.5 Graphs of Polar Equations 963 11.5.2 Answers 1. Circle: r = 6 sin(θ) y −6 6 −6 3. Rose: r = 2 sin(2θ) y −2 2 −2 2. Circle: r = 2 cos(θ) y 2 6 x −2 2 x −2 4. Rose: r = 4 cos(2θ) y 4 θ = 3π 4 θ = π 4 2 x −4 4 x 5. Rose: r = 5 sin(3θ) y 5 θ = 2π 3 θ = π 3 −5 5 x −4 6. Rose: r = cos(5θ) y 1 θ = 7π 10 θ = 3π 10 θ = 9π 10 −1 θ = π 10 1 x −5 −1 964 Applications of Trigonometry 7. Rose: r = sin(4θ) y 1 θ = 3π 4 θ = π 4 −1 1 x 8. Rose: r = 3 cos(4θ) y θ = 5π 8 3 θ = 3π 8 θ = 7π 8 −3 θ = π 8 3 x −1 −3 9. Cardioid: r = 3 − 3 cos(θ) 10. Cardioid: r = 5 + 5 sin(θ) y 6 3 y 10 5 −6 −3 3 6 x −10 −5 5 10 x −3 −6 −5 −10 11. Cardioid: r = 2 + 2 cos(θ) 12. Cardioid: r = 1 − sin(θ) y 4 2 y 2 1 −4 −2 2 4 x −2 −1 1 2 x −2 −4 −1 −2 11.5 Graphs of Polar Equations 965 13. Lima¸con: r = 1 − 2 cos(θ) 14. Lima¸con: r = 1 − 2 sin(θ = 5π 6 −3 −1 1 3 x −3 −1 1 3 x −1 −3 θ = 5π 3 15. Lima¸con: r = 2 √ 3 + 4 cos(θ) y √ 2 3 + 4 θ = 5π 6 √ 2 3 √ −2 3 − 4 θ = 7π 6 √ −2 3 √ −1 −3 16. Lima¸con: r = 3 − 5 cos(θ) y 8 3 θ = arccos 3 5 −8 −2 8 x −3 −8 θ = 2π − arccos 3 5 17. Lima¸con: r = 3 − 5 sin(θ) 18. Lima¸con: r = 2 + 7 sin(θ) y 8 θ = π − arcsin 3 5 θ = arcsin 3 5 y 9 5 −8 −3 3 −2 −8 8 x −9 θ = π + arcsin −2 2 7 2 9 θ = 2π − arcsin 2 7 x −9 966 Applications of Trigonometry 19. Lemniscate: r2 = sin(2θ) 20. Lemniscate: r2 = 4 cos(2θ) y 1 y 2 θ = 3π 4 θ = π 4 −1 1 x −2 2 x −1 −2 3 2 , π 3 , 3 2 , 5π 3 , pole √ 2 + 2 , √ 2 − 2 2 , 7π 4 2 , 3π 4 , pole 21. r = 3 cos(θ) and r = 1 + cos(θ) y 3 2 1 −3 −2 −1 1 2 3 −1 −2 −3 x 22. r = 1 + sin(θ) and r = 1 − cos(θ) y 2 1 −2 −1 1 2 x −1 −2 11.5 Graphs of Polar Equations 967 2, 7π 6 , 2, 11π 6 1, , π 2 1, 3π 2 , (−1, 0) √ 3, π 6 , pole 23. r = 1 − 2 sin(θ) and r = 2 y 3 1 −3 −1 1 3 x −1 −3 24. r = 1 − 2 cos(θ) and r = 1 y 3 1 −3 −1 1 3 x −1 −3 25. r = 2 cos(θ) and r = 2 √ 3 sin(θ) y 4 3 2 1 −3 −2 −1 1 2 3 x −1 −2 −3 −4 968 Applications of Trigonometry √ 3 10 10 , arctan(3) , pole √ 2, , π 6 √ , 5π 6 2, √ , 7π 6 2, √ 2, 11π 6 1, , π 12 1, , 5π 12 1, 13π 12 , 1, 17π 12 26. r = 3 cos(θ) and r = sin(θ) y 3 2 1 −3 −2 −1 1 2 3 x −1 −2 −3 27. r2 = 4 cos(2θ) and r = √ 2 y 2 −2 2 x −2 28. r2 = 2 sin(2θ) and r = 1 y √ 2 1 −1 11.5 Graphs of Polar Equations 969 29. r = 4 cos(2θ) and r = 2 y 4 2, 2, , π 6 11π 6 , 7π 6 2, 2, 5π 6 , −2, , π 3 , −2, , 2π 3 −2, 4π 3 , −2, 5π 3 −4 4 x −4 30. r = 2 sin(2θ) and r = 1 y 2 −2 2 x −2 , π 12 , 17π 12 1, 1, 1, −1, , 19π 12 , 5π 12 1, , 7π 12 −1, −1, 23π 12 , 13π 12 −1, , 11π 12 970 Applications of Trigonometry 31. {(r, θ) \| 0 ≤ r ≤ 3, 0 ≤ θ ≤ 2π} 32. {(r, θ) \| 0 ≤ r ≤ 4 sin(θ), 0 ≤ θ ≤ π3 −2 −1 1 2 3 −1 −2 −3 x x −4 −3 −2 −1 1 2 3 4 −1 −2 −3 −4 33. (r, θ) \| 0 ≤ r ≤ 3 cos(θ), − π 2 ≤ θ ≤ π 2 34. (r, θ) \| 0 ≤ r ≤ 2 sin(2θ), 3 −2 −1 1 2 3 −1 −2 −3 x −2 2 x −2 35. (r, θ) \| 0 ≤ r ≤ 4 cos(2θ), − π 4 ≤ θ ≤ π 4 36. (r, θ) \| 1 ≤ r ≤ 1 − 2 cos(θ), π 2 ≤ θ ≤ 3π 2 y 4 y 3 1 −4 4 x −3 −1 1 3 x −4 −1 −3 11.5 Graphs of Polar Equations 971 37. (r, θ) \| 1 + cos(θ) ≤ r ≤ 3 cos(θ), − 3 −2 −1 1 2 3 −1 −2 −3 x 38. (r, θ) \| 1 ≤ r ≤ y 2 sin(2θ), 13π 12 ≤ θ ≤ 17π 12 √ − 2 −1 √ 2 1 −r, θ) \| 0 ≤ r ≤ 2 cos(θ), π 6 ≤ θ ≤ π 2 39. (r, θ) \| 0 ≤ r ≤ 2 3 sin(θ), 3 −2 −1 1 2 3 x −1 −2 −3 −4 972 Applications of Trigonometry ∪ (r, θ) \| 0 ≤ r ≤ 1, π 12 ≤ θ ≤ π 4 40. (r, θ) \| 0 ≤ r ≤ 2 sin(2θ), 0 ≤ θ ≤ π 12 y −2 2 −2 2 x 41. {(r, θ) \| 0 ≤ r ≤ 5, 0 ≤ θ ≤ 2π} 42. (r, θ) \| 0 ≤ r ≤ 5, π ≤ θ ≤ 3π 2 43. (r, θ) \| 0 ≤ r ≤ 6 sin(θ), π 44. (r, θ) \| 4 cos(θ) ≤ r ≤ 0 45. {(r, θ) \| 0 ≤ r ≤ 3 − 3 cos(θ), 0 ≤ θ ≤ π} 46. (r, θ) \| 0 ≤ r ≤ 2 − 2 sin(θ), 0 ≤ θ ≤ π 2 or (r, θ) \| 0 ≤ r ≤ 2 − 2 sin(θ), 3π 47. (r, θ) \| 0 ≤ r ≤ 3 cos(4θ), 0 ≤ θ ≤ π 8 or (r, θ) \| 0 ≤ r ≤ 3 cos(4θ), − ≤ 5π 2 ∪ (r, θ) \| 0 ≤ r ≤ 2 − 2 sin(θ), 3π 2 ≤ θ ≤ 2π ∪ (r, θ) \| 0 ≤ r ≤ 3 cos(4θ), 15π 8 ≤ θ ≤ 2π 48. {(r, θ) \| 3 ≤ r ≤ 5, 0 ≤ θ ≤ 2π} 49. (r, θ) \| 0 ≤ r ≤ 3 cos(θ), − π 50. (r, θ) \| 0 ≤ r ≤ 6 sin(2θ), 0 ≤ θ ≤ π 12 12 ≤ θ ≤ π (r, θ) \| 0 ≤ r ≤ 6 sin(2θ), 5π 2 2 ≤ θ ≤ 0 ∪ {(r, θ) \| sin(θ) ≤ r ≤ 3 cos(θ), 0 ≤ θ ≤ arctan(3)} ∪ (r, θ) \| 0 ≤ r ≤ 3, π 12 ≤ θ ≤ 5π 12 ∪ 11.6 Hooked on Conics Again 973 11.6 Hooked on Conics Again In this section, we revisit our friends the Conic Sections which we began studying in Chapter 7. Our ﬁrst task is to formalize the notion of rotating axes so this subsection is actually a follow-up to Example 8.3.3 in Section 8.3. In that example, we saw that the graph of y = 2 x is actually a hyperbola. More speciﬁcally, it is the hyperbola obtained by rotating the graph of x2 − y2 = 4 counter-clockwise through a 45◦ angle. Armed with polar coordinates, we can generalize the process of rotating axes as shown below. 11.6.1 Rotation of Axes Consider the x- and y-axes below along with the dashed x- and y-axes obtained by rotating the xand y-axes counter-clockwise through an angle θ and consider the point P (x, y). The coordinates (x, y) are rectangular coordinates and are based on the x- and y-axes. Suppose we wished to ﬁnd rectangular coordinates based on the x- and y-axes. That is, we wish to determine P (x, y). While this seems like a formidable challenge, it is nearly trivial if we use polar coordinates. Consider the angle φ whose initial side is the positive x-axis and whose terminal side contains the point P . y y P (x, y) = P (x, y) x θ φ θ x We relate P (x, y) and P (x, y) by converting them to polar coordinates. Converting P (x, y) to polar coordinates with r > 0 yields x = r cos(θ + φ) and y = r sin(θ + φ). To convert the point P (x, y) into polar coordinates, we ﬁrst match the polar axis with the positive x-axis, choose the same r > 0 (since the origin is the same in both systems) and get x = r cos(φ) and y = r sin(φ). Using the sum formulas for sine and cosine, we have x = r cos(θ + φ) = r cos(θ) cos(φ) − r sin(θ) sin(φ) = (r cos(φ)) cos(θ) − (r sin(φ)) sin(θ) = x cos(θ) − y sin(θ) Sum formula for cosine Since x = r cos(φ) and y = r sin(φ) 974 Applications of Trigonometry Similarly, using the sum formula for sine we get y = x sin(θ) + y cos(θ). These equations enable us to easily convert points with xy-coordinates back into xy-coordinates. They also enable us to easily convert equations in the variables x and y into equations in the variables in terms of x and y.1 If we want equations which enable us to convert points with xy-coordinates into xy-coordinates, we need to solve the system x cos(θ) − y sin(θ) = x x sin(θ) + y cos(θ) = y for x and y. Perhaps the cleanest way2 to solve this system is to write it as a matrix equation. Using the machinery developed in Section 8.4, we write the above system as the matrix equation AX = X where A = cos(θ) − sin(θ) cos(θ) sin(θ Since det(A) = (cos(θ))(cos(θ)) − (− sin(θ))(sin(θ)) = cos2(θ) + sin2(θ) = 1, the determinant of A is not zero so A is invertible and X = A−1X. Using the formula given in Equation 8.2 with det(A) = 1, we ﬁnd so that A−1 = cos(θ) − sin(θ) sin(θ) cos(θ) X = A−1X x y x y = = cos(θ) − sin(θ) sin(θ) cos(θ) x y x cos(θ) + y sin(θ) −x sin(θ) + y cos(θ) From which we get x = x cos(θ) + y sin(θ) and y = −x sin(θ) + y cos(θ). To summarize, Theorem 11.9. Rotation of Axes: Suppose the positive x and y axes are rotated counterclockwise through an angle θ to produce the axes x and y, respectively. Then the coordinates P (x, y) and P (x, y) are related by the following systems of equations x = x cos(θ) − y sin(θ) y = x sin(θ) + y cos(θ) and x = x cos(θ) + y sin(θ) y = −x sin(θ) + y cos(θ) We put the formulas in Theorem 11.9 to good use in the following example. 1Sound familiar? In Section 11.4, the equations x = r cos(θ) and y = r sin(θ) make it easy to convert points from polar coordinates into rectangular coordinates, and they make it easy to convert equations from rectangular coordinates into polar coordinates. 2We could, of course, interchange the roles of x and x, y and y and replace φ with −
φ to get x and y in terms of x and y, but that seems like cheating. The matrix A introduced here is revisited in the Exercises. 11.6 Hooked on Conics Again 975 Example 11.6.1. Suppose the x- and y- axes are both rotated counter-clockwise through the angle θ = π 3 to produce the x- and y- axes, respectively. 1. Let P (x, y) = (2, −4) and ﬁnd P (x, y). Check your answer algebraically and graphically. 2. Convert the equation 21x2 + 10xy √ 3 + 31y2 = 144 to an equation in x and y and graph. Solution. √ 3 , Theorem 11.9 gives x = x cos(θ) + y sin(θ) = 2 cos π √ 1. If P (x, y) = (2, −4) then x = 2 and y = −4. Using these values for x and y along with which simpliﬁes 3. Similarly, y = −x sin(θ) + y cos(θ) = (−2) sin π + (−4) cos π which √ 3 3 3. To check our answer 3. Hence P (x, y to x = 1 − 2 √ gives y = − algebraically, we use the formulas in Theorem 11.9 to convert P (x, y) = 1 − 2 back into x and y coordinates. We get + (−4) sin π 3 3 − 2 = −2 − 3, −2 − 3, −2 − √ √ 3 x = x cos(θ) − y sin(θ) √ − (−2 − 3) cos ) sin π 3 = ( Similarly, using y = x sin(θ) + y cos(θ), we obtain y = −4 as required. To check our answer graphically, we sketch in the x-axis and y-axis to see if the new coordinates P (x, y) = 3 ≈ (−2.46, −3.73) seem reasonable. Our graph is below. 1 − 2 3, −x, y) = (2, −4) P (x, y) ≈ (−2.46, −3.73) 2. To convert the equation 21x2 +10xy we substitute x = x cos π 3 − y sin π 3 √ 3+31y2 = 144 to an equation in the variables x and y, = x 2 + y and y = x sin π 3 + y cos 976 Applications of Trigonometry and simplify. While this is by no means a trivial task, it is nothing more than a hefty dose of Beginning Algebra. We will not go through the entire computation, but rather, the reader should take the time to do it. Start by verifying that x2 = (x)2 4 − √ xy 2 3 + 3(y)2 4 , √ (x)2 4 3 − xy 2 − xy = √ 3 , y2 = 3(x)2 4 + √ xy 2 3 + (y)2 4 To our surprise and delight, the equation 21x2 + 10xy 3 + 31y2 = 144 in xy-coordinates 4 + (y)2 reduces to 36(x)2 + 16(y)2 = 144, or (x)2 9 = 1 in xy-coordinates. The latter is an ellipse centered at (0, 0) with vertices along the y-axis with (xy-coordinates) (0, ±3) and whose minor axis has endpoints with (xy-coordinates) (±2, 0). We graph it below. (y) 21x2 + 10xy √ 3 + 31y2 = 144 √ The elimination of the troublesome ‘xy’ term from the equation 21x2 + 10xy 3 + 31y2 = 144 in Example 11.6.1 number 2 allowed us to graph the equation by hand using what we learned in Chapter 7. It is natural to wonder if we can always do this. That is, given an equation of the form Ax2 +Bxy +Cy2 +Dx+Ey +F = 0, with B = 0, is there an angle θ so that if we rotate the x and yaxes counter-clockwise through that angle θ, the equation in the rotated variables x and y contains no xy term? To explore this conjecture, we make the usual substitutions x = x cos(θ) − y sin(θ) and y = x sin(θ) + y cos(θ) into the equation Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 and set the coeﬃcient of the xy term equal to 0. Terms containing xy in this expression will come from the ﬁrst three terms of the equation: Ax2, Bxy and Cy2. We leave it to the reader to verify that x2 = (x)2 cos2(θ) − 2xy cos(θ) sin(θ) + (y)2 sin(θ) xy = (x)2 cos(θ) sin(θ) + xy cos2(θ) − sin2(θ) − (y)2 cos(θ) sin(θ) y2 = (x)2 sin2(θ) + 2xy cos(θ) sin(θ) + (y)2 cos2(θ) 11.6 Hooked on Conics Again 977 The contribution to the xy-term from Ax2 is −2A cos(θ) sin(θ), from Bxy it is B cos2(θ) − sin2(θ), and from Cy2 it is 2C cos(θ) sin(θ). Equating the xy-term to 0, we get −2A cos(θ) sin(θ) + B cos2(θ) − sin2(θ) + 2C cos(θ) sin(θ) = 0 −A sin(2θ) + B cos(2θ) + C sin(2θ) = 0 Double Angle Identities From this, we get B cos(2θ) = (A − C) sin(2θ), and our goal is to solve for θ in terms of the coeﬃcients A, B and C. Since we are assuming B = 0, we can divide both sides of this equation by B. To solve for θ we would like to divide both sides of the equation by sin(2θ), provided of course that we have assurances that sin(2θ) = 0. If sin(2θ) = 0, then we would have B cos(2θ) = 0, and since B = 0, this would force cos(2θ) = 0. Since no angle θ can have both sin(2θ) = 0 and cos(2θ) = 0, we can safely assume3 sin(2θ) = 0. We get cos(2θ) B . We have just proved the following theorem. B , or cot(2θ) = A−C sin(2θ) = A−C Theorem 11.10. The equation Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 with B = 0 can be transformed into an equation in variables x and y without any xy terms by rotating the xand y- axes counter-clockwise through an angle θ which satisﬁes cot(2θ) = A−C B . We put Theorem 11.10 to good use in the following example. Example 11.6.2. Graph the following equations. 1. 5x2 + 26xy + 5y2 − 16x √ 2 + 16y √ 2 − 104 = 0 2. 16x2 + 24xy + 9y2 + 15x − 20y = 0 Solution. 1. Since the equation 5x2 + 26xy + 5y2 − 16x 2 − 104 = 0 is already given to us in the form required by Theorem 11.10, we identify A = 5, B = 26 and C = 5 so that cot(2θ) = A−C 26 = 0. This means cot(2θ) = 0 which gives θ = π 2 k for integers k. √ √ 2 + y 2 2 We choose θ = π 2 . The reader should verify that 4 so that our rotation equations are x = x 4 + π and y = x B = 5−5 2 − y 2 + 16y √ √ 2 2 2 √ √ x2 = (x)2 2 − xy + (y)2 2 , xy = (x)2 2 − (y)2 2 , y2 = (x)2 2 + xy + (y)2 2 Making the other substitutions, we get that 5x2 + 26xy + 5y2 − 16x 2 − 104 = 0 2 + 16y reduces to 18(x)2 − 8(y)2 + 32y − 104 = 0, or (x)2 9 = 1. The latter is the equation of a hyperbola centered at the xy-coordinates (0, 2) opening in the x direction with vertices (±2, 2) (in xy-coordinates) and asymptotes y = ± 3 4 − (y−2)2 2 x + 2. We graph it below. √ √ 3The reader is invited to think about the case sin(2θ) = 0 geometrically. What happens to the axes in this case? 978 Applications of Trigonometry 24 2. From 16x2 + 24xy + 9y2 + 15x − 20y = 0, we get A = 16, B = 24 and C = 9 so that cot(2θ) = 7 24 . Since this isn’t one of the values of the common angles, we will need to use inverse functions. Ultimately, we need to ﬁnd cos(θ) and sin(θ), which means we have two If we use the arccotangent function immediately, after the usual calculations we options. . To get cos(θ) and sin(θ) from this, we would need to use half angle 2 arccot 7 get θ = 1 identities. Alternatively, we can start with cot(2θ) = 7 24 , use a double angle identity, and then go after cos(θ) and sin(θ). We adopt the second approach. From cot(2θ) = 7 24 , we have tan(2θ) = 24 7 , which 7 . Using the double angle identity for tangent, we have gives 24 tan2(θ) + 14 tan(θ) − 24 = 0. Factoring, we get 2(3 tan(θ) + 4)(4 tan(θ) − 3) = 0 which gives tan(θ) = − 4 4 . While either of these values of tan(θ) satisﬁes the equation . To 4 , since this produces an acute angle,4 θ = arctan 3 cot(2θ) = 7 . ﬁnd the rotation equations, we need cos(θ) = cos arctan 3 4 Using the techniques developed in Section 10.6 we get cos(θ) = 4 5 . Our rotation 5 + 4y 5 − 3y equations are x = x cos(θ) − y sin(θ) = 4x 5 . As usual, we now substitute these quantities into 16x2 + 24xy + 9y2 + 15x − 20y = 0 and simplify. As a ﬁrst step, the reader can verify 3 or tan(θ) = 3 24 , we choose tan(θ) = 3 5 and y = x sin(θ) + y cos(θ) = 3x and sin(θ) = sin arctan 3 4 5 and sin(θ) = 3 1−tan2(θ) = 24 2 tan(θ) 4 x2 = 16(x)2 25 − 24xy 25 + 9(y)2 25 , xy = 12(x)2 25 + 7xy 25 − 12(y)2 25 , y2 = 9(x)2 25 + 24xy 25 + 16(y)2 25 Once the dust settles, we get 25(x)2 − 25y = 0, or y = (x)2, whose graph is a parabola opening along the positive y-axis with vertex (0, 0). We graph this equation below = arctan 3 4 x 5x2 + 26xy + 5y2 − 16x √ 2 + 16y √ 2 − 104 = 0 16x2 + 24xy + 9y2 + 15x − 20y = 0 4As usual, there are inﬁnitely many solutions to tan(θ) = 3 . The 4 reader is encouraged to think about why there is always at least one acute answer to cot(2θ) = A−C B and what this means geometrically in terms of what we are trying to accomplish by rotating the axes. The reader is also encouraged to keep a sharp lookout for the angles which satisfy tan(θ) = − 4 4 . We choose the acute angle θ = arctan 3 3 in our ﬁnal graph. (Hint: 3 = −1.) − 4 3 4 11.6 Hooked on Conics Again 979 We note that even though the coeﬃcients of x2 and y2 were both positive numbers in parts 1 and 2 of Example 11.6.2, the graph in part 1 turned out to be a hyperbola and the graph in part 2 worked out to be a parabola. Whereas in Chapter 7, we could easily pick out which conic section we were dealing with based on the presence (or absence) of quadratic terms and their coeﬃcients, Example 11.6.2 demonstrates that all bets are oﬀ when it comes to conics with an xy term which require rotation of axes to put them into a more standard form. Nevertheless, it is possible to determine which conic section we have by looking at a special, familiar combination of the coeﬃcients of the quadratic terms. We have the following theorem. Theorem 11.11. Suppose the equation Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 describes a non-degenerate conic section.a If B2 − 4AC > 0 then the graph of the equation is a hyperbola. If B2 − 4AC = 0 then the graph of the equation is a parabola. If B2 − 4AC < 0 then the graph of the equation is an ellipse or circle. aRecall that this means its graph is either a circle, parabola, ellipse or hyperbola. See page 497. As you may expect, the quantity B2 −4AC mentioned in Theorem 11.11 is called the discriminant of the conic section. While we will not attempt to explain the deep Mathematics which produces this ‘coincidence’, we will at least work through the proof of Theorem 11.11 mechanically to show that it is true.5 First note that if the coeﬃcient B = 0 in the equation Ax2 +Bxy +Cy2 +Dx+Ey +F = 0, Theorem 11.11 reduces to the result presented in Exercise 34 in Section 7.5, so we proceed here under the assumption that B = 0. We rotate the xy-axes counter-clockwise through an angle θ which satisﬁes cot(2θ) = A−C to produce an equation with no xy-term in accordance with B Theorem 11.10: A(x)2 + C(y)2 + Dx + Ey + F = 0. In this form, we can invoke Exercise 34 in Section 7.5 once more using the product AC. Our goal is to ﬁnd the product AC in terms of the coeﬃcients A, B
and C in the original equation. To that end, we make the usual substitutions x = x cos(θ) − y sin(θ) y = x sin(θ) + y cos(θ) into Ax2 + Bxy + Cy2 + Dx + Ey + F = 0. We leave it to the reader to show that, after gathering like terms, the coeﬃcient A on (x)2 and the coeﬃcient C on (y)2 are A = A cos2(θ) + B cos(θ) sin(θ) + C sin2(θ) C = A sin2(θ) − B cos(θ) sin(θ) + C cos2(θ) In order to make use of the condition cot(2θ) = A−C the power reduction formulas. After some regrouping, we get B , we rewrite our formulas for A and C using 2A = [(A + C) + (A − C) cos(2θ)] + B sin(2θ) 2C = [(A + C) − (A − C) cos(2θ)] − B sin(2θ) Next, we try to make sense of the product (2A)(2C) = {[(A + C) + (A − C) cos(2θ)] + B sin(2θ)} {[(A + C) − (A − C) cos(2θ)] − B sin(2θ)} 5We hope that someday you get to see why this works the way it does. 980 Applications of Trigonometry We break this product into pieces. First, we use the diﬀerence of squares to multiply the ‘ﬁrst’ quantities in each factor to get [(A + C) + (A − C) cos(2θ)] [(A + C) − (A − C) cos(2θ)] = (A + C)2 − (A − C)2 cos2(2θ) Next, we add the product of the ‘outer’ and ‘inner’ quantities in each factor to get −B sin(2θ) [(A + C) + (A − C) cos(2θ)] +B sin(2θ) [(A + C) − (A − C) cos(2θ)] = −2B(A − C) cos(2θ) sin(2θ) The product of the ‘last’ quantity in each factor is (B sin(2θ))(−B sin(2θ)) = −B2 sin2(2θ). Putting all of this together yields 4AC = (A + C)2 − (A − C)2 cos2(2θ) − 2B(A − C) cos(2θ) sin(2θ) − B2 sin2(2θ) From cot(2θ) = A−C sin(2θ) = A−C twice along with the Pythagorean Identity cos2(2θ) = 1 − sin2(2θ) to get B , we get cos(2θ) B , or (A−C) sin(2θ) = B cos(2θ). We use this substitution 4AC = (A + C)2 − (A − C)2 cos2(2θ) − 2B(A − C) cos(2θ) sin(2θ) − B2 sin2(2θ) = (A + C)2 − (A − C)2 1 − sin2(2θ) − 2B cos(2θ)B cos(2θ) − B2 sin2(2θ) = (A + C)2 − (A − C)2 + (A − C)2 sin2(2θ) − 2B2 cos2(2θ) − B2 sin2(2θ) = (A + C)2 − (A − C)2 + [(A − C) sin(2θ)]2 − 2B2 cos2(2θ) − B2 sin2(2θ) = (A + C)2 − (A − C)2 + [B cos(2θ)]2 − 2B2 cos2(2θ) − B2 sin2(2θ) = (A + C)2 − (A − C)2 + B2 cos2(2θ) − 2B2 cos2(2θ) − B2 sin2(2θ) = (A + C)2 − (A − C)2 − B2 cos2(2θ) − B2 sin2(2θ) = (A + C)2 − (A − C)2 − B2 cos2(2θ) + sin2(2θ) = (A + C)2 − (A − C)2 − B2 = A2 + 2AC + C2 − A2 − 2AC + C2 − B2 = 4AC − B2 Hence, B2 − 4AC = −4AC, so the quantity B2 − 4AC has the opposite sign of AC. The result now follows by applying Exercise 34 in Section 7.5. Example 11.6.3. Use Theorem 11.11 to classify the graphs of the following non-degenerate conics. 1. 21x2 + 10xy √ 3 + 31y2 = 144 2. 5x2 + 26xy + 5y2 − 16x √ 2 + 16y √ 2 − 104 = 0 3. 16x2 + 24xy + 9y2 + 15x − 20y = 0 Solution. This is a straightforward application of Theorem 11.11. 11.6 Hooked on Conics Again 981 1. We have A = 21, B = 10 3)2 − 4(21)(31) = −2304 < 0. Theorem 11.11 predicts the graph is an ellipse, which checks with our work from Example 11.6.1 number 2. 3 and C = 31 so B2 − 4AC = (10 √ √ 2. Here, A = 5, B = 26 and C = 5, so B2 − 4AC = 262 − 4(5)(5) = 576 > 0. Theorem 11.11 classiﬁes the graph as a hyperbola, which matches our answer to Example 11.6.2 number 1. 3. Finally, we have A = 16, B = 24 and C = 9 which gives 242 − 4(16)(9) = 0. Theorem 11.11 tells us that the graph is a parabola, matching our result from Example 11.6.2 number 2. 11.6.2 The Polar Form of Conics In this subsection, we start from scratch to reintroduce the conic sections from a more uniﬁed perspective. We have our ‘new’ deﬁnition below. Deﬁnition 11.1. Given a ﬁxed line L, a point F not on L, and a positive number e, a conic section is the set of all points P such that the distance from P to F the distance from P to L = e The line L is called the directrix of the conic section, the point F is called a focus of the conic section, and the constant e is called the eccentricity of the conic section. We have seen the notions of focus and directrix before in the deﬁnition of a parabola, Deﬁnition 7.3. There, a parabola is deﬁned as the set of points equidistant from the focus and directrix, giving an eccentricity e = 1 according to Deﬁnition 11.1. We have also seen the concept of eccentricity before. It was introduced for ellipses in Deﬁnition 7.5 in Section 7.4, and later extended to hyperbolas in Exercise 31 in Section 7.5. There, e was also deﬁned as a ratio of distances, though in these cases the distances involved were measurements from the center to a focus and from the center to a vertex. One way to reconcile the ‘old’ ideas of focus, directrix and eccentricity with the ‘new’ ones presented in Deﬁnition 11.1 is to derive equations for the conic sections using Deﬁnition 11.1 and compare these parameters with what we know from Chapter 7. We begin by assuming the conic section has eccentricity e, a focus F at the origin and that the directrix is the vertical line x = −d as in the ﬁgure below. y d r cos(θ) P (r, θ) r θ x = −d O = F x 982 Applications of Trigonometry Using a polar coordinate representation P (r, θ) for a point on the conic with r > 0, we get e = the distance from P to F the distance from P to L = r d + r cos(θ) so that r = e(d + r cos(θ)). Solving this equation for r, yields r = ed 1 − e cos(θ) At this point, we convert the equation r = e(d + r cos(θ)) back into a rectangular equation in the variables x and y. If e > 0, but e = 1, the usual conversion process outlined in Section 11.4 gives6 1 − e22 e2d2 x − e2d 1 − e2 2 1 − e2 e2d2 + y2 = 1 1−e2 , so the major axis has length 2ed 1−e2 and the minor axis has length 2ed√ e2d 1−e2 , 0 We leave it to the reader to show if 0 < e < 1, this is the equation of an ellipse centered at with major axis along the x-axis. Using the notation from Section 7.4, we have a2 = e2d2 (1−e2)2 and b2 = e2d2 1−e2 . Moreover, we ﬁnd that one focus is (0, 0) and working through the formula given in Deﬁnition 7.5 gives the eccentricity 1−e2 , 0 to be e, as required. If e > 1, then the equation generates a hyperbola with center whose transverse axis lies along the x-axis. Since such hyperbolas have the form (x−h)2 b2 = 1, we need to take the opposite reciprocal of the coeﬃcient of y2 to ﬁnd b2. We get7 a2 = e2d2 (e2−1)2 and b2 = − e2d2 e2−1 and the conjugate axis has length 2ed√ . e2−1 Additionally, we verify that one focus is at (0, 0), and the formula given in Exercise 31 in Section ed 1−e cos(θ) reduces to 7.5 gives the eccentricity is e in this case as well. If e = 1, the equation r = . This is a parabola with vertex r = − d 2 , the focus is (0, 0), the focal diameter is 2d and the directrix is x = −d, as required. Hence, we have shown that in all cases, our ‘new’ understanding of ‘conic section’, ‘focus’, ‘eccentricity’ and ‘directrix’ as presented in Deﬁnition 11.1 correspond with the ‘old’ deﬁnitions given in Chapter 7. 1−cos(θ) which gives the rectangular equation y2 = 2d x + d 2 , 0 opening to the right. In the language of Section 7.3, 4p = 2d so p = d e2−1 , so the transverse axis has length 2ed a2 − y2 (1−e2)2 = e2d2 1−e2 = e2d2 e2d d 2 ed Before we summarize our ﬁndings, we note that in order to arrive at our general equation of a conic 1−e cos(θ) , we assumed that the directrix was the line x = −d for d > 0. We could have just as r = easily chosen the directrix to be x = d, y = −d or y = d. As the reader can verify, in these cases 1+e sin(θ) , respectively. The key thing to 1−e sin(θ) and r = we obtain the forms r = remember is that in any of these cases, the directrix is always perpendicular to the major axis of an ellipse and it is always perpendicular to the transverse axis of the hyperbola. For parabolas, knowing the focus is (0, 0) and the directrix also tells us which way the parabola opens. We have established the following theorem. 1+e cos(θ) , r = ed ed ed 6Turn r = e(d + r cos(θ)) into r = e(d + x) and square both sides to get r2 = e2(d + x)2. Replace r2 with x2 + y2, expand (d + x)2, combine like terms, complete the square on x and clean things up. 7Since e > 1 in this case, 1 − e2 < 0. Hence, we rewrite 1 − e22 = e2 − 12 to help simplify things later on. 11.6 Hooked on Conics Again 983 Theorem 11.12. Suppose e and d are positive numbers. Then the graph of r = ed 1−e cos(θ) is the graph of a conic section with directrix x = −d. the graph of r = ed 1+e cos(θ) is the graph of a conic section with directrix x = d. the graph of r = ed 1−e sin(θ) is the graph of a conic section with directrix y = −d. the graph of r = ed 1+e sin(θ) is the graph of a conic section with directrix y = d. In each case above, (0, 0) is a focus of the conic and the number e is the eccentricity of the conic. If 0 < e < 1, the graph is an ellipse whose major axis has length 2ed 1−e2 and whose minor axis has length 2ed√ 1−e2 If e = 1, the graph is a parabola whose focal diameter is 2d. If e > 1, the graph is a hyperbola whose transverse axis has length 2ed e2−1 and whose conjugate axis has length 2ed√ e2−1 . We test out Theorem 11.12 in the next example. Example 11.6.4. Sketch the graphs of the following equations. 1. r = 4 1 − sin(θ) Solution. 2. r = 12 3 − cos(θ) 3. r = 6 1 + 2 sin(θ) 4 1. From r = 1−sin(θ) , we ﬁrst note e = 1 which means we have a parabola on our hands. Since ed = 4, we have d = 4 and considering the form of the equation, this puts the directrix at y = −4. Since the focus is at (0, 0), we know that the vertex is located at the point (in rectangular coordinates) (0, −2) and must open upwards. With d = 4, we have a focal diameter of 2d = 8, so the parabola contains the points (±4, 0). We graph r = 1−sin(θ) below. 4 12 2. We ﬁrst rewrite r = 4 1−(1/3) cos(θ) . 3−cos(θ) in the form found in Theorem 11.12, namely r = Since e = 1 3 satisﬁes 0 < e < 1, we know that the graph of this equation is an ellipse. Since ed = 4, we have d = 12 and, based on the form of the equation, the directrix is x = −12. This means that the ellipse has its major axis along the x-axis. We can ﬁnd the vertices of the ellipse by ﬁnding the points of the ellipse which lie on the x-axis. We ﬁnd r(0) = 6 and r(π) = 3 which correspond to the rectangular point
s (−3, 0) and (6, 0), so these are our 2 , 0.8 vertices. The center of the ellipse is the midpoint of the vertices, which in this case is 3 2 , 0 and this allows us to ﬁnd the We know one focus is (0, 0), which is 3 1−e2 = (2)(4) 2 from the center 3 8As a quick check, we have from Theorem 11.12 the major axis should have length 2ed 1−(1/3)2 = 9. 984 Applications of Trigonometry other focus (3, 0), even though we are not asked to do so. Finally, we know from Theorem 2ed√ 3 which means the endpoints 1−e2 = 11.12 that the length of the minor axis is of the minor axis are 3 1−(1/3)2 = 6 4√ 2. We now have everything we need to graph r = 12 √ √ 3−cos(θ) . 2 , ±3 3 2 1 −4 −3 −2 −1 −1 1 2 3 4 −2 −3 y = −4 r = 4 1−sin(θ) y 4 3 2 1 −3 −2 −1 −1 1 2 3 4 5 6 x −2 −3 −4 x = −12 r = 12 3−cos(θ) 3. From r = 6 1+2 sin(θ) we get e = 2 > 1 so the graph is a hyperbola. Since ed = 6, we get d = 3, and from the form of the equation, we know the directrix is y = 3. This means the transverse axis of the hyperbola lies along the y-axis, so we can ﬁnd the vertices by looking where the hyperbola intersects the y-axis. We ﬁnd r π = −6. These two 2 points correspond to the rectangular points (0, 2) and (0, 6) which puts the center of the hyperbola at (0, 4). Since one focus is at (0, 0), which is 4 units away from the center, we know the other focus is at (0, 8). According to Theorem 11.12, the conjugate axis has a length of 3. Putting this together with the location of the vertices, we get that √ 3 the asymptotes of the hyperbola have slopes ± 2 3 . Since the center of the hyperbola √ 2 = 2 and r 3π 2 = (2)(6) √ 22−1 = ± 2ed√ = 4 e2−1 √ 3 is (0, 4), the asymptotes are y = ± √ 3 3 x + 4. We graph the hyperbola below5 −4 −3 −2 −+2 sin(θ) 11.6 Hooked on Conics Again 985 In light of Section 11.6.1, the reader may wonder what the rotated form of the conic sections would look like in polar form. We know from Exercise 65 in Section 11.5 that replacing θ with (θ − φ) in an expression r = f (θ) rotates the graph of r = f (θ) counter-clockwise by an angle φ. For instance, to graph r = 1−sin(θ) , which we obtained in Example 11.6.4 number 1, counter-clockwise by π all we need to do is rotate the graph of r = 4 1−sin(θ− π 4 ) 4 radians, as shown below4 −3 −2 −1 −1 −2 −3 r = 4 1−sin(θ− π 4 ) Using rotations, we can greatly simplify the form of the conic sections presented in Theorem 11.12, since any three of the forms given there can be obtained from the fourth by rotating through some multiple of π 2 . Since rotations do not aﬀect lengths, all of the formulas for lengths Theorem 11.12 remain intact. In the theorem below, we also generalize our formula for conic sections to include circles centered at the origin by extending the concept of eccentricity to include e = 0. We conclude this section with the statement of the following theorem. Theorem 11.13. Given constants > 0, e ≥ 0 and φ, the graph of the equation r = 1 − e cos(θ − φ) is a conic section with eccentricity e and one focus at (0, 0). If e = 0, the graph is a circle centered at (0, 0) with radius . If e = 0, then the conic has a focus at (0, 0) and the directrix contains the point with polar coordinates (−d, φ) where d = e . minor axis has length 2ed√ 1−e2 – If 0 < e < 1, the graph is an ellipse whose major axis has length 2ed 1−e2 and whose – If e = 1, the graph is a parabola whose focal diameter is 2d. – If e > 1, the graph is a hyperbola whose transverse axis has length 2ed e2−1 and whose conjugate axis has length 2ed√ e2−1 . 986 Applications of Trigonometry 11.6.3 Exercises Graph the following equations. 1. x2 + 2xy + y2 − x √ √ 2 − 6 = 0 2. 7x2 − 4xy √ 3 + 3y2 − 2x − 2y √ 3 − 5 = 0 3. 5x2 + 6xy + 5y2 − 4 2x + 4 √ 2y = 0 2 + y √ 5. 13x2 − 34xy √ 3 + 47y2 − 64 = 0 7. x2 − 4xy + 4y2 − 2x √ 5 − y √ 5 = 0 Graph the following equations. 9. r = 11. r = 2 1 − cos(θ) 3 2 − cos(θ) 13. r = 4 1 + 3 cos(θ) 15. r = 2 1 + sin(θ − π 3 ) √ √ 4. x2 + 2 6. x2 − 2 3xy + 3y2 + 2 √ 3x − 2y − 16 = 0 3xy − y2 + 8 = 0 8. 8x2 + 12xy + 17y2 − 20 = 0 10. r = 12. r = 3 2 + sin(θ) 2 1 + sin(θ) 14. r = 2 1 − 2 sin(θ) 16. r = 6 3 − cos θ + π 4 The matrix A(θ) = cos(θ) − sin(θ) cos(θ) sin(θ) is called a rotation matrix. We’ve seen this matrix most recently in the proof of used in the proof of Theorem 11.9. 17. Show the matrix from Example 8.3.3 in Section 8.3 is none other than A π 4 . 18. Discuss with your classmates how to use A(θ) to rotate points in the plane. 19. Using the even / odd identities for cosine and sine, show A(θ)−1 = A(−θ). Interpret this geometrically. 11.6 Hooked on Conics Again 987 11.6.4 Answers 1. x2 + 2xy + y2 − becomes (x)2 = −(y − 3) after rotating counter-clockwise through x2 + 2xy + y2 − x √ 2 + y √ 3. 5x2 + 6xy + 5y2 − 4 2x + 4 √ 2 − 6 = 0 √ 2y = 0 becomes (x)2 + (y+2)2 counter-clockwise through θ = π 4 . 4 = 1 after rotating . 7x2 − 4xy 3 + 3y2 − 2x − 2y 3 − 5 = 0 9 + (y)2 = 1 after rotating becomes (x−2)2 counter-clockwise through √ 7x2 − 4xy 3 + 3y2 − 2x − 2y √ 3 − 5 = 0 √ 4. x2 + 2 3xy + 3y2 + 2 √ 3x − 2y − 16 = 0 becomes(x)2 = y + 4 after rotating counter-clockwise through 5x2 + 6xy + 5y2 − 4 √ 2x + 4 √ 2y = 0 √ x2 + 2 3xy + 3y2 + 2 √ 3x − 2y − 16 = 0 988 Applications of Trigonometry √ 5. 13x2 − 34xy 3 + 47y2 − 64 = 0 becomes (y)2 − (x)2 counter-clockwise through θ = π 6 . 16 = 1 after rotating √ 6. x2 − 2 3xy − y2 + 8 = 0 4 − (y)2 becomes (x)2 counter-clockwise through θ = π 3 4 = 1 after rotating 13x2 − 34xy √ 3 + 47y2 − 64 = 0 √ x2 − 2 3xy − y2 + 8 = 0 7. x2 − 4xy + 4y2 − 2x √ 5 − y √ 5 = 0 becomes (y)2 = x after rotating counter-clockwise through θ = arctan 1 2 y . y 8. 8x2 + 12xy + 17y2 − 20 = 0 becomes (x)2 + (y)2 4 = 1 after rotating counter-clockwise through θ = arctan(2) y x y x θ = arctan 1 2 x θ = arctan(2) x x2 − 4xy + 4y2 − 2x √ 5 − y √ 5 = 0 8x2 + 12xy + 17y2 − 20 = 0 11.6 Hooked on Conics Again 989 2 9. r = 1−cos(θ) is a parabola directrix x = −2 , vertex (−1, 0) focus (0, 0), focal diameter 4 10. r = 3 2+sin(θ) = 3 2 1+ 1 2 sin(θ) is an ellipse directrix y = 3 , vertices (0, 1), (0, −3) center (0, −2) , foci (0, 0), (0, −2) minor axis length 4 −3 −2 −1 1 2 3 4 x −1 −2 −3 −4 −4 −3 −2 −1 1 2 3 4 x −1 −2 −4 11. r = 3 2−cos(θ) = 3 2 1− 1 2 cos(θ) is an ellipse directrix x = −3 , vertices (−1, 0), (3, 0) center (1, 0) , foci (0, 0), (2, 0) minor axis length 2 √ 3 y 4 3 2 1 −4 −3 −2 −1 1 2 3 4 x −1 −2 −4 12. r = 2 1+sin(θ) is a parabola directrix y = 2 , vertex (0, 1) focus (0, 0), focal diameter 4 y 4 3 2 1 −4 −3 −2 −1 1 2 3 4 x −1 −2 −3 −4 990 Applications of Trigonometry 13. r = 4 1+3 cos(θ) is a hyperbola directrix x = 4 center 3 conjugate axis length 2 3 , vertices (1, 0), (2, 0) √ 2 2 , 0, foci (0, 0), (3, 0) 14. r = 2 1−2 sin(θ) is a hyperbola directrix y = −1, vertices 0, − 2 3 center 0, − 4 , foci (0, 0), 0, − 8 3 3 √ conjugate axis length 2 3 3 , (0, −24 −3 −2 −1 1 2 3 4 x −4 −3 −2 −1 1 2 3 4 x −1 −2 −3 −4 15. r = 2 3 ) is 1+sin(θ− π 2 the parabola r = 1+sin(θ) rotated through 1 −2 −3 −4 is the ellipse 16. r = 6 3−cos(θ+ π 4 ) 6 3−cos(θ) = r = 3 cos(θ) rotated through φ = − π 4 14 −3 −2 −1 1 2 3 4 x −1 −2 −3 −4 φ = − π 4 x 11.7 Polar Form of Complex Numbers 991 11.7 Polar Form of Complex Numbers √ In this section, we return to our study of complex numbers which were ﬁrst introduced in Section 3.4. Recall that a complex number is a number of the form z = a + bi where a and b are real −1. The number a is called the real part of numbers and i is the imaginary unit deﬁned by i = z, denoted Re(z), while the real number b is called the imaginary part of z, denoted Im(z). From Intermediate Algebra, we know that if z = a + bi = c + di where a, b, c and d are real numbers, then a = c and b = d, which means Re(z) and Im(z) are well-deﬁned.1 To start oﬀ this section, we associate each complex number z = a + bi with the point (a, b) on the coordinate plane. In this case, the x-axis is relabeled as the real axis, which corresponds to the real number line as usual, and the y-axis is relabeled as the imaginary axis, which is demarcated in increments of the imaginary unit i. The plane determined by these two axes is called the complex plane. Imaginary Axis (3, 0) ←→ z = 3 0 1 2 3 4 Real Axis (−4, 2) ←→ z = −4 + 2i 4i 3i 2i i −4 −3 −2 −1 −i −2i −3i (0, −3) ←→ z = −3i −4i The Complex Plane Since the ordered pair (a, b) gives the rectangular coordinates associated with the complex number z = a + bi, the expression z = a + bi is called the rectangular form of z. Of course, we could just as easily associate z with a pair of polar coordinates (r, θ). Although it is not as straightforward as the deﬁnitions of Re(z) and Im(z), we can still give r and θ special names in relation to z. Deﬁnition 11.2. The Modulus and Argument of Complex Numbers: Let z = a + bi be a complex number with a = Re(z) and b = Im(z). Let (r, θ) be a polar representation of the point with rectangular coordinates (a, b) where r ≥ 0. The modulus of z, denoted \|z\|, is deﬁned by \|z\| = r. The angle θ is an argument of z. The set of all arguments of z is denoted arg(z). If z = 0 and −π < θ ≤ π, then θ is the principal argument of z, written θ = Arg(z). 1‘Well-deﬁned’ means that no matter how we express z, the number Re(z) is always the same, and the number Im(z) is always the same. In other words, Re and Im are functions of complex numbers. 992 Applications of Trigonometry Some remarks about Deﬁnition 11.2 are in order. We know from Section 11.4 that every point in the plane has inﬁnitely many polar coordinate representations (r, θ) which means it’s worth our time to make sure the quantities ‘modulus’, ‘argument’ and ‘principal argument’ are well-deﬁned. Concerning the modulus, if z = 0 then the point associated with z is the origin. In this case, the only r-value which can be used here is r = 0. Hence for z = 0, \|z\| = 0 is well-deﬁned. If z = 0, then the point associated with z is not the origin, and there are two possibilities for r: one positive and one negative. However, we stipulated r ≥ 0 in our deﬁnition so this p
ins down the value of \|z\| to one and only one number. Thus the modulus is well-deﬁned in this case, too.2 Even with the requirement r ≥ 0, there are inﬁnitely many angles θ which can be used in a polar representation of a point (r, θ). If z = 0 then the point in question is not the origin, so all of these angles θ are coterminal. Since coterminal angles are exactly 2π radians apart, we are guaranteed that only one of them lies in the interval (−π, π], and this angle is what we call the principal argument of z, Arg(z). In fact, the set arg(z) of all arguments of z can be described using set-builder notation as arg(z) = {Arg(z) + 2πk \| k is an integer}. Note that since arg(z) is a set, we will write ‘θ ∈ arg(z)’ to mean ‘θ is in3 the set of arguments of z’. If z = 0 then the point in question is the origin, which we know can be represented in polar coordinates as (0, θ) for any angle θ. In this case, we have arg(0) = (−∞, ∞) and since there is no one value of θ which lies (−π, π], we leave Arg(0) undeﬁned.4 It is time for an example. Example 11.7.1. For each of the following complex numbers ﬁnd Re(z), Im(z), \|z\|, arg(z) and Arg(z). Plot z in the complex plane. 1. z = √ 3 − i Solution. 2. z = −2 + 4i 3. z = 3i 4. z = −117 √ √ √ 1. For z = 3 − i = 3 + (−1)i, we have Re(z) = 3 and Im(z) = −1. To ﬁnd \|z\|, arg(z) and Arg(z), we need to ﬁnd a polar representation (r, θ) with r ≥ 0 for the point P ( 3, −1) associated with z. We know r2 = ( 3)2 + (−1)2 = 4, so r = ±2. Since we require r ≥ 0, we choose r = 2, so \|z\| = 2. Next, we ﬁnd a corresponding angle θ. Since r > 0 and P lies in Quadrant IV, θ is a Quadrant IV angle. We know tan(θ) = −1√ 6 + 2πk 3 6 + 2πk \| k is an integer. Of these values, only θ = − π for integers k. Hence, arg(z) = − π satisﬁes the requirement that −π < θ ≤ π, hence Arg(z) = − π 6 . 3 , so . The complex number z = −2 + 4i has Re(z) = −2, Im(z) = 4, and is associated with the point P (−2, 4). Our next task is to ﬁnd a polar representation (r, θ) for P where r ≥ 0. 5. To ﬁnd θ, we get Running through the usual calculations gives r = 2 tan(θ) = −2, and since r > 0 and P lies in Quadrant II, we know θ is a Quadrant II angle. We ﬁnd θ = π + arctan(−2) + 2πk, or, more succinctly θ = π − arctan(2) + 2πk for integers k. Hence arg(z) = {π − arctan(2) + 2πk \| k is an integer}. Only θ = π − arctan(2) satisﬁes the requirement −π < θ ≤ π, so Arg(z) = π − arctan(2). 5, so \|z\| = 2 √ √ 2In case you’re wondering, the use of the absolute value notation \|z\| for modulus will be explained shortly. 3Recall the symbol being used here, ‘∈,’ is the mathematical symbol which denotes membership in a set. 4If we had Calculus, we would regard Arg(0) as an ‘indeterminate form.’ But we don’t, so we won’t. 11.7 Polar Form of Complex Numbers 993 3. We rewrite z = 3i as z = 0 + 3i to ﬁnd Re(z) = 0 and Im(z) = 3. The point in the plane which corresponds to z is (0, 3) and while we could go through the usual calculations to ﬁnd the required polar form of this point, we can almost ‘see’ the answer. The point (0, 3) lies 3 units away from the origin on the positive y-axis. Hence, r = \|z\| = 3 and θ = π 2 + 2πk for integers k. We get arg(z) = π 2 + 2πk \| k is an integer and Arg(z) = π 2 . 4. As in the previous problem, we write z = −117 = −117 + 0i so Re(z) = −117 and Im(z) = 0. The number z = −117 corresponds to the point (−117, 0), and this is another instance where we can determine the polar form ‘by eye’. The point (−117, 0) is 117 units away from the origin along the negative x-axis. Hence, r = \|z\| = 117 and θ = π + 2π = (2k + 1)πk for integers k. We have arg(z) = {(2k + 1)π \| k is an integer}. Only one of these values, θ = π, just barely lies in the interval (−π, π] which means and Arg(z) = π. We plot z along with the other numbers in this example below. Imaginary Axis z = 3i 4i 3i 2i i z = −2 + 4i z = −117 −117 −2 −1 − Real Axis Now that we’ve had some practice computing the modulus and argument of some complex numbers, it is time to explore their properties. We have the following theorem. Theorem 11.14. Properties of the Modulus: Let z and w be complex numbers. \|z\| is the distance from z to 0 in the complex plane \|z\| ≥ 0 and \|z\| = 0 if and only if z = 0 \|z\| = Re(z)2 + Im(z)2 Product Rule: \|zw\| = \|z\|\|w\| Power Rule: \|zn\| = \|z\|n for all natural numbers, n Quotient Rule: z w = \|z\| \|w\| , provided w = 0 To prove the ﬁrst three properties in Theorem 11.14, suppose z = a + bi where a and b are real numbers. To determine \|z\|, we ﬁnd a polar representation (r, θ) with r ≥ 0 for the point (a, b). From √ Section 11.4, we know r2 = a2 + b2 so that r = ± a2 + b2. Since we require r ≥ 0, then it must be a2 + b2. Using the distance formula, we ﬁnd the distance a2 + b2, which means \|z\| = that r = √ √ 994 Applications of Trigonometry √ a2 + b2, establishing the ﬁrst property.5 For the second property, note from (0, 0) to (a, b) is also that since \|z\| is a distance, \|z\| ≥ 0. Furthermore, \|z\| = 0 if and only if the distance from z to 0 is 0, and the latter happens if and only if z = 0, which is what we were asked to show.6 For the third property, we note that since a = Re(z) and b = Im(z), z = To prove the product rule, suppose z = a + bi and w = c + di for real numbers a, b, c and d. Then zw = (a + bi)(c + di). After the usual arithmetic7 we get zw = (ac − bd) + (ad + bc)i. Therefore, Re(z)2 + Im(z)2. a2 + b2 = √ \|zwac − bd)2 + (ad + bc)2 a2c2 − 2abcd + b2d2 + a2d2 + 2abcd + b2c2 Expand a2c2 + a2d2 + b2c2 + b2d2 a2 (c2 + d2) + b2 (c2 + d2) (a2 + b2) (c2 + d2) a2 + b2 c2 + d2 Factor Factor √ Rearrange terms = = \|z\|\|w\| Product Rule for Radicals Deﬁnition of \|z\| and \|w\| Hence \|zw\| = \|z\|\|w\| as required. Now that the Product Rule has been established, we use it and the Principle of Mathematical Induction8 to prove the power rule. Let P (n) be the statement \|zn\| = \|z\|n. Then P (1) is true since z1 = \|z\|k for some k ≥ 1. Our job = \|z\|k+1. As is customary with induction proofs, is to show that P (k + 1) is true, namely we ﬁrst try to reduce the problem in such a way as to use the Induction Hypothesis. = \|z\| = \|z\|1. Next, assume P (k) is true. That is, assume zk+1 zk zk+1 Properties of Exponents \|z\| Product Rule zkz = zk = = \|z\|k\|z\| = \|z\|k+1 Properties of Exponents Induction Hypothesis Hence, P (k + 1) is true, which means \|zn\| = \|z\|n is true for all natural numbers n. Like the Power Rule, the Quotient Rule can also be established with the help of the Product Rule. We assume w = 0 (so \|w\| = 0) and we get (z) 1 w Product Rule. = = \|z\| z w 1 w 5Since the absolute value \|x\| of a real number x can be viewed as the distance from x to 0 on the number line, this ﬁrst property justiﬁes the notation \|z\| for modulus. We leave it to the reader to show that if z is real, then the deﬁnition of modulus coincides with absolute value so the notation \|z\| is unambiguous. 6This may be considered by some to be a bit of a cheat, so we work through the underlying Algebra to see this is a2 + b2 = 0 if and only if a2 + b2 = 0, which is true if and only if a = b = 0. √ true. We know \|z\| = 0 if and only if The latter happens if and only if z = a + bi = 0. There. 7See Example 3.4.1 in Section 3.4 for a review of complex number arithmetic. 8See Section 9.3 for a review of this technique. 11.7 Polar Form of Complex Numbers 995 1 Hence, the proof really boils down to showing w Next, we characterize the argument of a complex number in terms of its real and imaginary parts. \|w\| . This is left as an exercise. = 1 Theorem 11.15. Properties of the Argument: Let z be a complex number. If Re(z) = 0 and θ ∈ arg(z), then tan(θ) = Im(z) Re(z) . 2 + 2πk \| k is an integer. 2 + 2πk \| k is an integer. If Re(z) = 0 and Im(z) > 0, then arg(z) = π If Re(z) = 0 and Im(z) < 0, then arg(z) = − π If Re(z) = Im(z) = 0, then z = 0 and arg(z) = (−∞, ∞). To prove Theorem 11.15, suppose z = a + bi for real numbers a and b. By deﬁnition, a = Re(z) and b = Im(z), so the point associated with z is (a, b) = (Re(z), Im(z)). From Section 11.4, we know that if (r, θ) is a polar representation for (Re(z), Im(z)), then tan(θ) = Im(z) Re(z) , provided Re(z) = 0. If Re(z) = 0 and Im(z) > 0, then z lies on the positive imaginary axis. Since we take r > 0, we have that θ is coterminal with π 2 , and the result follows. If Re(z) = 0 and Im(z) < 0, then z lies on the negative imaginary axis, and a similar argument shows θ is coterminal with − π 2 . The last property in the theorem was already discussed in the remarks following Deﬁnition 11.2. Our next goal is to completely marry the Geometry and the Algebra of the complex numbers. To that end, consider the ﬁgure below. Imaginary Axis bi (a, b) ←→ z = a + bi ←→ (r, θ ∈ arg(z) 0 a Real Axis Polar coordinates, (r, θ) associated with z = a + bi with r ≥ 0. We know from Theorem 11.7 that a = r cos(θ) and b = r sin(θ). Making these substitutions for a and b gives z = a + bi = r cos(θ) + r sin(θ)i = r [cos(θ) + i sin(θ)]. The expression ‘cos(θ) + i sin(θ)’ is abbreviated cis(θ) so we can write z = rcis(θ). Since r = \|z\| and θ ∈ arg(z), we get Deﬁnition 11.3. A Polar Form of a Complex Number: Suppose z is a complex number and θ ∈ arg(z). The expression: is called a polar form for z. \|z\|cis(θ) = \|z\| [cos(θ) + i sin(θ)] 996 Applications of Trigonometry Since there are inﬁnitely many choices for θ ∈ arg(z), there inﬁnitely many polar forms for z, so we used the indeﬁnite article ‘a’ in Deﬁnition 11.3. It is time for an example. Example 11.7.2. 1. Find the rectangular form of the following complex numbers. Find Re(z) and Im(z). (a) z = 4cis 2π 3 (b) z = 2cis − 3π 4 (c) z = 3cis(0) (d) z = cis π 2 2. Use the results from Example 11.7.1 to ﬁnd a polar form of the following complex numbers. (a) z = √ 3 − i Solution. (b) z = −2 + 4i (c) z = 3i (d) z = −117 1. The key to this problem is to write out cis(θ) as cos(θ) + i sin(θ). (a) By deﬁnition, z = 4cis 2π √ 3 z = −2 + 2i + i sin 2π = 4 cos 2π √ 3 3 3, so that Re(z) = −2 and Im(z) = 2 = 2 cos − 3π 4 . After some simplifying, we get 3. + i
sin − 3π 4 . From this, we ﬁnd (b) Expanding, we get z = 2cis − 3π √ 4 2 = Im(z). 2, so Re(zc) We get z = 3cis(0) = 3 [cos(0) + i sin(0)] = 3. Writing 3 = 3 + 0i, we get Re(z) = 3 and (d) Lastly, we have z = cis π 2 Im(z) = 0, which makes sense seeing as 3 is a real number. = cos π 2 = i. Since i = 0 + 1i, we get Re(z) = 0 and Im(z) = 1. Since i is called the ‘imaginary unit,’ these answers make perfect sense. + i sin π 2 2. To write a polar form of a complex number z, we need two pieces of information: the modulus \|z\| and an argument (not necessarily the principal argument) of z. We shamelessly mine our solution to Example 11.7.1 to ﬁnd what we need. (a) For z = √ 3 − i, \|z\| = 2 and θ = − π 6 , so z = 2cis − π 6 . We can check our answer by √ converting it back to rectangular form to see that it simpliﬁes to z = 3 − i. √ √ (b) For z = −2 + 4i, \|z\| = 2 5 and θ = π − arctan(2). Hence, z = 2 5cis(π − arctan(2)). It is a good exercise to actually show that this polar form reduces to z = −2 + 4i. (c) For z = 3i, \|z\| = 3 and θ = π 2 . In this case, z = 3cis π 2 geometrically. Head out 3 units from 0 along the positive real axis. Rotating π counter-clockwise lands you exactly 3 units above 0 on the imaginary axis at z = 3i. . This can be checked 2 radians (d) Last but not least, for z = −117, \|z\| = 117 and θ = π. We get z = 117cis(π). As with the previous problem, our answer is easily checked geometrically. 11.7 Polar Form of Complex Numbers 997 The following theorem summarizes the advantages of working with complex numbers in polar form. Theorem 11.16. Products, Powers and Quotients Complex Numbers in Polar Form: Suppose z and w are complex numbers with polar forms z = \|z\|cis(α) and w = \|w\|cis(β). Then Product Rule: zw = \|z\|\|w\|cis(α + β) Power Rule (DeMoivre’s Theorem) : zn = \|z\|ncis(nθ) for every natural number n Quotient Rule: z w = \|z\| \|w\| cis(α − β), provided \|w\| = 0 The proof of Theorem 11.16 requires a healthy mix of deﬁnition, arithmetic and identities. We ﬁrst start with the product rule. zw = [\|z\|cis(α)] [\|w\|cis(β)] = \|z\|\|w\| [cos(α) + i sin(α)] [cos(β) + i sin(β)] We now focus on the quantity in brackets on the right hand side of the equation. [cos(α) + i sin(α)] [cos(β) + i sin(β)] = cos(α) cos(β) + i cos(α) sin(β) + i sin(α) cos(β) + i2 sin(α) sin(β) = cos(α) cos(β) + i2 sin(α) sin(β) Rearranging terms + i sin(α) cos(β) + i cos(α) sin(β) = (cos(α) cos(β) − sin(α) sin(β)) Since i2 = −1 + i (sin(α) cos(β) + cos(α) sin(β)) Factor out i = cos(α + β) + i sin(α + β) Sum identities = cis(α + β) Deﬁnition of ‘cis’ Putting this together with our earlier work, we get zw = \|z\|\|w\|cis(α + β), as required. Moving right along, we next take aim at the Power Rule, better known as DeMoivre’s Theorem.9 We proceed by induction on n. Let P (n) be the sentence zn = \|z\|ncis(nθ). Then P (1) is true, since z1 = z = \|z\|cis(θ) = \|z\|1cis(1 · θ). We now assume P (k) is true, that is, we assume zk = \|z\|kcis(kθ) for some k ≥ 1. Our goal is to show that P (k + 1) is true, or that zk+1 = \|z\|k+1cis((k + 1)θ). We have zk+1 = zkz = \|z\|kcis(kθ) (\|z\|cis(θ)) = \|z\|k\|z\| cis(kθ + θ) = \|z\|k+1cis((k + 1)θ) Properties of Exponents Induction Hypothesis Product Rule 9Compare this proof with the proof of the Power Rule in Theorem 11.14. 998 Applications of Trigonometry Hence, assuming P (k) is true, we have that P (k + 1) is true, so by the Principle of Mathematical Induction, zn = \|z\|ncis(nθ) for all natural numbers n. The last property in Theorem 11.16 to prove is the quotient rule. Assuming \|w\| = 0 we have z w = = \|z\|cis(α) \|w\|cis(β) \|z\| \|w\| cos(α) + i sin(α) cos(β) + i sin(β) Next, we multiply both the numerator and denominator of the right hand side by (cos(β) − i sin(β)) which is the complex conjugate of (cos(β) + i sin(β)) to get z w = \|z\| \|w\| cos(α) + i sin(α) cos(β) + i sin(β) · cos(β) − i sin(β) cos(β) − i sin(β) If we let the numerator be N = [cos(α) + i sin(α)] [cos(β) − i sin(β)] and simplify we get N = [cos(α) + i sin(α)] [cos(β) − i sin(β)] = cos(α) cos(β) − i cos(α) sin(β) + i sin(α) cos(β) − i2 sin(α) sin(β) Expand = [cos(α) cos(β) + sin(α) sin(β)] + i [sin(α) cos(β) − cos(α) sin(β)] Rearrange and Factor = cos(α − β) + i sin(α − β) = cis(α − β) Diﬀerence Identities Deﬁnition of ‘cis’ If we call the denominator D then we get D = [cos(β) + i sin(β)] [cos(β) − i sin(β)] = cos2(β) − i cos(β) sin(β) + i cos(β) sin(β) − i2 sin2(β) Expand = cos2(β) − i2 sin2(β) Simplify = cos2(β) + sin2(β) Again, i2 = −1 Pythagorean Identity = 1 Putting it all together, we get · cos(β) − i sin(β) cos(β) − i sin(β) z w = = = \|z\| \|w\| \|z\| \|w\| cos(α) + i sin(α) cos(β) + i sin(β) cis(α − β) 1 \|z\| \|w\| cis(α − β) and we are done. The next example makes good use of Theorem 11.16. 11.7 Polar Form of Complex Numbers 999 Example 11.7.3. Let z = 2 √ 3 + 2i and w = −1 + i √ 3. Use Theorem 11.16 to ﬁnd the following. 1. zw 2. w5 Write your ﬁnal answers in rectangular form. 3. z w √ √ √ 3)2 + (2)2 = Solution. In order to use Theorem 11.16, we need to write z and w in polar form. For z = 2 3+2i, √ 3 3 . Since 3, we ﬁnd \|z\| = (2 z lies in Quadrant I, we have θ = π 16 = 4. If θ ∈ arg(z), we know tan(θ) = Im(z) 6 + 2πk for integers k. Hence, z = 4cis π 3)2 = 2. For an argument θ of w, we have tan(θ) = 3 + 2πk for integers k and w = 2cis 2π = 8cis π 2cis 2π √ 3 After simplifying, we get zw = −4 3 + 4i. = 3 . For w = −1 + i √ 3 −1 = − we have \|w\| = w lies in Quadrant II, θ = 2π 1. We get zw = 4cis π 6 . We can now proceed. + i sin 5π 6 . = 8 cos 5π 6 = 8cis 5π 6 Re(z) = 2 (−1)2 + ( 6 + 2π 3. Since √ √ √ √ 6 2 3 3 2. We use DeMoivre’s Theorem which yields w5 = 2cis 2π 3 + i sin 4π 3 is coterminal with 4π Since 10π 3 = 25cis 5 · 2π 3 = −16 − 16i = 32cis 10π √ 3 5 3. 3 . 3. Last, but not least, we have = 4 2 is a quadrantal angle, we can ‘see’ the rectangular form by moving out 2 units along the positive real axis, then rotating π 2 radians clockwise to arrive at the point 2 units below 0 on the imaginary axis. The long and short of it is that z 6 − 2π = 2cis − π 2 . Since − π 3 w = −2i. 3 , we get w5 = 32 cos 4π z 4cis( π 6 ) 2cis( 2π 3 ) w 2 cis π = Some remarks are in order. First, the reader may not be sold on using the polar form of complex numbers to multiply complex numbers – especially if they aren’t given in polar form to begin with. Indeed, a lot of work was needed to convert the numbers z and w in Example 11.7.3 into polar form, compute their product, and convert back to rectangular form – certainly more work than is required to multiply out zw = (2 3) the old-fashioned way. However, Theorem 11.16 pays huge dividends when computing powers of complex numbers. Consider how we computed w5 above and compare that to using the Binomial Theorem, Theorem 9.4, to accomplish the same feat by 3)5. Division is tricky in the best of times, and we saved ourselves a lot of expanding (−1 + i time and eﬀort using Theorem 11.16 to ﬁnd and simplify z w using their polar forms as opposed to starting with 2 , rationalizing the denominator, and so forth. 3 + 2i)(−1 + i √ √ √ √ 3+2i √ 3 −1+i There is geometric reason for studying these polar forms and we would be derelict in our duties if we did not mention the Geometry hidden in Theorem 11.16. Take the product rule, for instance. If z = \|z\|cis(α) and w = \|w\|cis(β), the formula zw = \|z\|\|w\|cis(α + β) can be viewed geometrically as a two step process. The multiplication of \|z\| by \|w\| can be interpreted as magnifying10 the distance \|z\| from z to 0, by the factor \|w\|. Adding the argument of w to the argument of z can be interpreted geometrically as a rotation of β radians counter-clockwise.11 Focusing on z and w from Example 10Assuming \|w\| > 1. 11Assuming β > 0. 1000 Applications of Trigonometry 11.7.3, we can arrive at the product zw by plotting z, doubling its distance from 0 (since \|w\| = 2), and rotating 2π 3 radians counter-clockwise. The sequence of diagrams below attempts to describe this process geometrically. Imaginary Axis Imaginary Axis 6i 5i 4i 3i 2i i z\|w\| = 8cis π 6 z = 4cis π 6 zw = 8cis π 6 + 2π 3 6i 5i 4i 3i 2i i z\|w\| = 8cis Real Axis −7 −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 7 Real Axis Multiplying z by \|w\| = 2. Rotating counter-clockwise by Arg(w) = 2π 3 radians. Visualizing zw for z = 4cis π 6 and w = 2cis 2π 3 . We may also visualize division similarly. Here, the formula z \|w\| cis(α − β) may be interpreted as shrinking12 the distance from 0 to z by the factor \|w\|, followed up by a clockwise 13 rotation of β radians. In the case of z and w from Example 11.7.3, we arrive at z w by ﬁrst halving the distance from 0 to z, then rotating clockwise 2π w = \|z\| 3 radians. Imaginary Axis Imaginary Axis 3i 2i i z = 4cis π 6 1 \|w\| z = 2cis π 6 0 1 2 3 Real Axis i −i −2i 1 \|w\| z = 2cis π 6 0 1 2 3 Real Axis zw = 2cis π 6 2π 3 Dividing z by \|w\| = 2. Rotating clockwise by Arg(w) = 2π 3 radians. Visualizing z w for z = 4cis π 6 and w = 2cis 2π 3 . Our last goal of the section is to reverse DeMoivre’s Theorem to extract roots of complex numbers. Deﬁnition 11.4. Let z and w be complex numbers. If there is a natural number n such that wn = z, then w is an nth root of z. Unlike Deﬁnition 5.4 in Section 5.3, we do not specify one particular prinicpal nth root, hence the use of the indeﬁnite article ‘an’ as in ‘an nth root of z’. Using this deﬁnition, both 4 and −4 are 12Again, assuming \|w\| > 1. 13Again, assuming β > 0. 11.7 Polar Form of Complex Numbers 1001 √ √ 16 means the principal square root of 16 as in square roots of 16, while 16 = 4. Suppose we wish to ﬁnd all complex third (cube) roots of 8. Algebraically, we are trying to solve w3 = 8. We √ know that there is only one real solution to this equation, namely w = 3 8 = 2, but if we take the time to rewrite this equation as w3 − 8 = 0 and factor, we get (w − 2) w2 + 2w + 4 = 0. The quadratic factor gives two more cube roots w = −1 ± i 3, for a total of three cube roots of 8. In accordance with Theorem 3.14, since the degree of p(w) = w3 − 8 is three, there are three complex zeros, counting multiplic
ity. Since we have found three distinct zeros, we know these are all of the zeros, so there are exactly three distinct cube roots of 8. Let us now solve this same problem using the machinery developed in this section. To do so, we express z = 8 in polar form. Since z = 8 lies 8 units away on the positive real axis, we get z = 8cis(0). If we let w = \|w\|cis(α) be a polar form of w, the equation w3 = 8 becomes √ w3 = 8 (\|w\|cis(α))3 = 8cis(0) \|w\|3cis(3α) = 8cis(0) DeMoivre’s Theorem The complex number on the left hand side of the equation corresponds to the point with polar coordinates \|w\|3, 3α, while the complex number on the right hand side corresponds to the point with polar coordinates (8, 0). Since \|w\| ≥ 0, so is \|w\|3, which means \|w\|3, 3α and (8, 0) are two polar representations corresponding to the same complex number, both with positive r values. From Section 11.4, we know \|w\|3 = 8 and 3α = 0 + 2πk for integers k. Since \|w\| is a real number, √ we solve \|w\|3 = 8 by extracting the principal cube root to get \|w\| = 3 8 = 2. As for α, we get α = 2πk for integers k. This produces three distinct points with polar coordinates corresponding to 3 k = 0, 1 and 2: speciﬁcally (2, 0), 2, 2π . These correspond to the complex numbers 3 w0 = 2cis(0), w1 = 2cis 2π , respectively. Writing these out in rectangular form 3 yields w0 = 2, w1 = −1 + i 3. While this process seems a tad more involved than our previous factoring approach, this procedure can be generalized to ﬁnd, for example, all of the ﬁfth roots of 32. (Try using Chapter 3 techniques on that!) If we start with a generic complex number in polar form z = \|z\|cis(θ) and solve wn = z in the same manner as above, we arrive at the following theorem. and w2 = 2cis 4π √ and 2, 4π 3 3 and w2 = −1 − i √ 3 Theorem 11.17. The nth roots of a Complex Number: Let z = 0 be a complex number with polar form z = rcis(θ). For each natural number n, z has n distinct nth roots, which we denote by w0, w1, . . . , wn − 1, and they are given by the formula √ wk = n rcis θ n + 2π n k The proof of Theorem 11.17 breaks into to two parts: ﬁrst, showing that each wk is an nth root, and second, showing that the set {wk \| k = 0, 1, . . . , (n − 1)} consists of n diﬀerent complex numbers. To show wk is an nth root of z, we use DeMoivre’s Theorem to show (wk)n = z. 1002 Applications of Trigonometry √ (wk)n = n √ = ( n rcis θ n + 2π r)n cis n · θ n kn n + 2π n k DeMoivre’s Theorem = rcis (θ + 2πk) Since k is a whole number, cos(θ + 2πk) = cos(θ) and sin(θ + 2πk) = sin(θ). Hence, it follows that cis(θ + 2πk) = cis(θ), so (wk)n = rcis(θ) = z, as required. To show that the formula in Theorem 11.17 generates n distinct numbers, we assume n ≥ 2 (or else there is nothing to prove) and note √ that the modulus of each of the wk is the same, namely n r. Therefore, the only way any two of these polar forms correspond to the same number is if their arguments are coterminal – that is, if the arguments diﬀer by an integer multiple of 2π. Suppose k and j are whole numbers between 0 and (n − 1), inclusive, with k = j. Since k and j are diﬀerent, let’s assume for the sake of argument that k > j. Then θ . For this to be an integer multiple of 2π, (k − j) must be a multiple of n. But because of the restrictions on k and j. (Think this through.) Hence, (k − j) is a positive number less than n, so it cannot be a multiple of n. As a result, wk and wj are diﬀerent complex numbers, and we are done. By Theorem 3.14, we know there at most n distinct solutions to wn = z, and we have just found all of them. We illustrate Theorem 11.17 in the next example. n j = 2π n k − θ k−j n n + 2π n + 2π Example 11.7.4. Use Theorem 11.17 to ﬁnd the following: 1. both square roots of z = −2 + 2i √ 3 2. the four fourth roots of z = −16 3. the three cube roots of z = √ √ 2 + i 2 4. the ﬁve ﬁfth roots of z = 1. Solution. √ 1. We start by writing z = −2 + 2i 3 = 4cis 2π 3 θ = 2π in Theorem 11.17, we’ll call them w0 and w1. We get w0 = . To use Theorem 11.17, we identify r = 4, 3 and n = 2. We know that z has two square roots, and in keeping with the notation = 2cis π 3 . In rectangular form, the two square roots of 3. We can check our answers by squaring them and 2 + 2π 2 (1) 3 and w1 = −1 − i = 2cis 4π 3 √ (2π/3) √ 2 + 2π 2 (0) and w1 = (2π/3) 4cis 4cis √ √ z are w0 = 1 + i showing that we get z = −2 + 2i √ 3. 2. Proceeding as above, we get z = −16 = 16cis(π). With r = 16, θ = π and n = 4, we get the √ √ 4 (1) = 4 (0) = 2cis π 16cis π 16cis π , w1 = 4 4 + 2π 4 + 2π four fourth roots of z to be w0 = 4 √ √ 4 . and w3 = 4 , w2 = 4 16cis π 2cis 3π 4 (3) = 2cis 7π 4 (2) = 2cis 5π 16cis −i 2, w2 = − Converting these to rectangular form gives w0 = 2 √ and w3 = 4 + 2π √ 2+i 2, w1 = − 4 + 2π 2 − i 2+i √ 2. 11.7 Polar Form of Complex Numbers 1003 √ √ 3. For z = 2cis π 12 2, we have z = 2cis π √ 4 , w1 = 3 2cis 9π 12 4 and n = 3 the usual computations 2+i √ 2cis 17π yield w0 = 3 If we were 12 to convert these to rectangular form, we would need to use either the Sum and Diﬀerence Identities in Theorem 10.16 or the Half-Angle Identities in Theorem 10.19 to evaluate w0 and w2. Since we are not explicitly told to do so, we leave this as a good, but messy, exercise. . With r = 2, θ = π √ = 3 √ and w2 = 3 2cis 3π 4 . 1 = 1, the roots are w0 = cis(0) = 1, w1 = cis 2π 4. To ﬁnd the ﬁve ﬁfth roots of 1, we write 1 = 1cis(0). We have r = 1, θ = 0 and n = 5. and . The situation here is even graver than in the previous example, since we have 5 . At this stage, we √ Since 5 w4 = cis 8π not developed any identities to help us determine the cosine or sine of 2π could approximate our answers using a calculator, and we leave this as an exercise. , w3 = cis 6π , w2 = cis 4π 5 5 5 5 Now that we have done some computations using Theorem 11.17, we take a step back to look at things geometrically. Essentially, Theorem 11.17 says that to ﬁnd the nth roots of a complex number, we ﬁrst take the nth root of the modulus and divide the argument by n. This gives the ﬁrst root w0. Each succeessive root is found by adding 2π n to the argument, which amounts to rotating w0 by 2π n radians. This results in n roots, spaced equally around the complex plane. As an example of this, we plot our answers to number 2 in Example 11.7.4 below. Imaginary Axis w1 2i i w0 −2 −1 0 1 2 Real Axis w2 −i −2i w3 The four fourth roots of z = −16 equally spaced 2π 4 = π 2 around the plane. We have only glimpsed at the beauty of the complex numbers in this section. The complex plane is without a doubt one of the most important mathematical constructs ever devised. Coupled with Calculus, it is the venue for incredibly important Science and Engineering applications.14 For now, the following exercises will have to suﬃce. 14For more on this, see the beautifully written epilogue to Section 3.4 found on page 294. 1004 Applications of Trigonometry 11.7.1 Exercises In Exercises 1 - 20, ﬁnd a polar representation for the complex number z and then identify Re(z), Im(z), \|z\|, arg(z) and Arg(z). 2. z = 5 + 5i √ 3 3. z = 6i 4. z = −3 √ √ 2 2+3i 1. z = 9 + 9i 5. z = −6 √ 3 + 6i 9. z = −5i 13. z = 3 + 4i 6. z = −2 7. z = − √ √ 2 − 2i 2 11. z = 6 10. z = 2 √ 14. z = 2 + i 15. z = −7 + 24i 16. z = −2 + 6i √ 3 2 − 1 2 i 8. z = −3 − 3i √ 12. z = i 3 7 17. z = −12 − 5i 18. z = −5 − 2i 19. z = 4 − 2i 20. z = 1 − 3i In Exercises 21 - 40, ﬁnd the rectangular form of the given complex number. Use whatever identities are necessary to ﬁnd the exact values. 21. z = 6cis(0) 22. z = 2cis π 6 3π 4 3π 2 √ 26. z = 6cis √ 30. z = 13cis √ 23. z = 7 2cis π 4 24. z = 3cis 27. z = 9cis (π) 28. z = 3cis π 2 4π 3 31. z = 1 2 cis 32. z = 12cis − π 3 7π 4 7π 8 34. z = 2cis 10cis arctan √ 3 arctan − √ 1 3 2 5 12 4 3 √ 36. z = 37. z = 15cis (arctan (−2)) 38. z = 39. z = 50cis π − arctan 7 24 40. z = 1 2 cis π + arctan For Exercises 41 - 52, use z = − your answers in polar form using the principal argument. i and w = 3 + 2 2 − 3i √ 3 3 3 2 √ √ 2 to compute the quantity. Express 41. zw 45. w3 42. z w 46. z5w2 43. w z 47. z3w2 44. z4 48. z2 w 25. z = 4cis 2π 3 29. z = 7cis − 3π 4 33. z = 8cis π 12 35. z = 5cis arctan 11.7 Polar Form of Complex Numbers 1005 49. w z2 50. z3 w2 51. w2 z3 52. 6 w z In Exercises 53 - 64, use DeMoivre’s Theorem to ﬁnd the indicated power of the given complex number. Express your ﬁnal answers in rectangular form. 53. −2 + 2i √ 33 √ 54. (− 3 − i)3 57 + 61 58. 62. (2 + 2i)5 55. (−3 + 3i)4 59)5 63. ( √ 56. ( 3 + i)4 4 √ 3 3 − 1 3 i 60. 64. (1 − i)8 In Exercises 65 - 76, ﬁnd the indicated complex roots. Express your answers in polar form and then convert them into rectangular form. 65. the two square roots of z = 4i 66. the two square roots of z = −25i 67. the two square roots of z = 1 + i √ 3 68. the two square roots of 5 2 − 5 √ 3 2 i 69. the three cube roots of z = 64 70. the three cube roots of z = −125 71. the three cube roots of z = i 72. the three cube roots of z = −8i 73. the four fourth roots of z = 16 74. the four fourth roots of z = −81 75. the six sixth roots of z = 64 76. the six sixth roots of z = −729 77. Use the Sum and Diﬀerence Identities in Theorem 10.16 or the Half Angle Identities in 2 in rectangular form. (See 2 + i Theorem 10.19 to express the three cube roots of z = Example 11.7.4, number 3.) √ √ 78. Use a calculator to approximate the ﬁve ﬁfth roots of 1. (See Example 11.7.4, number 4.) 79. According to Theorem 3.16 in Section 3.4, the polynomial p(x) = x4 + 4 can be factored into the product linear and irreducible quadratic factors. In Exercise 28 in Section 8.7, we showed you how to factor this polynomial into the product of two irreducible quadratic factors using a system of non-linear equations. Now that we can compute the complex fourth roots of −4 directly, we can simply apply the Complex Factorization Theorem, Theorem 3.14, to obtain the linear factorization p(x) = (x − (1 + i))(x − (1 − i))(x − (−1 + i))(x − (−1 − i)). By multiplying the ﬁrst two factors together and then the second two factors
together, thus pairing up the complex conjugate pairs of zeros Theorem 3.15 told us we’d get, we have that p(x) = (x2 − 2x + 2)(x2 + 2x + 2). Use the 12 complex 12th roots of 4096 to factor p(x) = x12 − 4096 into a product of linear and irreducible quadratic factors. 1006 Applications of Trigonometry 80. Complete the proof of Theorem 11.14 by showing that if w = 0 than 1 w = 1 \|w\| . 81. Recall from Section 3.4 that given a complex number z = a+bi its complex conjugate, denoted z, is given by z = a − bi. (a) Prove that \|z\| = \|z\|. √ (b) Prove that \|z\| = (c) Show that Re(z) = zz z + z 2 and Im(z) = z − z 2i (d) Show that if θ ∈ arg(z) then −θ ∈ arg (z). Interpret this result geometrically. (e) Is it always true that Arg (z) = −Arg(z)? 82. Given any natural number n ≥ 2, the n complex nth roots of the number z = 1 are called the In the following exercises, assume that n is a ﬁxed, but arbitrary, nth Roots of Unity. natural number such that n ≥ 2. (a) Show that w = 1 is an nth root of unity. (b) Show that if both wj and wk are nth roots of unity then so is their product wjwk. (c) Show that if wj is an nth root of unity then there exists another nth root of unity wj such that wjwj = 1. Hint: If wj = cis(θ) let wj = cis(2π − θ). You’ll need to verify that wj = cis(2π − θ) is indeed an nth root of unity. 83. Another way to express the polar form of a complex number is to use the exponential function. For real numbers t, Euler’s Formula deﬁnes eit = cos(t) + i sin(t). (a) Use Theorem 11.16 to show that eixeiy = ei(x+y) for all real numbers x and y. (b) Use Theorem 11.16 to show that eixn = ei(nx) for any real number x and any natural number n. (c) Use Theorem 11.16 to show that eix eiy = ei(x−y) for all real numbers x and y. (d) If z = rcis(θ) is the polar form of z, show that z = reit where θ = t radians. (e) Show that eiπ + 1 = 0. (This famous equation relates the ﬁve most important constants in all of Mathematics with the three most fundamental operations in Mathematics.) (f) Show that cos(t) = eit + e−it 2 and that sin(t) = eit − e−it 2i for all real numbers t. 11.7 Polar Form of Complex Numbers 1007 11.7.2 Answers √ 2cis π 4 , Re(z) = 9, 1. z = 9 + 9i = 9 4 + 2πk \| k is an integer and Arg(z) = π arg(z) = π 4 . √ √ , Re(z) = 5, 3 = 10cis π 3, 3 3 + 2πk \| k is an integer and Arg(z) = π 3 . arg(z) = π 2. z = 5 + 5i Im(z) = 9, Im(z) = 5 √ \|z\| = 9 2 \|z\| = 10 3. z = 6i = 6cis π 2 arg(z) = π , Re(z) = 0, \|z\| = 6 2 + 2πk \| k is an integer and Arg(z) = π 2 . Im(z) = 6, 4. z = −3 √ √ 2 + 3i 2 = 6cis 3π 4 , Re(z) = −3 √ 2, Im(z) = 3 √ 2, \|z\| = 6 arg(z) = 3π √ 5. z = −6 arg(z) = 5π 4 + 2πk \| k is an integer and Arg(z) = 3π 4 . , Re(z) = −6 6 + 2πk \| k is an integer and Arg(z) = 5π 6 . 3 + 6i = 12cis 5π 6 √ 3, Im(z) = 6, \|z\| = 12 6. z = −2 = 2cis (π), Re(z) = −2, Im(z) = 0, \|z\| = 2 arg(z) = {(2k + 1)π \| k is an integer} and Arg(z) = π. √ 3 7. z = − 2 − 1 arg(z) = 7π 8. z = −3 − 3i = 3 arg(z) = 5π √ 3 2 , 6 , Re(z) = − 2 i = cis 7π Im(z) = − 1 2 , 6 + 2πk \| k is an integer and Arg(z) = − 5π 6 . , Re(z) = −3, Im(z) = −3, 4 + 2πk \| k is an integer and Arg(z) = − 3π 4 . 2cis 5π 4 √ \|z\| = 1 √ \|z\| = 3 2 9. z = −5i = 5cis 3π 2 arg(z) = 3π √ , Re(z) = 0, Im(z) = −5, \|z\| = 5 2 + 2πk \| k is an integer and Arg(z) = − π 2 . √ , Re(z) = 2 4 + 2πk \| k is an integer and Arg(z) = − π 4 . 2 = 4cis 7π 4 √ 2, 10. z = 2 2 − 2i arg(z) = 7π √ Im(z) = −2 2, \|z\| = 4 11. z = 6 = 6cis (0), Re(z) = 6, Im(z) = 0, \|z\| = 6 arg(z) = {2πk \| k is an integer} and Arg(z) = 0. √ Im(z) = 3 , Re(z) = 0, √ 12. z = i 3 7, 7cis π 2 √ \|z\| = 3 7 2 + 2πk \| k is an integer and Arg(z) = π 2 . √ 7 = 3 arg(z) = π 13. z = 3 + 4i = 5cis arctan 4 3 , Re(z) = 3, Im(z) = 4, \|z\| = 5 arg(z) = arctan 4 3 + 2πk \| k is an integer and Arg(z) = arctan 4 3 . 1008 14. z = √ 2 + i = √ arctan √ 2 2 , Re(z) = √ 2, Im(z) = 1, \|z\| = arg(z) = arctan + 2πk \| k is an integer and Arg(z) = arctan 3cis √ 2 2 √ 3 √ 2 2 . Applications of Trigonometry 15. z = −7 + 24i = 25cis π − arctan 24 7 , Re(z) = −7, Im(z) = 24, \|z\| = 25 arg(z) = π − arctan 24 7 + 2πk \| k is an integer and Arg(z) = π − arctan 24 7 . 16. z = −2 + 6i = 2 √ 10cis (π − arctan (3)), Re(z) = −2, Im(z) = 6, √ \|z\| = 2 10 arg(z) = {π − arctan (3) + 2πk \| k is an integer} and Arg(z) = π − arctan (3). 17. z = −12 − 5i = 13cis π + arctan 5 12 , Re(z) = −12, Im(z) = −5, \|z\| = 13 arg(z) = π + arctan 5 12 + 2πk \| k is an integer and Arg(z) = arctan 5 12 18. z = −5 − 2i = √ 29cis π + arctan 2 5 , Re(z) = −5, Im(z) = −2, arg(z) = π + arctan 2 5 5cis arctan − 1 2 19. z = 4 − 2i = 2 √ + 2πk \| k is an integer and Arg(z) = arctan 2 5 √ , Re(z) = 4, Im(z) = −2, \|z\| = 2 arg(z) = arctan − 1 2 + 2πk \| k is an integer and Arg(z) = arctan − 1 2 − π. √ \|z\| = 29 − π. 5 = − arctan 1 2 . 20. z = 1 − 3i = √ 10cis (arctan (−3)), Re(z) = 1, Im(z) = −3, \|z\| = √ 10 arg(z) = {arctan (−3) + 2πk \| k is an integer} and Arg(z) = arctan (−3) = − arctan(3). 21. z = 6cis(0) = 6 √ 23. z = 7 25. z = 4cis 2π 3 2cis π 4 = −2 + 2i = 7 + 7i √ 3 27. z = 9cis (π) = −9 √ 3 + i = 22. z = 2cis π 6 24. z = 3cis π 2 6cis 3π 4 26. z = √ = 3i √ = − √ 3 + i 3 28. z = 3cis 4π 3 √ 30. z = = − 3 2 − 3i √ √ 3 2 13 13cis 3π 2 = −i 32. z = 12cis − π 3 = 6 − 6i √ 3 29. z = 7cis − 3π 4 31. z = 1 2 cis 7π 33. z = 8cis π 12 35. z = 5cis arctan + 4i 2 − √ 3 = 3 + 4i √ 37. z = 15cis (arctan (−2)) = 3 5 − 6i √ 5 √ 38. z = 34. z = 2cis 7π 8 √ 36 10cis arctan 1 3 √ 3cis arctan − 2 = 1 − i √ 2 39. z = 50cis π − arctan 7 24 = −48 + 14i 40. z = 1 2 cis π + arctan 5 12 = − 6 13 − 5i 26 11.7 Polar Form of Complex Numbers 1009 In Exercises 41 - 52, we have that z = − 3 we get the following. √ 3 2 + 3 2 i = 3cis 5π 6 41. zw = 18cis 7π 12 44. z4 = 81cis − 2π 3 47. z3w2 = 972cis(0) 50. z3 w2 = 3 4 cis(π) √ 53. −2 + 2i 33 = 64 and w = 3 √ √ 2 − 3i 2 = 6cis − π 4 so 43. w z = 2cis 11π 46. z5w2 = 8748cis − π 3 12 42. z w = 1 2 cis − 11π 12 45. w3 = 216cis − 3π 4 48. z2 w = 3 2 cis − π 12 49. w z2 = 2 3 cis π 12 51. w2 z3 = 4 √ 3 cis(π) 54. (− 3 − i)3 = −8i 52. w z 6 = 64cis − π 2 55. (−3 + 3i)4 = −324 √ 56. ( 3 + i)4 = −8 + 8i √ 3 57. 5 2 + 5 2 i3 = − 125 4 + 125 58. 59. 3 2 − 3 2 i3 = − 27 4 − 27 4 i 60. 62. (2 + 2i)5 = −128 − 128i 63. ( 3 − i)5 = −16 3 − 16i 81 − 8i √ √ 3 81 √ 2 2 + 4 √ 2 2 i 61. = −1 64. (1 − i)8 = 16 65. Since z = 4i = 4cis π 2 w0 = 2cis π 4 = √ 2 + i 2 we have √ 66. Since z = −25i = 25cis 3π 2 we have w0 = 5cis 3π w1 = 2cis 5π 4 √ = − √ 2 − i 2 w1 = 5cis 7π 67. Since z = 1 + i √ 3 = 2cis π 3 we have w0 = √ 2cis w1 = √ 2cis 7π 68. Since z = 5 2 − 5 3 2 i = 5cis 5π 3 we have √ w0 = √ 5cis 5π 6 = − √ 15 2 + √ 5 2 i w1 = √ 5cis 11π 6 = √ 15 2 − √ 5 2 i 69. Since z = 64 = 64cis (0) we have w0 = 4cis (0) = 4 w1 = 4cis 2π 3 = −2 + 2i √ 3 w2 = 4cis 4π 3 = −2 − 2i √ 3 1010 Applications of Trigonometry 70. Since z = −125 = 125cis (π) we have w0 = 5cis w1 = 5cis (π) = −5 w2 = 5cis 5π 71. Since z = i = cis π 2 we have w0 = cis w1 = cis 5π w2 = cis 3π 2 = −i 72. Since z = −8i = 8cis 3π 2 we have w0 = 2cis π 2 = 2i w1 = 2cis 7π 6 √ = − 3 − i w2 = cis 11π 6 = √ 3 − i 73. Since z = 16 = 16cis (0) we have w0 = 2cis (0) = 2 w2 = 2cis (π) = −2 74. Since z = −81 = 81cis (π) we have w1 = 2cis π w3 = 2cis 3π = 2i = −2i 2 2 w0 = 3cis π 4 = 3 2 2 + 3 √ w2 = 3cis 5π w1 = 3cis 3π 4 w3 = 3cis 7π 75. Since z = 64 = 64cis(0) we have w0 = 2cis(0) = 2 w3 = 2cis (π) = −2 √ w1 = 2cis π w4 = 2cis − 2π 3 3 = 1 + 3i = −1 − √ w2 = 2cis 2π 3i w5 = 2cis − π = −1 + √ = 1 − 3 3i 3 √ 3i 76. Since z = −729 = 729cis(π) we have w0 = 3cis w1 = 3cis π 2 = 3i w2 = 3cis 5π w3 = 3cis 7π w4 = 3cis − 3π 2 = −3i w5 = 3cis − 11π 77. Note: In the answers for w0 and w2 the ﬁrst rectangular form comes from applying the 3 − π appropriate Sum or Diﬀerence Identity ( π 4 , respectively) and the second comes from using the Half-Angle Identities. √ √ = 3 2 √ w0 = 3 4 and 17π 12 = 2π 3 + 3π 12 = 2cis π 12 2+ 2 2− 2 6− 6 4 + i − √ 2 √ w1 = 3 2cis 3π 4 √ w2 = 3 2cis 17π 12 √ − √ 6 √ = 3 2 2− 4 √ √ 3 2− 2 + i √ 3 √ 2+ 2 11.7 Polar Form of Complex Numbers 1011 5 78. w0 = cis(0) = 1 w1 = cis 2π w2 = cis 4π w3 = cis 6π w4 = cis 8π ≈ 0.309 + 0.951i ≈ −0.809 + 0.588i ≈ −0.809 − 0.588i ≈ 0.309 − 0.951i 5 5 5 79. p(x) = x12 −4096 = (x−2)(x+2)(x2 +4)(x2 −2x+4)(x2 +2x+4)(x2 −2 √ 3x+4)(x2 +2 √ 3+4) 1012 Applications of Trigonometry 11.8 Vectors As we have seen numerous times in this book, Mathematics can be used to model and solve real-world problems. For many applications, real numbers suﬃce; that is, real numbers with the appropriate units attached can be used to answer questions like “How close is the nearest Sasquatch nest?” There are other times though, when these kinds of quantities do not suﬃce. Perhaps it is important to know, for instance, how close the nearest Sasquatch nest is as well as the direction in which it lies. (Foreshadowing the use of bearings in the exercises, perhaps?) To answer questions like these which involve both a quantitative answer, or magnitude, along with a direction, we use the mathematical objects called vectors.1 A vector is represented geometrically as a directed line segment where the magnitude of the vector is taken to be the length of the line segment and the direction is made clear with the use of an arrow at one endpoint of the segment. When referring to vectors in this text, we shall adopt2 the ‘arrow’ notation, so the symbol v is read as ‘the vector v’. Below is a typical vector v with endpoints P (1, 2) and Q (4, 6). The point P is called the initial point or tail of v and the point Q is called the terminal point or head of v. Since we can reconstruct −−→ v completely from P and Q, we write v = P Q, where the order of points P (initial point) and Q (terminal point) is important. (Think about this before moving on.) Q (4, 6) P (1, 2) −−→ P Q v = While it is true that P and Q completely determine v, it is important to note that since vectors are deﬁned in terms of their two characteristics, magnitude and direction, any directed line segment with the same length and direction as v is considered to be the same vector as v, reg
ardless of its initial point. In the case of our vector v above, any vector which moves three units to the right and four up3 from its initial point to arrive at its terminal point is considered the same vector as v. The notation we use to capture this idea is the component form of the vector, v = 3, 4, where the ﬁrst number, 3, is called the x-component of v and the second number, 4, is called the y-component of v. If we wanted to reconstruct v = 3, 4 with initial point P (−2, 3), then we would ﬁnd the terminal point of v by adding 3 to the x-coordinate and adding 4 to the y-coordinate to obtain the terminal point Q(1, 7), as seen below. 1The word ‘vector’ comes from the Latin vehere meaning ‘to convey’ or ‘to carry.’ 2Other textbook authors use bold vectors such as v. We ﬁnd that writing in bold font on the chalkboard is inconvenient at best, so we have chosen the ‘arrow’ notation. 3If this idea of ‘over’ and ‘up’ seems familiar, it should. The slope of the line segment containing v is 4 3 . 11.8 Vectors 1013 Q (1, 7) up 4 P (−2, 3) over 3 v = 3, 4 with initial point P (−2, 3). The component form of a vector is what ties these very geometric objects back to Algebra and ultimately Trigonometry. We generalize our example in our deﬁnition below. Deﬁnition 11.5. Suppose v is represented by a directed line segment with initial point P (x0, y0) and terminal point Q (x1, y1). The component form of v is given by −−→ P Q = x1 − x0, y1 − y0 v = Using the language of components, we have that two vectors are equal if and only if their corresponding components are equal. That is, v1, v2 = v 1 and v2 = v 2. (Again, think about this before reading on.) We now set about deﬁning operations on vectors. Suppose we are given two vectors v and w. The sum, or resultant vector v + w is obtained as follows. First, plot v. Next, plot w so that its initial point is the terminal point of v. To plot the vector v + w we begin at the initial point of v and end at the terminal point of w. It is helpful to think of the vector v + w as the ‘net result’ of moving along v then moving along w. 2 if and only if v1 = v 1, v v + w w v v, w, and v + w Our next example makes good use of resultant vectors and reviews bearings and the Law of Cosines.4 Example 11.8.1. A plane leaves an airport with an airspeed5 of 175 miles per hour at a bearing of N40◦E. A 35 mile per hour wind is blowing at a bearing of S60◦E. Find the true speed of the plane, rounded to the nearest mile per hour, and the true bearing of the plane, rounded to the nearest degree. 4If necessary, review page 905 and Section 11.3. 5That is, the speed of the plane relative to the air around it. If there were no wind, plane’s airspeed would be the same as its speed as observed from the ground. How does wind aﬀect this? Keep reading! 1014 Applications of Trigonometry Solution: For both the plane and the wind, we are given their speeds and their directions. Coupling speed (as a magnitude) with direction is the concept of velocity which we’ve seen a few times before in this textbook.6 We let v denote the plane’s velocity and w denote the wind’s velocity in the diagram below. The ‘true’ speed and bearing is found by analyzing the resultant vector, v + w. From the vector diagram, we get a triangle, the lengths of whose sides are the magnitude of v, which is 175, the magnitude of w, which is 35, and the magnitude of v + w, which we’ll call c. From the given bearing information, we go through the usual geometry to determine that the angle between the sides of length 35 and 175 measures 100◦. N 40◦ w 60◦ v v + w N 35 100◦ 175 α c 40◦ E E 60◦ 31850 − 12250 cos(100◦) ≈ 184, which means the From the Law of Cosines, we determine c = true speed of the plane is (approximately) 184 miles per hour. To determine the true bearing of the plane, we need to determine the angle α. Using the Law of Cosines once more,7 we ﬁnd cos(α) = c2+29400 so that α ≈ 11◦. Given the geometry of the situation, we add α to the given 40◦ and ﬁnd the true bearing of the plane to be (approximately) N51◦E. 350c Our next step is to deﬁne addition of vectors component-wise to match the geometric action.8 Deﬁnition 11.6. Suppose v = v1, v2 and w = w1, w2. The vector v + w is deﬁned by v + w = v1 + w1, v2 + w2 Example 11.8.2. Let v = 3, 4 and suppose w = and interpret this sum geometrically. −−→ P Q where P (−3, 7) and Q(−2, 5). Find v + w Solution. Before can add the vectors using Deﬁnition 11.6, we need to write w in component form. Using Deﬁnition 11.5, we get w = −2 − (−3), 5 − 7 = 1, −2. Thus 6See Section 10.1.1, for instance. 7Or, since our given angle, 100◦, is obtuse, we could use the Law of Sines without any ambiguity here. 8Adding vectors ‘component-wise’ should seem hauntingly familiar. Compare this with how matrix addition was deﬁned in section 8.3. In fact, in more advanced courses such as Linear Algebra, vectors are deﬁned as 1 × n or n × 1 matrices, depending on the situation. 11.8 Vectors 1015 v + w = 3, 4 + 1, −2 = 3 + 1, 4 + (−2) = 4, 2 To visualize this sum, we draw v with its initial point at (0, 0) (for convenience) so that its terminal point is (3, 4). Next, we graph w with its initial point at (3, 4). Moving one to the right and two down, we ﬁnd the terminal point of w to be (4, 2). We see that the vector v + w has initial point (0, 0) and terminal point (4, 2) so its component form is 4, 2, as required In order for vector addition to enjoy the same kinds of properties as real number addition, it is necessary to extend our deﬁnition of vectors to include a ‘zero vector’, 0 = 0, 0. Geometrically, 0 represents a point, which we can think of as a directed line segment with the same initial and terminal points. The reader may well object to the inclusion of 0, since after all, vectors are supposed to have both a magnitude (length) and a direction. While it seems clear that the magnitude of 0 should be 0, it is not clear what its direction is. As we shall see, the direction of 0 is in fact undeﬁned, but this minor hiccup in the natural ﬂow of things is worth the beneﬁts we reap by including 0 in our discussions. We have the following theorem. Theorem 11.18. Properties of Vector Addition Commutative Property: For all vectors v and w, v + w = w + v. Associative Property: For all vectors u, v and w, (u + v) + w = u + (v + w). Identity Property: The vector 0 acts as the additive identity for vector addition. That is, for all vectors v. Inverse Property: Every vector v has a unique additive inverse, denoted −v. That is, for every vector v, there is a vector −v so that v + (−v) = (−v) + v = 0. 1016 Applications of Trigonometry The properties in Theorem 11.18 are easily veriﬁed using the deﬁnition of vector addition.9 For the commutative property, we note that if v = v1, v2 and w = w1, w2 then v + w = v1, v2 + w1, w2 = v1 + w1, v2 + w2 = w1 + v1, w2 + v2 = w + v Geometrically, we can ‘see’ the commutative property by realizing that the sums v + w and w + v are the same directed diagonal determined by the parallelogram below Demonstrating the commutative property of vector addition. The proofs of the associative and identity properties proceed similarly, and the reader is encouraged to verify them and provide accompanying diagrams. The existence and uniqueness of the additive inverse is yet another property inherited from the real numbers. Given a vector v = v1, v2, suppose we wish to ﬁnd a vector w = w1, w2 so that v + w = 0. By the deﬁnition of vector addition, we have v1 + w1, v2 + w2 = 0, 0, and hence, v1 + w1 = 0 and v2 + w2 = 0. We get w1 = −v1 and w2 = −v2 so that w = −v1, −v2. Hence, v has an additive inverse, and moreover, it is unique and can be obtained by the formula −v = −v1, −v2. Geometrically, the vectors v = v1, v2 and −v = −v1, −v2 have the same length, but opposite directions. As a result, when adding the vectors geometrically, the sum v + (−v) results in starting at the initial point of v and ending back at the initial point of v, or in other words, the net result of moving v then −v is not moving at all. v −v Using the additive inverse of a vector, we can deﬁne the diﬀerence of two vectors, v − w = v + (− w). If v = v1, v2 and w = w1, w2 then 9The interested reader is encouraged to compare Theorem 11.18 and the ensuing discussion with Theorem 8.3 in Section 8.3 and the discussion there. 11.8 Vectors 1017 v − w = v + (− w) = v1, v2 + −w1, −w2 = v1 + (−w1) , v2 + (−w2) = v1 − w1, v2 − w2 In other words, like vector addition, vector subtraction works component-wise. To interpret the vector v − w geometrically, we note w + (v − w) = w + (v + (− w)) Deﬁnition of Vector Subtraction = w + ((− w) + v) Commutativity of Vector Addition = ( w + (− w)) + v Associativity of Vector Addition = 0 + v = v Deﬁnition of Additive Inverse Deﬁnition of Additive Identity This means that the ‘net result’ of moving along w then moving along v − w is just v itself. From the diagram below, we see that v − w may be interpreted as the vector whose initial point is the terminal point of w and whose terminal point is the terminal point of v as depicted below. It is also worth mentioning that in the parallelogram determined by the vectors v and w, the vector v − w is one of the diagonals – the other being v + w Next, we discuss scalar multiplication – that is, taking a real number times a vector. We deﬁne scalar multiplication for vectors in the same way we deﬁned it for matrices in Section 8.3. Deﬁnition 11.7. If k is a real number and v = v1, v2, we deﬁne kv by kv = k v1, v2 = kv1, kv2 Scalar multiplication by k in vectors can be understood geometrically as scaling the vector (if k > 0) or scaling the vector and reversing its direction (if k < 0) as demonstrated below. 1018 Applications of Trigonometry v 2v 1 2 v −2v Note that, by deﬁnition 11.7, (−1)v = (−1) v1, v2 = (−1)v1, (−1)v2 = −v1, −v2 = −v. This, and other properties of scalar multiplication are summarized below. Theorem 11.19. Properties of Scalar Multiplication Associative
Property: For every vector v and scalars k and r, (kr)v = k(rv). Identity Property: For all vectors v, 1v = v. Additive Inverse Property: For all vectors v, −v = (−1)v. Distributive Property of Scalar Multiplication over Scalar Addition: For every vector v and scalars k and r, (k + r)v = kv + rv Distributive Property of Scalar Multiplication over Vector Addition: For all vectors v and w and scalars k, k(v + w) = kv + k w Zero Product Property: If v is vector and k is a scalar, then kv = 0 if and only if k = 0 or v = 0 The proof of Theorem 11.19, like the proof of Theorem 11.18, ultimately boils down to the deﬁnition of scalar multiplication and properties of real numbers. For example, to prove the associative property, we let v = v1, v2. If k and r are scalars then (kr)v = (kr) v1, v2 = (kr)v1, (kr)v2 Deﬁnition of Scalar Multiplication = k(rv1), k(rv2) Associative Property of Real Number Multiplication = k rv1, rv2 = k (r v1, v2) = k(rv) Deﬁnition of Scalar Multiplication Deﬁnition of Scalar Multiplication 11.8 Vectors 1019 The remaining properties are proved similarly and are left as exercises. Our next example demonstrates how Theorem 11.19 allows us to do the same kind of algebraic manipulations with vectors as we do with variables – multiplication and division of vectors notwithstanding. If the pedantry seems familiar, it should. This is the same treatment we gave Example 8.3.1 in Section 8.3. As in that example, we spell out the solution in excruciating detail to encourage the reader to think carefully about why each step is justiﬁed. Example 11.8.3. Solve 5v − 2 (v + 1, −2) = 0 for v. Solution. 5v − 2 (v + 1, −2) = 0 5v + (−1) [2 (v + 1, −2)] = 0 5v + [(−1)(2)] (v + 1, −2) = 0 5v + (−2) (v + 1, −2) = 0 5v + [(−2)v + (−2) 1, −2] = 0 5v + [(−2)v + (−2)(1), (−2)(−2)] = 0 [5v + (−2)v] + −2, 4 = 0 (5 + (−2))v + −2, 4 = 0 3v + −2, 4 = 0 (3v + −2, 4) + (− −2, 4) = 0 + (− −2, 4) 3v + [−2, 4 + (− −2, 4)] = 0 + (−1) −2, 4 3v + 0 = 0 + (−1)(−2), (−1)(4) 3v = 2, −4 1 3 1 3 (3v) = 1 (3) v = 1 3 1v = 2 v = 2 3 (2, −4) (2), (−4) A vector whose initial point is (0, 0) is said to be in standard position. If v = v1, v2 is plotted in standard position, then its terminal point is necessarily (v1, v2). (Once more, think about this before reading on.) y (v1, v2) v = v1, v2 in standard position. x 1020 Applications of Trigonometry Plotting a vector in standard position enables us to more easily quantify the concepts of magnitude and direction of the vector. We can convert the point (v1, v2) in rectangular coordinates to a pair (r, θ) in polar coordinates where r ≥ 0. The magnitude of v, which we said earlier was length 2 and is denoted by v. From Section 11.4, we of the directed line segment, is r = know v1 = r cos(θ) = v cos(θ) and v2 = r sin(θ) = v sin(θ). From the deﬁnition of scalar multiplication and vector equality, we get 1 + v2 v2 v = v1, v2 = v cos(θ), v sin(θ) = v cos(θ), sin(θ) This motivates the following deﬁnition. Deﬁnition 11.8. Suppose v is a vector with component form v = v1, v2. Let (r, θ) be a polar representation of the point with rectangular coordinates (v1, v2) with r ≥ 0. The magnitude of v, denoted v, is given by v = r = 1 + v2 v2 2 If v = 0, the (vector) direction of v, denoted ˆv is given by ˆv = cos(θ), sin(θ) Taken together, we get v = v cos(θ), v sin(θ). A few remarks are in order. First, we note that if v = 0 then even though there are inﬁnitely many angles θ which satisfy Deﬁnition 11.8, the stipulation r > 0 means that all of the angles are coterminal. Hence, if θ and θ both satisfy the conditions of Deﬁnition 11.8, then cos(θ) = cos(θ) and sin(θ) = sin(θ), and as such, cos(θ), sin(θ) = cos(θ), sin(θ) making ˆv is well-deﬁned.10 If v = 0, then v = 0, 0, and we know from Section 11.4 that (0, θ) is a polar representation for the origin for any angle θ. For this reason, ˆ0 is undeﬁned. The following theorem summarizes the important facts about the magnitude and direction of a vector. Theorem 11.20. Properties of Magnitude and Direction: Suppose v is a vector. v ≥ 0 and v = 0 if and only if v = 0 For all scalars k, k v = \|k\|v. If v = 0 then v = vˆv, so that ˆv = 1 v v. 1 + v2 v2 v2 1 + v2 1 + v2 2 = 0 if and only of v2 The proof of the ﬁrst property in Theorem 11.20 is a direct consequence of the deﬁnition of v. If v = v1, v2, then v = 2 which is by deﬁnition greater than or equal to 0. Moreover, 2 = 0 if and only if v1 = v2 = 0. Hence, v = 0 if and only if v = 0, 0 = 0, as required. The second property is a result of the deﬁnition of magnitude and scalar multiplication along with a propery of radicals. If v = v1, v2 and k is a scalar then 10If this all looks familiar, it should. The interested reader is invited to compare Deﬁnition 11.8 to Deﬁnition 11.2 in Section 11.7. 11.8 Vectors 1021 k v = k v1, v2 = kv1, kv2 Deﬁnition of scalar multiplication 2 = = = √ (kv1)2 + (kv2)2 Deﬁnition of magnitude 1 + k2v2 k2v2 1 + v2 k2(v2 2 ) k2 v2 1 + v2 = = \|k\| 1 + v2 v2 = \|k\|v Product Rule for Radicals k2 = \|k\| Since √ 2 2 The equation v = vˆv in Theorem 11.20 is a consequence of the deﬁnitions of v and ˆv and was worked out in the discussion just prior to Deﬁnition 11.8 on page 1020. In words, the equation v = vˆv says that any given vector is the product of its magnitude and its direction – an important v is a result of concept to keep in mind when studying and using vectors. The equation ˆv = solving v = vˆv for ˆv by multiplying11 both sides of the equation by 1 v and using the properties of Theorem 11.19. We are overdue for an example. 1 v Example 11.8.4. 1. Find the component form of the vector v with v = 5 so that when v is plotted in standard position, it lies in Quadrant II and makes a 60◦ angle12 with the negative x-axis. 2. For v = 3, −3 √ 3, ﬁnd v and θ, 0 ≤ θ < 2π so that v = v cos(θ), sin(θ). 3. For the vectors v = 3, 4 and w = 1, −2, ﬁnd the following. (a) ˆv (b) v − 2 w (c) v − 2 w (d) ˆw Solution. 1. We are told that v = 5 and are given information about its direction, so we can use the formula v = vˆv to get the component form of v. To determine ˆv, we appeal to Deﬁnition 11.8. We are told that v lies in Quadrant II and makes a 60◦ angle with the negative x-axis, so the polar form of the terminal point of v, when plotted in standard position is (5, 120◦). (See the diagram below.) Thus ˆv = cos (120◦) , sin (120◦) = , so v = vˆv = √ − . 11Of course, to go from v = vˆv to ˆv = 1 v v, we are essentially ‘dividing both sides’ of the equation by the scalar v. The authors encourage the reader, however, to work out the details carefully to gain an appreciation of the properties in play. 12Due to the utility of vectors in ‘real-world’ applications, we will usually use degree measure for the angle when giving the vector’s direction. However, since Carl doesn’t want you to forget about radians, he’s made sure there are examples and exercises which use them. 1022 Applications of Trigonometry y 5 4 3 2 1 v 60◦ θ = 120◦ −3 −2 −1 1 2 3 x 2. For v = 3, −3 √ 3, we get v = (3)2 + (−3 √ ﬁnd the θ we’re after by converting the point with rectangular coordinates (3, −3 form (r, θ) where r = v > 0. From Section 11.4, we have tan(θ) = −3 (3, −3 may check our answer by verifying v = 3, −3 3) is a point in Quadrant IV, θ is a Quadrant IV angle. Hence, we pick θ = 5π √ 3)2 = 6. In light of Deﬁnition 11.8, we can 3) to polar √ 3. Since 3 . We √ 3 = − . 3 = 6 cos 5π 3 , sin 5π 3 √ √ 3 3. (a) Since we are given the component form of v, we’ll use the formula ˆv = 5 3, 4, we have v = 25 = 5. Hence, ˆv = 1 32 + 42 = √ √ 5 1 v . v. For (b) We know from our work above that v = 5, so to ﬁnd v−2 w, we need only ﬁnd w. 12 + (−2)2 = Since w = 1, −2, we get w = 5. Hence. (c) In the expression v −2 w, notice that the arithmetic on the vectors comes ﬁrst, then the magnitude. Hence, our ﬁrst step is to ﬁnd the component form of the vector v − 2 w. We get v − 2 w = 3, 4 − 2 1, −2 = 1, 8. Hence, v − 2 w = 1, 8 = 65. √ 12 + 82 = √ √ (d) To ﬁnd ˆw, we ﬁrst need ˆw. Using the formula ˆw = 1 w which we found the in the previous problem, we get ˆw = 1√ 5 √ √ 5 5 . Hence, ˆw = 25 + 20 . 25 = w along with w = 5, 1, −2 = 1√ 5 , − 2√ 5 = The process exempliﬁed by number 1 in Example 11.8.4 above by which we take information about the magnitude and direction of a vector and ﬁnd the component form of a vector is called resolving a vector into its components. As an application of this process, we revisit Example 11.8.1 below. Example 11.8.5. A plane leaves an airport with an airspeed of 175 miles per hour with bearing N40◦E. A 35 mile per hour wind is blowing at a bearing of S60◦E. Find the true speed of the plane, rounded to the nearest mile per hour, and the true bearing of the plane, rounded to the nearest degree. Solution: We proceed as we did in Example 11.8.1 and let v denote the plane’s velocity and w denote the wind’s velocity, and set about determining v + w. If we regard the airport as being 11.8 Vectors 1023 at the origin, the positive y-axis acting as due north and the positive x-axis acting as due east, we see that the vectors v and w are in standard position and their directions correspond to the angles 50◦ and −30◦, respectively. Hence, the component form of v = 175 cos(50◦), sin(50◦) = 175 cos(50◦), 175 sin(50◦) and the component form of w = 35 cos(−30◦), 35 sin(−30◦). Since we have no convenient way to express the exact values of cosine and sine of 50◦, we leave both vectors in terms of cosines and sines.13 Adding corresponding components, we ﬁnd the resultant vector v + w = 175 cos(50◦) + 35 cos(−30◦), 175 sin(50◦) + 35 sin(−30◦). To ﬁnd the ‘true’ speed of the plane, we compute the magnitude of this resultant vector v + w = (175 cos(50◦) + 35 cos(−30◦))2 + (175 sin(50◦) + 35 sin(−30◦))2 ≈ 184 Hence, the ‘true’ speed of the plane is approximately 184 miles per hour. To ﬁnd the true bearing, we need to ﬁnd the angle θ which corresponds to the polar form (r, θ), r > 0, of the point (x, y) = (175 cos(50◦) + 35 cos(−30◦), 175 sin(
50◦) + 35 sin(−30◦)). Since both of these coordinates are positive,14 we know θ is a Quadrant I angle, as depicted below. Furthermore, tan(θ) = y x = 175 sin(50◦) + 35 sin(−30◦) 175 cos(50◦) + 35 cos(−30◦) , so using the arctangent function, we get θ ≈ 39◦. Since, for the purposes of bearing, we need the angle between v + w and the positive y-axis, we take the complement of θ and ﬁnd the ‘true’ bearing of the plane to be approximately N51◦E. y (N) y (N) v v v + w 40◦ 50◦ −30◦ w 60◦ x (E) θ w x (E) In part 3d of Example 11.8.4, we saw that ˆw = 1. Vectors with length 1 have a special name and are important in our further study of vectors. Deﬁnition 11.9. Unit Vectors: Let v be a vector. If v = 1, we say that v is a unit vector. 13Keeping things ‘calculator’ friendly, for once! 14Yes, a calculator approximation is the quickest way to see this, but you can also use good old-fashioned inequalities and the fact that 45◦ ≤ 50◦ ≤ 60◦. 1024 Applications of Trigonometry 1 v If v is a unit vector, then necessarily, v = vˆv = 1 · ˆv = ˆv. Conversely, we leave it as an exercise15 In practice, if v is a unit v is a unit vector for any nonzero vector v. to show that ˆv = vector we write it as ˆv as opposed to v because we have reserved the ‘ˆ’ notation for unit vectors. 1 The process of multiplying a nonzero vector by the factor v to produce a unit vector is called ‘normalizing the vector,’ and the resulting vector ˆv is called the ‘unit vector in the direction of v’. The terminal points of unit vectors, when plotted in standard position, lie on the Unit Circle. (You should take the time to show this.) As a result, we visualize normalizing a nonzero vector v as shrinking16 its terminal point, when plotted in standard position, back to the Unit Circle. y v 1 ˆv −1 1 x −1 Visualizing vector normalization ˆv = 1 v v Of all of the unit vectors, two deserve special mention. Deﬁnition 11.10. The Principal Unit Vectors: The vector ˆı is deﬁned by ˆı = 1, 0 The vector ˆ is deﬁned by ˆı = 0, 1 We can think of the vector ˆı as representing the positive x-direction, while ˆ represents the positive y-direction. We have the following ‘decomposition’ theorem.17 Theorem 11.21. Principal Vector Decomposition Theorem: Let v be a vector with component form v = v1, v2. Then v = v1ˆı + v2ˆ. The proof of Theorem 11.21 is straightforward. Since ˆı = 1, 0 and ˆ = 0, 1, we have from the deﬁnition of scalar multiplication and vector addition that v1ˆı + v2ˆ = v1 1, 0 + v2 0, 1 = v1, 0 + 0, v2 = v1, v2 = v 15One proof uses the properties of scalar multiplication and magnitude. If v = 0, consider ˆv = 1 v . Use v the fact that v ≥ 0 is a scalar and consider factoring. 16. . . if v > 1 . . . 17We will see a generalization of Theorem 11.21 in Section 11.9. Stay tuned! 11.8 Vectors 1025 Geometrically, the situation looks like this: y v2ˆ v = v1, v2 ˆ ˆı v1ˆı x v = v1, v2 = v1ˆı + v2ˆ. We conclude this section with a classic example which demonstrates how vectors are used to model forces. A ‘force’ is deﬁned as a ‘push’ or a ‘pull.’ The intensity of the push or pull is the magnitude of the force, and is measured in Netwons (N) in the SI system or pounds (lbs.) in the English system.18 The following example uses all of the concepts in this section, and should be studied in great detail. Example 11.8.6. A 50 pound speaker is suspended from the ceiling by two support braces. If one of them makes a 60◦ angle with the ceiling and the other makes a 30◦ angle with the ceiling, what are the tensions on each of the supports? Solution. We represent the problem schematically below and then provide the corresponding vector diagram. 30◦ 60◦ 30◦ 60◦ T1 T2 30◦ 60◦ w 50 lbs. We have three forces acting on the speaker: the weight of the speaker, which we’ll call w, pulling the speaker directly downward, and the forces on the support rods, which we’ll call T1 and T2 (for ‘tensions’) acting upward at angles 60◦ and 30◦, respectively. We are looking for the tensions on the support, which are the magnitudes T1 and T2. In order for the speaker to remain stationary,19 we require w + T1 + T2 = 0. Viewing the common initial point of these vectors as the 18See also Section 11.1.1. 19This is the criteria for ‘static equilbrium’. 1026 Applications of Trigonometry origin and the dashed line as the x-axis, we use Theorem 11.20 to get component representations for the three vectors involved. We can model the weight of the speaker as a vector pointing directly downwards with a magnitude of 50 pounds. That is, w = 50 and ˆw = −ˆ = 0, −1. Hence, w = 50 0, −1 = 0, −50. For the force in the ﬁrst support, we get T1 = T1 cos (60◦) , sin (60◦) T1 2 , √ T1 2 3 = For the second support, we note that the angle 30◦ is measured from the negative x-axis, so the angle needed to write T2 in component form is 150◦. Hence T2 = T2 cos (150◦) , sin (150◦) √ = − T2 2 3 , T2 2 The requirement w + T1 + T2 = 0 gives us this vector equation. 0, −50 + , T1 2 T1 2 − T1 2 T2 2 √ 3 + √ 3 , T1 2 − √ , 3 T2 2 T2 2 + 3 w + T1 + T2 = 0 √ T2 2 = 0, 0 − 50 = 0, 0 Equating the corresponding components of the vectors on each side, we get a system of linear equations in the variables T1 and T2.    (E1) (E2) T1 2 3 √ + T1 2 √ T2 2 3 − = 0 T2 2 − 50 = 0 From (E1), we get T1 = T2 which yields 2 T2 − 50 = 0. Hence, T2 = 25 pounds and T1 = T2 3. Substituting that into (E2) gives ( T2 √ 2 3 = 25 + T2 3 √ 3 pounds. 2 − 50 = 0 √ √ √ 3) 11.8 Vectors 11.8.1 Exercises 1027 In Exercises 1 - 10, use the given pair of vectors v and w to ﬁnd the following quantities. State whether the result is a vector or a scalar. • v + w • w − 2v • − wv • wˆv Finally, verify that the vectors satisfy the Parallelogram Law v2 + w2 = 1 2 v + w2 + v − w2 1. v = 12, −5, w = 3, 4 2. v = −7, 24, w = −5, −12 3. v = 2, −1, w = −2, 4 √ √ 3, 1, w = 2 5. v = − √ 7. v = 3. v = 4. v = 10, 4, w = −2, 5 61, − 3 2 √ 3 9. v = 3ˆı + 4ˆ, w = −2ˆ 10. v = 1 2 (ˆı + ˆ), w = 1 2 (ˆı − ˆ) In Exercises 11 - 25, ﬁnd the component form of the vector v using the information given about its magnitude and direction. Give exact values. 11. v = 6; when drawn in standard position v lies in Quadrant I and makes a 60◦ angle with the positive x-axis 12. v = 3; when drawn in standard position v lies in Quadrant I and makes a 45◦ angle with the positive x-axis 13. v = 2 3 ; when drawn in standard position v lies in Quadrant I and makes a 60◦ angle with the positive y-axis 14. v = 12; when drawn in standard position v lies along the positive y-axis 15. v = 4; when drawn in standard position v lies in Quadrant II and makes a 30◦ angle with the negative x-axis √ 16. v = 2 3; when drawn in standard position v lies in Quadrant II and makes a 30◦ angle with the positive y-axis 17. v = 7 2 ; when drawn in standard position v lies along the negative x-axis √ 18. v = 5 6; when drawn in standard position v lies in Quadrant III and makes a 45◦ angle with the negative x-axis 19. v = 6.25; when drawn in standard position v lies along the negative y-axis 1028 20. v = 4 √ 3; when drawn in standard position v lies in Quadrant IV and makes a 30◦ angle Applications of Trigonometry with the positive x-axis 21. v = 5 √ 2; when drawn in standard position v lies in Quadrant IV and makes a 45◦ angle with the negative y-axis 22. v = 2 √ 5; when drawn in standard position v lies in Quadrant I and makes an angle measuring arctan(2) with the positive x-axis √ 23. v = 10; when drawn in standard position v lies in Quadrant II and makes an angle measuring arctan(3) with the negative x-axis 24. v = 5; when drawn in standard position v lies in Quadrant III and makes an angle measuring arctan 4 3 with the negative x-axis 25. v = 26; when drawn in standard position v lies in Quadrant IV and makes an angle measuring arctan 5 12 with the positive x-axis In Exercises 26 - 31, approximate the component form of the vector v using the information given about its magnitude and direction. Round your approximations to two decimal places. 26. v = 392; when drawn in standard position v makes a 117◦ angle with the positive x-axis 27. v = 63.92; when drawn in standard position v makes a 78.3◦ angle with the positive x-axis 28. v = 5280; when drawn in standard position v makes a 12◦ angle with the positive x-axis 29. v = 450; when drawn in standard position v makes a 210.75◦ angle with the positive x-axis 30. v = 168.7; when drawn in standard position v makes a 252◦ angle with the positive x-axis 31. v = 26; when drawn in standard position v makes a 304.5◦ angle with the positive x-axis In Exercises 32 - 52, for the given vector v, ﬁnd the magnitude v and an angle θ with 0 ≤ θ < 360◦ so that v = v cos(θ), sin(θ) (See Deﬁnition 11.8.) Round approximations to two decimal places. 32. v = 1, √ 35. v = − √ 2, 3 √ 2 33. v = 5, 5 36 38. v = 6, 0 41. v = −10ˆ 39. v = −2.5, 0 40. v = 0, 42. v = 3, 4 43. v = 12, 5 34. v = −2 √ 37. v = − 1 3 11.8 Vectors 1029 44. v = −4, 3 45. v = −7, 24 46. v = −2, −1 47. v = −2, −6 48. v = ˆı + ˆ 49. v = ˆı − 4ˆ 50. v = 123.4, −77.05 51. v = 965.15, 831.6 52. v = −114.1, 42.3 53. A small boat leaves the dock at Camp DuNuthin and heads across the Nessie River at 17 miles per hour (that is, with respect to the water) at a bearing of S68◦W. The river is ﬂowing due east at 8 miles per hour. What is the boat’s true speed and heading? Round the speed to the nearest mile per hour and express the heading as a bearing, rounded to the nearest tenth of a degree. 54. The HMS Sasquatch leaves port with bearing S20◦E maintaining a speed of 42 miles per hour (that is, with respect to the water). If the ocean current is 5 miles per hour with a bearing of N60◦E, ﬁnd the HMS Sasquatch’s true speed and bearing. Round the speed to the nearest mile per hour and express the heading as a bearing, rounded to the nearest tenth of a degree. 55. If the captain of the HMS Sasquatch in Exercise 54 wishes to reach Chupacabra Cove, an island 100 miles away at a bearing of S20◦E from port, in three hours, what speed an
d heading should she set to take into account the ocean current? Round the speed to the nearest mile per hour and express the heading as a bearing, rounded to the nearest tenth of a degree. HINT: If v denotes the velocity of the HMS Sasquatch and w denotes the velocity of the current, what does v + w need to be to reach Chupacabra Cove in three hours? 56. In calm air, a plane ﬂying from the Pedimaxus International Airport can reach Cliﬀs of Insanity Point in two hours by following a bearing of N8.2◦E at 96 miles an hour. (The distance between the airport and the cliﬀs is 192 miles.) If the wind is blowing from the southeast at 25 miles per hour, what speed and bearing should the pilot take so that she makes the trip in two hours along the original heading? Round the speed to the nearest hundredth of a mile per hour and your angle to the nearest tenth of a degree. 57. The SS Bigfoot leaves Yeti Bay on a course of N37◦W at a speed of 50 miles per hour. After traveling half an hour, the captain determines he is 30 miles from the bay and his bearing back to the bay is S40◦E. What is the speed and bearing of the ocean current? Round the speed to the nearest mile per hour and express the heading as a bearing, rounded to the nearest tenth of a degree. 58. A 600 pound Sasquatch statue is suspended by two cables from a gymnasium ceiling. If each cable makes a 60◦ angle with the ceiling, ﬁnd the tension on each cable. Round your answer to the nearest pound. 59. Two cables are to support an object hanging from a ceiling. If the cables are each to make a 42◦ angle with the ceiling, and each cable is rated to withstand a maximum tension of 100 pounds, what is the heaviest object that can be supported? Round your answer down to the nearest pound. 1030 Applications of Trigonometry 60. A 300 pound metal star is hanging on two cables which are attached to the ceiling. The left hand cable makes a 72◦ angle with the ceiling while the right hand cable makes a 18◦ angle with the ceiling. What is the tension on each of the cables? Round your answers to three decimal places. 61. Two drunken college students have ﬁlled an empty beer keg with rocks and tied ropes to it in order to drag it down the street in the middle of the night. The stronger of the two students pulls with a force of 100 pounds at a heading of N77◦E and the other pulls at a heading of S68◦E. What force should the weaker student apply to his rope so that the keg of rocks heads due east? What resultant force is applied to the keg? Round your answer to the nearest pound. 62. Emboldened by the success of their late night keg pull in Exercise 61 above, our intrepid young scholars have decided to pay homage to the chariot race scene from the movie ‘Ben-Hur’ by tying three ropes to a couch, loading the couch with all but one of their friends and pulling it due west down the street. The ﬁrst rope points N80◦W, the second points due west and the third points S80◦W. The force applied to the ﬁrst rope is 100 pounds, the force applied to the second rope is 40 pounds and the force applied (by the non-riding friend) to the third rope is 160 pounds. They need the resultant force to be at least 300 pounds otherwise the couch won’t move. Does it move? If so, is it heading due west? 63. Let v = v1, v2 be any non-zero vector. Show that 1 v v has length 1. 64. We say that two non-zero vectors v and w are parallel if they have same or opposite directions. That is, v = 0 and w = 0 are parallel if either ˆv = ˆw or ˆv = − ˆw. Show that this means v = k w for some non-zero scalar k and that k > 0 if the vectors have the same direction and k < 0 if they point in opposite directions. 65. The goal of this exercise is to use vectors to describe non-vertical lines in the plane. To that end, consider the line y = 2x − 4. Let v0 = 0, −4 and let s = 1, 2. Let t be any real number. Show that the vector deﬁned by v = v0 + ts, when drawn in standard position, has its terminal point on the line y = 2x − 4. (Hint: Show that v0 + ts = t, 2t − 4 for any real number t.) Now consider the non-vertical line y = mx + b. Repeat the previous analysis with v0 = 0, b and let s = 1, m. Thus any non-vertical line can be thought of as a collection of terminal points of the vector sum of 0, b (the position vector of the y-intercept) and a scalar multiple of the slope vector s = 1, m. 66. Prove the associative and identity properties of vector addition in Theorem 11.18. 67. Prove the properties of scalar multiplication in Theorem 11.19. 11.8 Vectors 11.8.2 Answers 1. v + w = 15, −1, vector √ v + w = 226, scalar 1031 w − 2v = −21, 14, vector v + w = 18, scalar v w − wv = −21, 77, vector wˆv = 60 13 , − 25 13 , vector 2. v + w = −12, 12, vector w − 2v = 9, −60, vector v + w = 12 2, scalar √ v w − wv = −34, −612, vector v + w = 38, scalar wˆv = − 91 25 , 312 25 , vector 3. v + w = 0, 3, vector v + w = 3, scalar w − 2v = −6, 6, vector v + w = 3 √ 5, scalar v w − wv = −6 √ √ 5, 6 5, vector wˆv = 4, −2, vector 4. v + w = 8, 9, vector 5. 6. v + w = √ v w − wv = −14 v + w = √ 145, scalar √ 3, 3, vector √ 3, scalar v + w = 2 √ 29, vector 29, 6 v w − wv = 8 √ 3, 0, vector , scalar , vector v w − wv = − 7 5 , − 1 5 , vector w − 2v = −22, −3, vector v + w = 3 √ 29, scalar wˆv = 5, 2, vector w − 2v = 4 √ 3, 0, vector v + w = 6, scalar wˆv = −2 √ 3, 2, vector w − 2v = −2, −1, vector v + w = 2, scalar wˆv = 3 5 5 , 4 , vector √ √ − 3 2 2 , 3 2 2 7. v + w = 0, 0, vector w − 2v = , vector v + w = 0, scalar v w − wv = − √ √ 2, 2, vector v + w = 2, scalar wˆv = √ 2 2 , − √ 2 2 , vector 1032 Applications of Trigonometry 8 , vector w − 2v = −2, −2 √ 3, vector v + w = 1, scalar v w − wv = −2, −2 √ 3, vector v + w = 3, scalar wˆv = 1, √ 3, vector 9. v + w = 3, 2, vector v + w = √ 13, scalar w − 2v = −6, −10, vector v + w = 7, scalar v w − wv = −6, −18, vector 10. v + w = 1, 0, vector v + w = 1, scalar v w − wv = 0, − √ 2 2 , vector , vector wˆv = 6 5 , 8 5 w − 2v = − , scalar , vector wˆv = 1 2 , 1 2 , vector 11. v = 3, 3 √ 3 14. v = 0, 12 17. v = − 7 2 , 0 √ 20. v = 6, −2 3 12. v = √ 3 2 2 , 3 √ 15. v = −2 3, 2 √ 2 2 13. v = √ 3 3 3 , 1 √ 16. v = − 3, 3 18. v = −5 √ √ 3 3, −5 19. v = 0, −6.25 21. v = 5, −5 22. v = 2, 4 23. v = −1, 3 24. v = −3, −4 25. v = 24, −10 26. v ≈ −177.96, 349.27 27. v ≈ 12.96, 62.59 28. v ≈ 5164.62, 1097.77 29. v ≈ −386.73, −230.08 32. v = 2, θ = 60◦ 30. v ≈ −52.13, −160.44 √ 33. v = 5 2, θ = 45◦ 31. v ≈ 14.73, −21.43 34. v = 4, θ = 150◦ 35. v = 2, θ = 135◦ 36. v = 1, θ = 225◦ 38. v = 6, θ = 0◦ 39. v = 2.5, θ = 180◦ 41. v = 10, θ = 270◦ 42. v = 5, θ ≈ 53.13◦ 44. v = 5, θ ≈ 143.13◦ 45. v = 25, θ ≈ 106.26◦ 37. v = 1, θ = 240◦ √ 40. v = 7, θ = 90◦ 43. v = 13, θ ≈ 22.62◦ √ 46. v = 5, θ ≈ 206.57◦ 11.8 Vectors 1033 47. v = 2 √ 10, θ ≈ 251.57◦ 48. v = √ 2, θ ≈ 45◦ √ 49. v = 17, θ ≈ 284.04◦ 50. v ≈ 145.48, θ ≈ 328.02◦ 51. v ≈ 1274.00, θ ≈ 40.75◦ 52. v ≈ 121.69, θ ≈ 159.66◦ 53. The boat’s true speed is about 10 miles per hour at a heading of S50.6◦W. 54. The HMS Sasquatch’s true speed is about 41 miles per hour at a heading of S26.8◦E. 55. She should maintain a speed of about 35 miles per hour at a heading of S11.8◦E. 56. She should ﬂy at 83.46 miles per hour with a heading of N22.1◦E 57. The current is moving at about 10 miles per hour bearing N54.6◦W. 58. The tension on each of the cables is about 346 pounds. 59. The maximum weight that can be held by the cables in that conﬁguration is about 133 pounds. 60. The tension on the left hand cable is 285.317 lbs. and on the right hand cable is 92.705 lbs. 61. The weaker student should pull about 60 pounds. The net force on the keg is about 153 pounds. 62. The resultant force is only about 296 pounds so the couch doesn’t budge. Even if it did move, the stronger force on the third rope would have made the couch drift slightly to the south as it traveled down the street. 1034 Applications of Trigonometry 11.9 The Dot Product and Projection In Section 11.8, we learned how add and subtract vectors and how to multiply vectors by scalars. In this section, we deﬁne a product of vectors. We begin with the following deﬁnition. Deﬁnition 11.11. Suppose v and w are vectors whose component forms are v = v1, v2 and w = w1, w2. The dot product of v and w is given by v · w = v1, v2 · w1, w2 = v1w1 + v2w2 For example, let v = 3, 4 and w = 1, −2. Then v · w = 3, 4 · 1, −2 = (3)(1) + (4)(−2) = −5. Note that the dot product takes two vectors and produces a scalar. For that reason, the quantity v· w is often called the scalar product of v and w. The dot product enjoys the following properties. Theorem 11.22. Properties of the Dot Product Commutative Property: For all vectors v and w, v · w = w · v. Distributive Property: For all vectors u, v and w, u · (v + w) = u · v + u · w. Scalar Property: For all vectors v and w and scalars k, (kv) · w = k(v · w) = v · (k w). Relation to Magnitude: For all vectors v, v · v = v2. Like most of the theorems involving vectors, the proof of Theorem 11.22 amounts to using the deﬁnition of the dot product and properties of real number arithmetic. To show the commutative property for instance, let v = v1, v2 and w = w1, w2. Then v · w = v1, v2 · w1, w2 = v1w1 + v2w2 Deﬁnition of Dot Product = w1v1 + w2v2 = w1, w2 · v1, v2 Deﬁnition of Dot Product = w · v Commutativity of Real Number Multiplication The distributive property is proved similarly and is left as an exercise. For the scalar property, assume that v = v1, v2 and w = w1, w2 and k is a scalar. Then (kv) · w = (k v1, v2) · w1, w2 = kv1, kv2 · w1, w2 = (kv1)(w1) + (kv2)(w2) Deﬁnition of Dot Product Deﬁnition of Scalar Multiplication = k(v1w1) + k(v2w2) Associativity of Real Number Multiplication = k(v1w1 + v2w2) = k v1, v2 · w1, w2 = k(v · w) Distributive Law of Real Numbers Deﬁnition of Dot Product We leave the proof of k(v · w) = v · (k w) as an exercise. 11.9 The Dot Product and Projection 1035 For the last property, we note that if v = v1, v2, then v · v = v1, v2 · v1, v2 = v2 where the last equality comes c
ourtesy of Deﬁnition 11.8. The following example puts Theorem 11.22 to good use. As in Example 11.8.3, we work out the problem in great detail and encourage the reader to supply the justiﬁcation for each step. 2 = v2, 1 + v2 Example 11.9.1. Prove the identity: v − w2 = v2 − 2(v · w) + w2. Solution. We begin by rewriting v − w2 in terms of the dot product using Theorem 11.22. v − w2 = (v − w) · (v − w) = (v + [− w]) · (v + [− w]) = (v + [− w]) · v + (v + [− w]) · [− w] = v · (v + [− w]) + [− w] · (v + [− w]) = v · v + v · [− w] + [− w] · v + [− w] · [− w] = v · v + v · [(−1) w] + [(−1) w] · v + [(−1) w] · [(−1) w] = v · v + (−1)(v · w) + (−1)( w · v) + [(−1)(−1)]( w · w) = v · v + (−1)(v · w) + (−1)(v · w(v · w) + w · w = v2 − 2(v · w) + w2 Hence, v − w2 = v2 − 2(v · w) + w2 as required. If we take a step back from the pedantry in Example 11.9.1, we see that the bulk of the work is needed to show that (v − w)·(v − w) = v ·v −2(v · w)+ w· w. If this looks familiar, it should. Since the dot product enjoys many of the same properties enjoyed by real numbers, the machinations required to expand (v − w) · (v − w) for vectors v and w match those required to expand (v − w)(v − w) for real numbers v and w, and hence we get similar looking results. The identity veriﬁed in Example 11.9.1 plays a large role in the development of the geometric properties of the dot product, which we now explore. Suppose v and w are two nonzero vectors. If we draw v and w with the same initial point, we deﬁne the angle between v and w to be the angle θ determined by the rays containing the vectors v and w, as illustrated below. We require 0 ≤ θ ≤ π. (Think about why this is needed in the deﬁnition.) The following theorem gives us some insight into the geometric role the dot product plays. Theorem 11.23. Geometric Interpretation of Dot Product: vectors then v · w = v w cos(θ), where θ is the angle between v and w. If v and w are nonzero 1036 Applications of Trigonometry We prove Theorem 11.23 in cases. If θ = 0, then v and w have the same direction. It follows1 that there is a real number k > 0 so that w = kv. Hence, v · w = v · (kv) = k(v · v) = kv2 = kvv. Since k > 0, k = \|k\|, so kv = \|k\|v = kv by Theorem 11.20. Hence, kvv = v(kv) = vkv = v w. Since cos(0) = 1, we get v · w = kvv = v w = v w cos(0), proving that the formula holds for θ = 0. If θ = π, we repeat the argument with the diﬀerence being w = kv where k < 0. In this case, \|k\| = −k, so kv = −\|k\|v = −kv = − w. Since cos(π) = −1, we get v · w = −v w = v w cos(π), as required. Next, if 0 < θ < π, the vectors v, w and v − w determine a triangle with side lengths v, w and v − w, respectively, as seen below The Law of Cosines yields v − w2 = v2 + w2 − 2v w cos(θ). From Example 11.9.1, we know v − w2 = v2 − 2(v · w) + w2. Equating these two expressions for v − w2 gives v2 + w2 −2v w cos(θ) = v2 −2(v· w)+ w2 which reduces to −2v w cos(θ) = −2(v· w), or v · w = v w cos(θ), as required. An immediate consequence of Theorem 11.23 is the following. Theorem 11.24. Let v and w be nonzero vectors and let θ the angle between v and w. Then θ = arccos v · w v w = arccos(ˆv · ˆw) We obtain the formula in Theorem 11.24 by solving the equation given in Theorem 11.23 for θ. Since v and w are nonzero, so are v and w. Hence, we may divide both sides of v · w = v w cos(θ) by v w to get cos(θ) = v· w v w . Since 0 ≤ θ ≤ π by deﬁnition, the values of θ exactly match the . Using Theorem 11.22, we can rewrite v· w range of the arccosine function. Hence, θ = arccos v w = ˆv · ˆw, giving us the alternative formula θ = arccos(ˆv · ˆw). v· w v w = 1 vv 1 w w · We are overdue for an example. Example 11.9.2. Find the angle between the following pairs of vectors. 1. v = 3, −3 √ 3, and w = − 3, 1 √ 2. v = 2, 2, and w = 5, −5 3. v = 3, −4, and w = 2, 1 Solution. We use the formula θ = arccos v· w v w from Theorem 11.24 in each case below. 1Since v = vˆv and w = w ˆw, if ˆv = ˆw then w = wˆv = w v (vˆv) = w v v. In this case, k = w v > 0. 1037 √ 3)2 = 11.9 The Dot Product and Projection 1. We have v· w = 3, −3 √ 3·− √ 3, 1 = −3 √ 36 = 6 and w = (− 3)2 + 12 = √ √ √ √ 3−3 3 = −6 √ 4 = 2, θ = arccos 3. Since v = −6 12 = arccos √ 3 32 + (−3 √ − 3 2 = 5π 6 . 2. For v = 2, 2 and w = 5, −5, we ﬁnd v · w = 2, 2 · 5, −5 = 10 − 10 = 0. Hence, it doesn’t v· w matter what v and w are,2 θ = arccos = arccos(0) = π 2 . v w 3. We ﬁnd v · w = 3, −4 · 2, 1 = 6 − 4 = 2. Also v = √ √ w = 22 + 12 = 5, so θ = arccos = arccos isn’t the cosine of one 32 + (−4)2 = √ 25 = 5 and 2 √ 5 5 of the common angles, we leave our answer as θ = arccos 2 √ 5 25 √ 5 25 . Since 2 2 √ 5 25 . The vectors v = 2, 2, and w = 5, −5 in Example 11.9.2 are called orthogonal and we write v ⊥ w, because the angle between them is π 2 radians = 90◦. Geometrically, when orthogonal vectors are sketched with the same initial point, the lines containing the vectors are perpendicular. w v v and w are orthogonal, v ⊥ w We state the relationship between orthogonal vectors and their dot product in the following theorem. Theorem 11.25. The Dot Product Detects Orthogonality: Let v and w be nonzero vectors. Then v ⊥ w if and only if v · w = 0. To prove Theorem 11.25, we ﬁrst assume v and w are nonzero vectors with v ⊥ w. By deﬁnition, = 0. Conversely, the angle between v and w is π v· w if v and w are nonzero vectors and v · w = 0, then Theorem 11.24 gives θ = arccos = 2 . By Theorem 11.23, v · w = v w cos π 2 v w arccos to provide a diﬀerent proof about the relationship between the slopes of perpendicular lines.3 2 , so v ⊥ w. We can use Theorem 11.25 in the following example v w = arccos(0) = π 0 Example 11.9.3. Let L1 be the line y = m1x + b1 and let L2 be the line y = m2x + b2. Prove that L1 is perpendicular to L2 if and only if m1 · m2 = −1. Solution. Our strategy is to ﬁnd two vectors: v1, which has the same direction as L1, and v2, which has the same direction as L2 and show v1 ⊥ v2 if and only if m1m2 = −1. To that end, we substitute x = 0 and x = 1 into y = m1x + b1 to ﬁnd two points which lie on L1, namely P (0, b1) 2Note that there is no ‘zero product property’ for the dot product since neither v nor w is 0, yet v · w = 0. 3See Exercise 2.1.1 in Section 2.1. 1038 Applications of Trigonometry −−→ P Q = 1 − 0, (m1 + b1) − b1 = 1, m1, and note that since v1 is and Q(1, m1 + b1). We let v1 = determined by two points on L1, it may be viewed as lying on L1. Hence it has the same direction as L1. Similarly, we get the vector v2 = 1, m2 which has the same direction as the line L2. Hence, L1 and L2 are perpendicular if and only if v1 ⊥ v2. According to Theorem 11.25, v1 ⊥ v2 if and only if v1 · v2 = 0. Notice that v1 · v2 = 1, m1 · 1, m2 = 1 + m1m2. Hence, v1 · v2 = 0 if and only if 1 + m1m2 = 0, which is true if and only if m1m2 = −1, as required. While Theorem 11.25 certainly gives us some insight into what the dot product means geometrically, there is more to the story of the dot product. Consider the two nonzero vectors v and w drawn with a common initial point O below. For the moment, assume that the angle between v and w, which we’ll denote θ, is acute. We wish to develop a formula for the vector p, indicated below, which is called the orthogonal projection of v onto w. The vector p is obtained geometrically as follows: drop a perpendicular from the terminal point T of v to the vector w and call the point −−→ of intersection R. The vector p is then deﬁned as p = OR. Like any vector, p is determined by its magnitude p and its direction ˆp according to the formula p = pˆp. Since we want ˆp to have the same direction as w, we have ˆp = ˆw. To determine p, we make use of Theorem 10.4 as applied to the right triangle ORT . We ﬁnd cos(θ) = p v , or p = v cos(θ). To get things in terms of just v and w, we use Theorem 11.23 to get p = v cos(θ) = v w cos(θ) w . Using Theorem 11.22, we rewrite v· w = v · ˆw. Hence, p = v · ˆw, and since ˆp = ˆw, we now have a formula for p completely in terms of v and w, namely p = pˆp = (v · ˆw) ˆw. w = v · = v −−→ OR p = v w θ R p O Now suppose that the angle θ between v and w is obtuse, and consider the diagram below. In this case, we see that ˆp = − ˆw and using the triangle ORT , we ﬁnd p = v cos(θ). Since θ +θ = π, it follows that cos(θ) = − cos(θ), which means p = v cos(θ) = −v cos(θ). Rewriting this last equation in terms of v and w as before, we get p = −(v · ˆw). Putting this together with ˆp = − ˆw, we get p = pˆp = −(v · ˆw)(− ˆw) = (v · ˆw) ˆw in this case as well. 11.9 The Dot Product and Projection 1039 T v w θ O θ −−→ OR p = R If the angle between v and w is π v · w = 0. It follows that v · ˆw = 0 and p = 0 = 0 ˆw = (v · ˆw) ˆw in this case, too. This gives us 2 then it is easy to show4 that p = 0. Since v ⊥ w in this case, Deﬁnition 11.12. Let v and w be nonzero vectors. The orthogonal projection of v onto w, denoted proj w(v) is given by proj w(v) = (v · ˆw) ˆw. Deﬁnition 11.12 gives us a good idea what the dot product does. The scalar v · ˆw is a measure of how much of the vector v is in the direction of the vector w and is thus called the scalar projection of v onto w. While the formula given in Deﬁnition 11.12 is theoretically appealing, because of the presence of the normalized unit vector ˆw, computing the projection using the formula proj w(v) = (v · ˆw) ˆw can be messy. We present two other formulas that are often used in practice. Theorem 11.26. Alternate Formulas for Vector Projections: If v and w are nonzero vectors then proj w(v) = (v · ˆw) ˆw = v · w w2 w = v · w w · w w The proof of Theorem 11.26, which we leave to the reader as an exercise, amounts to using the formula ˆw = w and properties of the dot product. It is time for an example. 1 w Example 11.9.4. Let v = 1, 8 and w = −1, 2. Find p = proj w(v), and plot v, w and p in standard position. Solution. We ﬁnd v · w = 1, 8 · −1, 2 = (−1) + 16 = 15 and w · w = −1, 2 · −1, 2 = 1 + 4 = 5. Hence, p = v· w 5 −1, 2 = −3, 6. We plot v, w and p below. w· w w = 15 4In this case
, the point R coincides with the point O, so p = −→ OR = −−→ OO = 0. 1040 Applications of Trigonometry 3 −2 −1 1 Suppose we wanted to verify that our answer p in Example 11.9.4 is indeed the orthogonal projection of v onto w. We ﬁrst note that since p is a scalar multiple of w, it has the correct direction, so what remains to check is the orthogonality condition. Consider the vector q whose initial point is the terminal point of p and whose terminal point is the terminal point of v3 −2 −1 1 From the deﬁnition of vector arithmetic, p + q = v, so that q = v − p. In the case of Example 11.9.4, v = 1, 8 and p = −3, 6, so q = 1, 8−−3, 6 = 4, 2. Then q· w = 4, 2·−1, 2 = (−4)+4 = 0, which shows q ⊥ w, as required. This result is generalized in the following theorem. Theorem 11.27. Generalized Decomposition Theorem: Let v and w be nonzero vectors. There are unique vectors p and q such that v = p + q where p = k w for some scalar k, and q · w = 0. Note that if the vectors p and q in Theorem 11.27 are nonzero, then we can say p is parallel 5 to w and q is orthogonal to w. In this case, the vector p is sometimes called the ‘vector component of v parallel to w’ and q is called the ‘vector component of v orthogonal to w.’ To prove Theorem 11.27, we take p = proj w(v) and q = v − p. Then p is, by deﬁnition, a scalar multiple of w. Next, we compute q · w. 5See Exercise 64 in Section 11.8. 11.9 The Dot Product and Projection 1041 q · w = (v − p) · w Deﬁnition of q. = v · w − p · w Properties of Dot Product · w Since p = proj w(v). ( w · w) Properties of Dot Product Hence, q · w = 0, as required. At this point, we have shown that the vectors p and q guaranteed by Theorem 11.27 exist. Now we need to show that they are unique. Suppose v = p + q = p + q where the vectors p and q satisfy the same properties described in Theorem 11.27 as p and q. Then p − p = q − q, so w · (p − p ) = w · (q − q. Hence, w · (p − p ) = 0. Now there are scalars k and k so that p = k w and p = k w. This means w · (p − p ) = w · (k w − k w) = w · ([k − k ] w) = (k − k )( w · w) = (k − k ) w2. Since w = 0, w2 = 0, which means the only way w · (p − p ) = (k − k ) w2 = 0 is for k − k = 0, or k = k . This means p = k w = k w = p . With , it must be that q = q as well. Hence, we have shown there is only one way to write v as a sum of vectors as described in Theorem 11.27. We close this section with an application of the dot product. In Physics, if a constant force F is exerted over a distance d, the work W done by the force is given by W = F d. Here, we assume the force is being applied in the direction of the motion. If the force applied is not in the direction of the motion, we can use the dot product to ﬁnd the work done. Consider the scenario below where the constant force F is applied to move an object from the point P to the point Q. F F θ θ P Q To ﬁnd the work W done in this scenario, we need to ﬁnd how much of the force F is in the −−→ P Q. This is precisely what the dot product F · P Q represents. Since direction of the motion −−→ the distance the object travels is P QP Q, −−→ −−→ W = ( F · P Q) P Q = F · ( P Q cos(θ), where θ is the angle between the applied force F and the trajectory of the motion −−→ P Q, we get W = ( F · P Q) −−→ P Q = F −−→ P Q. We have proved the following. −−→ P QP Q) = F · −−→ P Q. Since −−→ P Q = 1042 Applications of Trigonometry Theorem 11.28. Work as a Dot Product: Suppose a constant force F is applied along the vector −−→ P Q. The work W done by F is given by where θ is the angle between F and −−→ P Q = F −−→ P Q cos(θ), W = F · −−→ P Q. Example 11.9.5. Taylor exerts a force of 10 pounds to pull her wagon a distance of 50 feet over level ground. If the handle of the wagon makes a 30◦ angle with the horizontal, how much work did Taylor do pulling the wagon? Assume Taylor exerts the force of 10 pounds at a 30◦ angle for the duration of the 50 feet. 30◦ −−→ Solution. There are two ways to attack this problem. One way is to ﬁnd the vectors F and P Q −−→ mentioned in Theorem 11.28 and compute W = F · P Q. To do this, we assume the origin is at the point where the handle of the wagon meets the wagon and the positive x-axis lies along the dashed line in the ﬁgure above. Since the force applied is a constant 10 pounds, we have F = 10. Since it is being applied at a constant angle of θ = 30◦ with respect to the positive x-axis, Deﬁnition √ 11.8 gives us F = 10 cos(30◦, sin(30◦) = 5 3, 5. Since the wagon is being pulled along 50 −−→ P Q = 50ˆı = 50 1, 0 = 50, 0. We get feet in the positive direction, the displacement vector is W = F · 3. Since force is measured in pounds and distance is measured in feet, we get W = 250 3 foot-pounds. Alternatively, we can use the formulation W = F −−→ P Q cos(θ) to get W = (10 pounds)(50 feet) cos (30◦) = 250 3, 5 · 50, 0 = 250 3 foot-pounds of work. −−→ P Q = 5 √ √ √ √ 11.9 The Dot Product and Projection 1043 11.9.1 Exercises In Exercises 1 - 20, use the pair of vectors v and w to ﬁnd the following quantities. v · w The angle θ (in degrees) between v and w proj w(v) q = v − proj w(v) (Show that q · w = 0.) 1. v = −2, −7 and w = 5, −9 2. v = −6, −5 and w = 10, −12 3. v = 1, √ 3 and w = 1, − √ 3 5. v = −2, 1 and w = 3, 6 4. v = 3, 4 and w = −6, −8 6. v = −3 √ 3, 3 and w = − 3, −1 √ 7. v = 1, 17 and w = −1, 0 8. v = 3, 4 and w = 5, 12 9. v = −4, −2 and w = 1, −5 10. v = −5, 6 and w = 4, −7 11. v = −8, 3 and w = 2, 6 12. v = 34, −91 and w = 0, 1 13. v = 3ˆı − ˆ and w = 4ˆ 14. v = −24ˆı + 7ˆ and w = 2ˆı 15. v = 3 17. v = 19. v = 2ˆı + 3 1 2 , 2 ˆ and w = ˆı − ˆ √ √ 2 2 , and and 16. v = 5ˆı + 12ˆ and w = −3ˆı + 4ˆ 18. v = 20 and w = 1 2 , − √ 3 2 and 21. A force of 1500 pounds is required to tow a trailer. Find the work done towing the trailer along a ﬂat stretch of road 300 feet. Assume the force is applied in the direction of the motion. 22. Find the work done lifting a 10 pound book 3 feet straight up into the air. Assume the force of gravity is acting straight downwards. 23. Suppose Taylor ﬁlls her wagon with rocks and must exert a force of 13 pounds to pull her wagon across the yard. If she maintains a 15◦ angle between the handle of the wagon and the horizontal, compute how much work Taylor does pulling her wagon 25 feet. Round your answer to two decimal places. 24. In Exercise 61 in Section 11.8, two drunken college students have ﬁlled an empty beer keg with rocks which they drag down the street by pulling on two attached ropes. The stronger of the two students pulls with a force of 100 pounds on a rope which makes a 13◦ angle with the direction of motion. (In this case, the keg was being pulled due east and the student’s heading was N77◦E.) Find the work done by this student if the keg is dragged 42 feet. 1044 Applications of Trigonometry 25. Find the work done pushing a 200 pound barrel 10 feet up a 12.5◦ incline. Ignore all forces acting on the barrel except gravity, which acts downwards. Round your answer to two decimal places. HINT: Since you are working to overcome gravity only, the force being applied acts directly upwards. This means that the angle between the applied force in this case and the motion of the object is not the 12.5◦ of the incline! 26. Prove the distributive property of the dot product in Theorem 11.22. 27. Finish the proof of the scalar property of the dot product in Theorem 11.22. 28. Use the identity in Example 11.9.1 to prove the Parallelogram Law v2 + w2 = 1 2 v + w2 + v − w2 29. We know that \|x + y\| ≤ \|x\| + \|y\| for all real numbers x and y by the Triangle Inequality established in Exercise 36 in Section 2.2. We can now establish a Triangle Inequality for vectors. In this exercise, we prove that u + v ≤ u + v for all pairs of vectors u and v. (a) (Step 1) Show that u + v2 = u2 + 2u · v + v2. (b) (Step 2) Show that \|u · v\| ≤ uv. This is the celebrated Cauchy-Schwarz Inequality.6 (Hint: To show this inequality, start with the fact that \|u · v\| = \| uv cos(θ) \| and use the fact that \| cos(θ)\| ≤ 1 for all θ.) (c) (Step 3) Show that u + v2 = u2 + 2u · v + v2 ≤ u2 + 2\|u · v\| + v2 ≤ u2 + 2uv + v2 = (u + v)2. (d) (Step 4) Use Step 3 to show that u + v ≤ u + v for all pairs of vectors u and v. (e) As an added bonus, we can now show that the Triangle Inequality \|z + w\| ≤ \|z\| + \|w\| holds for all complex numbers z and w as well. Identify the complex number z = a + bi with the vector u = a, b and identify the complex number w = c + di with the vector v = c, d and just follow your nose! 6It is also known by other names. Check out this site for details. 11.9 The Dot Product and Projection 1045 11.9.2 Answers 1. v = −2, −7 and w = 5, −9 2. v = −6, −5 and w = 10, −12 v · w = 53 θ = 45◦ proj w(v = 90◦ proj w(v) = 0, 0 q = −6, −5 3. v = 1, √ 3 and w = 1, − √ 3 4. v = 3, 4 and w = −6, −8 v · w = −2 θ = 120◦ proj w(v50 θ = 180◦ proj w(v) = 3, 4 q = 0, 0 5. v = −2, 1 and w = 3, 6 v · w = 0 θ = 90◦ proj w(v) = 0, 0 q = −2, 1 3, 3 and w = − 3, −1 √ √ 6. v = −3 v · w = 6 θ = 60◦ proj w(v. v = 1, 17 and w = −1, 0 8. v = 3, 4 and w = 5, 12 v · w = −1 θ ≈ 93.37◦ proj w(v) = 1, 0 q = 0, 17 v · w = 63 θ ≈ 14.25◦ proj w(v) = 315 169 , 756 q = 192 169 , − 80 169 169 9. v = −4, −2 and w = 1, −5 10. v = −5, 6 and w = 4, −7 v · w = 6 θ ≈ 74.74◦ proj w(v) = 3 q = − 55 13 , − 11 13 13 , − 15 13 v · w = −62 θ ≈ 169.94◦ proj w(v) = − 248 q = − 77 65 , − 44 65 65 , 434 65 1046 Applications of Trigonometry 11. v = −8, 3 and w = 2, 6 12. v = 34, −91 and w = 0, 1 v · w = 2 θ ≈ 87.88◦ proj w(v) = 1 10 , 3 q = − 81 10 , 27 10 10 v · w = −91 θ ≈ 159.51◦ proj w(v) = 0, −91 q = 34, 0 13. v = 3ˆı − ˆ and w = 4ˆ 14. v = −24ˆı + 7ˆ and w = 2ˆı v · w = −4 θ ≈ 108.43◦ proj w(v) = 0, −1 q = 3, 0 15. v = 3 2 ˆ and w = ˆı − ˆ 2ˆı + 3 v · w = 0 θ = 90◦ proj w(v) = 048 θ ≈ 163.74◦ proj w(v) = −24, 0 q = 0, 7 16. v = 5ˆı + 12ˆ and w = −3ˆı + 4ˆ v · w = 33 θ ≈ 59.49◦ proj w(v) = − 99 q = 224 25 , 168 25 25 , 132 25 17 = 75◦ and − 4 √ √ 3−1 4 3 , proj w(v) = 1+ 4 q = √ 3 1− 4 √ , 1+ 4 3 18 = 105◦ and − 4 6 proj
w(v) = √ 3 √ q = 2+ 8 √ √ 2− 8 6 √ √ 2+ 8 6 , √ 6 2 20. v = 1 and 19. v = √ 3 2 , 1 2 √ and w = √ v · w = − θ = 165◦ 2 6+ 4 √ proj w(v) = √ q = 3−1 4 , 1− 4 3+1 4 √ 3 √ , 3++ 4 v · w = θ = 15◦ √ proj w(v) = 1− 4 q = √ 3 , 1− 4 3+1 4 √ 3 √ 3+1 4 , − 21. (1500 pounds)(300 feet) cos (0◦) = 450, 000 foot-pounds 22. (10 pounds)(3 feet) cos (0◦) = 30 foot-pounds 11.9 The Dot Product and Projection 1047 23. (13 pounds)(25 feet) cos (15◦) ≈ 313.92 foot-pounds 24. (100 pounds)(42 feet) cos (13◦) ≈ 4092.35 foot-pounds 25. (200 pounds)(10 feet) cos (77.5◦) ≈ 432.88 foot-pounds 1048 Applications of Trigonometry 11.10 Parametric Equations As we have seen in Exercises 53 - 56 in Section 1.2, Chapter 7 and most recently in Section 11.5, there are scores of interesting curves which, when plotted in the xy-plane, neither represent y as a function of x nor x as a function of y. In this section, we present a new concept which allows us to use functions to study these kinds of curves. To motivate the idea, we imagine a bug crawling across a table top starting at the point O and tracing out a curve C in the plane, as shown below. y 5 4 3 2 1 P (x, y) = (f (t), g(t)) Q O x 1 2 3 4 5 The curve C does not represent y as a function of x because it fails the Vertical Line Test and it does not represent x as a function of y because it fails the Horizontal Line Test. However, since the bug can be in only one place P (x, y) at any given time t, we can deﬁne the x-coordinate of P as a function of t and the y-coordinate of P as a (usually, but not necessarily) diﬀerent function of t. (Traditionally, f (t) is used for x and g(t) is used for y.) The independent variable t in this case is called a parameter and the system of equations x = f (t) y = g(t) is called a system of parametric equations or a parametrization of the curve C.1 The parametrization of C endows it with an orientation and the arrows on C indicate motion in the direction of increasing values of t. In this case, our bug starts at the point O, travels upwards to the left, then loops back around to cross its path2 at the point Q and ﬁnally heads oﬀ into the ﬁrst quadrant. It is important to note that the curve itself is a set of points and as such is devoid of any orientation. The parametrization determines the orientation and as we shall see, diﬀerent parametrizations can determine diﬀerent orientations. If all of this seems hauntingly familiar, it should. By deﬁnition, the system of equations {x = cos(t), y = sin(t) parametrizes the Unit Circle, giving it a counter-clockwise orientation. More generally, the equations of circular motion {x = r cos(ωt), y = r sin(ωt) developed on page 732 in Section 10.2.1 are parametric equations which trace out a circle of radius r centered at the origin. If ω > 0, the orientation is counterclockwise; if ω < 0, the orientation is clockwise. The angular frequency ω determines ‘how fast’ the 1Note the use of the indeﬁnite article ‘a’. As we shall see, there are inﬁnitely many diﬀerent parametric represen- tations for any given curve. 2Here, the bug reaches the point Q at two diﬀerent times. While this does not contradict our claim that f (t) and g(t) are functions of t, it shows that neither f nor g can be one-to-one. (Think about this before reading on.) 11.10 Parametric Equations 1049 object moves around the circle. In particular, the equations x = 2960 cos π 12 t that model the motion of Lakeland Community College as the earth rotates (see Example 10.2.7 in Section 10.2) parameterize a circle of radius 2960 with a counter-clockwise rotation which completes one revolution as t runs through the interval [0, 24). It is time for another example. 12 t , y = 2960 sin π Example 11.10.1. Sketch the curve described by x = t2 − 3 y = 2t − 1 for t ≥ −2. Solution. We follow the same procedure here as we have time and time again when asked to graph anything new – choose friendly values of t, plot the corresponding points and connect the results in a pleasing fashion. Since we are told t ≥ −2, we start there and as we plot successive points, we draw an arrow to indicate the direction of the path for increasing values of t. t −2 −1 0 1 2 3 x(t) y(t) 1 −5 −2 −3 −3 −1 −2 1 3 1 5 6 (x(t), y(t)) (1, −5) (−2, −3) (−3, −1) (−2, 1) (1, 3) (6, 5) y 5 4 3 2 1 −2−1 −1 1 2 3 4 5 6 x −2 −3 −5 2 . Substituting this into the equation x = t2 − 3 yields x = The curve sketched out in Example 11.10.1 certainly looks like a parabola, and the presence of the t2 term in the equation x = t2 − 3 reinforces this hunch. Since the parametric equations x = t2 − 3, y = 2t − 1 given to describe this curve are a system of equations, we can use the technique of substitution as described in Section 8.7 to eliminate the parameter t and get an equation involving just x and y. To do so, we choose to solve the equation y = 2t − 1 for t to get t = y+1 − 3 or, after some rearrangement, (y + 1)2 = 4(x + 3). Thinking back to Section 7.3, we see that the graph of this equation is a parabola with vertex (−3, −1) which opens to the right, as required. Technically speaking, the equation (y + 1)2 = 4(x + 3) describes the entire parabola, while the parametric equations x = t2 − 3, y = 2t − 1 for t ≥ −2 describe only a portion of the parabola. In this case,3 we can remedy this situation by restricting the bounds on y. Since the portion of the parabola we want is exactly the part where y ≥ −5, the equation (y + 1)2 = 4(x + 3) coupled with the restriction y ≥ −5 describes the same curve as the given parametric equations. The one piece of information we can never recover after eliminating the parameter is the orientation of the curve. y+1 2 2 Eliminating the parameter and obtaining an equation in terms of x and y, whenever possible, can be a great help in graphing curves determined by parametric equations. If the system of parametric equations contains algebraic functions, as was the case in Example 11.10.1, then the usual techniques of substitution and elimination as learned in Section 8.7 can be applied to the 3We will have an example shortly where no matter how we restrict x and y, we can never accurately describe the curve once we’ve eliminated the parameter. 1050 Applications of Trigonometry system {x = f (t), y = g(t) to eliminate the parameter. If, on the other hand, the parametrization In this case, it is often best involves the trigonometric functions, the strategy changes slightly. to solve for the trigonometric functions and relate them using an identity. We demonstrate these techniques in the following example. Example 11.10.2. Sketch the curves described by the following parametric equations. x = t3 y = 2t2 x = e−t y = e−2t 1. 2. for −1 ≤ t ≤ 1 for t ≥ 0 Solution. 3. 4. for 0 < t < π x = sin(t) y = csc(t) x = 1 + 3 cos(t) y = 2 sin(t) for 0 ≤ t ≤ 3π 2 1. To get a feel for the curve described by the system x = t3, y = 2t2 we ﬁrst sketch the graphs of x = t3 and y = 2t2 over the interval [−1, 1]. We note that as t takes on values in the interval [−1, 1], x = t3 ranges between −1 and 1, and y = 2t2 ranges between 0 and 2. This means that all of the action is happening on a portion of the plane, namely {(x, y) \| − 1 ≤ x ≤ 1, 0 ≤ y ≤ 2}. Next, we plot a few points to get a sense of the position and orientation of the curve. Certainly, t = −1 and t = 1 are good values to pick since these are the extreme values of t. We also choose t = 0, since that corresponds to a relative minimum4 on the graph of y = 2t2. Plugging in t = −1 gives the point (−1, 2), t = 0 gives (0, 0) and t = 1 gives (1, 2). More generally, we see that x = t3 is increasing over the entire interval [−1, 1] whereas y = 2t2 is decreasing over the interval [−1, 0] and then increasing over [0, 1]. Geometrically, this means that in order to trace out the path described by the parametric equations, we start at (−1, 2) (where t = −1), then move to the right (since x is increasing) and down (since y is decreasing) to (0, 0) (where t = 0). We continue to move to the right (since x is still increasing) but now move upwards (since y is now increasing) until we reach (1, 2) (where t = 1). Finally, to get a good sense of the shape of the curve, we √ eliminate the parameter. Solving x = t3 for t, we get t = 3 x. Substituting this into y = 2t2 √ x)2 = 2x2/3. Our experience in Section 5.3 yields the graph of our ﬁnal answer gives y = 2( 3 below. x 1 −1 1 t −1 y 2 1 y 2 1 −1 1 t −1 1 x x = t3, −1 ≤ t ≤ 1 y = 2t2, −1 ≤ t ≤ 1 x = t3, y = 2t2 , −1 ≤ t ≤ 1 4You should review Section 1.6.1 if you’ve forgotten what ‘increasing’, ‘decreasing’ and ‘relative minimum’ mean. 11.10 Parametric Equations 1051 9 3 , 1 and for t = ln(3) we get 2 2. For the system x = 2e−t, y = e−2t for t ≥ 0, we proceed as in the previous example and graph x = 2e−t and y = e−2t over the interval [0, ∞). We ﬁnd that the range of x in this case is (0, 2] and the range of y is (0, 1]. Next, we plug in some friendly values of t to get a sense of the orientation of the curve. Since t lies in the exponent here, ‘friendly’ values of t involve natural logarithms. Starting with t = ln(1) = 0 we get5 (2, 1), for t = ln(2) we get 1, 1 . Since t is ranging over the unbounded interval [0, ∞), 4 we take the time to analyze the end behavior of both x and y. As t → ∞, x = 2e−t → 0+ and y = e−2t → 0+ as well. This means the graph of x = 2e−t, y = e−2t approaches the point (0, 0). Since both x = 2e−t and y = e−2t are always decreasing for t ≥ 0, we know that our ﬁnal graph will start at (2, 1) (where t = 0), and move consistently to the left (since x is decreasing) and down (since y is decreasing) to approach the origin. To eliminate the parameter, one way to proceed is to solve x = 2e−t for t to get t = − ln x . Substituting 2 = x2 this for t in y = e−2t gives y = e−2(− ln(x/2)) = e2 ln(x/2) = eln(x/2)2 4 . Or, we could recognize that y = e−2t = e−t2 2 2 , we get y = x = x2 4 this way as well. Either way, the graph of x = 2e−t, y = e−2t for t ≥ 0 is a portion of the parabola y = x2 4 which starts
at the point (2, 1) and heads towards, but never reaches,6 (0, 0). = x 2 , and since x = 2e−t means e− = 2e−t, t ≥ 0 y = e−2t, t ≥ 0 x = 2e−t, y = e−2t , t ≥ 0 6 gives the point 1 3. For the system {x = sin(t), y = csc(t) for 0 < t < π, we start by graphing x = sin(t) and y = csc(t) over the interval (0, π). We ﬁnd that the range of x is (0, 1] while the range of 2 , 2, t = π y is [1, ∞). Plotting a few friendly points, we see that t = π 2 , 2. Since t = 0 and t = π aren’t included in the 6 returns us to 1 gives (1, 1) and t = 5π domain for t, (because y = csc(t) is undeﬁned at these t-values), we analyze the behavior of the system as t approaches 0 and π. We ﬁnd that as t → 0+ as well as when t → π−, we get x = sin(t) → 0+ and y = csc(t) → ∞. Piecing all of this information together, we get that for t near 0, we have points with very small positive x-values, but very large positive y-values. As t ranges through the interval 0, π , x = sin(t) is increasing and y = csc(t) is decreasing. 2 This means that we are moving to the right and downwards, through 1 6 to (1, 1) when t = π 2 , the orientation reverses, and we start to head to the left, since x = sin(t) is now decreasing, and up, since y = csc(t) is now increasing. We pass back through 1 6 back to the points with small positive x-coordinates and large 2 , 2 when t = 5π 2 , 2 when t = π 2 . Once t = π 2 5The reader is encouraged to review Sections 6.1 and 6.2 as needed. 6Note the open circle at the origin. See the solution to part 3 in Example 1.2.1 on page 22 and Theorem 4.1 in Section 4.1 for a review of this concept. 1052 Applications of Trigonometry positive y-coordinates. To better explain this behavior, we eliminate the parameter. Using a reciprocal identity, we write y = csc(t) = 1 sin(t) . Since x = sin(t), the curve traced out by this parametrization is a portion of the graph of y = 1 x . We now can explain the unusual behavior as t → 0+ and t → π− – for these values of t, we are hugging the vertical asymptote x = 0 of the graph of y = 1 x . We see that the parametrization given above traces out the portion of y = 1 x for 0 < x ≤ 1 twice as t runs through the interval (0, π). = sin(t), 0 < t < π y = csc(t), 0 < t < π {x = sin(t), y = csc(t) , 0 < t < π 4. Proceeding as above, we set about graphing {x = 1 + 3 cos(t), y = 2 sin(t) for 0 ≤ t ≤ 3π 2 by . We see that x ranges ﬁrst graphing x = 1 + 3 cos(t) and y = 2 sin(t) on the interval 0, 3π 2 from −2 to 4 and y ranges from −2 to 2. Plugging in t = 0, π 2 , π and 3π 2 gives the points (4, 0), (1, 2), (−2, 0) and (1, −2), respectively. As t ranges from 0 to π 2 , x = 1 + 3 cos(t) is decreasing, while y = 2 sin(t) is increasing. This means that we start tracing out our answer at (4, 0) and continue moving to the left and upwards towards (1, 2). For π 2 ≤ t ≤ π, x is decreasing, as is y, so the motion is still right to left, but now is downwards from (1, 2) to (−2, 0). On the interval π, 3π , x begins to increase, while y continues to decrease. Hence, the motion becomes left 2 to right but continues downwards, connecting (−2, 0) to (1, −2). To eliminate the parameter here, we note that the trigonometric functions involved, namely cos(t) and sin(t), are related by the Pythagorean Identity cos2(t) + sin2(t) = 1. Hence, we solve x = 1 + 3 cos(t) for cos(t) 3 , and we solve y = 2 sin(t) for sin(t) to get sin(t) = y to get cos(t) = x−1 2 . Substituting these expressions into cos2(t)+sin2(t) = 1 gives x−1 9 + y2 4 = 1. From Section 3 7.4, we know that the graph of this equation is an ellipse centered at (1, 0) with vertices at (−2, 0) and (4, 0) with a minor axis of length 4. Our parametric equations here are tracing out three-quarters of this ellipse, in a counter-clockwise direction. = 1, or (x−1)1 π 2 π t 3π 2 −2 x = 1 + 3 cos(t), 0 ≤ t ≤ 3π 2 y 2 1 π 2 π t 3π 2 −1 1 2 3 x 4 −1 −2 y 2 1 −1 −2 y = 2 sin(t), 0 ≤ t ≤ 3π 2 {x = 1 + 3 cos(t), y = 2 sin(t) , 0 ≤ t ≤ 3π 2 11.10 Parametric Equations 1053 Now that we have had some good practice sketching the graphs of parametric equations, we turn to the problem of ﬁnding parametric representations of curves. We start with the following. Parametrizations of Common Curves To parametrize y = f (x) as x runs through some interval I, let x = t and y = f (t) and let t run through I. To parametrize x = g(y) as y runs through some interval I, let x = g(t) and y = t and let t run through I. To parametrize a directed line segment with initial point (x0, y0) and terminal point (x1, y1), let x = x0 + (x1 − x0)t and y = y0 + (y1 − y0)t for 0 ≤ t ≤ 1. To parametrize (x−h)2 a2 + (y−k)2 b2 = 1 where a, b > 0, let x = h + a cos(t) and y = k + b sin(t) for 0 ≤ t < 2π. (This will impart a counter-clockwise orientation.) The reader is encouraged to verify the above formulas by eliminating the parameter and, when indicated, checking the orientation. We put these formulas to good use in the following example. Example 11.10.3. Find a parametrization for each of the following curves and check your answers. 1. y = x2 from x = −3 to x = 2 2. y = f −1(x) where f (x) = x5 + 2x + 1 3. The line segment which starts at (2, −3) and ends at (1, 5) 4. The circle x2 + 2x + y2 − 4y = 4 4 + y2 5. The left half of the ellipse x2 9 = 1 Solution. 1. Since y = x2 is written in the form y = f (x), we let x = t and y = f (t) = t2. Since x = t, the bounds on t match precisely the bounds on x so we get x = t, y = t2 for −3 ≤ t ≤ 2. The check is almost trivial; with x = t we have y = t2 = x2 as t = x runs from −3 to 2. 2. We are told to parametrize y = f −1(x) for f (x) = x5 + 2x + 1 so it is safe to assume that f is one-to-one. (Otherwise, f −1 would not exist.) To ﬁnd a formula y = f −1(x), we follow the procedure outlined on page 384 – we start with the equation y = f (x), interchange x and y and solve for y. Doing so gives us the equation x = y5 + 2y + 1. While we could attempt to solve this equation for y, we don’t need to. We can parametrize x = f (y) = y5 + 2y + 1 by setting y = t so that x = t5 + 2t + 1. We know from our work in Section 3.1 that since f (x) = x5 + 2x + 1 is an odd-degree polynomial, the range of y = f (x) = x5 + 2x + 1 is (−∞, ∞). Hence, in order to trace out the entire graph of x = f (y) = y5 + 2y + 1, we need to let y run through all real numbers. Our ﬁnal answer to this problem is x = t5 + 2t + 1, y = t for −∞ < t < ∞. As in the previous problem, our solution is trivial to check.7 7Provided you followed the inverse function theory, of course. 1054 Applications of Trigonometry 3. To parametrize line segment which starts at (2, −3) and ends at (1, 5), we make use of the formulas x = x0 +(x1 −x0)t and y = y0 +(y1 −y0)t for 0 ≤ t ≤ 1. While these equations at ﬁrst glance are quite a handful,8 they can be summarized as ‘starting point + (displacement)t’. To ﬁnd the equation for x, we have that the line segment starts at x = 2 and ends at x = 1. This means the displacement in the x-direction is (1 − 2) = −1. Hence, the equation for x is x = 2 + (−1)t = 2 − t. For y, we note that the line segment starts at y = −3 and ends at y = 5. Hence, the displacement in the y-direction is (5 − (−3)) = 8, so we get y = −3 + 8t. Our ﬁnal answer is {x = 2 − t, y = −3 + 8t for 0 ≤ t ≤ 1. To check, we can solve x = 2 − t for t to get t = 2 − x. Substituting this into y = −3 + 8t gives y = −3 + 8t = −3 + 8(2 − x), or y = −8x + 13. We know this is the graph of a line, so all we need to check is that it starts and stops at the correct points. When t = 0, x = 2 − t = 2, and when t = 1, x = 2 − t = 1. Plugging in x = 2 gives y = −8(2) + 13 = −3, for an initial point of (2, −3). Plugging in x = 1 gives y = −8(1) + 13 = 5 for an ending point of (1, 5), as required. 9 + (y−2)2 9 + (y−2)2 4. In order to use the formulas above to parametrize the circle x2 +2x+y2 −4y = 4, we ﬁrst need to put it into the correct form. After completing the squares, we get (x + 1)2 + (y − 2)2 = 9, or (x+1)2 9 = 1. Once again, the formulas x = h + a cos(t) and y = k + b sin(t) can be a challenge to memorize, but they come from the Pythagorean Identity cos2(t) + sin2(t) = 1. In the equation (x+1)2 3 . Rearranging these last two equations, we get x = −1 + 3 cos(t) and y = 2 + 3 sin(t). In order to complete one revolution around the circle, we let t range through the interval [0, 2π). We get as our ﬁnal answer {x = −1 + 3 cos(t), y = 2 + 3 sin(t) for 0 ≤ t < 2π. To check our answer, we could eliminate the parameter by solving x = −1 + 3 cos(t) for cos(t) and y = 2 + 3 sin(t) for sin(t), invoking a Pythagorean Identity, and then manipulating the resulting equation in x and y into the original equation x2 + 2x + y2 − 4y = 4. Instead, we opt for a more direct approach. We substitute x = −1 + 3 cos(t) and y = 2 + 3 sin(t) into the equation x2 + 2x + y2 − 4y = 4 and show that the latter is satisﬁed for all t such that 0 ≤ t < 2π. 9 = 1, we identify cos(t) = x+1 3 and sin(t) = y−2 (−1 + 3 cos(t))2 + 2(−1 + 3 cos(t)) + (2 + 3 sin(t))2 − 4(2 + 3 sin(t)) 1 − 6 cos(t) + 9 cos2(t) − 2 + 6 cos(t) + 4 + 12 sin(t) + 9 sin2(t) − 8 − 12 sin(t) 9 cos2(t) + 9 sin2(t) − 5 9 cos2(t) + sin2(t) − 5 9 (1) − 5 x2 + 2x + y2 − 4y = Now that we know the parametric equations give us points on the circle, we can go through the usual analysis as demonstrated in Example 11.10.2 to show that the entire circle is covered as t ranges through the interval [0, 2π). 8Compare and contrast this with Exercise 65 in Section 11.8. 11.10 Parametric Equations 1055 5. In the equation x2 4 + y2 9 = 1, we can either use the formulas above or think back to the Pythagorean Identity to get x = 2 cos(t) and y = 3 sin(t). The normal range on the parameter in this case is 0 ≤ t < 2π, but since we are interested in only the left half of the ellipse, we restrict t to the values which correspond to Quadrant II and Quadrant III angles, namely 2 ≤ t ≤ 3π π 2 . Substituting x = 2 cos(t) and y = 3 sin(t) into x2 = 1, which reduces to the Pythagorean Identity cos2(t) + sin2(t) = 1. This proves that
the points generated by the parametric equations {x = 2 cos(t), y = 3 sin(t) lie on the ellipse x2 9 = 1. Employing 2 ≤ t ≤ 3π the techniques demonstrated in Example 11.10.2, we ﬁnd that the restriction π 2 generates the left half of the ellipse, as required. 2 . Our ﬁnal answer is {x = 2 cos(t), y = 3 sin(t) for π 4 + y2 9 = 1 gives 4 cos2(t) 2 ≤ t ≤ 3π + 9 sin2(t) 9 4 + y2 4 We note that the formulas given on page 1053 oﬀer only one of literally inﬁnitely many ways to parametrize the common curves listed there. At times, the formulas oﬀered there need to be altered to suit the situation. Two easy ways to alter parametrizations are given below. Adjusting Parametric Equations Reversing Orientation: Replacing every occurrence of t with −t in a parametric description for a curve (including any inequalities which describe the bounds on t) reverses the orientation of the curve. Shift of Parameter: Replacing every occurrence of t with (t − c) in a parametric description for a curve (including any inequalities which describe the bounds on t) shifts the start of the parameter t ahead by c units. We demonstrate these techniques in the following example. Example 11.10.4. Find a parametrization for the following curves. 1. The curve which starts at (2, 4) and follows the parabola y = x2 to end at (−1, 1). Shift the parameter so that the path starts at t = 0. 2. The two part path which starts at (0, 0), travels along a line to (3, 4), then travels along a line to (5, 0). 3. The Unit Circle, oriented clockwise, with t = 0 corresponding to (0, −1). Solution. 1. We can parametrize y = x2 from x = −1 to x = 2 using the formula given on Page 1053 as x = t, y = t2 for −1 ≤ t ≤ 2. This parametrization, however, starts at (−1, 1) and ends at (2, 4). Hence, we need to reverse the orientation. To do so, we replace every occurrence of t with −t to get x = −t, y = (−t)2 for −1 ≤ −t ≤ 2. After simplifying, we get x = −t, y = t2 for −2 ≤ t ≤ 1. We would like t to begin at t = 0 instead of t = −2. The problem here is that the parametrization we have starts 2 units ‘too soon’, so we need to introduce a ‘time delay’ of 2. Replacing every occurrence of t with (t − 2) gives x = −(t − 2), y = (t − 2)2 for −2 ≤ t − 2 ≤ 1. Simplifying yields x = 2 − t, y = t2 − 4t + 4 for 0 ≤ t ≤ 3. 1056 Applications of Trigonometry 2. When parameterizing line segments, we think: ‘starting point + (displacement)t’. For the ﬁrst part of the path, we get {x = 3t, y = 4t for 0 ≤ t ≤ 1, and for the second part we get {x = 3 + 2t, y = 4 − 4t for 0 ≤ t ≤ 1. Since the ﬁrst parametrization leaves oﬀ at t = 1, we shift the parameter in the second part so it starts at t = 1. Our current description of the second part starts at t = 0, so we introduce a ‘time delay’ of 1 unit to the second set of parametric equations. Replacing t with (t − 1) in the second set of parametric equations gives {x = 3 + 2(t − 1), y = 4 − 4(t − 1) for 0 ≤ t − 1 ≤ 1. Simplifying yields {x = 1 + 2t, y = 8 − 4t for 1 ≤ t ≤ 2. Hence, we may parametrize the path as {x = f (t), y = g(t) for 0 ≤ t ≤ 2 where f (t) = 3t, 1 + 2t, for 0 ≤ t ≤ 1 for 1 ≤ t ≤ 2 and g(t) = 4t, 8 − 4t, for 0 ≤ t ≤ 1 for 1 ≤ t ≤ 2 3. We know that {x = cos(t), y = sin(t) for 0 ≤ t < 2π gives a counter-clockwise parametrization of the Unit Circle with t = 0 corresponding to (1, 0), so the ﬁrst order of business is to reverse the orientation. Replacing t with −t gives {x = cos(−t), y = sin(−t) for 0 ≤ −t < 2π, which simpliﬁes9 to {x = cos(t), y = − sin(t) for −2π < t ≤ 0. This parametrization gives a clockwise orientation, but t = 0 still corresponds to the point (1, 0); the point (0, −1) is reached when t = − 3π 2 . Our strategy is to ﬁrst get the parametrization to ‘start’ at the point (0, −1) and then shift the parameter accordingly so the ‘start’ coincides with t = 0. We know that any interval of length 2π will parametrize the entire circle, so we keep the equations {x = cos(t), y = − sin(t) , but start the parameter t at − 3π 2 , and ﬁnd the upper bound by adding 2π so − 3π 2 . The reader can verify that {x = cos(t), y = − sin(t) for − 3π 2 traces out the Unit Circle clockwise starting at the point (0, −1). We now shift the parameter by introducing a ‘time delay’ of 3π 2 units by replacing every occurrence of t with t − 3π 2 . This 2 simpliﬁes10 to {x = − sin(t), y = − cos(t) for 0 ≤ t < 2π, as required. . We get x = cos t − 3π 2 , y = − sin t − 3π 2 2 ≤ t < π 2 ≤ t − 3π 2 ≤ t < π for − 3π 2 < π We put our answer to Example 11.10.4 number 3 to good use to derive the equation of a cycloid. Suppose a circle of radius r rolls along the positive x-axis at a constant velocity v as pictured below. Let θ be the angle in radians which measures the amount of clockwise rotation experienced by the radius highlighted in the ﬁgure. y r P (x, y) θ 9courtesy of the Even/Odd Identities 10courtesy of the Sum/Diﬀerence Formulas x 11.10 Parametric Equations 1057 Our goal is to ﬁnd parametric equations for the coordinates of the point P (x, y) in terms of θ. From our work in Example 11.10.4 number 3, we know that clockwise motion along the Unit Circle starting at the point (0, −1) can be modeled by the equations {x = − sin(θ), y = − cos(θ) for 0 ≤ θ < 2π. (We have renamed the parameter ‘θ’ to match the context of this problem.) To model this motion on a circle of radius r, all we need to do11 is multiply both x and y by the factor r which yields {x = −r sin(θ), y = −r cos(θ) . We now need to adjust for the fact that the circle isn’t stationary with center (0, 0), but rather, is rolling along the positive x-axis. Since the velocity v is constant, we know that at time t, the center of the circle has traveled a distance vt down the positive x-axis. Furthermore, since the radius of the circle is r and the circle isn’t moving vertically, we know that the center of the circle is always r units above the x-axis. Putting these two facts together, we have that at time t, the center of the circle is at the point (vt, r). From Section 10.1.1, we know v = rθ t , or vt = rθ. Hence, the center of the circle, in terms of the parameter θ, is (rθ, r). As a result, we need to modify the equations {x = −r sin(θ), y = −r cos(θ) by shifting the x-coordinate to the right rθ units (by adding rθ to the expression for x) and the y-coordinate up r units12 (by adding r to the expression for y). We get {x = −r sin(θ) + rθ, y = −r cos(θ) + r , which can be written as {x = r(θ − sin(θ)), y = r(1 − cos(θ)) . Since the motion starts at θ = 0 and proceeds indeﬁnitely, we set θ ≥ 0. We end the section with a demonstration of the graphing calculator. Example 11.10.5. Find the parametric equations of a cycloid which results from a circle of radius 3 rolling down the positive x-axis as described above. Graph your answer using a calculator. Solution. We have r = 3 which gives the equations {x = 3(t − sin(t)), y = 3(1 − cos(t)) for t ≥ 0. (Here we have returned to the convention of using t as the parameter.) Sketching the cycloid by hand is a wonderful exercise in Calculus, but for the purposes of this book, we use a graphing utility. Using a calculator to graph parametric equations is very similar to graphing polar equations on a calculator.13 Ensuring that the calculator is in ‘Parametric Mode’ and ‘radian mode’ we enter the equations and advance to the ‘Window’ screen. As always, the challenge is to determine appropriate bounds on the parameter, t, as well as for x and y. We know that one full revolution of the circle occurs over the interval 0 ≤ t < 2π, so 11If we replace x with x 2 = 1 which reduces to x2 + y2 = r2. In the language of Section 1.7, we are stretching the graph by a factor of r in both the x- and y-directions. Hence, we multiply both the x- and y-coordinates of points on the graph by r. r in the equation for the Unit Circle x2 + y2 = 1, we obtain x r and y with y 2 + y r r 12Does this seem familiar? See Example 11.1.1 in Section 11.1. 13See page 959 in Section 11.5. 1058 Applications of Trigonometry it seems reasonable to keep these as our bounds on t. The ‘Tstep’ seems reasonably small – too large a value here can lead to incorrect graphs.14 We know from our derivation of the equations of the cycloid that the center of the generating circle has coordinates (rθ, r), or in this case, (3t, 3). Since t ranges between 0 and 2π, we set x to range between 0 and 6π. The values of y go from the bottom of the circle to the top, so y ranges between 0 and 6. Below we graph the cycloid with these settings, and then extend t to range from 0 to 6π which forces x to range from 0 to 18π yielding three arches of the cycloid. (It is instructive to note that keeping the y settings between 0 and 6 messes up the geometry of the cycloid. The reader is invited to use the Zoom Square feature on the graphing calculator to see what window gives a true geometric perspective of the three arches.) 14Again, see page 959 in Section 11.5. 11.10 Parametric Equations 1059 11.10.1 Exercises In Exercises 1 - 20, plot the set of parametric equations by hand. Be sure to indicate the orientation imparted on the curve by the parametrization. 1. 3. 5. 7. 9. 11. 13. 15. 17. 19. x = 4t − 3 y = 6t − 2 for 0 ≤ t ≤ 1 x = 2t y = t2 for − 1 ≤ t ≤ 2 x = t2 + 2t + 1 y = t + 1 for t ≤ 1 x = t y = t3 for − ∞ < t < ∞ x = cos(t) y = sin(t) for − 1 + 3 cos(t) y = 4 sin(t) for 0 ≤ t ≤ 2π x = 2 cos(t) y = sec(t) for 0 ≤ t < π 2 x = sec(t) y = tan(t) for − π 2 < t < π 2 x = tan(t) y = 2 sec(t) for − π 2 < t < π 2 x = cos(t) y = t for 0 ≤ t ≤ π 2. 4. 6. 8. 10. 12. 14. 16. 18. 20. x = 4t − 1 y = 3 − 4t for + 2t − t2 18 − t2 x = 1 9 y = 1 3 t for 0 ≤ t ≤ 3 for t ≥ −3 x = t3 y = t for − ∞ < t < ∞ x = 3 cos(t) y = 3 sin(t) for 0 ≤ t ≤ π x = 3 cos(t) y = 2 sin(t) + 1 for π 2 ≤ t ≤ 2π x = 2 tan(t) y = cot(t) for 0 < t < π 2 x = sec(t) y = tan(t) for π 2 < t < 3π 2 x = tan(t) y = 2 sec(t) for π 2 < t < 3π 2 x = sin(t) y = t for − π 2 ≤ t ≤ π 2 In Exercises 21 - 24, plot the set of parametric equations with the help of a graphin
g utility. Be sure to indicate the orientation imparted on the curve by the parametrization. 21. 23. x = t3 − 3t y = t2 − 4 x = et + e−t y = et − e−t for − 2 ≤ t ≤ 2 for − 2 ≤ t ≤ 2 22. 24. x = 4 cos3(t) y = 4 sin3(t) x = cos(3t) y = sin(4t) for 0 ≤ t ≤ 2π for 0 ≤ t ≤ 2π 1060 Applications of Trigonometry In Exercises 25 - 39, ﬁnd a parametric description for the given oriented curve. 25. the directed line segment from (3, −5) to (−2, 2) 26. the directed line segment from (−2, −1) to (3, −4) 27. the curve y = 4 − x2 from (−2, 0) to (2, 0). 28. the curve y = 4 − x2 from (−2, 0) to (2, 0) (Shift the parameter so t = 0 corresponds to (−2, 0).) 29. the curve x = y2 − 9 from (−5, −2) to (0, 3). 30. the curve x = y2 − 9 from (0, 3) to (−5, −2). (Shift the parameter so t = 0 corresponds to (0, 3).) 31. the circle x2 + y2 = 25, oriented counter-clockwise 32. the circle (x − 1)2 + y2 = 4, oriented counter-clockwise 33. the circle x2 + y2 − 6y = 0, oriented counter-clockwise 34. the circle x2 + y2 − 6y = 0, oriented clockwise (Shift the parameter so t begins at 0.) 35. the circle (x − 3)2 + (y + 1)2 = 117, oriented counter-clockwise 36. the ellipse (x − 1)2 + 9y2 = 9, oriented counter-clockwise 37. the ellipse 9x2 + 4y2 + 24y = 0, oriented counter-clockwise 38. the ellipse 9x2 + 4y2 + 24y = 0, oriented clockwise (Shift the parameter so t = 0 corresponds to (0, 0).) 39. the triangle with vertices (0, 0), (3, 0), (0, 4), oriented counter-clockwise (Shift the parameter so t = 0 corresponds to (0, 0).) 40. Use parametric equations and a graphing utility to graph the inverse of f (x) = x3 + 3x − 4. 41. Every polar curve r = f (θ) can be translated to a system of parametric equations with parameter θ by {x = r cos(θ) = f (θ) cos(θ), y = r sin(θ) = f (θ) sin(θ) . Convert r = 6 cos(2θ) to a system of parametric equations. Check your answer by graphing r = 6 cos(2θ) by hand using the techniques presented in Section 11.5 and then graphing the parametric equations you found using a graphing utility. 42. Use your results from Exercises 3 and 4 in Section 11.1 to ﬁnd the parametric equations which model a passenger’s position as they ride the London Eye. 11.10 Parametric Equations 1061 Suppose an object, called a projectile, is launched into the air. Ignoring everything except the force gravity, the path of the projectile is given by15    x = v0 cos(θ) t 1 2 y = − gt2 + v0 sin(θ) t + s0 for 0 ≤ t ≤ T where v0 is the initial speed of the object, θ is the angle from the horizontal at which the projectile is launched,16 g is the acceleration due to gravity, s0 is the initial height of the projectile above the ground and T is the time when the object returns to the ground. (See the ﬁgure below.) y θ s0 (x(T ), 0) x 43. Carl’s friend Jason competes in Highland Games Competitions across the country. In one event, the ‘hammer throw’, he throws a 56 pound weight for distance. If the weight is released 6 feet above the ground at an angle of 42◦ with respect to the horizontal with an initial speed of 33 feet per second, ﬁnd the parametric equations for the ﬂight of the hammer. (Here, use g = 32 ft. s2 .) When will the hammer hit the ground? How far away will it hit the ground? Check your answer using a graphing utility. 44. Eliminate the parameter in the equations for projectile motion to show that the path of the projectile follows the curve y = − g sec2(θ) 2v2 0 x2 + tan(θ)x + s0 Use the vertex formula (Equation 2.4) to show the maximum height of the projectile is y = 0 sin2(θ) v2 2g + s0 when x = v2 0 sin(2θ) 2g 15A nice mix of vectors and Calculus are needed to derive this. 16We’ve seen this before. It’s the angle of elevation which was deﬁned on page 753. 1062 Applications of Trigonometry 45. In another event, the ‘sheaf toss’, Jason throws a 20 pound weight for height. If the weight is released 5 feet above the ground at an angle of 85◦ with respect to the horizontal and the sheaf reaches a maximum height of 31.5 feet, use your results from part 44 to determine how fast the sheaf was launched into the air. (Once again, use g = 32 ft. s2 .) 46. Suppose θ = π formula given for y(t) above using g = 9.8 m in Exercise 25 in Section 2.3. What is x(t) in this case? 2 . (The projectile was launched vertically.) Simplify the general parametric s2 and compare that to the formula for s(t) given In Exercises 47 - 52, we explore the hyperbolic cosine function, denoted cosh(t), and the hyperbolic sine function, denoted sinh(t), deﬁned below: cosh(t) = et + e−t 2 and sinh(t) = et − e−t 2 47. Using a graphing utility as needed, verify that the domain of cosh(t) is (−∞, ∞) and the range of cosh(t) is [1, ∞). 48. Using a graphing utility as needed, verify that the domain and range of sinh(t) are both (−∞, ∞). 49. Show that {x(t) = cosh(t), y(t) = sinh(t) parametrize the right half of the ‘unit’ hyperbola x2 − y2 = 1. (Hence the use of the adjective ‘hyperbolic.’) 50. Compare the deﬁnitions of cosh(t) and sinh(t) to the formulas for cos(t) and sin(t) given in Exercise 83f in Section 11.7. 51. Four other hyperbolic functions are waiting to be deﬁned: the hyperbolic secant sech(t), the hyperbolic cosecant csch(t), the hyperbolic tangent tanh(t) and the hyperbolic cotangent coth(t). Deﬁne these functions in terms of cosh(t) and sinh(t), then convert them to formulas involving et and e−t. Consult a suitable reference (a Calculus book, or this entry on the hyperbolic functions) and spend some time reliving the thrills of trigonometry with these ‘hyperbolic’ functions. 52. If these functions look familiar, they should. Enjoy some nostalgia and revisit Exercise 35 in Section 6.5, Exercise 47 in Section 6.3 and the answer to Exercise 38 in Section 6.4. 11.10 Parametric Equations 1063 11.10.2 Answers x = 4t − 3 y = 6t − 2 1. for 3 −2 −1 −1 −2 2. x = 4t − 1 y = 3 − 4t y for 0 ≤ t ≤ 1 3 2 1 −1 −1 1 2 3 x x = 2t y = t2 3. for − + 2t − t2 4. for 3 −2 −1 1 2 3 4 x −1 1 2 x x = t2 + 2t + 1 y = t + 1 5. for t ≤ 1 6. 18 − t2 x = 1 9 y = 1 3 t for t ≥ −3 1 2 3 4 5 x −3 −2 −1 −1 −2 Applications of Trigonometry x = t3 y = t 8. for − ∞ < t < ∞ y 1 −4 −3 −2 −1 −1 1 2 3 4 x 1064 7. x = t y = t3 y for − ∞ < t < ∞ 4 3 2 1 −1 −1 −2 −3 −4 1 x 9. x = cos(t) y = sin(t) y 1 for − π 2 ≤ t ≤ π 2 −1 x 1 −1 10. x = 3 cos(t) y = 3 sin(t) y for 0 ≤ t ≤ π 3 2 1 −3 −2 −1 1 2 3 x 11. x = −1 + 3 cos(t) y = 4 sin(t) for 0 ≤ t ≤ 2π 12. x = 3 cos(t) y = 2 sin(t) + 1 for π 2 ≤ t ≤ 2π y 4 3 2 1 1 2 x −4 −3 −2 −1 −1 −2 −3 −4 y 3 2 1 −3 −1 −1 1 3 x 11.10 Parametric Equations 1065 13. x = 2 cos(t) y = sec(t) y for 0 ≤ t < π 2 14. x = 2 tan(t) y = cot(t) y for 15. x = sec(t) y = tan(t) for − π 2 < t < π 2 y 16. x = sec(t) y = tan(t) < t < 3π 2 π 2 for y 4 3 2 1 −1 −2 −3 −4 −3 −2 −1 −1 −2 −3 −4 1066 Applications of Trigonometry 17. x = tan(t) y = 2 sec(t) π 2 for − y < t < π 2 18. x = tan(t) y = 2 sec(t) < t < 3π 2 π 2 for y 4 3 2 1 −2 −1 1 2 x −2 −1 1 2 x −1 −2 −3 −4 19. x = cos(t) y = t for 0 < t < π 20. x = sin(t) y = t for − 1 1 x y π 2 −1 1 x − π 2 21. x = t3 − 3t y = t2 − 4 for − 2 ≤ t ≤ 2 22. x = 4 cos3(t) y = 4 sin3(t) y for 0 ≤ t ≤ 2π y 1 2 x −2 −1 −1 −2 −3 −4 4 3 2 1 −4 −3 −2 −1 −1 1 2 3 4 x −2 −3 −4 11.10 Parametric Equations 1067 23. x = et + e−t y = et − e−t for − 2 ≤ t ≤ 2 24. x = cos(3t) y = sin(4t) for 0 ≤ t ≤ 2π 1 1 x y 7 5 3 1 −1 −3 −5 −7 26. 28. 30. 32. 34. −1 x = 5t − 2 y = −1 − 3t x = t − 2 y = 4t − t2 x = t2 − 6t y = 3 − t for 0 ≤ t ≤ 1 for 0 ≤ t ≤ 4 for cos(t) y = 2 sin(t) x = 3 cos(t) y = 3 − 3 sin(t) x = 1 + 3 cos(t) y = sin(t) for 0 ≤ t < 2π for 0 ≤ t < 2π for 0 ≤ t < 2π x = 3 − 5t y = −5 + 7t for 0 ≤ t ≤ 1 for − 2 ≤ t ≤ 2 for − 2 ≤ t ≤ 3 for 0 ≤ t < 2π x = t y = 4 − t2 x = t2 − 9 y = t x = 5 cos(t) y = 5 sin(t) x = 3 cos(t) x = 3 + √ y = −1 + x = 2 cos(t) 25. 27. 29. 31. 33. 35. 37. 38. y = 3 + 3 sin(t) for 0 ≤ t < 2π 117 cos(t) √ 117 sin(t) for 0 ≤ t < 2π 36. for 0 ≤ t < 2π y = 3 sin(t) − 3    x = 2 cos t − y = −3 − 3 sin π 2 = 2 sin(t) π 2 t − = −3 + 3 cos(t) for 0 ≤ t < 2π 39. {x(t), y(t) where: x(t) =    3t, 0 ≤ t ≤ 1 6 − 3t, 1 ≤ t ≤ 2 0, 2 ≤ t ≤ 3 y(t) =    0, 0 ≤ t ≤ 1 4t − 4, 1 ≤ t ≤ 2 12 − 4t, 2 ≤ t ≤ 3 1068 Applications of Trigonometry 40. The parametric equations for the inverse are x = t3 + 3t − 4 y = t for − ∞ < t < ∞ 41. r = 6 cos(2θ) translates to x = 6 cos(2θ) cos(θ) y = 6 cos(2θ) sin(θ) for 0 ≤ θ < 2π. 42. The parametric equations which describe the locations of passengers on the London Eye are x = 67.5 cos π y = 67.5 sin π 15 t − π 15 t − π 2 = 67.5 sin π 15 t 2 + 67.5 = 67.5 − 67.5 cos π 15 t for − ∞ < t < ∞ x = 33 cos(42◦)t 43. The parametric equations for the hammer throw are y = −16t2 + 33 sin(42◦)t + 6 for t ≥ 0. To ﬁnd when the hammer hits the ground, we solve y(t) = 0 and get t ≈ −0.23 or 1.61. Since t ≥ 0, the hammer hits the ground after approximately t = 1.61 seconds after it was launched into the air. To ﬁnd how far away the hammer hits the ground, we ﬁnd x(1.61) ≈ 39.48 feet from where it was thrown into the air. 45. We solve y = 0 sin2(θ) v2 2g + s0 = 0 sin2(85◦) v2 2(32) + 5 = 31.5 to get v0 = ±41.34. The initial speed of the sheaf was approximately 41.34 feet per second. Index nth root of a complex number, 1000, 1001 principal, 397 nth Roots of Unity, 1006 u-substitution, 273 x-axis, 6 x-coordinate, 6 x-intercept, 25 y-axis, 6 y-coordinate, 6 y-intercept, 25 abscissa, 6 absolute value deﬁnition of, 173 inequality, 211 properties of, 173 acidity of a solution pH, 432 acute angle, 694 adjoint of a matrix, 622 alkalinity of a solution pH, 432 amplitude, 794, 881 angle acute, 694 between two vectors, 1035, 1036 central angle, 701 complementary, 696 coterminal, 698 decimal degrees, 695 deﬁnition, 693 degree, 694 DMS, 695 initial side, 698 measurement, 693 negative, 698 obtuse, 694 of declination, 761 of depression, 761 of elevation, 753 of inclination, 753 oriented, 697 positive, 698 quadrantal, 698 radian measure, 701 reference, 721 right, 694 standard position, 698 straight, 693 supplementary, 696 terminal side, 698 vertex, 693 angle sid
e opposite pairs, 896 angular frequency, 708 annuity annuity-due, 667 ordinary deﬁnition of, 666 future value, 667 applied domain of a function, 60 arccosecant calculus friendly deﬁnition of, 831 graph of, 830 properties of, 831 1069 Index 1070 trigonometry friendly deﬁnition of, 828 graph of, 827 properties of, 828 arccosine deﬁnition of, 820 graph of, 819 properties of, 820 arccotangent deﬁnition of, 824 graph of, 824 properties of, 824 arcsecant calculus friendly deﬁnition of, 831 graph of, 830 properties of, 831 trigonometry friendly deﬁnition of, 828 graph of, 827 properties of, 828 arcsine deﬁnition of, 820 graph of, 820 properties of, 820 arctangent deﬁnition of, 824 graph of, 823 properties of, 824 argument of a complex number deﬁnition of, 991 properties of, 995 of a function, 55 of a logarithm, 425 of a trigonometric function, 793 arithmetic sequence, 654 associative property for function composition, 366 matrix addition, 579 matrix multiplication, 585 scalar multiplication, 581 vector addition, 1015 scalar multiplication, 1018 asymptote horizontal formal deﬁnition of, 304 intuitive deﬁnition of, 304 location of, 308 of a hyperbola, 531 slant determination of, 312 formal deﬁnition of, 311 slant (oblique), 311 vertical formal deﬁnition of, 304 intuitive deﬁnition of, 304 location of, 306 augmented matrix, 568 average angular velocity, 707 average cost, 346 average cost function, 82 average rate of change, 160 average velocity, 706 axis of symmetry, 191 back substitution, 560 bearings, 905 binomial coeﬃcient, 683 Binomial Theorem, 684 Bisection Method, 277 BMI, body mass index, 355 Boyle’s Law, 350 buﬀer solution, 478 cardioid, 951 Cartesian coordinate plane, 6 Cartesian coordinates, 6 Cauchy’s Bound, 269 center of a circle, 498 of a hyperbola, 531 of an ellipse, 516 Index 1071 central angle, 701 change of base formulas, 442 characteristic polynomial, 626 Charles’s Law, 355 circle center of, 498 deﬁnition of, 498 from slicing a cone, 495 radius of, 498 standard equation, 498 standard equation, alternate, 519 circular function, 744 cis(θ), 995 coeﬃcient of determination, 226 cofactor, 616 Cofunction Identities, 773 common base, 420 common logarithm, 422 commutative property function composition does not have, 366 matrix addition, 579 vector addition, 1015 dot product, 1034 complementary angles, 696 Complex Factorization Theorem, 290 complex number nth root, 1000, 1001 nth Roots of Unity, 1006 argument deﬁnition of, 991 properties of, 995 conjugate deﬁnition of, 288 properties of, 289 deﬁnition of, 2, 287, 991 imaginary part, 991 imaginary unit, i, 287 modulus deﬁnition of, 991 properties of, 993 polar form cis-notation, 995 principal argument, 991 real part, 991 rectangular form, 991 set of, 2 complex plane, 991 component form of a vector, 1013 composite function deﬁnition of, 360 properties of, 367 compound interest, 470 conic sections deﬁnition, 495 conjugate axis of a hyperbola, 532 conjugate of a complex number deﬁnition of, 288 properties of, 289 Conjugate Pairs Theorem, 291 consistent system, 553 constant function as a horizontal line, 156 formal deﬁnition of, 101 intuitive deﬁnition of, 100 constant of proportionality, 350 constant term of a polynomial, 236 continuous, 241 continuously compounded interest, 472 contradiction, 549 coordinates Cartesian, 6 polar, 919 rectangular, 919 correlation coeﬃcient, 226 cosecant graph of, 801 of an angle, 744, 752 properties of, 802 cosine graph of, 791 of an angle, 717, 730, 744 properties of, 791 1072 cost average, 82, 346 ﬁxed, start-up, 82 variable, 159 cost function, 82 cotangent graph of, 805 of an angle, 744, 752 properties of, 806 coterminal angle, 698 Coulomb’s Law, 355 Cramer’s Rule, 619 curve orientated, 1048 cycloid, 1056 decibel, 431 decimal degrees, 695 decreasing function formal deﬁnition of, 101 intuitive deﬁnition of, 100 degree measure, 694 degree of a polynomial, 236 DeMoivre’s Theorem, 997 dependent system, 554 dependent variable, 55 depreciation, 420 Descartes’ Rule of Signs, 273 determinant of a matrix deﬁnition of, 614 properties of, 616 Diﬀerence Identity for cosine, 771, 775 for sine, 773, 775 for tangent, 775 diﬀerence quotient, 79 dimension of a matrix, 567 direct variation, 350 directrix of a conic section in polar form, 981 of a parabola, 505 discriminant Index of a conic, 979 of a quadratic equation, 195 trichotomy, 195 distance deﬁnition, 10 distance formula, 11 distributive property matrix matrix multiplication, 585 scalar multiplication, 581 vector dot product, 1034 scalar multiplication, 1018 DMS, 695 domain applied, 60 deﬁnition of, 45 implied, 58 dot product commutative property of, 1034 deﬁnition of, 1034 distributive property of, 1034 geometric interpretation, 1035 properties of, 1034 relation to orthogonality, 1037 relation to vector magnitude, 1034 work, 1042 Double Angle Identities, 776 earthquake Richter Scale, 431 eccentricity, 522, 981 eigenvalue, 626 eigenvector, 626 ellipse center, 516 deﬁnition of, 516 eccentricity, 522 foci, 516 from slicing a cone, 496 guide rectangle, 519 major axis, 516 minor axis, 516 Index 1073 reﬂective property, 523 standard equation, 519 vertices, 516 ellipsis (. . . ), 31, 651 empty set, 2 end behavior of f (x) = axn, n even, 240 of f (x) = axn, n odd, 240 of a function graph, 239 polynomial, 243 entry in a matrix, 567 equation contradiction, 549 graph of, 23 identity, 549 linear of n variables, 554 linear of two variables, 549 even function, 95 Even/Odd Identities, 770 exponential function algebraic properties of, 437 change of base formula, 442 common base, 420 deﬁnition of, 418 graphical properties of, 419 inverse properties of, 437 natural base, 420 one-to-one properties of, 437 solving equations with, 448 extended interval notation, 756 Factor Theorem, 258 factorial, 654, 681 ﬁxed cost, 82 focal diameter of a parabola, 507 focal length of a parabola, 506 focus of a conic section in polar form, 981 focus (foci) of a hyperbola, 531 of a parabola, 505 of an ellipse, 516 free variable, 552 frequency angular, 708, 881 of a sinusoid, 795 ordinary, 708, 881 function (absolute) maximum, 101 (absolute, global) minimum, 101 absolute value, 173 algebraic, 399 argument, 55 arithmetic, 76 as a process, 55, 378 average cost, 82 circular, 744 composite deﬁnition of, 360 properties of, 367 constant, 100, 156 continuous, 241 cost, 82 decreasing, 100 deﬁnition as a relation, 43 dependent variable of, 55 diﬀerence, 76 diﬀerence quotient, 79 domain, 45 even, 95 exponential, 418 Fundamental Graphing Principle, 93 identity, 168 increasing, 100 independent variable of, 55 inverse deﬁnition of, 379 properties of, 379 solving for, 384 uniqueness of, 380 linear, 156 local (relative) maximum, 101 local (relative) minimum, 101 logarithmic, 422 1074 Index notation, 55 odd, 95 one-to-one, 381 periodic, 790 piecewise-deﬁned, 62 polynomial, 235 price-demand, 82 product, 76 proﬁt, 82 quadratic, 188 quotient, 76 range, 45 rational, 301 revenue, 82 smooth, 241 sum, 76 transformation of graphs, 120, 135 zero, 95 fundamental cycle of y = cos(x), 791 Fundamental Graphing Principle for equations, 23 for functions, 93 for polar equations, 938 Fundamental Theorem of Algebra, 290 Gauss-Jordan Elimination, 571 Gaussian Elimination, 557 geometric sequence, 654 geometric series, 669 graph hole in, 305 horizontal scaling, 132 horizontal shift, 123 of a function, 93 of a relation, 20 of an equation, 23 rational function, 321 reﬂection about an axis, 126 transformations, 135 vertical scaling, 130 vertical shift, 121 greatest integer function, 67 growth model limited, 475 logistic, 475 uninhibited, 472 guide rectangle for a hyperbola, 532 for an ellipse, 519 Half-Angle Formulas, 779 harmonic motion, 885 Henderson-Hasselbalch Equation, 446 Heron’s Formula, 914 hole in a graph, 305 location of, 306 Hooke’s Law, 350 horizontal asymptote formal deﬁnition of, 304 intuitive deﬁnition of, 304 location of, 308 horizontal line, 23 Horizontal Line Test (HLT), 381 hyperbola asymptotes, 531 branch, 531 center, 531 conjugate axis, 532 deﬁnition of, 531 foci, 531 from slicing a cone, 496 guide rectangle, 532 standard equation horizontal, 534 vertical, 534 transverse axis, 531 vertices, 531 hyperbolic cosine, 1062 hyperbolic sine, 1062 hyperboloid, 542 identity function, 367 matrix, additive, 579 Index 1075 matrix, multiplicative, 585 statement which is always true, 549 imaginary axis, 991 imaginary part of a complex number, 991 imaginary unit, i, 287 implied domain of a function, 58 inconsistent system, 553 increasing function interval deﬁnition of, 3 notation for, 3 notation, extended, 756 inverse matrix, additive, 579, 581 matrix, multiplicative, 602 of a function formal deﬁnition of, 101 intuitive deﬁnition of, 100 independent system, 554 independent variable, 55 index of a root, 397 induction base step, 673 induction hypothesis, 673 inductive step, 673 inequality absolute value, 211 graphical interpretation, 209 non-linear, 643 quadratic, 215 sign diagram, 214 inﬂection point, 477 information entropy, 477 initial side of an angle, 698 instantaneous rate of change, 161, 472, 707 integer deﬁnition of, 2 greatest integer function, 67 set of, 2 intercept deﬁnition of, 25 location of, 25 interest compound, 470 compounded continuously, 472 simple, 469 Intermediate Value Theorem polynomial zero version, 241 interrobang, 321 intersection of two sets, 4 deﬁnition of, 379 properties of, 379 solving for, 384 uniqueness of, 380 inverse variation, 350 invertibility function, 382 invertible function, 379 matrix, 602 irrational number deﬁnition of, 2 set of, 2 irreducible quadratic, 291 joint variation, 350 Kepler’s Third Law of Planetary Motion, 355 Kirchhoﬀ’s Voltage Law, 605 latus rectum of a parabola, 507 Law of Cosines, 910 Law of Sines, 897 leading coeﬃcient of a polynomial, 236 leading term of a polynomial, 236 Learning Curve Equation, 315 least squares regression line, 225 lemniscate, 950 lima¸con, 950 line horizontal, 23 least squares regressio
n, 225 linear function, 156 of best ﬁt, 225 parallel, 166 1076 Index perpendicular, 167 point-slope form, 155 slope of, 151 slope-intercept form, 155 vertical, 23 linear equation n variables, 554 two variables, 549 linear function, 156 local maximum formal deﬁnition of, 102 intuitive deﬁnition of, 101 local minimum formal deﬁnition of, 102 intuitive deﬁnition of, 101 logarithm algebraic properties of, 438 change of base formula, 442 common, 422 general, “base b”, 422 graphical properties of, 423 inverse properties of, 437 natural, 422 one-to-one properties of, 437 solving equations with, 459 logarithmic scales, 431 logistic growth, 475 LORAN, 538 lower triangular matrix, 593 main diagonal, 585 major axis of an ellipse, 516 Markov Chain, 592 mathematical model, 60 matrix addition associative property, 579 commutative property, 579 deﬁnition of, 578 properties of, 579 additive identity, 579 additive inverse, 579 adjoint, 622 augmented, 568 characteristic polynomial, 626 cofactor, 616 deﬁnition, 567 determinant deﬁnition of, 614 properties of, 616 dimension, 567 entry, 567 equality, 578 invertible, 602 leading entry, 569 lower triangular, 593 main diagonal, 585 matrix multiplication associative property of, 585 deﬁnition of, 584 distributive property, 585 identity for, 585 properties of, 585 minor, 616 multiplicative inverse, 602 product of row and column, 584 reduced row echelon form, 570 rotation, 986 row echelon form, 569 row operations, 568 scalar multiplication associative property of, 581 deﬁnition of, 580 distributive properties, 581 identity for, 581 properties of, 581 zero product property, 581 size, 567 square matrix, 586 sum, 578 upper triangular, 593 maximum formal deﬁnition of, 102 intuitive deﬁnition of, 101 measure of an angle, 693 1077 Index midpoint deﬁnition of, 12 midpoint formula, 13 minimum formal deﬁnition of, 102 intuitive deﬁnition of, 101 minor, 616 minor axis of an ellipse, 516 model mathematical, 60 modulus of a complex number deﬁnition of, 991 properties of, 993 multiplicity eﬀect on the graph of a polynomial, 245, 249 of a zero, 244 natural base, 420 natural logarithm, 422 natural number deﬁnition of, 2 set of, 2 negative angle, 698 Newton’s Law of Cooling, 421, 474 Newton’s Law of Universal Gravitation, 351 oblique asymptote, 311 obtuse angle, 694 odd function, 95 Ohm’s Law, 350, 605 one-to-one function, 381 ordered pair, 6 ordinary frequency, 708 ordinate, 6 orientation, 1048 oriented angle, 697 oriented arc, 704 origin, 7 orthogonal projection, 1038 orthogonal vectors, 1037 overdetermined system, 554 parabola axis of symmetry, 191 deﬁnition of, 505 directrix, 505 focal diameter, 507 focal length, 506 focus, 505 from slicing a cone, 496 graph of a quadratic function, 188 latus rectum, 507 reﬂective property, 510 standard equation horizontal, 508 vertical, 506 vertex, 188, 505 vertex formulas, 194 paraboloid, 510 parallel vectors, 1030 parameter, 1048 parametric equations, 1048 parametric solution, 552 parametrization, 1048 partial fractions, 628 Pascal’s Triangle, 688 password strength, 477 period circular motion, 708 of a function, 790 of a sinusoid, 881 periodic function, 790 pH, 432 phase, 795, 881 phase shift, 795, 881 pi, π, 700 piecewise-deﬁned function, 62 point of diminishing returns, 477 point-slope form of a line, 155 polar coordinates conversion into rectangular, 924 deﬁnition of, 919 equivalent representations of, 923 polar axis, 919 pole, 919 1078 Index polar form of a complex number, 995 polar rose, 950 polynomial division dividend, 258 divisor, 258 factor, 258 quotient, 258 remainder, 258 synthetic division, 260 polynomial function completely factored over the complex numbers, 291 over the real numbers, 291 constant term, 236 deﬁnition of, 235 degree, 236 end behavior, 239 leading coeﬃcient, 236 leading term, 236 variations in sign, 273 zero lower bound, 274 multiplicity, 244 upper bound, 274 positive angle, 698 Power Reduction Formulas, 778 power rule for absolute value, 173 for complex numbers, 997 for exponential functions, 437 for logarithms, 438 for radicals, 398 for the modulus of a complex number, 993 price-demand function, 82 principal, 469 principal nth root, 397 principal argument of a complex number, 991 principal unit vectors, ˆı, ˆ, 1024 Principle of Mathematical Induction, 673 product rule for absolute value, 173 for complex numbers, 997 for exponential functions, 437 for logarithms, 438 for radicals, 398 for the modulus of a complex number, 993 Product to Sum Formulas, 780 proﬁt function, 82 projection x−axis, 45 y−axis, 46 orthogonal, 1038 Pythagorean Conjugates, 751 Pythagorean Identities, 749 quadrantal angle, 698 quadrants, 8 quadratic formula, 194 quadratic function deﬁnition of, 188 general form, 190 inequality, 215 irreducible quadratic, 291 standard form, 190 quadratic regression, 228 Quotient Identities, 745 quotient rule for absolute value, 173 for complex numbers, 997 for exponential functions, 437 for logarithms, 438 for radicals, 398 for the modulus of a complex number, 993 radian measure, 701 radical properties of, 398 radicand, 397 radioactive decay, 473 radius of a circle, 498 range deﬁnition of, 45 rate of change average, 160 Index 1079 instantaneous, 161, 472 slope of a line, 154 rational exponent, 398 rational functions, 301 rational number deﬁnition of, 2 set of, 2 Rational Zeros Theorem, 269 ray deﬁnition of, 693 initial point, 693 real axis, 991 Real Factorization Theorem, 292 real number deﬁnition of, 2 set of, 2 real part of a complex number, 991 Reciprocal Identities, 745 rectangular coordinates also known as Cartesian coordinates, 919 conversion into polar, 924 rectangular form of a complex number, 991 recursion equation, 654 reduced row echelon form, 570 reference angle, 721 Reference Angle Theorem for cosine and sine, 722 for the circular functions, 747 reﬂection of a function graph, 126 of a point, 10 regression coeﬃcient of determination, 226 correlation coeﬃcient, 226 least squares line, 225 quadratic, 228 total squared error, 225 relation algebraic description, 23 deﬁnition, 20 Fundamental Graphing Principle, 23 Remainder Theorem, 258 revenue function, 82 Richter Scale, 431 right angle, 694 root index, 397 radicand, 397 Roots of Unity, 1006 rotation matrix, 986 rotation of axes, 974 row echelon form, 569 row operations for a matrix, 568 scalar multiplication matrix associative property of, 581 deﬁnition of, 580 distributive properties of, 581 properties of, 581 vector associative property of, 1018 deﬁnition of, 1017 distributive properties of, 1018 properties of, 1018 scalar projection, 1039 secant graph of, 800 of an angle, 744, 752 properties of, 802 secant line, 160 sequence nth term, 652 alternating, 652 arithmetic common diﬀerence, 654 deﬁnition of, 654 formula for nth term, 656 sum of ﬁrst n terms, 666 deﬁnition of, 652 geometric common ratio, 654 deﬁnition of, 654 formula for nth term, 656 sum of ﬁrst n terms, 666 1080 recursive, 654 series, 668 set deﬁnition of, 1 empty, 2 intersection, 4 roster method, 1 set-builder notation, 1 sets of numbers, 2 union, 4 verbal description, 1 set-builder notation, 1 Side-Angle-Side triangle, 910 Side-Side-Side triangle, 910 sign diagram algebraic function, 399 for quadratic inequality, 214 polynomial function, 242 rational function, 321 simple interest, 469 sine graph of, 792 of an angle, 717, 730, 744 properties of, 791 sinusoid amplitude, 794, 881 baseline, 881 frequency angular, 881 ordinary, 881 graph of, 795, 882 period, 881 phase, 881 phase shift, 795, 881 properties of, 881 vertical shift, 881 slant asymptote, 311 slant asymptote determination of, 312 formal deﬁnition of, 311 slope deﬁnition, 151 Index of a line, 151 rate of change, 154 slope-intercept form of a line, 155 smooth, 241 sound intensity level decibel, 431 square matrix, 586 standard position of a vector, 1019 standard position of an angle, 698 start-up cost, 82 steady state, 592 stochastic process, 592 straight angle, 693 Sum Identity for cosine, 771, 775 for sine, 773, 775 for tangent, 775 Sum to Product Formulas, 781 summation notation deﬁnition of, 661 index of summation, 661 lower limit of summation, 661 properties of, 664 upper limit of summation, 661 supplementary angles, 696 symmetry about the x-axis, 9 about the y-axis, 9 about the origin, 9 testing a function graph for, 95 testing an equation for, 26 synthetic division tableau, 260 system of equations back-substitution, 560 coeﬃcient matrix, 590 consistent, 553 constant matrix, 590 deﬁnition, 549 dependent, 554 free variable, 552 Gauss-Jordan Elimination, 571 Gaussian Elimination, 557 Index 1081 inconsistent, 553 independent, 554 leading variable, 556 linear n variables, 554 two variables, 550 linear in form, 646 non-linear, 637 overdetermined, 554 parametric solution, 552 triangular form, 556 underdetermined, 554 unknowns matrix, 590 tangent graph of, 804 of an angle, 744, 752 properties of, 806 terminal side of an angle, 698 Thurstone, Louis Leon, 315 total squared error, 225 transformation non-rigid, 129 rigid, 129 transformations of function graphs, 120, 135 transverse axis of a hyperbola, 531 Triangle Inequality, 183 triangular form, 556 underdetermined system, 554 uninhibited growth, 472 union of two sets, 4 Unit Circle deﬁnition of, 501 important points, 724 unit vector, 1023 Upper and Lower Bounds Theorem, 274 upper triangular matrix, 593 variable dependent, 55 independent, 55 variable cost, 159 variation constant of proportionality, 350 direct, 350 inverse, 350 joint, 350 variations in sign, 273 vector x-component, 1012 y-component, 1012 addition associative property, 1015 commutative property, 1015 deﬁnition of, 1014 properties of, 1015 additive identity, 1015 additive inverse, 1015, 1018 angle between two, 1035, 1036 component form, 1012 Decomposition Theorem Generalized, 1040 Principal, 1024 deﬁnition of, 1012 direction deﬁnition of, 1020 properties of, 1020 dot product commutative property of, 1034 deﬁnition of, 1034 distributive prope
rty of, 1034 geometric interpretation, 1035 properties of, 1034 relation to magnitude, 1034 relation to orthogonality, 1037 work, 1042 head, 1012 initial point, 1012 magnitude deﬁnition of, 1020 properties of, 1020 relation to dot product, 1034 normalization, 1024 orthogonal projection, 1038 1082 Index multiplicity of, 244 of a function, 95 upper and lower bounds, 274 orthogonal vectors, 1037 parallel, 1030 principal unit vectors, ˆı, ˆ, 1024 resultant, 1013 scalar multiplication associative property of, 1018 deﬁnition of, 1017 distributive properties, 1018 identity for, 1018 properties of, 1018 zero product property, 1018 scalar product deﬁnition of, 1034 properties of, 1034 scalar projection, 1039 standard position, 1019 tail, 1012 terminal point, 1012 triangle inequality, 1044 unit vector, 1023 velocity average angular, 707 instantaneous, 707 instantaneous angular, 707 vertex of a hyperbola, 531 of a parabola, 188, 505 of an angle, 693 of an ellipse, 516 vertical asymptote formal deﬁnition of, 304 intuitive deﬁnition of, 304 location of, 306 vertical line, 23 Vertical Line Test (VLT), 43 whole number deﬁnition of, 2 set of, 2 work, 1041 wrapping function, 704 zero
ferent times and see how they differ (i.e., how they have changed). The two times of comparison are usually called the ﬁnal time and the initial time. We really need to be precise about this, so here is what we mean: ∆ quantity = (value of quantity at ﬁnal time) − (value of quantity at initial time) ∆ time = (ﬁnal time) − (initial time). For example, suppose that on June 4 we measure that the temperature at 8:00 am is 65◦ F and at 10:00 am it is 71◦ F. So, the ﬁnal time is 10:00 am, the initial time is 8:00 am and the temperature is changing according to the rate = ∆ quantity ∆ time = = ﬁnal value of quantity − initial value of quantity ﬁnal time − initial time 71 − 65 degrees 10:00 − 8:00 hours = 3 deg/hr. As a second example, suppose on June 5 the temperature at 8:00 am is 71◦ and at 10:00 am it is 65◦. So, the ﬁnal time is 10:00 am, the initial 1.2. TOTAL CHANGE = RATE TIME × 5 time is 8:00 am and the temperature is changing according to the rate = ∆ quantity ∆ time = = ﬁnal value of quantity − initial value of quantity ﬁnal time − initial time 65 − 71 degrees 10:00 − 8:00 hours = −3 deg/hr. These two examples illustrate that a rate can be either a positive or a negative number. More importantly, it highlights that we really need to be careful when making a rate computation. In both examples, the initial and ﬁnal times are the same and the two temperatures involved are the same, BUT whether they occur at the initial or ﬁnal time is interchanged. If we accidently mix this up, we will end up being off by a minus sign. There are many situations where the rate is the same for all time periods. In a case like this, we say we have a constant rate. For example, imagine you are driving down the freeway at a constant speed of 60 mi/hr. The fact that the speedometer needle indicates a steady speed of 60 mi/hr means the rate your distance is changing is constant. In cases when we have a constant rate, we often want to ﬁnd the total amount of change in the quantity over a speciﬁc time period. The key principle in the background is this: Total Change in some Quantity = Rate Time × (1.2) It is important to mention that this formula only works when we have a constant rate, but that will be the only situation we encounter in this course. One of the main goals of calculus is to develop a version of (1.2) that works for non-constant rates. Here is another example; others will occur throughout the text. It is Example 1.2.1. A water pipe mounted to the ceiling has a leak. dripping onto the ﬂoor below and creates a circular puddle of water. The surface area of this puddle is increasing at a constant rate of 4 cm2/hour. Find the surface area and dimensions of the puddle after 84 minutes. Solution. The quantity changing is “surface area” and we are given a “rate” and “time.” Using (1.2) with time t = 84 minutes, Total Surface Area = Rate Time × cm2 hr = 4 = 5.6 cm2. 84 60 hr × The formula for the area of a circular region of radius r is given at the 5.6 π = 1.335 cm at back of this text. Using this, the puddle has radius r = time t = 84 minutes. q 6 CHAPTER 1. WARMING UP 1.3 The Modeling Process Modeling is a method used in disciplines ranging from architecture to zoology. This mathematical technique will crop up any time we are problem solving and consciously trying to both “describe” and “predict.” Inevitably, mathematics is introduced to add structure to the model, but the clean equations and formulas only arise after some (or typically a lot) of preliminary work. A model can be thought of as a caricature in that it will pick out certain features (like a nose or a face) and focus on those at the expense of others. It takes a lot of experience to know which models are “good” and “bad,” in the sense of isolating the right features. In the beginning, modeling will lead to frustration and confusion, but by the end of this course our comfort level will dramatically increase. Let’s look at an illustration of the problem solving process. Example 1.3.1. How much time do you anticipate studying precalculus each week? Solution. One possible response is simply to say “a little” or “way too much!” You might not think these answers are the result of modeling, but they are. They are a consequence of modeling the total amount of study time in terms of categories such as “a little,” “some,” “lots,” “way too much,” etc. By drawing on your past experiences with math classes and using this crude model you arrived at a preliminary answer to the question. Let’s put a little more effort into the problem and try to come up with a numerical estimate. If T is the number of hours spent on precalculus a given week, it is certainly the case that: T = (hours in class) + (hours reading text) + (hours doing homework) Our time in class each week is known to be 5 hours. However, the other two terms require a little more thought. For example, if we can comfortably read and digest a page of text in (on average) 15 minutes and there are r pages of text to read during the week, then (hours reading text) = 15 60 r hours. As for homework, if a typical homework problem takes (on average) 25 minutes and there are h homework problems for the week, then (hours doing homework) = 25 60 h hours. We now have a mathematical model for the weekly time commitment to precalculus: T = 5 + 15 60 r + 25 60 h hours. 1.3. THE MODELING PROCESS 7 Is this a good model? Well, it is certainly more informative than our original crude model in terms of categories like “a little” or “lots.” But, the real plus of this model is that it clearly isolates the features being used to make our estimated time commitment and it can be easily modiﬁed as the amount of reading or homework changes. So, this is a pretty good model. However, it isn’t perfect; some homework problems will take a lot more than 25 minutes! 8 CHAPTER 1. WARMING UP 1.4 Exercises Problem 1.1. (a) Verify that 7685.33 seconds is 2 hours 8 minutes 5.33 seconds. (b) Allyson has a pace of 6 min/mile; what is her speed? (b) Which is faster: 100 mph or 150 ft/s? (c) Gina’s salary is 1 cent/second for a 40 hour work week. Tiare’s salary is $1400 for a 40 hour work week. Who has a higher salary? (d) Suppose it takes 180 credits to get a You accumubaccalaureate degree. late credit at the rate of one credit per quarter for each hour that the class meets per week. For instance, a class that meets three hours each week of the quarter will count for three credits. In addition, suppose that you spend 2.5 hours of study outside of class for each hour in class. A quarter is 10 weeks long. How many total hours, including time spent in class and time spent studying out of class, must you invest to get a degree? Problem 1.2. Sarah can bicycle a loop around the north part of Lake Washington in 2 hours and 40 minutes. If she could increase her average speed by 1 km/hr, it would reduce her time around the loop by 6 minutes. How many kilometers long is the loop? Problem 1.3. The is 11.34 g/cm3 and the density of aluminum is 2.69 g/cm3. Find the radius of lead and aluminum spheres each having a mass of 50 kg. density lead of Problem 1.4. The Eiffel Tower has a mass of 7.3 million kilograms and a height of 324 meIts base is square with a side length of ters. 125 meters. The steel used to make the Tower occupies a volume of 930 cubic meters. Air has a density of 1.225 kg per cubic meter. Suppose the Tower was contained in a cylinder. Find the mass of the air in the cylinder. Is this more or less than the mass of the Tower? Problem 1.5. Marathon runners keep track of their speed using units of pace = minutes/mile. (a) Lee has a speed of 16 ft/sec; what is his pace? (c) Adrienne and Dave are both running a race. Adrienne has a pace of 5.7 min/mile and Dave is running 10.3 mph. Who is running faster? Problem 1.6. Convert each of the following sentences into “pseudo-equations.” For example, suppose you start with the sentence: “The cost of the book was more than $10 and the cost of the magazine was $4.” A ﬁrst step would be these “pseudo-equations”: (Book cost) > $10 and (Magazine cost) = $4. (a) John’s salary is $56,000 a year and he pays no taxes. (b) John’s salary is at most $56,000 a year and he pays 15% of his salary in taxes. (c) John’s salary is at least $56,000 a year and he pays more than 28% of his salary in taxes. (d) The number of students taking Math 120 at the UW is somewhere between 1500 and 1800 each year. (e) The cost of a new red Porsche is more than three times the cost of a new Ford F-150 pickup truck. (f) Each week, students spend at least two but no more than three hours studying for each credit hour. (g) Twice the number of happy math students exceeds ﬁve times the number of happy chemistry students. However, all of the happy math and chemistry students combined is less than half the total number of cheerful biology students. (h) The difference between Cady’s high and low midterm scores was 10%. Her ﬁnal exam score was 97%. (i) The vote tally for Gov. Tush was within one-hundredth of one percent of onehalf the total number of votes cast. Problem 1.7. Which is a better deal: A 10 inch diameter pizza for $8 or a 15 inch diameter pizza for $16? 1.4. EXERCISES 9 Problem 1.8. The famous theory of relativity predicts that a lot of weird things will happen when you approach the speed of light 108 m/sec. For example, here is a forc = 3 mula that relates the mass mo (in kg) of an object at rest and its mass when it is moving at a speed v: × m = mo 1 − v2 c2 . q (a) Suppose the object moving is Dave, who has a mass of mo = 66 kg at rest. What is Dave’s mass at 90% of the speed of light? At 99% of the speed of light? At 99.9% of the speed of light? (b) How fast should Dave be moving to have a mass of 500 kg? Problem 1.9. During a typical evening in Seattle, Pagliacci receives phone orders for pizza delivery at a constant rate: 18 orders in a typical 4 minute period. How many pies are sold in 4 hours? Assume Pagliacci starts taking orders at 5 : 00
pm and the proﬁt is a constant rate of $11 on 10 orders. When will phone order proﬁt exceed $1,000? Problem 1.10. Aleko’s Pizza has delivered a beautiful 16 inch diameter pie to Lee’s dorm room. The pie is sliced into 8 equal sized pieces, but Lee is such a non-conformist he cuts off an edge as pictured. John then takes one of the remaining triangular slices. Who has more pizza and by how much? John’s part Lee’s part Problem 1.11. A typical cell in the human body contains molecules of deoxyribonucleic acid, referred to as DNA for short. In the cell, this DNA is all twisted together in a tight little packet. But imagine unwinding (straightening out) all of the DNA from a single typical cell and laying it “end-to-end”; then the sum total length will be approximately 2 meters. isolate DNA from nucleus cell nucleus lay out end−to−end 2 m Assume the human body has 1014 cells containing DNA. How many times would the sum total length of DNA in your body wrap around the equator of the earth? Problem 1.12. A water pipe mounted to the ceiling has a leak and is dripping onto the ﬂoor below, creating a circular puddle of water. The area of the circular puddle is increasing at a constant rate of 11 cm2/hour. (a) Find the area and radius of the puddle after 1 minute, 92 minutes, 5 hours, 1 day. (b) Is the radius of the puddle increasing at a constant rate? Problem 1.13. During the 1950s, Seattle was dumping an average of 20 million gallons of sewage into Lake Washington each day. (a) How much sewage went into Lake Wash- ington in a week? In a year? (b) In order to illustrate the amounts inimagine a rectangular prism volved, whose base is the size of a football ﬁeld (100 yards 50 yards) with height h yards. What are the dimensions of such a rectangular prism containing the × 10 CHAPTER 1. WARMING UP sewage dumped into Lake Washington in a single day? (Note: There are 7.5 gallons in one cubic foot. Dumping into Lake Washington has stopped; now it goes into the Puget Sound.) ter 1, 2, 5 and 20 years. Would tax rates increase or decrease over time? Congress claims that this law would ultimately cut peoples’ tax rates by 75 %. Do you believe this claim? Problem 1.14. Dave has inherited an apple orchard on which 60 trees are planted. Under these conditions, each tree yields 12 bushels of apples. According to the local WSU extension agent, each time Dave removes a tree the yield per tree will go up 0.45 bushels. Let x be the number of trees in the orchard and N the yield per tree. (a) Find a formula for N in terms of the unknown x. (Hint: Make a table of data with one column representing various values of x and the other column the corresponding values of N. After you complete the ﬁrst few rows of the table, you need to discover the pattern.) (b) What possible reason(s) might explain why the yield goes up when you remove trees? Problem 1.15. Congress is debating a proposed law to reduce tax rates. If the current tax rate is r %, then the proposed rate after x years is given by this formula Rewrite this formula as a simple fraction. Use your formula to calculate the new tax rate af- Problem 1.16. (a) The temperature at 7:00 am is 44◦F and the temperature at 10:00 am is 50◦F. What are the initial time, the ﬁnal time, the initial temperature and the ﬁnal temperature? What is the rate of change in the temperature between 7:00 am and 10:00 am? (b) Assume it is 50◦F at 10:00 am and the rate of change in the temperature between 10:00 am and 2:00 pm is the same as the rate in part (a). What is the temperature at 2:00 pm? (c) The temperature at 4:30 pm is 54◦F and the temperature at 6:15 pm is 26◦F. What are the initial time, the ﬁnal time, the initial temperature and the ﬁnal temperature? What is the rate of change in the temperature between 4:30 pm and 6:15 pm? Problem 1.17. (a) Solve for t: 3t −7 = 11+ t. (b) Solve for a: 1 + 1 a = 3. q (c) Solve for x: √x2 + a2 = 2a + x. (d) Solve for t: 1 − t > 4 − 2t. (e) Write as a single fraction: 2 x − 1 x + 1 Chapter 2 Imposing Coordinates You ﬁnd yourself visiting Spangle, WA and dinner time is approaching. A friend has recommended Tiff’s Diner, an excellent restaurant; how will you ﬁnd it? Of course, the solution to this simple problem amounts to locating a “point” on a two-dimensional map. This idea will be important in many problem solving situations, so we will quickly review the key ideas. Q P Figure 2.1: Two points in a plane. 2.1 The Coordinate System If we are careful, we can develop the ﬂow of ideas underlying two-dimensional coordinate systems in such a way that it easily generalizes to three-dimensions. Suppose we start with a blank piece of paper and mark two points; let’s label these two points “P ” and “Q.” This presents the basic problem of ﬁnding a foolproof method to reconstruct the picture. The basic idea is to introduce a coordinate system for the plane (analogous to the city map grid of streets), allowing us to catalog points in the plane using pairs of real numbers (analogous to the addresses of locations in the city). Here are the details. Start by drawing two perpendicular lines, called the horizontal axis and the vertical axis, each of which looks like a copy of the real number line. We refer to the intersection point of these two lines as the origin. Given P in the plane, the plan is to use these two axes to obtain a pair of real numbers (x,y) that will give us the exact location of P. With this in mind, the horizontal axis is often called the x-axis and the vertical axis is often called the y-axis. Remember, a typical real number line (like the x-axis or the y-axis) is divided into three parts: the positive numbers, the negative numbers, and the number zero (see Figure 2.2(a)). This allows us to specify positive and negative portions of the x-axis and y-axis. Unless we say otherwise, we will always adopt the convention that the positive x-axis consists of those numbers to the right of the origin on 11 12 CHAPTER 2. IMPOSING COORDINATES the x-axis and the positive y-axis consists of those numbers above the origin on the y-axis. We have just described the xy-coordinate system for the plane: Negative real numbers Positive real numbers Positive y-axis Origin Zero Negative x-axis Positive x-axis Negative y-axis (a) Number line. (b) xy-coordinate system. Figure 2.2: Coordinates. 2.1.1 Going from P to a Pair of Real Numbers. y-axis y P ℓ x-axis x ℓ∗ Figure 2.3: Coordinate pairs. Imagine a coordinate system had been drawn on our piece of paper in Figure 2.1. Let’s review the procedure of going from a point P to a pair of real numbers: 1. First, draw two new lines passing through P, one parallel to the x-axis and the other parallel to the y-axis; call these ℓ and ℓ∗, as pictured in Figure 2.3. 2. Notice that ℓ will cross the y-axis exactly once; the point on the y-axis where these two lines cross will be called “y.” Likewise, the line ℓ∗ will cross the x-axis exactly once; the point on the x-axis where these two lines cross will be called “x.” 3. If you begin with two different points, like P and Q in Figure 2.1, you will see that the two pairs of points you obtain will be different; i.e., if Q gives you the pair (x∗,y∗), then either x = y∗. This shows that two different points in the plane give two different pairs of real numbers and describes the process of assigning a pair of real numbers to the point P. = x∗ or y The great thing about the procedure we just described is that it is reversible! In other words, suppose you start with a pair of real numbers, say (x,y). Locate the number x on the x-axis and the number y on the y-axis. Now draw two lines: a line ℓ parallel to the x-axis passing through the number y on the y-axis and a line ℓ∗ parallel to the y-axis passing through the number x on the x-axis. The two lines ℓ and ℓ∗ will intersect 6 6 2.2. THREE FEATURES OF A COORDINATE SYSTEM 13 in exactly one point in the plane, call it P. This procedure describes how to go from a given pair of real numbers to a point in the plane. In addition, if you start with two different pairs of real numbers, then the corresponding two points in the plane are going to be different. In the future, we will constantly be going back and forth between points in the plane and pairs of real numbers using these ideas. Deﬁnition 2.1.1. Coordinate System: Every point P in the xy-plane corresponds to a unique pair of real numbers (x, y), where x is a number on the horizontal x-axis and y is a number on the vertical y-axis; for this reason, we commonly use the notation “P = (x,y).” Having speciﬁed positive and negative directions on the horizontal and vertical axes, we can now divide our two dimensional plane into four quadrants. The ﬁrst quadrant corresponds to all the points where both coordinates are positive, the second quadrant consists of points with the ﬁrst coordinate negative and the second coordinate positive, etc. Every point in the plane will lie in one of these four quadrants or on one of the two axes. This quadrant terminology is useful to give a rough sense of location, just as we use the terminology “Northeast, Northwest, Southwest and Southeast” when discussing locations on a map. y-axis Second Quadrant First Quadrant Third Quadrant Fourth Quadrant x-axis Figure 2.4: Quadrants in the xy-plane. 2.2 Three Features of a Coordinate System A coordinate system involves scaling, labeling and units on each of the axes. 2.2.1 Scaling Sketch two xy coordinate systems. In the ﬁrst, make the scale on each axis the same. In the second, assume “one unit” on the x axis has the same length as “two units” on the y axis. Plot the points (1,1), (−1,1), 5 , 16 − 4 , (1,1). , (0,0), Both pictures illustrate how the points lie on a parabola in the xy- coordinate system, but the aspect ratio has changed. The aspect ratio is deﬁned by this fraction: 5, 16 5 , 4 5, 9 5, 1 5, 1 5, 4 5, 9 − 1 − 2 − 3 25 25 25 25 25 25 25 25 , , , , , , 1 2 3 4 aspect ratio def= length of one unit on the vertical axis length of o
ne unit on the horizontal axis . Figure 2.5(a) has aspect ratio 1, whereas Figure 2.5(b) has aspect ratio 1 2. In problem solving, you will often need to make a rough assumption about the relative axis scaling. This scaling will depend entirely on the 14 CHAPTER 2. IMPOSING COORDINATES y-axis 1.0 0.8 0.6 0.4 0.2 −1.0 −0.5 0.0 0.5 x-axis 1.0 y-axis 1.0 0.8 0.6 0.4 0.2 x-axis −1.0 −0.5 0.0 0.5 1.0 (a) Aspect ratio = 1. (b) Aspect ratio = 1 2 . Figure 2.5: Coordinates. information given in the problem. Most graphing devices will allow you to specify the aspect ratio. 2.2.2 Axes Units Sometimes we are led to coordinate systems where each of the two axes involve different types of units (labels). Here is a sample, that illustrates the power of using pictures. Example 2.2.1. As the marketing director of Turbowebsoftware, you have been asked to deliver a brief message at the annual stockholders meeting on the performance of your product. Your staff has assembled this tabular collection of data; how can you convey the content of this table most clearly? TURBOWEB SALES (in $1000’s) week 1 2 3 4 5 6 7 8 9 10 sales 11.0517 12.214 13.4986 14.9182 16.4872 18.2212 20.1375 22.2554 24.596 27.1828 week 11 12 13 14 15 16 17 18 19 20 sales 30.0417 33.2012 36.693 40.552 44.8169 49.5303 54.7395 60.4965 66.8589 73.8906 week 21 22 23 24 25 26 27 28 29 30 sales 81.6617 90.2501 99.7418 110.232 121.825 134.637 148.797 164.446 181.741 200.855 week 31 32 33 34 35 36 37 38 39 40 sales 221.98 245.325 271.126 299.641 331.155 365.982 404.473 447.012 494.024 545.982 week 41 42 43 44 45 46 47 48 49 50 sales 603.403 666.863 736.998 814.509 900.171 994.843 1099.47 1215.1 1342.9 1,484.131 One idea is to simply ﬂash an overhead slide of this data to the audience; this can be deadly! A better idea is to use a visual aid. Suppose we let the variable x represent the week and the variable y represent the gross sales (in thousands of dollars) in week x. We can then plot the points (x,y) in the xy-coordinate system; see Figure 2.6. Notice, the units on the two axes are very different: y-axis units are “thousands of dollars” and x-axis units are “weeks.” In addition, the aspect ratio of this coordinate system is not 1. The beauty of this picture is the visual impact it gives your audience. From the coordinate plot we can get a sense of how the sales ﬁgures are dramatically increasing. In fact, this plot is good evidence you deserve a big raise! 2.3. A KEY STEP IN ALL MODELING PROBLEMS 15 Mathematical modeling is all about relating concrete phenomena and symbolic equations, so we want to embrace the idea of visualization. Most typically, visualization will involve plotting a collection of points in the plane. This can be achieved by providing a “list” or a “prescription” for plotting the points. The material we review in the next couple of sections makes the transition from symbolic mathematics to visual pictures go more smoothly. y-axis (Thousands of Dollars) 1,400 1,200 1,000 800 600 400 200 x-axis (Weeks) 10 20 30 40 50 Figure 2.6: Turboweb sales. 2.3 A Key Step in all Modeling Problems The initial problem solving or modeling step of deciding on a choice of xy-coordinate system is called imposing a coordinate system: There will often be many possible choices; it takes problem solving experience to develop intuition for a “natural” choice. This is a key step in all modeling problems. Example 2.3.1. Return to the tossed ball scenario on page 1. How do we decide where to draw a coordinate system in the picture? Figure 2.7 on page 16 shows four natural choices of xy-coordinate system. To choose a coordinate system we must specify the origin. The four logical choices for the origin are either the top of the cliff, the bottom of the cliff, the launch point of the ball or the landing point of the ball. So, which choice do we make? The answer is that any of these choices will work, but one choice may be more natural than another. For example, Figure 2.7(b) is probably the most natural choice: in this coordinate system, the motion of the ball takes place entirely in the ﬁrst quadrant, so the x and y coordinates of any point on the path of the ball will be non-negative. Example 2.3.2. Michael and Aaron are running toward each other, beginning at opposite ends of a 10,000 ft. airport runway, as pictured in Figure 2.8 on page 17. Where and when will these guys collide? Solution. This problem requires that we ﬁnd the “time” and “location” of the collision. Our ﬁrst step is to impose a coordinate system. We choose the coordinate system so that Michael is initially located at the point M = (0, 0) (the origin) and Aaron is initially located at the point A = (10,000, 0). To ﬁnd the coordinates of Michael after t seconds, we need to think about how distance and time are related. Since Michael is moving at the rate of 15 ft/second, then after one second he is located 15 feet right of the origin; i.e., at the point (15, 0). After 2 seconds, Michael has moved an additional 15 feet, for a total of 30 feet; so he is located at the point (30, 0), etc. Conclude Michael has traveled 15t ft. to the right after t seconds; i.e., his location is the point 16 CHAPTER 2. IMPOSING COORDINATES y-axis Path of tossed ball. x-axis Cliff. y-axis Path of tossed ball. Cliff. x-axis (a) Origin at the top of the ledge. (b) Origin at the bottom of the ledge. Path of tossed ball. y-axis y-axis Path of tossed ball. Cliff. x-axis Cliff. x-axis (c) Origin at the landing point. (d) Origin at the launch point. Figure 2.7: Choices when imposing an xy-coordinate system. M(t) = (15t, 0). Similarly, Aaron is located 8 ft. left of his starting location after 1 second (at the point (9,992, 0)), etc. Conclude Aaron has traveled 8t ft. to the left after t seconds; i.e., his location is the point A(t) = (10,000 − 8t, 0). The key observation required to solve the problem is that the point of collision occurs when the coordinates of Michael and Aaron are equal. Because we are moving along the horizontal axis, this amounts to ﬁnding where and when the x-coordinates of M(t) and A(t) agree. This is a straight forward algebra problem: 15t = 10,000 − 8t 23t = 10,000 t = 434.78 (2.1) To the nearest tenth of a second, the runners collide after 434.8 seconds. Plugging t = 434.78 into either expression for the position: M(434.8) = (15(434.8), 0) = (6,522, 0). 2.4. DISTANCE Michael: 15 ft sec Aaron: 8 ft sec y-axis 17 x-axis 10,000 ft M = (0,0) A = (10,000, 0) (a) The physical picture. (b) The xy-coordinate picture. y-axis NOT TO SCALE! Michael starts here. Aaron starts here. (0,0) M(t) = (15t, 0) A(t) = (10,000 − 8t, 0) M after t seconds. A after t seconds. x-axis (c) Building a visual model Michael: 10 mph. Aaron: 8 mph. 6,522 feet to collision point. (d) Michael and Aaron’s collision point. Figure 2.8: Michael and Aaron running head-on. 2.4 Distance We end this Chapter with a discussion of direction and distance in the plane. To set the stage, think about the following analogy: Example 2.4.1. You are in an airplane ﬂying from Denver to New York. How far will you ﬂy? To what extent will you travel north? To what extent will you travel east? Consider two points P = (x1,y1) and Q = (x2,y2) in the xy coordinate system, where we assume that the units on each axis are the same; for example, both in units of “feet.” Imagine starting at the location P (Denver) and ﬂying to the location Q (New York) along a straight line segment; see Figure 2.9(a). Now ask yourself this question: To what overall extent have the x and y coordinates changed? To answer this, we introduce visual and notational aides into this ﬁgure. We have inserted an “arrow” pointing from the starting position P to the ending position Q; see Figure 2.9(b). To simplify things, introduce the notation ∆x to keep track of the change in the x-coordinate and ∆y 18 CHAPTER 2. IMPOSING COORDINATES y-axis y-axis ?? ?? P Q End (stop) here. Begin (start) here. x-axis ?? ∆y y2 y1 Q = (x2, y2) Ending point. d P = (x1, y1) Beginning point. x1 x2 ∆x x-axis (a) Starting and stopping points. (b) Coordinates for P and Q. Figure 2.9: The meaning of ∆x and ∆y. to keep track of the change in the y-coordinate, as we move from P to Q. Each of these quantities can now be computed: ∆x = change in x-coordinate going from P to Q (2.2) = (x-coord of ending point) − (x-coord of beginning point) = x2 − x1 ∆y = change in y-coordinate going from P to Q = (y-coord of ending point) − (y-coord of beginning point) = y2 − y1. We can interpret ∆x and ∆y using the right triangle in Figure 2.9(b). This means we can use the Pythagorean Theorem to write: d2 = (∆x)2 + (∆y)2; that is, d = (∆x)2 + (∆y)2, p which tells us the distance d from P to Q. In other words, d is the distance we would ﬂy if we had ﬂown along that line segment connecting the two points. As an example, if P = (1, 1) and Q = (5, 4), then ∆x = 5 − 1 = 4, ∆y = 4 − 1 = 3 and d = 5. There is a subtle idea behind the way we deﬁned ∆x and ∆y: You need to specify the “beginning” and “ending” points used to do the calculation in Equations 2.2. What happens if we had reversed the choices in Figure 2.9? Then the quantities ∆x and ∆y will both be negative and the lengths of the sides of the right triangle are computed by taking the absolute value 2.4. DISTANCE 19 of ∆x and ∆y. As far as a distance calculation is concerned, the previous formula still works because of this algebra equality: d = = (∆x)2 + (∆y)2 (\|∆x\|)2 + (\|∆y\|)2. p p We will sometimes refer to ∆x and ∆y as directed distances in the x and y directions. The notion of directed distance becomes important in our discussion of lines in Chapter 4 and later when you learn about vectors; it is also very important in calculus. For example, if P = (5, 4) and Q = (1, 1), then ∆x = 1 − 5 = −4, ∆y = 1 − 4 = −3 and d = 5. Important Fact 2.4.2 (Distance formula). If P = (x1, y1) and Q = (x2, y2) are two points in the plane, then the straight line distance between the points (in the same units as th
e two axes) is given by the formula (∆x)2 + (∆y)2 (x2 − x1)2 + (y2 − y1)2. d = = p p (2.3) y-axis \|∆y\| y1 y2 P = (x1, y1) Beginning point. d Q = (x2, y2) Ending point. x2 x1 x-axis \|∆x\| Figure 2.10: A different direction. If your algebra is a little rusty, a very common mistake may crop up when you are using the distance formula. For example, 32 + 42 ? = √32 + √42 ? p √9 + 16 = 3 + 4 = 7. 5 Notice, you have an impossible situation: 5 is never equal to 7. !!! CAUTION !!! 6 20 CHAPTER 2. IMPOSING COORDINATES Example 2.4.3. Two cars depart from a four way intersection at the same time, one heading East and the other heading North. Both cars are traveling at the constant speed of 30 ft/sec. Find the distance (in miles) between the two cars after 1 hour 12 minutes. In addition, determine when the two cars would be exactly 1 mile apart. (0,b) y-axis (North) d x-axis (East) (a,0) Figure 2.11: Two departing cars. Solution. Begin with a picture of the situation. We have indicated the locations of the two vehicles after t seconds and the distance d between them at time t. By the distance formula, the distance between them is d = (a − 0)2 + (0 − b)2 = a2 + b2. p p This formula is a ﬁrst step; the difﬁculty is that we have traded the mystery distance d for two new unknown numbers a and b. To ﬁnd the coordinate a for the Eastbound car, we know the car is moving at the rate of 30 ft/sec, so it will travel 30t feet after t seconds; i.e., a = 30t. Similarly, we ﬁnd that b = 30t. Substituting into the formula for d we arrive at d = (30t)2 + (30t)2 2t2(30)2 = p = 30t√2. p First, we need to convert 1 hour and 12 minutes into seconds so that our formula can be used: 1 hr 12 min = 1 + 12/60 hr = 1.2 hr = (1.2hr) = 4,320 sec. 60 min hr 60 sec min Substituting t = 4,320 sec and recalling that 1 mile = 5,280 feet, we arrive at d = 129,600√2 feet = 183,282 feet = 34.71 miles. For the second question, we specify the distance being 1 mile and want to ﬁnd when this occurs. The idea is to set d equal to 1 mile and solve for 2.4. DISTANCE 21 t. However, we need to be careful, since the units for d are feet: 30t√2 = d = 5,280 Solving for t: t = 5,280 30√2 = 124.45 seconds = 2 minutes 4 seconds. The two cars will be 1 mile apart in 2 minutes, 4 seconds. 22 CHAPTER 2. IMPOSING COORDINATES 2.5 Exercises Problem 2.1. In the following four cases, let P be the initial (starting) point and Q the ending point; recall Equation 2.2 and Figure 2.10 on Page 19. Compute d = the distance from P to Q, ∆x and ∆y. Give your answer in exact form; eg. √2 is an exact answer, whereas 1.41 is an approximation of √2. (a) P = (0,0), Q = (1,1). (b) P = (2,1), Q = (1, − 1). (c) P = (−1,2), Q = (4, − 1). (d) P = (1,2), Q = (1 + 3t,3 + t), where t is a constant. Problem 2.2. Start with two points M = (a,b) and N = (s,t) in the xy-coordinate system. Let d be the distance between these two points. Answer these questions and make sure you can justify your answers: (a) TRUE or FALSE: d = (a − s)2 + (b − t)2. (b) TRUE or FALSE: d = p (a − s)2 + (t − b)2. (c) TRUE or FALSE: d = p (s − a)2 + (t − b)2. (d) Suppose M is the beginning point and N is the ending point; recall Equation 2.2 and Figure 2.10 on Page 19. What is ∆x? What is ∆y? p (e) Suppose N is the beginning point and M is the ending point; recall Equation 2.2 and Figure 2.10 on Page 19. What is ∆x? What is ∆y? (f) If ∆x=0, what can you say about the relationship between the positions of the two points M and N ? If ∆y=0, what can you say about the relationship between the positions of the two points M and N? (Hint: Use some speciﬁc values for the coordinates and draw some pictures to see what is going on.) Problem 2.3. Steve and Elsie are camping in the desert, but have decided to part ways. Steve heads North, at 6 AM, and walks steadily at 3 miles per hour. Elsie sleeps in, and starts walking West at 3.5 miles per hour starting at 8 AM. When will the distance between them be 25 miles? Problem 2.4. Erik’s disabled sailboat is ﬂoating at a stationary location 3 miles East and 2 miles North of Kingston. A ferry leaves Kingston heading due East toward Edmonds at 12 mph. At the same time, Erik leaves the sailboat in a dinghy heading due South at 10 ft/sec (hoping to intercept the ferry). Edmonds is 6 miles due East of Kingston. sailboat Kingston Edmonds North Ballard UDub (a) Compute Erik’s speed in mph and the Ferry speed in ft/sec. (b) Impose a coordinate system and complete this table of data concerning locations (i.e., coordinates) of Erik and the ferry. Insert into the picture the locations of the ferry and Erik after 7 minutes. Time Ferry Erik Distance Between 0 sec 30 sec 7 min t hr (c) Explain why Erik misses the ferry. (d) After 10 minutes, a Coast Guard boat leaves Kingston heading due East at a speed of 25 ft/sec. Will the Coast Guard boat catch the ferry before it reaches Edmonds? Explain. Problem 2.5. Suppose two cars depart from a four way intersection at the same time, one heading north and the other heading west. The car heading north travels at the steady speed of 30 ft/sec and the car heading west travels at the steady speed of 58 ft/sec. 2.5. EXERCISES 23 (a) Find an expression for the distance between the two cars after t seconds. Brooke ocean (b) Find the distance in miles between the two cars after 3 hours 47 minutes. 5 mi (c) When are the two cars 1 mile apart? kayak reaches shore here Problem 2.6. Allyson and Adrian have decided to connect their ankles with a bungee cord; one end is tied to each person’s ankle. The cord is 30 feet long, but can stretch up to 90 feet. They both start from the same location. Allyson moves 10 ft/sec and Adrian moves 8 ft/sec in the directions indicated. 20 ft Building 30 ft Allyson Adrian start (a) Where are the two girls located after 2 seconds? (b) After 2 seconds, will the slack in the bungee cord be used up? (c) Determine when the bungee cord ﬁrst becomes tight; there is no slack i.e. in the line. Where are the girls located when this occurs? (d) When will the bungee cord ﬁrst touch the corner of the building? (Hint: Use a fact about “similar triangles”.) Problem 2.7. Brooke is located 5 miles out from the nearest point A along a straight shoreline in her seakayak. Hunger strikes and she wants to make it to Kono’s for lunch; see picture. Brooke can paddle 2 mph and walk 4 mph. shore A Kono’s 6 mi (a) If she paddles along a straight line course to the shore, ﬁnd an expression that computes the total time to reach lunch in terms of the location where Brooke beaches the boat. (b) Determine the total time to reach Kono’s if she paddles directly to the point “A”. (c) Determine the total time to reach Kono’s if she paddles directly to Kono’s. (d) Do you think your answer to (b) or (c) is the minimum time required for Brooke to reach lunch? (e) Determine the total time to reach Kono’s if she paddles directly to a point on the shore half way between point “A” and Kono’s. How does this time compare to the times in parts (b) and (c)? Do you need to modify your answer to part (d)? Problem 2.8. A spider is located at the position (1,2) in a coordinate system, where the units on each axis are feet. An ant is located at the position (15,0) in the same coordinate system. Assume the location of the spider after t minutes is s(t) = (1 + 2t,2 + t) and the location of the ant after t minutes is a(t) = (15 − 2t,2t). (a) Sketch a picture of the situation, indicating the locations of the spider and ant at times t = 0,1,2,3,4,5 minutes. Label the locations of the bugs in your picture, using the notation s(0), s(1),...,s(5), a(0), a(1), ..., a(5). (b) When will the x-coordinate of the spider equal 5? When will the y-coordinate of the ant equal 5? (c) Where is the spider located when its y-coordinate is 3? (d) Where is each bug located when the y-coordinate of the spider is twice as large as the y-coordinate of the ant? 24 CHAPTER 2. IMPOSING COORDINATES (e) How far apart are the bugs when their x-coordinates coincide? Draw a picture, indicating the locations of each bug when their x-coordinates coincide. (f) A sugar cube is located at the position (9,6). Explain why each bug will pass through the position of the sugar cube. Which bug reaches the sugar cube ﬁrst? (g) Find the speed of each bug along its line of motion; which bug is moving faster? Problem 2.9. A Ferrari is heading south at a constant speed on Broadway (a north/south street) at the same time a Mercedes is heading west on Aloha Avenue (an east/west street). The Ferrari is 624 feet north of the intersection of Broadway and Aloha, at the same time that the Mercedes is 400 feet east of the intersection. Assume the Mercedes is traveling at the constant speed of 32 miles/hour. Find the speed of the Ferrari so that a collision occurs in the intersection of Broadway and Aloha. Problem 2.10. Two planes ﬂying opposite directions (North and South) pass each other 80 miles apart at the same altitude. The Northbound plane is ﬂying 200 mph (miles per hour) and the Southbound plane is ﬂying 150 mph. How far apart are the planes in 20 minutes? When are the planes 300 miles apart? Problem 2.11. Here is a list of some algebra problems with ”solutions.” Some of the solutions are correct and some are wrong. For each problem, determine: (i) if the answer is correct, (ii) if the steps are correct, (iii) identify any incorrect steps in the solution (noting that the answer may be correct but some steps may not be correct). (a) If x = 1, x2 − 1 x + 1 = x2 + (−1)1 x + 1 −1 1 + = x2 x = x − 1 (b) (x + y)2 − (x − y)2 = (x2 + y2) − x2 − y2 = 0 (c) If x = 4, 9(x − 4)2 3x − 12 = 32(x − 4)2 3x − 12 (3x − 12)2 3x − 12 = 3x − 12. = Problem 2.12. Assume α, β are nonzero constants. Solve for x. (a) αx + β = 1 αx−β (b) 1 α + 1 (c Problem 2.13. Simplify as far as possible. (a) (1 − t)2 + (2 + 2t)2 (b) (t + 1)2 + (−t − 1)2 − 2 (c) (d) 1 t−1 − 1 t+1 (write as a single fraction) (2 + t)2 + 4t2 p 6 6 Chapter 3 Three Simple Curves Before we discuss graphing, we ﬁ
rst want to become acquainted with the sorts of pictures that will arise. This is surprisingly easy to accomplish: Impose an xy-coordinate system on a blank sheet of paper. Take a sharp pencil and begin moving it around on the paper. The resulting picture is what we will call a curve. For example, here is a sample of the sort of “artwork” we are trying to visualize. A number of examples in the text will involve basic curves in the plane. When confronted with a curve in the plane, the fundamental question we always try to answer is this: Can we give a condition (think of it as a “test”) that will tell us preciselywhen a pointin the plane lies on a curve? y-axis A typical curve. x-axis Figure 3.1: A typical curve. Typically, the kind of condition we will give involves an equation in two variables (like x and y). We consider the three simplest situations in this chapter: horizontal lines, vertical lines and circles. 3.1 The Simplest Lines Undoubtedly, the simplest curves in the plane are the horizontal and vertical lines. For example, sketch a line parallel to the x-axis passing through 2 on the y-axis; the result is a horizontal line ℓ, as pictured. This means the line ℓ passes through the point (0, 2) in our coordinate system. A concise symbolic prescription for ALL of the points on ℓ can be given using “set notation”: ℓ = {(x, 2)\|x is any real number}. y-axis (−1, 2) (0, 2) (2, 2) (x, 2), a typical point. ℓ x-axis Figure 3.2: The points (x, y). We read the right-hand side of this expression as “the set of all points (x, 2) where x = any real number.” Notice, the points (x, y) on the line ℓ are EXACTLY the ones that lead to solutions of the equation y = 2; i.e., take any point on this line, plug the coordinates into the equation y = 2 25 26 CHAPTER 3. THREE SIMPLE CURVES and you get a true statement. Because the equation does not involve the variable x and only constrains y to equal 2, we see that x can take on any real value. In short, we see that plotting all of the solutions (x, y) to the equation y = 2 gives the line ℓ. We usually refer to the set of all solutions of the equation y = 2 as the graph of the equation y = 2. y-axis m (3, y), a typical point. (3, 3) As a second example, sketch the vertical line m passing through 3 on the x-axis; this means the line m passes through the point (3, 0) in our coordinate system. A concise symbolic prescription for ALL of the points on m can be given using “set notation”: m = {(3, y)\|y is any real number}. (3, 0) (3, −2) Figure 3.3: Stacked points. x-axis Notice, the points (x, y) on the line m are EXACTLY the ones that lead to solutions of the equation x = 3; i.e., take any point on this line, plug the coordinates into the equation x = 3 and you get a true statement. Because the equation does not involve the variable y and only speciﬁes that x = 3, y can take on any real number value. Plotting all of the solutions (x, y) to the equation x = 3 gives the line m. We usually refer to the set of all solutions of the equation x = 3 as the graph of the equation x = 3. These two simple examples highlight our ﬁrst clear connection between a geometric ﬁgure and an equation; the link is achieved by plotting all of the solutions (x, y) of the equation in the xy-coordinate system. These observations work for any horizontal or vertical line. Deﬁnition 3.1.1. Horizontal and Vertical Lines: A horizontal line ℓ passing through k on the y-axis is precisely a plot of all solutions (x, y) of the equation y = k; i.e., ℓ is the graph of y = k. A vertical line m passing through h on the x-axis is precisely a plot of all solutions (x, y) of the equation x = h; i.e., m is the graph of x = h. 3.2 Circles Another common curve in the plane is a circle. Let’s see how to relate a circle and an equation involving the variables x and y. As a special case of the distance formula (2.3), suppose P = (0, 0) is the origin and Q = (x, y) is any point in the plane; then distance from P to Q = (x − 0)2 + (y − 0)2 x2 + y2. This calculation tells us that a point (x, y) is of distance r from the origin x2 + y2 or, squaring each side, that x2 + y2 = r2. This if and only if r = shows p p = {(x, y)\|distance (x, y) to origin is r} = {(x, y)\|x2 + y2 = r2 }. (3.1) p 3.2. CIRCLES What is the left-hand side of Equation 3.1? To picture all points in the plane of distance r from the origin, fasten a pencil to one end of a non-elastic string (a string that will not stretch) of length r and tack the other end to the origin. Holding the string tight, the pencil point will locate a point of distance r from the origin. We could visualize all such points by simply moving the pencil around the origin, all the while keeping the string tight. 27 Pencil. r Start. r Draw with a tight string. Figure 3.4: Drawing a circle. What is the right-hand side of Equation 3.1? A point (x, y) in the right-hand set is a solution to the equation x2 + y2 = r2; i.e., if we plug in the coordinates we get a true statement. For example, in Figure 3.5 we plot eight solutions (r, 0), (−r, 0), (0, r), (0, −r), A = , , B = , r √2 r √2 r √2 , −r √2 C = −r √2 , −r √2 , D = −r √2 , r √2 , of the equation. To see that the last point is a solution, here is the sample calculation: 2 + −r √2 2 r √2 = r2 2 + r2 2 = r2. Since the two sides of Equation 3.1 are equal, drawing the circle of radius r is the same as plotting all of the solutions of the equation x2 + y2 = r2. The same reasoning can be used to show that drawing a circle of radius r centered at a point (h, k) is the same as plotting all of the solutions of the equation: (x − h)2 + (y − k)2 = r2. We usually refer to the set of all solutions of the equation as the graph of the equation. (−r, 0) D C (0, r) A (r, 0) B (0, −r) Figure 3.5: Computing points. Deﬁnition 3.2.1 (Circles). Let (h, k) be a given point in the xy-plane and r > 0 a given positive real number. The circle of radius r centered at (h, k) is precisely all of the solutions (x,y) of the equation (x − h)2 + (y − k)2 = r2; i.e., the circle is the graph of this equation. y-axis (x, y) r (h, k) x-axis We refer to the equation in the box as the standard form of the equation of a circle. From this equation you know both the center and radius of the circle described. Figure 3.6: Deﬁning a circle. 28 CHAPTER 3. THREE SIMPLE CURVES !!! CAUTION !!! Be very careful with the minus signs “−” in the standard form for a circle equation. For example, the equation (x + 3)2 + (y − 1)2 = 7 is NOT in standard form. We can rewrite it in standard form: (x − (−3))2 + (y − 1)2 = (√7)2; so, this equation describes a circle of radius √7 centered at (−3, 1). Examples 3.2.2. Here are some of the ways we can discuss circles: 1. The circle of radius 1 centered at the origin is the graph of the equation x2 + y2 = 1. This circle is called the unit circle and will be used extensively. 2. A circle of radius 3 centered at the point (h, k) = (1, −1) is the graph of the equation (x−1)2 +(y−(−1))2 = 32; or, equivalently (x−1)2+(y+1)2 = 32; or, equivalently x2 + y2 − 2x + 2y = 7. 3. The circle of radius √5 centered at (2, −3) does not pass through the origin; this is because (0, 0) is not a solution of the equation (x − 2)2 + (y + 3)2 = 5. 3.3 Intersecting Curves I In many problem solving situations, we will have two curves in the plane and need to determine where the curves intersect one another. Before we discuss a general procedure, let’s make sure we really understand the meaning of the word “intersect.” From Latin, the word “inter” means “within or in between” and the word “sectus” means “to cut.” So, the intersection of two curves is the place where the curves “cut into” each other; in other words, where the two curves cross one another. If the pictures of two curves are given to us up front, we can often visually decide whether or not they intersect. This is one good reason for drawing a picture of any physical problem we are trying to solve. We will need a small bag of tricks used for ﬁnding intersections of curves. We begin with intersections involving the curves studied in this section. Two different horizontal lines (or two different vertical lines) will never intersect. However, a horizontal line always intersects a vertical line exactly once; Figure 3.7(a). Given a circle and a horizontal or vertical line, we may or may not have an intersection. Looking at Figure 3.7(b), you can convince yourself a given horizontal or vertical line will intersect a circle in either two points, one point or no points. This analysis is all pictorial; how do you ﬁnd the explicit coordinates of an intersection point? Let’s look at a sample problem to isolate the procedure used. 3.3. INTERSECTING CURVES I 29 y-axis Point: (h, k). Horizontal line: y = k. x-axis Vertical line: x = h. (a) Line equations. (b) Possible intersections. Figure 3.7: Circles and lines. Example 3.3.1. Glo-Tek Industries has designed a new halogen street light ﬁxture for the city of Seattle. According to the product literature, when placed on a 50 foot light pole, the resulting useful illuminated area is a circular disc 120 feet in diameter. Assume the light pole is located 20 feet east and 40 feet north of the intersection of Parkside Ave. (a north/south street) and Wilson St. (an east/west street). What portion of each street is illuminated? y-axis (Parkside) P Illuminated zone. x-axis (Wilson) R S Q Figure 3.8: Illuminated street. Solution. The illuminated area is a circular disc whose diameter and center are both known. Consequently, we really need to study the intersection of this circle with the two streets. Begin by imposing the pictured coordinate system; we will use units of feet for each axis. The illuminated region will be a circular disc centered at the point (20, 40) in the coordinate system; the radius of the disc will be r = 60 feet. We need to ﬁnd the points of intersection P, Q, R, and S of the circle with the x-axis and the y-axis. The equation for the circle with r = 60 and center (h, k) = (20, 40)
is (x − 20)2 + (y − 40)2 = 3600. To ﬁnd the circular disc intersection with the y-axis, we have a system of two equations to work with: (x − 20)2 + (y − 40)2 = 3,600; x = 0. To ﬁnd the intersection points we simultaneously solve both equations. To do this, we replace x = 0 in the ﬁrst equation (i.e., we impose the 30 CHAPTER 3. THREE SIMPLE CURVES conditions of the second equation on the ﬁrst equation) and arrive at (0 − 20)2 + (y − 40)2 = 3,600; 400 + (y − 40)2 = 3,600; (y − 40)2 = 3,200; (y − 40) = √3,200 ± y = 40 √3,200 ± = −16.57 or 96.57. Notice, we have two solutions. This means that the circle and y-axis intersect at the points P = (0,96.57) and Q = (0, − 16.57). Similarly, to ﬁnd the circular disc intersection with the x-axis, we have a system of two equations to work with: (x − 20)2 + (y − 40)2 = 3,600; y = 0. Replace y = 0 in the ﬁrst equation (i.e., we impose the conditions of the second equation on the ﬁrst equation) and arrive at (x − 20)2 + (0 − 40)2 = 3,600; (x − 20)2 = 2,000; (x − 20) = √2,000; ± x = 20 √2,000 ± = −24.72 or 64.72. Conclude the circle and x-axis intersect at the points S = (64.72,0) and R = (−24.72,0). The procedure we used in the solution of Example 3.3.1 gives us a general approach to ﬁnding the intersection points of circles with horizontal and vertical lines; this will be important in the exercises. 3.4 Summary • • • Every horizontal line has equation of the form y = c. Every vertical line has equation of the form x = c. Every circle has equation of the form (x − h)2 + (y − k)2 = r2 where (h,k) is the center of the circle, and r is the circle’s radius. 3.5. EXERCISES 3.5 Exercises Problem 3.1. This exercise emphasizes the “mechanical aspects” of circles and their equations. (a) Find an equation whose graph is a circle of radius 3 centered at (−3,4). (b) Find an equation whose graph is a cir2 centered at the point cle of diameter 1 (3, − 11 3 ). (c) Find four different equations whose graphs are circles of radius 2 through (1,1). (d) Consider the equation (x − 1)2 + (y + 1)2 = 4. Which of the following points lie on the graph of this equation: (1,1), (1, − 1), (1, − 3), (1 + √3,0), (0, − 1 − √3), (0,0). Problem 3.2. Find the center and radius of each of the following circles. (a) x2 − 6x + y2 + 2y − 2 = 0 (b) x2 + 4x + y2 + 6y + 9 = 0 (c) x2 + 1 3 y = 127 3 x + y2 − 10 9 (d) x2 + y2 = 3 2 x − y + 35 16 Problem 3.3. Water is ﬂowing from a major broken water main at the intersection of two streets. The resulting puddle of water is circular and the radius r of the puddle is given by the equation r = 5t feet, where t represents time in seconds elapsed since the the main broke. (a) When the main broke, a runner was located 6 miles from the intersection. The runner continues toward the intersection at the constant speed of 17 feet per second. When will the runner’s feet get wet? (b) Suppose, instead, that when the main broke, the runner was 6 miles east, and 5000 feet north of the intersection. The runner runs due west at 17 feet per second. When will the runner’s feet get wet? 31 rider 100 feet ground level 62 ft. tower 60 feet operator 24 feet (a) Impose a coordinate system. (b) Suppose a rider is located at the point in the picture, 100 feet above the ground. If the rider drops an ice cream cone straight down, where will it land on the ground? (c) The ride operator is standing 24 feet to one side of the support tower on the level ground at the location in the picture. Determine the location(s) of a rider on the Ferris Wheel so that a dropped ice cream cone lands on the operator. (Note: There are two answers.) Problem 3.5. A crawling tractor sprinkler is located as pictured below, 100 feet South of a sidewalk. Once the water is turned on, the sprinkler waters a circular disc of radius 20 feet and moves North along the hose at the rate of 1 2 inch/second. The hose is perpendicular to the 10 ft. wide sidewalk. Assume there is grass on both sides of the sidewalk. hose W N S tractor sprinkler E sidewalk Problem 3.4. An amusement park Ferris Wheel has a radius of 60 feet. The center of the wheel is mounted on a tower 62 feet above the ground (see picture). For these questions, the wheel is not turning. (a) Impose a coordinate system. Describe the initial coordinates of the sprinkler and ﬁnd equations of the lines forming the North and South boundaries of the sidewalk. 32 CHAPTER 3. THREE SIMPLE CURVES (b) When will the water ﬁrst strike the side- (e) How long does the ferry spend inside the walk? radar zone? (c) When will the water from the sprinkler fall completely North of the sidewalk? (d) Find the total amount of time water from the sprinkler falls on the sidewalk. (e) Sketch a picture of the situation after 33 minutes. Draw an accurate picture of the watered portion of the sidewalk. (f) Find the area of GRASS watered after one hour. Problem 3.6. Erik’s disabled sailboat is ﬂoating stationary 3 miles East and 2 miles North of Kingston. A ferry leaves Kingston heading toward Edmonds at 12 mph. Edmonds is 6 miles due east of Kingston. After 20 minutes the ferry turns heading due South. Ballard is 8 miles South and 1 mile West of Edmonds. Impose coordinates with Ballard as the origin. sailboat Kingston Edmonds North Ballard UDub (a) Find the equations for the lines along which the ferry is moving and draw in these lines. (b) The sailboat has a radar scope that will detect any object within 3 miles of the sailboat. Looking down from above, as in the picture, the radar region looks like a circular disk. The boundary is the ”edge” or circle around this disc, the interior is the inside of the disk, and the exterior is everything outside of the disk (i.e. outside of the circle). Give a mathematical (equation) description of the boundary, interior and exterior of the radar zone. Sketch an accurate picture of the radar zone by determining where the line connecting Kingston and Edmonds would cross the radar zone. (c) When does the ferry enter the radar zone? (d) Where and when does the ferry exit the radar zone? Problem 3.7. Nora spends part of her summer driving a combine during the wheat harvest. Assume she starts at the indicated position heading east at 10 ft/sec toward a circular wheat ﬁeld of radius 200 ft. The combine cuts a swath 20 feet wide and begins when the corner of the machine labeled “a” is 60 feet north and 60 feet west of the western-most edge of the ﬁeld. N W E S 20 ft combine a swath cut center wheat field (a) When does Nora’s rig ﬁrst start cutting the wheat? (b) When does Nora’s rig ﬁrst start cutting a swath 20 feet wide? (c) Find the total amount of time wheat is being cut during this pass across the ﬁeld. (d) Estimate the area of the swath cut dur- ing this pass across the ﬁeld. Problem 3.8. (a) Solve for x: x2 − 2x + 1 x + 5 = x − 2 (b) Solve for xc) If x = −2, ﬁnd ALL solutions of the equa- tion (x + 1)2 + (y − 1)2 = 10 (d) If y = 3, ﬁnd ALL solutions of the equa- tion 5(x + 1)2 + 2(y − 1)2 = 10 Chapter 4 Linear Modeling Sometimes, we will begin a section by looking at a speciﬁc problem which highlights the topic to be studied; this section offers the ﬁrst such vista. View these problems as illustrations of precalculus in action, rather than confusing examples. Don’t panic, the essential algebraic skills will be reviewed once the motivation is in place. 4.1 The Earning Power Problem The government likes to gather all kinds of data. For example, Table 4.1 contains some data on the average annual income for full-time workers; these data were taken from the 1990 Statistical Abstract of the U.S. Given this information, a natural question would be: How can we predict the future earning power of women and men? One way to answer this (a) Women. (b) Men. YEAR MEN (dollars) YEAR WOMEN (dollars) 1970 $9,521 $5,616 1970 $18,531 $28,313 1987 1987 Table 4.1: Earning power data. question would be to use the data in the table to construct two different mathematical models that predict the future (or past) earning power for women or men. In order to do that, we would need to make some kind of initial assumption about the type of mathematical model expected. Let’s begin by drawing two identical xy-coordinate systems, where the x-axis has units of “year” and the y-axis has units of “dollars;” see Figure 4.1. In each coordinate system, the data in our table gives us two points to plot: In the case of women, the data table gives us the points P = (1970, 5,616) and Q = (1987, 18,531). Likewise, for the men, the data table gives us the points R = (1970, 9,521) and S = (1987, 28,313). To study the future earning power of men and women, we are going to make an assumption: For women, if the earning power in year x is $y, 33 34 PSfrag CHAPTER 4. LINEAR MODELING y-axis (dollars) Q = (1987,18531) P = (1970,5616) x-axis (year) 30000 25000 20000 15000 10000 5000 aca 30000 25000 20000 15000 10000 5000 y-axis (dollars) S = (1987,28313) R = (1970,5616) x-axis (yeara) Data points for women. (b) Data points for men. Figure 4.1: Visualizing the data. then the point (x, y) lies on the line connecting P and Q. Likewise, for men, if the earning power in year x is $y, then the point (x, y) lies on the line connecting R and S. In the real world, the validity of this kind of assumption would involve a lot of statistical analysis. This kind of assumption leads us to what is called a linear model, since we are demanding that the data points predicted by the model (i.e., the points (x, y)) lie on a straight line in a coordinate system. Now that we have made this assumption, our job is to ﬁnd a way to mathematically describe when a point (x, y) lies on one of the two lines pictured in Figure 4.2. Our goal in the next subsection is to review the mathematics necessary to show that the lines in Figure 4.2 are the so-called graphs of Equations 4.1 and 4.2. 30000 25000 20000 15000 10000 y y-axis (dollars) Q = (1987,18531) (x,y) on the line: This means women earn y dollars in year x. 5000 P = (1970,5616) x-axis (y
ear 30000 25000 20000 15000 10000 5000 y y-axis (dollars) S = (1987,28313) (x,y) on the line: This means men earn y dollars in year x. R = (1970,9521) x-axis (yeara) Linear model for women. (b) Linear model for men. Figure 4.2: Linear models of earning power. 4.2. RELATING LINES AND EQUATIONS ywomen = = ymen = = 18,531 − 5,616 1987 − 1970 12,915 17 28,313 − 9,521 1987 − 1970 18,792 17 (x − 1970) + 5,616 (x − 1970) + 5,616 (x − 1970) + 9,521 (x − 1970) + 9,521 35 (4.1) (4.2) 4.2 Relating Lines and Equations A systematic approach to studying equations and their graphs would begin with the simple cases, gradually working toward the more complicated. Thinking visually, the simplest curves in the plane would be straight lines. As we discussed in Chapter 3, a point on the vertical line in Figure 4.3(a) will always have the same x-coordinate; we refer to this line as the graph of the equation x = h. Likewise, a point on the horizontal line in Figure 4.3(b) will always have the same y-coordinate; we refer to this line as the graph of the equation y = k. Figure 4.3(c) is different, in the sense that neither the x nor the y coordinate is constant; i.e., as you move a point along the line, both coordinates of the point are changing. It is reasonable to guess that this line is the graph of some equation involving both x and y. The question is: What is the equation? y-axis graph of x = h y-axis (x,k) is a typical point on this line y-axis (h,y) is a typical point on this line. x-axis location h on x-axis graph of y = k location k on y-axis x-axis (x,y) is a typical point on this line graph some of equation involving x and y x-axis (a) Vertical line. (b) Horizontal line. (c) Sloped line. Figure 4.3: Lines in a plane. 36 CHAPTER 4. LINEAR MODELING Here is the key geometric fact needed to model lines by mathematical equations: Important Fact 4.2.1. Two different points completely determine a straight line. This fact tells us that if you are given two different points on a line, you can reconstruct the line in a coordinate system by simply lining a ruler up with the two points. In our discussion, we will need to pay special attention to the difference between vertical and non-vertical lines. 4.3 Non-vertical Lines Assume in this section that ℓ is a non-vertical line in the plane; for exIf we are given two points P = (x1, y1) ample, the line in Figure 4.3(c). and Q = (x2, y2) on a line ℓ, then Equation (2.2) on page 18 deﬁned two quantities we can calculate: ∆x = change in x going from P to Q = x2 − x1. ∆y = change in y going from P to Q = y2 − y1. We deﬁne the slope of the line ℓ to be the ratio of ∆y by ∆x, which is usually denoted by m: m def= = slope of ℓ = ∆y ∆x y2 − y1 x2 − x1 change in y change in x (4.3) Notice, we are using the fact that the line is non-vertical to know that this ratio is always deﬁned; i.e., we will never have ∆x = 0 (which would lead to illegal division by zero). There is some additional terminology that goes along with the deﬁnition of the slope. The term ∆y is sometimes called the rise of ℓ and ∆x is called the run of ℓ. For this reason, people often refer to the slope of a line ℓ as “the rise over the run,” meaning slope of ℓ def= rise of ℓ run of ℓ = ∆y ∆x = m. In addition, notice that the calculation of ∆y involves taking the difference of two numbers; likewise, the calculation of ∆x involves taking the difference of two numbers. For this reason, the slope of a line ℓ is sometimes called a difference quotient. 4.3. NON-VERTICAL LINES For example, suppose P = (1, 1) and Q = (4, 5) lie on a line ℓ. In this case, the rise = ∆y = 4 and the run = ∆x = 3, so m = 4 3 is the slope of ℓ. When computing ∆x, pay special attention that it is the x-coordinate of the destination point Q minus the xcoordinate of the starting point P; likewise, when computing ∆y, it is the y-coordinate of the destination point Q minus the y-coordinate of the starting point P. We can reverse this order in both calculations and get the same slope: m = ∆y ∆x = y2 − y1 x2 − x1 = −(y2 − y1) −(x2 − x1) = y1 − y2 x1 − x2 = −∆y −∆x . 37 Q = (4,5) 5 1 P = (1,1) ∆y = 4 ∆x = 3 1 4 Figure 4.4: Computing the slope of a line. We CANNOT reverse the order in just one of the calculations and get the same slope: m = y2 − y1 x2 − x1 6 = y2 − y1 x1 − x2 , and m = y2 − y1 x2 − x1 6 = y1 − y2 x2 − x1 . !!! CAUTION !!! It is very important to notice that the calculation of the slope of a line does not depend on the choice of the two points P and Q. This is a real windfall, since we are then always at liberty to pick our favorite two points on the line to determine the slope. The reason for this freedom of choice is pretty easy to see by looking at a picture. If we were to choose two other points P∗ = (x∗1, y∗1) and Q∗ = (x∗2, y∗2) on ℓ, then we would get two similar right triangles: See Figure 4.5. Basic properties of similar triangles tell us ratios of lengths of common sides are equal, so that m = y2 − y1 x2 − x1 = y∗2 − y∗1 x∗2 − x∗1 ; ∗ 2,y ∗ 2 = Q ∗ x (x2,y2) = Q y2 − y1 (x1,y1) = P x2 − x1 ∗ , Figure 4.5: Using similar triangles. but this just says the calculation of the slope is the same for any pair of distinct points on ℓ. For example, lets redo the slope calculation when P∗ = P = (x1, y1) and Q∗ = (x, y) represents an arbitrary point on the line. Then the two ratios of lengths of common sides give us the equation m = y − y1 x − x1 , y − y1 = m(x − x1). 38 CHAPTER 4. LINEAR MODELING This can be rewritten as y = m(x − x1) + y1 or y = y2 − y1 x2 − x1 (x − x1) + y1. (4.4) (4.5) Equation 4.4 is usually called the point slope formula for the line ℓ (since the data required to write the equation amounts to a point (x1,y1) on the line and the slope m), whereas Equation 4.5 is called the two point formula for the line ℓ (since the data required amounts to the coordinates of the points P and Q). In any event, we now see that Important Fact 4.3.1. (x,y) lies on ℓ if and only if (x,y) is a solution to y = m(x − x1) + y1. We can plot the collection of ALL solutions to the equation in Fact 4.3.1, which we refer to as the graph of the equation. As a subset of the xycoordinate system, the line Important Fact 4.3.2. ℓ = {(x, m(x − x1) + y1) \|x is any real number}. graph of y = 4 3 x − 1 3 (6, 23 3 ) (4,5) (−1, − 5 3 ) (1,1) (0, − 1 3 ) Figure 4.6: Verifying points on a line. Example 4.3.3. Consider the line ℓ, in Figure 4.6, through the two points P = (1,1) and Q = (4,5). Then the slope of ℓ is m = 4/3 and ℓ consists of all pairs of points (x,y) such that the coordinates x and y satisfy the equation y = 4 3(x − 1) + 1. Letting x = 0, 1, 6 and −1, we conclude that the following four points lie on the line ℓ: (0, 4 3(0 − 1) + 1) = (0, −1 3(6 − 1) + 1) = (6, 23 3(1 − 1) + 1) = (1,1), (6, 4 3 ) and (−1, 4 3 ). By the same reasoning, the point (0,0) does not lie on the line ℓ. As a set of points in the plane, we have 3 ), (1, 4 3(−1 − 1) + 1) = (−1, −5 ℓ = x, 4 3 (x − 1) + 1 \|x is any real number Returning to the general situation, we can obtain a third general equation for a non-vertical line. To emphasize what is going on here, plug the speciﬁc value x = 0 into Equation 4.4 and obtain the point R = (0,b) on the line, where b = m(0 − x1) + y1 = −mx1 + y1. But, Equation (4.4) can be written y = m(x − x1) + y1 = mx − mx1 + y1 = mx + b. 4.4. GENERAL LINES 39 The point R is important; it is precisely the point where the line ℓ crosses the y-axis, usually called the y-intercept. The slope intercept equation of the line is the form y = mx + b, where the slope of the line is m and b is the y-intercept of the line. Summary 4.3.4. Non-vertical Lines: Let ℓ be a non-vertical line in the xy-plane. There are three ways to obtain an equation whose graph is ℓ, depending on the data provided for ℓ: 1. If P = (x1,y1), Q = (x2,y2) are two different points on the line, then y2 − y1 x2 − x1 (x − x1) + y1 gives an equation the two-point formula y = whose graph is ℓ. 2. If P = (x1,y1) is a point on the line and m is the slope of ℓ, then the point-slope formula y = m(x − x1) + y1 gives an equation whose graph is ℓ. 3. If the line ℓ intersects the y-axis at the point (0,b) and m is the slope of the line ℓ, then the slope-intercept formula y = mx + b gives an equation whose graph is ℓ. 4.4 General Lines Summarizing, any line in the plane is the graph of an equation involving x and y and the equation always has the form Ax + By + C = 0, for some constants A, B, C. Equations like this are called linear equaIn general, non-vertical lines will be of the most interest to us, tions. since these are the lines that can be viewed as the graphs of functions; we will discuss this in Chapter 5. 4.5 Lines and Rate of Change If we draw a non-vertical line in the xy coordinate system, then its slope will be the rate of change of y with respect to x: slope = = ∆y ∆x change in y change in x def= rate of change of y with respect to x. 40 CHAPTER 4. LINEAR MODELING We should emphasize that this rate of change is a constant; in other words, this rate is the same no matter where we compute the slope on the line. The point-slope formula for a line can now be interpreted as follows: A line is determined by a point on the line and the rate of change of y with respect to x. An interesting thing to notice is how the units for x and y ﬁgure into the rate of change calculation. For example, suppose that we have the equation y = 10,000 x+200,000, which relates the value y of a house (in dollars) to the number of years x you own it. For example, after 5 years, x = 5 and the value of the house would be y = 10,000 (5) + 200,000 = $250,000. In this case, the equation y = 10,000 x + 200,000 is linear and already written in slope-intercept form, so the slope can be read as m = 10,000. If we carry along the units in the calculation of , then the numerator involves “dollar” units and the denominator “years” units. That means that carrying along units, the slope is actually m = 10,000 dollars/year. In other words, the va
lue of the house is changing at a rate of 10,000 dollars/year. ∆y ∆x At the other extreme, if the units for both x and y are the same, then the units cancel out in the rate of change calculation; in other words, the slope is a unit-less quantity, simply a number. This sort of thing will come up in the mathematics you see in chemistry and physics. s b speed m reference point initial location of the object location of the object at time t Figure 4.7: Motion along a line. One important type of rate encountered is the speed of a moving object. Suppose an object moves along a straight line at a constant speed m: See Figure 4.7. If we specify a reference point, we can let b be the starting location of the moving object, which is usually called the initial location of the object. We can write down an equation relating the initial location b, the time t, the constant speed m and the location s at time t: s = (location of object at time t) s = (initial location of object) + (distance object travels in time t) s = b + mt. where t is in the same time units used to deﬁne the rate m. Notice, both b and m would be constants given to us, so this is a linear equation involving the variables s and t. We can graph the equation in the tscoordinate system: See Figure 4.8. It is important to distinguish between this picture (the graph of s = mt+b) and the path of the object in Figure 4.7. The graph of the equation should be thought of as a visual aid attached to the equation s = mt + b. The general idea is that using this visual aid can help answer various questions involving the equation, which in turn will tell us things about the motion of the object in Figure 4.7. 4.5. LINES AND RATE OF CHANGE Two other comments related to this discussion are important. First, concerning notation, the speed m is often symbolized by v to denote constant velocity and b is writ(the subscript “0” meaning “time zero”). With ten as s ◦ these changes, the equation becomes s = s + vt, which is ◦ the form in which it would be written in a typical physics text. As a second note, if you return to Figure 4.8, you will notice we only drew in the positive t axis. This was because t represented time, which is always a non-negative quantity. 41 s-axis graph of s = mt + b ∆s ∆t rate of change = b = s-intercept t-axis ∆s ∆t Figure 4.8: The graph of s = mt + b. Example 4.5.1. Linda, Asia and Mookie are all playing frisbee. Mookie is 10 meters in front of Linda and always runs 5 m/sec. Asia is 34 meters in front of Linda and always runs 4 m/sec. Linda yells “go!” and both Mookie and Asia start running directly away from Linda to catch a tossed frisbee. Find linear equations for the distances between Linda, Mookie and Asia after t seconds. Linda Mookie Asia Solution. Let sM be the distance between Linda and Mookie and sA the distance between Linda and Asia, after t seconds. An application of the above formula tells us sM = (initial distance between Linda and Mookie) + + (distance Mookie runs in t seconds) · · · sM = 10 + 5t. · · · Likewise, sA = (initial distance between Linda and Asia) + + (distance Asia runs in t seconds) · · · sA = 34 + 4t. · · · If sMA is the distance between Mookie and Asia after t seconds, we compute sMA = sA − sM = (34 + 4t) − (10 + 5t) = (24 − t) meters. In all cases, the distances we computed are given by linear equations of the form s = b + mt, for appropriate b and rate m. 42 CHAPTER 4. LINEAR MODELING 4.6 Back to the Earning Power Problem 40000 35000 30000 25000 20000 15000 10000 5000 y-axis (dollars) U T S y R (x,y) on the line: means men earn y dollars in year x x-axis (year We now return to the motivating problem at the start of this section. Recall the plot in Figure 4.1(b). We can model the men’s earning power using the ﬁrst and last data points, using the ideas we have discussed about linear equations. To do this, we should specify a “beginning point” and an “ending point” (recall Figure 2.9) and calculate the slope: Rbegin = (1970,9,521) and Send = (1987,28,313). Figure 4.9: Linear model of Men’s Earning Power. We ﬁnd that ∆y ∆x 28,313 − 9,521 1987 − 1970 18,792 17 m ∆y ∆x 18,792 17 If we apply the “point-slope formula” for the equation of a line, we arrive at the equation: y = 18,792 17 (x − 1970) + 9,521. (4.6) The graph of this line will pass through the two points R and S in Figure 4.2. We sketch the graph in Figure 4.9, indicating two new points T and U. We can use the model in (4.6) to make predictions of two different sorts: (i) predict earnings at some date, or (ii) predict when a desired value for earnings will occur. For example, let’s graphically discuss the earnings in 1995: • • Draw a vertical line x = 1995 up to the graph and label the intersection point U. Draw a horizontal line ℓ through U. The line ℓ crosses the y axis at the point 18,792 17 (1995 − 1970) + 9521 = 37,156. 4.7. WHAT’S NEEDED TO BUILD A LINEAR MODEL? 43 The coordinates of the point U = (1995, 37,156). • Conclude that $37,156 is the Men’s Earning Power in 1995. For another example, suppose we wanted to know when men’s earning power will equal $33,000? This means we seek a data point T on the men’s earning curve whose y-coordinate is 33,000. By (4.6), T has the form T = x, (x − 1970) + 9,521 . 18,792 17 We want this to be a data point of the form (x, 33,000). Setting these two points equal and equating the second coordinates leads to an algebra problem: 18,792 17 (x − 1970) + 9,521 = 33,000 x = 1991.24. This means men’s earning power will be $33,000 at the end of the ﬁrst quarter of 1991. Graphically, we interpret this reasoning as follows: • • Draw a horizontal line y = 33,000 and label the intersection point T on the model. Draw a vertical line ℓ through T . The line ℓ crosses the x axis at the point 1991.24. The coordinates of the point T = (1991.24,33,000). • In the exercises, you will be asked to show that the women’s earning power model is given by the equation y = 12,915 17 (x − 1970) + 5,616. Using the two linear models for the earning power of men and women, are women gaining on men? You will also be asked to think about this question in the exercises. 4.7 What’s Needed to Build a Linear Model? As we progress through this text, a number of different “types” of mathematical models will be discussed. We will want to think about the information needed to construct that particular kind of mathematical model. Why would we care? For example, in a laboratory context, if we knew a situation being studied was given by a linear model, this would effect the amount of data collected. In the case of linear models, we can now make this useful statement: 44 CHAPTER 4. LINEAR MODELING Important Fact 4.7.1. A linear model is completely determined by: 1. One data point and a slope (a rate of change), or 2. Two data points, or 3. An intercept and a slope (a rate of change). 4.8 Linear Application Problems Example 4.8.1. The yearly resident tuition at the University of Washington was $1827 in 1989 and $2907 in 1995. Assume that the tuition growth at the UW follows a linear model. What will be the tuition in the year 2000? When will yearly tuition at the University of Washington be $10,000? 10000 8000 6000 4000 2000 y-axis (dollars) P Q x-axis (year Solution. If we consider a coordinate system where the x-axis represents the year and the y-axis represents dollars, we are given two data points: P = (1989, 1,827) and Q = (1995, 2,907). Using the two-point formula for the equation of line through P and Q, we obtain the equation Figure 4.10: Linear tuition model. y = 180(x − 1989) + 1,827. The graph of this equation gives a line through the given points as pictured in Figure 4.10. If we let x = 2000, we get y = $3,807, which tells us the tuition in the year 2000. On the other hand, if we set the equation equal to $10,000, we can solve for x: 10,000 = 180(x − 1989) + 1,827 8,173 = 180(x − 1989) 2,034.4 = x. Conclude the tuition is $10,000 in the year 2035. 4.9 Perpendicular and Parallel Lines Here is a useful fact to keep in mind. Important Facts 4.9.1. Two non-vertical lines in the plane are parallel precisely when they both have the same slope. Two non-vertical lines are perpendicular precisely when their slopes are negative reciprocals of one another. 4.10. INTERSECTING CURVES II 45 Example 4.9.2. Let ℓ be a line in the plane passing through the points (1, 1) and (6, −1). Find a linear equation whose graph is a line parallel to ℓ passing through 5 on the y-axis. Find a linear equation whose graph is perpendicular to ℓ and passes through (4, 6). Solution. Letting P = (1, 1) and Q = (6, −1), apply the “two point formula”: y = −2 5 2 5 (x − 1) + 1 x + 7 5 . = − The graph of this equation will be ℓ. This equation is in slope intercept form and we can read off that the slope is m = −2 5 . The desired line a is parallel to ℓ; it must have slope m = − 2 5 and y-intercept 5. Plugging into the “slope intercept form”: y = −2 5 x + 5. The desired line b is a line perpendicular to ℓ (so its slope is m ′ = −1 = 5 2) and passes through the point (4, 6), so we can use the “point slope formula”: −2 5 y = 5 2 (x − 4) + 6. 4.10 Intersecting Curves II We have already encountered problems that require us to investigate the intersection of two curves in the plane. Ultimately, this reduces to solving a system of two (or more) equations in the variables x and y. A useful tool when working with equations involving squared terms (i.e., x2 or y2), is the quadratic formula. Important Fact 4.10.1. Quadratic Formula: Consider the equation az2 + bz + c = 0, where a,b,c are constants. The solutions for this equation are given by the formula z = −b ± √b2 − 4ac 2a . The solutions are real numbers if and only if b2 − 4ac 0. ≥ The next example illustrates a typical application of the quadratic forIn addition, we describe a very useful technique for ﬁnding the mula. shortest distance between a “line” and a “point.” 46 CHAPTER 4. LINEAR MODELING North ﬂight path irrigated ﬁeld West Q East s e l i m
2 crop duster P 1.5 miles South Figure 4.11: The ﬂight path of a crop duster. Example 4.10.2. A crop dusting airplane ﬂying a constant speed of 120 mph is spotted 2 miles South and 1.5 miles East of the center of a circular irrigated ﬁeld. The irrigated ﬁeld has a radius of 1 mile. Impose a coordinate system as pictured, with the center of the ﬁeld the origin (0,0). The ﬂight path of the duster is a straight line passing over the labeled points P and Q. Assume that the point Q where the plane exits the airspace above the ﬁeld is the Western-most location of the ﬁeld. Answer these questions: 1. Find a linear equation whose graph is the line along which the crop duster travels. 2. Find the location P where the crop duster enters airspace above the irrigated ﬁeld. 3. How much time does the duster spend ﬂying over the irrigated ﬁeld? 4. Find the shortest distance from the ﬂight path to the center of the irrigated ﬁeld. Solution. 1. Take Q = (−1, 0) and S = (1.5, −2) = duster spotting point. Construct a line through Q and S. The slope is −0.8 = m and the line equation becomes: y = −0.8x − 0.8. (4.7) 2. The equation of the boundary of the irrigated region is x2 + y2 = 1. We need to solve this equation AND the line equation y = −0.8x − 0.8 simultaneously. Plugging the line equation into the unit circle equation gives: x2 + (−0.8x − 0.8)2 = 1 x2 + 0.64x2 + 1.28x + 0.64 = 1 1.64x2 + 1.28x − 0.36 = 0 Apply the quadratic formula and ﬁnd x = −1, 0.2195. Conclude that the x coordinate of P is 0.2195. To ﬁnd the y coordinate, plug into the line equation and get y = −0.9756. Conclude that P = (0.2195, −0.9756) 4.11. UNIFORM LINEAR MOTION 47 3. Find the distance from P to Q by using the distance formula: d = (−1 − 0.2195)2 + (0 − (−0.9756))2 = 1.562 miles p Now, 1.562 miles 120 mph = 0.01302 hours = 47 seconds. 4. The idea is to construct a line perpendicular to the ﬂight path passing through the origin of the coordinate system. This line will have slope m = − 1 −0.8 = 1.25. So this perpendicular line has equation y = 1.25x. Intersecting this line with the ﬂight path gives us the point closest to the center of the ﬁeld. The x-coordinate of this point is found by setting the two line equations equal and solving: −0.8x − 0.8 = 1.25x x = −0.3902 This means that the closest point on the ﬂight path is (−0.39, −0.49). Apply the distance formula and the shortest distance to the ﬂight path is d = (−0.39)2 + (−0.49)2 = 0.6263. p 4.11 Uniform Linear Motion When an object moves along a line in the xy-plane at a constant speed, we say that the object exhibits uniform linear motion. Its location can be described using a pair of linear equations involving a variable which represents time. That is, we can ﬁnd constants a, b, c, and d such that, at any time t, the object’s location is given by (x,y), where x = a + bt and y = c + dt. Such equations are called parametric equations of motion. The motion is deﬁned in terms of the parameter t. Since there are four constants to be determined, one needs four pieces of information to determine these equations. Knowing the object’s location at two points in time is sufﬁcient. Example 4.11.1. Bob is running in the xy-plane. He runs in a straight line from the point (2,3) to the point (5, − 4), taking 6 seconds to do so. Find his equations of motion. 48 CHAPTER 4. LINEAR MODELING Solution. We begin by setting a reference for our time parameter. Let’s let t = 0 represent the instant when Bob is at the point (2,3). In this way, t will represent the time since Bob left the point (2,3). When t = 6, we know he will be at the point (5, − 4). This is enough information to determine his equations of motion. We seek constants a, b, c, and d so that at time t, Bob’s location is given by x = a + bt and y = c + dt. When t = 0, we know Bob’s location is (2,3). That is, x = 2 and y = 3. Thus, with t = 0, we have the two equations x = 2 = a + b(0) = a and y = 3 = c + d(0) = c. and so a = 2 and c = 3. We’re half-way done. When t = 6, we know Bob’s location is (5, − 4). Thus, with t = 6, we have the two equations x = 5 = a + b(6) = 2 + 6b and y = −4 = c + d(6) = 3 + 6d which we can solve to ﬁnd b = 1 2 and d = − 7 6 . So, we arrive at the equations of Bob’s motion x = 2 + 1 2 t and y = 3 − 7 6 t. Notice it is easy to check that these are correct. If we plug in t = 0, we ﬁnd x = 2, y = 3 as required. If we plug in t = 6, we ﬁnd x = 5, y = −4, as required. So we know we’ve done it right. Now that we have these equations of motion, it is very easy to calculate Bob’s location at any time. For instance, 30 seconds after leaving the point (2,3), we can ﬁnd that he is at the point (17, − 32) since x = 2 + 1 2 (30) = 17, y = 3 − 7 6 (30) = −32. Example 4.11.2. Olga is running in the xy-plane, and the coordinate are given in meters (so, for example, the point (1,0) is one meter from the origin (0,0)). She runs in a straight line, starting at the point (3,5) and running along the line y = − 1 3 x + 6 at a speed of 7 meters per second, heading away from the y-axis. What are her parametric equations of motion? 4.11. UNIFORM LINEAR MOTION 49 Solution. This example differs in some respects from the last example. In particular, instead of knowing where the runner is at two points in time, we only know one point, and have other information given to us about the speed and path of the runner. One approach is to use this new information to ﬁnd where the runner is at some other point in time: this will then give us exactly the same sort of information as we used in the last example, and so we may solve it in an identical manner. We know that Olga starts at the point (3,5). Letting t = 0 represent the time when she starts, we then know that when t = 0, x = 3 and y = 5. To get another point (and time), we can use the fact that we know what line she travels along, and which direction she runs. We may consider any point on the line in the correct direction: any will do. For instance, the point (6,4) is on the line. We then need to ﬁnd when Olga reaches this point. To do this, we ﬁnd the distance from her starting point to the point (6,4), and divide this by her speed. The time she takes to get to (6,4) is thus (6 − 3)2 + (4 − 5)2 7 p = 0.45175395 seconds . At this point, we are now in the same situation as in the last example. We know two facts: when t = 0, x = 3 and y = 5, and when t = 0.45175395, x = 6 and y = 4. As we saw in the last example, this is enough information to ﬁnd the parametric equations of motion. We seek a, b, c, and d such that Olga’s location t seconds after she starts is (x,y) where x = a + bt and y = c + dt. When t = 0, x = 3, and y = 5, so x = 3 = a + b(0) = a and y = 5 = c + d(0) = c and so a = 3 and c = 5. Also, when t = 0.45175395, x = 6 and y = 4, so x = 6 = a+b(0.45175395) = 3+b(0.45175395) and y = 4 = c+d(0.45175395) = 5+d(0.45175395). Solving these equations for b and d, we ﬁnd b = 3 0.45175395 = 6.64078311 and d = −1 0.45175395 = −2.21359436. Thus, Olga’s equations of motion are x = 3 + 6.64078311t, y = 5 − 2.21359436t. 50 CHAPTER 4. LINEAR MODELING 4.12 Summary • • • The equation of every non-vertical line can be expressed in the form y = m(x − h) + k (the point-slope form) and y = mx + b (the slope-intercept form) A vertical line has an equation of the form x = c. The shortest distance between a point, P, and a line, l, can be found by determining a line l2 which passes through P and is perpendicular to l. Then the point at which l and l2 intersect is the point on l which is closest to l. The distance from this point to P is the shortest distance between P and l. • The location of an object moving at constant speed along a line can be described using a pair of equations (parametric equations) x = a + bt, y = c + dt. 4.13. EXERCISES 4.13 Exercises 51 Problem 4.1. This exercise emphasizes the “mechanical aspects” of working with linear equations. Find the equation of a line: Problem 4.4. Complete Table 4.2 on page 52. In many cases there may be several possible correct answers. (a) Passing through the points (1, − 1) and (−2,4). (b) Passing through the point (−1, − 2) with slope m = 40. (c) With y-intercept b = −2 and slope m = −2. (d) Passing through the point (4,11) and having slope m = 0. (e) Perpendicular to the line in (a) and pass- ing through (1,1). (f) Parallel to the line in (b) and having y- intercept b = −14. (g) Having the equation 3x + 4y = 7. (h) Crossing the x-axis at x = 1 and having slope m = 1. Problem 4.2. Sketch an accurate picture of the line having equation y = 2 − 1 2 x. Let α be an unknown constant. (a) Find the point of intersection between the line you have graphed and the line y = 1 + αx; your answer will be a point in the xy plane whose coordinates involve the unknown α. (b) Find α so that the intersection point in (a) has x-coordinate 10. (c) Find α so that the intersection point in (a) lies on the x-axis. Problem 4.3. (a) What is the area of the tri2 x + angle determined by the lines y = − 1 5, y = 6x and the y-axis? (b) If b > 0 and m < 0, then the line y = mx + b cuts off a triangle from the ﬁrst quadrant. Express the area of that triangle in terms of m and b. (c) The lines y = mx + 5, y = x and the yaxis form a triangle in the ﬁrst quadrant. Suppose this triangle has an area of 10 square units. Find m. Problem 4.5. The (average) sale price for single family property in Seattle and Port Townsend is tabulated below: YEAR SEATTLE PORT TOWNSEND $8400 1970 $168,400 1990 $38,000 $175,000 (a) Find a linear model relating the year x and the sales price y for a single family property in Seattle. (b) Find a linear model relating the year x and the sales price y for a single family property in Port Townsend. (c) Sketch the graph of both modeling equations in a common coordinate system; restrict your attention to x 1970. ≥ (d) What is the sales price in Seattle and Port Townsend in 1983 and 1998? (e) When will the average sales price in Seattle and Port Townsend be equal and what is this p
rice? (f) When will the average sales price in Port Townsend be $15,000 less than the Seattle sales price? What are the two sales prices at this time? (g) When will the Port Townsend sales price be $15,000 more than the Seattle sales price? What are the two sales prices at this time? (h) When will the Seattle sales price be double the Port Townsend sales price? (i) Is the Port Townsend sales price ever double the Seattle sales price? Problem 4.6. The cup on the 9th hole of a golf course is located dead center in the middle of a circular green that is 70 feet in diameter. Your ball is located as in the picture below: 52 CHAPTER 4. LINEAR MODELING Equation Slope y-intercept Point on the line Point on the line y = 2x + 1 (3, −4) (−1, 7) 1 1,000 −2 1 2 0 (0, 1) (3, 3) (5, −9) (3, −2) Table 4.2: Linear equations table for Problem 4.4. 9 green cup ball path 50 feet rough ball 40 feet The ball follows a straight line path and exits the green at the right-most edge. Assume the ball travels a constant rate of 10 ft/sec. (a) Where does the ball enter the green? (b) When does the ball enter the green? (c) How long does the ball spend inside the green? (d) Where is the ball located when it is closest to the cup and when does this occur. Problem 4.7. Allyson and Adrian have decided to connect their ankles with a bungee cord; one end is tied to each person’s ankle. The cord is 30 feet long, but can stretch up to 90 feet. They both start from the same location. Allyson moves 10 ft/sec and Adrian moves 8 ft/sec in the directions indicated. Adrian stops moving at time t = 5.5 sec, but Allyson keeps on moving 10 ft/sec in the indicated direction. (a) Sketch an accurate picture of the situation at time t = 7 seconds. Make sure to label the locations of Allyson and Adrian; also, compute the length of the bungee cord at t = 7 seconds. (b) Where is Allyson when the bungee reaches its maximum length? 4.13. EXERCISES 53 20 ft Building 30 ft Allyson Adrian start Problem 4.8. Dave is going to leave academia and go into business building grain silos. A grain silo is a cylinder with a hemispherical top, used to store grain for farm animals. Here is a 3D view, a cross-section, and the top view: silo h 3D-view r cross-section y-axis blind spot line of sight #1 (12,0) x-axis Dave line of sight #2 TOP VIEW If Dave is standing next to a silo of crosssectional radius r = 8 feet at the indicated position, his vision will be partially obstructed. Find the portion of the y-axis that Dave cannot see. (Hint: Let a be the x-coordinate of the point where line of sight #1 is tangent to the silo; compute the slope of the line using two points (the tangent point and (12,0)). On the other hand, compute the slope of line of sight #1 by noting it is perpendicular to a radial line through the tangency point. Set these two calculations of the slope equal and solve for a.) Problem 4.9. While speaking on the phone to a friend in Oslo, Norway, you learned that the current temperature there was −23◦ Celsius (−23◦C). After the phone conversation, you wanted to convert this temperature to Fahrenheit degrees ◦F, but you could not ﬁnd a reference with the correct formulas. You then remembered that the relationship between ◦F and ◦C is linear. (a) Using this and the knowledge that 32◦F = 0◦C and 212◦F = 100◦C, ﬁnd an equation that computes Celsius temperature in terms of Fahrenheit temperai.e., an equation of the form C= ture; “an expression involving only the variable F .” (b) Likewise, ﬁnd an equation that computes Fahrenheit temperature in terms of Celsius temperature; i.e. an equation of the form F= “an expression involving only the variable C .” (c) How cold was it in Oslo in ◦F? Problem 4.10. Pam is taking a train from the town of Rome to the town of Florence. Rome is located 30 miles due West of the town of Paris. Florence is 25 miles East, and 45 miles North of Rome. On her trip, how close does Pam get to Paris? Problem 4.11. Angela, Mary and Tiff are all standing near the intersection of University and 42nd streets. Mary and Tiff do not move, but Angela runs toward Tiff at 12 ft/sec along a straight line, as pictured. Assume the roads are 50 feet wide and Tiff is 60 feet north of the nearest corner. Where is Angela located when she is closest to Mary and when does she reach this spot? 54 CHAPTER 4. LINEAR MODELING tiff 42 nd St. mary angela University Way Problem 4.12. The infamous crawling tractor sprinkler is located as pictured below, 100 feet South of a 10 ft. wide sidewalk; notice the hose and sidewalk are not perpendicular. Once the water is turned on, the sprinkler waters a circular disc of radius 20 feet and moves North along the hose at the rate of 1 2 inch/second. (a) Impose a coordinate system. Describe the initial coordinates of the sprinkler and ﬁnd the equation of the line forming the southern boundary of the sidewalk. (b) After 33 minutes, sketch a picture of the wet portion of the sidewalk; ﬁnd the length of the wet portion of the Southern edge of the sidewalk. (c) Find the equation of the line forming the northern boundary of the sidewalk. (Hint: You can use the properties of right triangles.) Problem 4.13. Margot is walking in a straight line from a point 30 feet due east of a statue in a park toward a point 24 feet due north of the statue. She walks at a constant speed of 4 feet per second. (a) Write parametric equations for Margot’s position t seconds after she starts walking. (b) Write an expression for the distance from Margot’s position to the statue at time t. (c) Find the times when Margot is 28 feet from the statue. Problem 4.14. Juliet and Mercutio are moving at constant speeds in the xy-plane. They start moving at the same time. Juliet starts at the point (0, − 6) and heads in a straight line toward the point (10,5), reaching it in 10 seconds. Mercutio starts at (9, − 14) and moves in a straight line. Mercutio passes through the same point on the x axis as Juliet, but 2 seconds after she does. How long does it take Mercutio to reach the y-axis? Problem 4.15. (a) Solve for x. hose N S E W 100 ft (b) Solve for t: 2 = (1 + t)2 + (1 − 2t)2. (c) Solve for t: 3 √5 p = (1 + t)2 + (1 − 2t)2. (d) Solve for t: 0 = p (1 + t)2 + (1 − 2t)2. p 20 ft Problem 4.16. (a) Solve for x: 100 ft. sidewalk circular watered zone x4 − 4x2 + 2 = 0 (b) Solve for y: y − 2√y = 4 Chapter 5 Functions and Graphs Pictures are certainly important in the work of an architect, but it is perhaps less evident that visual aids can be powerful tools for solving mathematical problems. If we start with an equation and attach a picture, then the mathematics can come to life. This adds a new dimension to both interpreting and solving problems. One of the real triumphs of modern mathematics is a theory connecting pictures and equations via the concept of a graph. This transition from “equation” to “picture” (called graphing) and its usefulness (called graphical analysis) are the theme of the next two sections. The importance of these ideas is HUGE and cannot be overstated. Every moment spent studying these ideas will pay back dividends in this course and in any future mathematics, science or engineering courses. 5.1 Relating Data, Plots and Equations Imagineyou are standinghigh atop an oceanside cliff and spot a seagull hovering in the air-current. Assuming the gull moves up and down along a vertical line of motion, how can we best describe its location at time t seconds? There are three different (but closely linked) ways to describe the location of the gull: • • • a table of data of the gull’s height above cliff level at various times t; a plot of the data in a “time” (seconds) vs. “height” (feet) coordinate system; an equation relating time t (seconds) and height s (feet). gull line of motion cliff level ocean Figure 5.1: Seagull’s height. To make sure we really understand how to pass back and forth between these three descriptive modes, imagine we have tabulated (Figure 5.2) the height of the gull above cliff level at one-second time intervals 55 56 CHAPTER 5. FUNCTIONS AND GRAPHS for a 10 second time period. Here, a “negative height” means the gull is below cliff level. We can try to visualize the meaning of this data by plotting these 11 data points (t, s) in a time (sec.) vs. height (ft.) coordinate system. t (sec) 0 1 2 3 Gull Height (feet above cliff level) t (sec) 8 9 10 t (sec) 4 5 6 7 s (ft) 20 6.88 -2.5 -8.12 s (ft) -10 -8.12 -2.5 6.88 Feet 50 40 30 20 10 −10 s (ft) 20 36.88 57.5 2 4 6 8 10 Seconds (a) Symbolic data. (b) Visual data. Figure 5.2: Symbolic versus visual view of data. We can improve the quality of this description by increasing the number of data points. For example, if we tabulate the height of the gull above cliff level at 1/2 second or 1/4 second time intervals (over the same 10 second time period), we might get these two plots: Feet 50 40 30 20 10 4 Seconds Feet 50 40 30 20 10 4 Seconds −10 2 6 8 10 −10 2 6 8 10 (a) 1 2 second intervals. (b) 1 4 second intervals. Figure 5.3: Shorter time intervals mean more data points. We have focused on how to go from data to a plot, but the reverse process is just as easy: A point (t, s) in any of these three plots is interpreted to mean that the gull is s feet above cliff level at time t seconds. Furthermore, increasing the amount of data, we see how the plotted points are “ﬁlling in” a portion of a parabola. Of course, it is way too tedious to create longer and longer tables of data. What we really want is a “formula” (think of it as a prescription) that tells us how to produce a data point for the gull’s height at any given time t. If we had such a formula, then we could completely dispense with the tables of data and just use the formula to crank out data points. For example, look at this equation involving the variables t and s: s = 15 8 (t − 4)2 − 10. 5.2. WHAT IS A FUNCTION? 57 If we plug in t = 0, 1, 2, 9, 10, then we get s = 20, 6.88, −2.5, 36.88, 57.5, respectively; this was some of our initial tab
ulated data. This same equation produces ALL of the data points for the other two plots, using 1/2 second and 1/4 second time intervals. (Granted, we have swept under the rug the issue of “...where the heck the equation comes from...”; that is a consequence of mathematically modeling the motion of this gull. Right now, we are focusing on how the equation relates to the data and the plot, assuming the equation is in front of us to start with.) In addition, it is very important to notice that having this equation produces an inﬁnite number of data points for our gull’s location, since we can plug in any t value between 0 and 10 and get out a corresponding height s. In other words, the equation is A LOT more powerful than a ﬁnite (usually called discrete) collection of tabulated data. 5.2 What is a Function? Our lives are chock full of examples where two changing quantities are related to one another: • • • • The cost of postage is related to the weight of the item. The value of an investment will depend upon the time elapsed. The population of cells in a growth medium will be related to the amount of time elapsed. The speed of a chemical reaction will be related to the temperature of the reaction vessel. In all such cases, it would be beneﬁcial to have a “procedure” whereby we can assign a unique output value to any acceptable input value. For example, given the time elapsed (an input value), we would like to predict Informally, a unique future value of an investment (the output value). this leads to the broadest (and hence most applicable) deﬁnition of what we will call a function: Deﬁnition 5.2.1. A function is a procedure for assigning a unique output to any allowable input. The key word here is “procedure.” Our discussion of the hovering seagull in 5.1 highlights three ways to produce such a “procedure” using data, plots of curves and equations. y-axis P = (x,y) y x x-axis • A table of data, by its very nature, will relate two columns of data: The output and input values are listed as column entries of the table and reading across each row is the “procedure” which relates an input with a unique output. Figure 5.4: Graph of a procedure. 58 • • CHAPTER 5. FUNCTIONS AND GRAPHS Given a curve in Figure 5.4, consider the “procedure” which associates to each x on the horizontal axis the y coordinate of the pictured point P on the curve. Given an equation relating two quantities x and y, plugging in a particular x value and going through the “procedure” of algebra often produces a unique output value y. 5.2.1 The deﬁnition of a function (equation viewpoint) Now we focus on giving a precise deﬁnition of a function, in the situation when the “procedure” relating two quantities is actually given by an equation. Keep in mind, this is only one of three possible ways to describe a function; we could alternatively use tables of data or the plot of a curve. We focus on the equation viewpoint ﬁrst, since it is no doubt the most familiar. If we think of x and y as related physical quantities (e.g. time and distance), then it is sometimes possible (and often desirable) to express one of the variables in terms of the other. For example, by simple arithmetic, the equations 3x + 2y = 4 x2 − x = 1 2 y − 4 y x2 + 1 = 1, p can be rewritten as equivalent equations y = 1 2 (4 − 3x) 2x2 − 2x + 8 = y y = 1 √x2 + 1 . This leads to THE MOST IMPORTANT MATH DEFINITION IN THE WORLD: Deﬁnition 5.2.2. A function is a package, consisting of three parts: An equation of the form • y = “a mathematical expression only involving the variable x,” which we usually indicate via the shorthand notation y = f(x). This equation has the very special property that each time we plug in an x value, it produces exactly one (a unique) y value. We call the mathematical expression f(x) ”the rule”. • A set D of x-values we are allowed to plug into f(x), called the ”domain” of the function. 5.2. WHAT IS A FUNCTION? 59 • The set R of output values f(x), where x varies over the domain, called the ”range” of the function. Any time we have a function y = f(x), we refer to x as the independent variable (the “input data”) and y as the dependent variable (the “output data”). This terminology emphasizes the fact that we have freedom in the values of x we plug in, but once we specify an x value, the y value is uniquely determined by the rule f(x). Examples 5.2.3. (i) The equation y = −2x + 3 is in the form y = f(x), where the rule is f(x) = −2x+3. Once we specify a domain of x values, we have a function. For example, we could let the domain be all real numbers. y-axis Graph of y = b b x-axis (ii) Take the same rule f(x) = −2x + 3 from (i) and let the domain be all non-negative real numbers. This describes a function. However, the functions f(x) = −2x+ 3 on the domain of all non-negative real numbers and f(x) = −2x + 3 on the domain of all real numbers (from (i)) are different, even though they share the same rule; this is because their domains differ! This example illustrates the idea of what is called a restricted domain. In other words, we started with the function in (i) on the domain of all real numbers, then we “restricted” to the subset of non-negative real numbers. Figure 5.5: Constant function. (iii) The equation y = b, where b is a constant, deﬁnes a function on the domain of all real numbers, where the rule is f(x) = b; we call these the constant functions. Recall, in Chapter 3, we observed that the solutions of the equation y = b, plotted in the xy coordinate system, will give a horizontal line. For example, if b = 0, you get the horizontal axis. (iv) Consider the equation y = 1 x , then the rule f(x) = 1 x deﬁnes a function, as long as we do not plug in x = 0. For example, take the domain to be the non-zero real numbers. (v) Consider the equation y = √1 − x2. Before we start plugging in x values, we want to know the expression under the radical symbol (square root symbol) is nonnegative; this insures the square root is a real number. This amounts to solving an inequality equation: 1. These remarks show that 0 the rule f(x) = √1 − x2 deﬁnes a function, where the domain of x values is −1 x 1 − x2; i.e., −1 1. ≤ ≤ ≤ x ≤ ≤ 60 CHAPTER 5. FUNCTIONS AND GRAPHS Typically, the domain of a function y = f(x) will either be the entire number line, an interval on the number line, or a ﬁnite union of such intervals. We summarize the notation used to represent intervals in Table 5.1. Common Intervals on the Number Line Description Symbolic Notation Picture All numbers x between a and b, x possibly equal to either a or b a b x ≤ ≤ All numbers x between a = a and x and b, x = b a < x < b All numbers x between a = b and x possiand b, x bly equal to a x < b a ≤ All numbers x between a = a and x possiand b, x bly equal to Table 5.1: Interval Notations We can interpret a function as a “prescription” that takes a given x value (in the domain) and produces a single unique y value (in the range). We need to be really careful and not fall into the trap of thinking that every equation in the world is a function. For example, if we look at this equation x + y2 = 1 and plug in x = 0, the equation becomes y2 = 1. ± This equation has two solutions, y = 1, so the conclusion is that plugging in x = 0 does NOT produce a single output value. This violates one of the conditions of our function deﬁnition, so the equation x + y2 = 1 is NOT a function in the independent variable x. Notice, if you were to try and solve this equation for y in terms of x, you’d ﬁrst write y2 = 1 − x and then take a square root (to isolate y); but the square root introduces TWO roots, which is just another way of reﬂecting the fact there can be two y values attached to a single x value. Alternatively, you can solve the equation for x in terms of y, getting x = 1 − y2; this shows the equation does deﬁne a function x = g(y) in the independent variable y. 6 6 6 6 5.2. WHAT IS A FUNCTION? 61 5.2.2 The deﬁnition of a function (conceptual viewpoint) Conceptually, you can think of a function as a “process”: An allowable input goes into a “black box” and out pops a unique new value denoted by the symbol f(x). Compare this with the machine making “hula-hoops” in Figure 5.6. While you are problem solving, you will ﬁnd this to be a useful viewpoint when a function is described in words. tube in hoop out in x out y tubes f(tube) domain f(x) (a) A hoop machine as a “process” taking “tubes” to “hoops.” (b) A function as a “process” which takes x to y. Figure 5.6: Viewing a function as a “process.” Examples 5.2.4. Here are four examples of relationships that are functions: (i) The total amount of water used by a household since midnight on a particular day. Let y be the total number of gallons of water used by a household between 12:00am and a particular time t; we will use time units of “hours.” Given a time t, the household will have used a speciﬁc (unique) amount of water, call it S(t). Then y = S(t) deﬁnes a function in the independent variable t with dependent variable y. The domain would be 0 24 and the largest possible value of S(t) on this domain is S(24). This tells us that the range would be the set of values 0 S(24). ≤ ≤ y t ≤ ≤ (ii) The height of the center of a basketball as you dribble, depending on time. Let s be the height of the basketball center at time t seconds after you start dribbling. Given a time t, if we freeze the action, the center of the ball has a single unique height above the ﬂoor, call it h(t). So, the height of the basketball center is given by a function s = h(t). The domain would be a given interval of time you are dribbling the ball; for example, maybe 0 2 (the ﬁrst 2 seconds). In this case, the range would be all of the possible heights attained by the center of the basketball during this 2 seconds. ≤ ≤ t (iii) The state sales tax due on a taxable item. Let T be the state tax (in dollars) due on a taxable item that sells for z dollars. Given a taxable item that costs z dollars, the state tax due is
a single unique 62 CHAPTER 5. FUNCTIONS AND GRAPHS amount, call it W(z). So, T = W(z) is a function, where the independent variable is z. The domain could be taken to be 0 1,000,000, which would cover all items costing up to one-million dollars. The range of the function would be the set of all values W(z), as z ranges over the domain. ≤ ≤ z (iv) The speed of a chemical reaction depending on the temperature. Let v be the speed of a particular chemical reaction and T the temperature in Celsius ◦C. Given a particular temperature T , one could experimentally measure the speed of the reaction; there will be a unique speed, call it r(T ). So, v = r(T ) is a function, where the independent variable is T . The domain could be taken to be 0 100, which would cover the range of temperatures between the freezing and boiling points of water. The range of the function would be the set of all speeds r(T ), as T ranges over the domain. ≤ ≤ T 5.3 The Graph of a Function Let’s start with a concrete example; the function f(x) = −2x + 3 on the domain of all real numbers. We discussed this Example 5.2.3. Plug in the speciﬁc x values, where x = −1, 0, 1, 2 and tabulate the resulting y values of the function: x -1 0 1 2 ... x y 5 3 1 -1 ... −2x + 3 point (x,y) (-1,5) (0,3) (1,1) (2,-1) ... (x, − 2x + 3) y-axis Graph of y = −2x + 3. x-axis (a) Tabulated data. (b) Visual data. Figure 5.7: Symbolic versus visual view of data. · This tells us that the points (0, 3), (1, 1), (2, −1), (−1, 5) are solutions of the equation y = −2x + 3. For example, if y = −2x + 3, x = 0, y = 3, 0 + 3 (which is true), or if y = −2x + 3, x = 2, y = −1, then then 3 = −2 −1 = −2 2 + 3 (which is true), etc. In general, if we plug in x we get out −2x + 3, so the point (x, − 2x + 3) is a solution to the function equation y = f(x). We can plot all of these solutions in the xy-coordinate system. The set of points we obtain, as we vary over all x in the domain, is called the set of solutions of the equation y = −2x + 3: · Solutions = { (x, −2x + 3) \| x any real number}. Notice that plotting these points produces a line of slope m = −2 with y-intercept 3. In other words, the graph of the function f(x) = −2x + 3 is 5.4. THE VERTICAL LINE TEST 63 the same as the graph of the equation y = −2x + 3, as we discussed in Chapter 4. In general, by deﬁnition, we say that a point (x,y) is a solution to the function equation y = f(x) if plugging x and y into the equation gives a true statement. HowcanweﬁndALLthesolutionsoftheequation y = f(x)? In general, the deﬁnition of a function is “rigged” so it is easy to describe all solutions of the equation y = f(x): Each time we specify an x value (in the domain), there is only one y value, namely f(x). This means the point P = (x, f(x)) is the ONLY solution to the equation y = f(x) with ﬁrst coordinate x. We deﬁne the graph of the function y = f(x) to be the plot of all solutions of this equation (in the xy coordinate system). It is common to refer to this as either the “graph of f(x)” or the “graph of f.” Graph = {(x,f(x)) \| x in the domain} (5.1) Important Procedure 5.3.1. Points on a graph. The description of the graph of a function gives us a procedure to produce points on the graph AND to test whether a given point is on the graph. On the one hand, if you are given u in the domain of a function y = f(x), then you immediately can plot the point (u, f(u)) on the graph. On the other hand, if someone gives you a point (u, v), it will be on the graph only if v = f(u) is true. We illustrate this in Example 5.3.2. t Example 5.3.2. The function s = h(t) = 15 8 (t − 4)2 − 10 deﬁnes a function in the independent variable t. If we restrict 10, then the discussion in Chapter 7 to the domain 0 tells us that the graph is a portion of a parabola: See Figure 5.8. Using the above procedure, you can verify that the data points discussed in the seagull example (in 5.1) all § lie on this parabola. On the other hand, the point (0,0) is NOT on the graph, since h(0) = 20 = 0. ≤ ≤ s-axis 60 50 40 30 20 10 4 t-axis −10 2 6 8 10 Figure 5.8: s = h(t). 5.4 The Vertical Line Test There is a pictorial aspect of the graph of a function that is very revealing: Since (x, f(x)) is the only point on the graph with ﬁrst coordinate equal to x, a vertical line passing through x on the x-axis (with x in the domain) crosses the graph of y = f(x) once and only once. This gives us a decisive way to test if a curve is the graph of a function. Important Procedure 5.4.1. The vertical line test. Draw a curve in the xy-plane and specify a set D of x-values. Suppose every vertical line through a value in D intersects the curve exactly once. Then the curve is 6 64 CHAPTER 5. FUNCTIONS AND GRAPHS the graph of some function on the domain D. If we can ﬁnd a single vertical line through some value in D that intersects the curve more than once, then the curve is not the graph of a function on the domain D. For example, draw any straight line m in the plane. By the vertical line test, if the line m is not vertical, m is the graph of a function. On the other hand, if the line m is vertical, then m is not the graph of a function. These two situations are illustrated in Figure 5.9. As another example, consider the equation x2 + y2 = 1, whose graph is the unit circle and specify the domain D to be −1 1; recall Example 3.2.2. The vertical line passing through the point will intersect the unit circle twice; by the vertical line test, the unit circle is not the graph of a function on the domain −1 x ≤ 1 2, 0 1. ≤ x ≤ ≤ y-axis y-axis m m y-axis l C l crosses curve twice x-axis x-axis x-axis Figure 5.9: Applying the vertical line test. 5.4.1 Imposed Constraints In physical problems, it might be natural to constrain (meaning to “limit” or “restrict”) the domain. As an example, suppose the height s (in feet) of a ball above the ground after t seconds is given by the function s = h(t) = −16t2 + 4. s-axis Physically interesting portion of graph. t-axis Figure 5.10: Restricting the domain. We could look at the graph of the function in the tsplane and we will review in Chapter 7 that the graph looks like a parabola. The physical context of this problem makes it natural to only consider the portion of the graph in the ﬁrst quadrant; why? One way of specifying this quadrant would be to restrict the domain of possible t values to lie between 0 and 1 2; notationally, we would 1 write this constraint as 0 2. t ≤ ≤ 5.5 Linear Functions A major goal of this course is to discuss several different kinds of functions. The work we did in Chapter 4 actually sets us up to describe one 5.6. PROFIT ANALYSIS 65 very useful type of function called a linear function. Back in Chapter 4, we discussed how lines in the plane can be described using equations in the variables x and y. One of the key conclusions was: Important Fact 5.5.1. A non-vertical line in the plane will be the graph of an equation y = mx + b, where m is the slope of the line and b is the y-intercept. Notice that any non-vertical line will satisfy the conditions of the vertical line test, which means it must be the graph of a function. What is the function? The answer is to use the equation in x and y we already obtained in Chapter 4: The rule f(x) = mx + b on some speciﬁed domain will have a line of slope m and y-intercept b as its graph. We call a function of this form a linear function. Example 5.5.2. You are driving 65 mph from the Kansas state line (mile marker 0) to Salina (mile marker 130) along I-35. Describe a linear function that calculates mile marker after t hours. Describe another linear function that will calculate your distance from Salina after t hours. Solution. Deﬁne a function d(t) to be the mile marker after t hours. Using “distance=rate time,” we conclude that 65t will be the distance traveled after t hours. Since we started at mile marker 0, d(t) = 65t is the rule for the ﬁrst function. A reasonable domain would be to take 0 2, since it takes 2 hours to reach Salina. × ≤ ≤ t For the second situation, we need to describe a different function, call it s(t), that calculates your distance from Salina after t hours. To describe the rule of s(t) we can use the previous work: s(t) = (mile marker Salina) − (your mile marker at t hrs.) = 130 − d(t) = 130 − 65t. y-axis 120 100 80 60 40 20 0.5 1 1.5 2 t-axis Figure 5.11: Distance functions. For the rule s(t), the best domain would again be 0 2. We have graphed these two functions in the same coordinate system: See Figure 5.11 (Which function goes with which graph?). ≤ ≤ t 5.6 Proﬁt Analysis Let’s give a ﬁrst example of how to interpret the graph of a function in the context of an application. 66 CHAPTER 5. FUNCTIONS AND GRAPHS Example 5.6.1. A software company plans to bring a new product to market. The sales price per unit is $15 and the expense to produce and market x units is $100(1 + √x). What is the proﬁt potential? Two functions control the proﬁt potential of the new software. The ﬁrst tells us the gross income, in dollars, on the sale of x units. All of the costs involved in developing, supporting, distributing and marketing x units are controlled by the expense equation (again in dollars): g(x) = 15x e(x) = 100(1 + √x) (gross income function) (expense function) A proﬁt will be realized on the sale of x units whenever the gross income exceeds expenses; i.e., this occurs when g(x) > e(x). A loss occurs on the sale of x units when expenses exceed gross income; i.e., when e(x) > g(x). Whenever the sale of x units yields zero proﬁt (and zero loss), we call x a break-even point; i.e., when e(x) = g(x). The above approach is “symbolic.” Let’s see how to study proﬁt and loss visually, by studying the graphs of the two functions g(x) and e(x). To begin with, plot the graphs of the two individual functions in the xycoordinate system. We will focus on the situation when the sales ﬁgures are between 0 to 100 units; so the domain of x values is the interval 100. Given any sales ﬁgure x, we ca
n graphically relate three 0 x ≤ ≤ y-axis (dollars) y-axis (dollars) 1500 1000 500 x, 15x) P x-axis (units sold) 20 40 60 80 100 e s n e p x e 1500 1000 500 (x, e(x)) Q x-axis (units sold) 20 40 60 80 100 (a) Gross income graph. (b) Expenses graph. Figure 5.12: Visualizing income and expenses. things: x on the horizontal axis; a point on the graph of the gross income or expense function; • • y on the vertical axis. • If x = 20 units sold, there is a unique point P = (20, g(20)) = (20, 300) on the gross income graph and a unique point Q = (20, e(20)) = (20, 547) on the expenses graph. Since the y-coordinates of P and Q are the function values at x = 20, the height of the point above the horizontal axis is controlled by the function. 5.6. PROFIT ANALYSIS 67 dollars y (x, g(x)) (x, e(x)) sold units x 80 100 Figure 5.13: Modelling proﬁt and loss. 1400 1200 1000 800 600 400 200 If we plot both graphs in the same coordinate system, we can visually study the distance between points on each graph above x on the horizontal axis. In the ﬁrst part of this plot, the expense graph is above the income graph, showing a loss is realized; the exact amount of the loss will be e(x) − g(x), which is the length of the pictured line segment. Further to the right, the two graphs cross at the point labeled “B”; this is the break-even point; i.e., expense and income agree, so there is zero proﬁt (and zero loss). Finally, to the right of B the income graph is above the expense graph, so there is a proﬁt; the exact amount of the proﬁt will be g(x) − e(x), which is the length of the right-most line segment. Our analysis will be complete once we pin down the break-even point B. This amounts to solving the equation g(x) = e(x). (x, g(x)) (x, e(x)) 20 40 B 60 15x = 100(1 + √x) 15x − 100 = 100√x 225x2 − 3000x + 10000 = 10000x 225x2 − 13000x + 10000 = 0. Applying the quadratic formula, we get two answers: x = 0.78 or 57. Now, we face a problem: Which of these two solutions is the answer to the original problem? We are going to argue that only the second solution x = 57 gives us the break even point. What about the other ”solution” at x = 0.78? Try plugging x = 0.78 into the original equation: = 100(1 + √0.78). What has happened? Well, when going from 15(0.78) the second to the third line, both sides of the equation were squared. Whenever we do this, we run the risk of adding extraneous solutions. What should you do? After solving any equation, look back at your steps and ask yourself whether or not you may have added (or lost) solutions. In particular, be wary when squaring or taking the square root of both sides of an equation. Always check your ﬁnal answer in the original equation. We can now compute the coordinates of the break-even point using either function: B = (57, g(57)) = (57, 855) = (57, e(57)). 6 68 CHAPTER 5. FUNCTIONS AND GRAPHS 5.7 Exercises Problem 5.1. For each of the following functions, ﬁnd the expression for padelford gould f(x + h) − f(x) h . Simplify each of your expressions far enough so that plugging in h = 0 would be allowed. (a) f(x) = x2 − 2x. (b) f(x) = 2x + 3 (c) f(x) = x2 − 3 (d) f(x) = 4 − x2 (e) f(x) = −πx2 − π2 (f) f(x) = √x − 1. (Hint: Rationalize the nu- merator) x ≤ Problem 5.2. Here are the graphs of two linear functions on the domain 0 20. Find the formula for each of the rules y = f(x) and y = g(x). Find the formula for a NEW function v(x) that calculates the vertical distance between the two lines at x. Explain in terms of the picture what v(x) is calculating. What is v(5)? What is v(20)? What are the smallest and largest values of v(x) on the domain 0 20? ≤ x ≤ ≤ y-axis 60 40 20 (0,4) g(x) (20,60) (0,24) f(x) (20,20) 10 20 x-axis Problem 5.3. Dave leaves his ofﬁce in Padelford Hall on his way to teach in Gould Hall. Below are several different scenarios. In each case, sketch a plausible (reasonable) graph of the function s = d(t) which keeps track of Dave’s distance s from Padelford Hall at time t. Take distance units to be “feet” and time units to be “minutes.” Assume Dave’s path to Gould Hall is along a straight line which is 2400 feet long. (a) Dave leaves Padelford Hall and walks at a constant speed until he reaches Gould Hall 10 minutes later. (b) Dave leaves Padelford Hall and walks at a constant speed. It takes him 6 minutes to reach the half-way point. Then he gets confused and stops for 1 minute. He then continues on to Gould Hall at the same constant speed he had when he originally left Padelford Hall. (c) Dave leaves Padelford Hall and walks at a constant speed. It takes him 6 minutes to reach the half-way point. Then he gets confused and stops for 1 minute to ﬁgure out where he is. Dave then continues on to Gould Hall at twice the constant speed he had when he originally left Padelford Hall. (d) Dave leaves Padelford Hall and walks at a constant speed. It takes him 6 minutes to reach the half-way point. Dave gets confused and stops for 1 minute to ﬁgure out where he is. Dave is totally lost, so he simply heads back to his ofﬁce, walking the same constant speed he had when he originally left Padelford Hall. (e) Dave leaves Padelford heading for Gould Hall at the same instant Angela leaves Gould Hall heading for Padelford Hall. Both walk at a constant speed, but Angela walks twice as fast as Dave. Indicate a plot of “distance from Padelford” vs. “time” for both Angela and Dave. (f) Suppose you want to sketch the graph of a new function s = g(t) that keeps track of Dave’s distance s from Gould Hall at time t. How would your graphs change in (a)-(e)? Problem 5.4. At 5 AM one day, a monk began a trek from his monastery by the sea to the monastery at the top of a mountain. He reached the mountain-top monastery at 11 AM, spent the rest of the day in meditation, 5.7. EXERCISES 69 and then slept the night there. In the morning, at 5 AM, he began walking back to the seaside monastery. Though walking downhill should have been faster, he dawdled in the beautiful sunshine, and ending up getting to the seaside monastery at exactly 11 AM. (a) Was there necessarily a time during each trip when the monk was in exactly the same place on both days? Why or why not? (b) Suppose the monk walked faster on the second day, and got back at 9 AM. What is your answer to part (a) in this case? (c) Suppose the monk started later, at 10 AM, and reached the seaside monastery at 3 PM. What is your answer to part (a) in this case? Problem 5.5. Sketch a reasonable graph for each of the following functions. Specify a reasonable domain and range and state any assumptions you are making. Finally, describe the largest and smallest values of your function. (a) Height of a person depending on age. (b) Height of the top of your head as you jump on a pogo stick for 5 seconds. (c) The amount of postage you must put on a ﬁrst class letter, depending on the weight of the letter. (d) Distance of your big toe from the ground as you ride your bike for 10 seconds. (e) Your height above the water level in a swimming pool after you dive off the high board. y-axis x-axis Recall the procedure 5.3.1 on page 63. (a) Find the x and y intercepts of the graph. (b) Find the exact coordinates of all points (x,y) on the graph which have ycoordinate equal to 5. (c) Find the coordinates of all points (x,y) on the graph which have y-coordinate equal to -3. (d) Which of these points is on the graph: (1, − 2), (−1,3), (2.4,8), (√3,7 − 3√3). (e) Find the exact coordinates of the point (x,y) on the graph with x = 1 + √2. p Problem 5.7. After winning the lottery, you decide to buy your own island. The island is located 1 km offshore from a straight portion of the mainland. There is currently no source of electricity on the island, so you want to run a cable from the mainland to the island. An electrical power sub-station is located 4 km from your island’s nearest location to the shore. It costs $50,000 per km to lay a cable in the water and $30,000 per km to lay a cable over the land. ocean power your island cable path 1 km x 4 km Problem 5.6. Here is a picture of the graph of the function f(x) = 3x2 − 3x − 2. (a) Explain why we can assume the cable follows the path indicated in the picture; 70 CHAPTER 5. FUNCTIONS AND GRAPHS explain why the path consists of i.e. two line segments, rather than a weird curved path AND why it is OK to assume the cable reaches shore to the right of the power station and the left of the island. (b) Let x be the distance downshore from the power sub-station to where the cable reaches the land. Find a function f(x) in the variable x that computes the cost to lay a cable out to your island. (c) Make a table of values of f(x), where x = 0, 1 2 ,4. Use these calculations to estimate the installation of minimal cost. 2 ,2,. . . , 7 2 ,1, 3 Problem 5.8. This problem deals with the “mechanical aspects” of working with the rule of a function. For each of the functions listed ), in (a)-(c), calculate: f(0), f(−2), f(x + 3), f( f( ♥ + ). △ ♥ (a) The function f(x) = 1 main of all real numbers. 2 (x − 3) on the do- Problem 5.9. Which of the curves in Figure 5.14 represent the graph of a function? If the curve is not the graph of a function, describe what goes wrong and how you might “ﬁx it.” When you describe how to “ﬁx” the graph, you are allowed to cut the curve into pieces and such that each piece is the graph of a function. Many of these problems have more than one correct answer. Problem 5.10. Find an EXACT answer for each problem. (a) Solve for = 30 x2 − 4x − 21 (b) Solve for x √5x − 4 = x 2 + 2 (c) Solve for x √x + √x − 20 = 10 (b) The function f(x) = 2x2 − 6x on the do- (d) Solve for t main of all real numbers. (c) The function f(x) = 4π2. √2t − 1 + √3t + 3 = 5 5.7. EXERCISES 71 (a) (e) (i) (b) (c) (d) (f) (g) (h) (j) (k) (l) (p) (m) (n) (o) Figure 5.14: Curves to consider for Problem 5.9 72 CHAPTER 5. FUNCTIONS AND GRAPHS Chapter 6 Graphical Analysis We ended the previous section with an in-depth look at a “Proﬁt Analysis Problem.”
In that discussion, we looked at the graphs of the relevant functions and used these as visual aids to help us answer the questions posed. This was a concrete illustration of what is typically called “graphical analysis of a function.” This is a fundamental technique we want to carry forward throughout the course. Let’s highlight the key ideas for future reference. 6.1 Visual Analysis of a Graph A variety of information can be visually read off of a function graph. To see this, we ask ourselves the following question: What is the most basic qualitative feature of a graph? To answer this, we need to return to the deﬁnition of the graph (see Equation (5.1) on page 63) and the surrounding discussion. The key thing about the graph of a function f(x) is that it keeps track of a particular set of points in the plane whose coordinates are related by the function rule. To be precise, a point P = (x,y) will be on the graph of the function f(x) exactly when y = f(x). 6.1.1 Visualizing the domain and range A function is a package that consists of a rule y = f(x), a domain of allowed x-values and a range of output y-values. The domain can be visualized as a subset of the x-axis and the range as a subset of the y-axis. If you are handed the domain, it is graphically easy to describe the range values obtained; here is the procedure: Important Procedure 6.1.1. Look at all points on the graph corresponding to domain values on the x-axis, then project these points to the y-axis. The collection of all values you obtain on the y-axis will be the range of the function. This idea of “projection” is illustrated in the two graphs be- 73 74 CHAPTER 6. GRAPHICAL ANALYSIS low. We use arrows “ graph, then over to the y-axis: ” to indicate going from a domain x-value, up to the → graph of f (x) = −2x + 3 domain = [3,5] range = [1,9] range = [−7, − 3] domain = [−3,1] graph of f (x) = −2x + 3 (a) Domain: [3, 5]. (b) Domain: [−3, 1]. Figure 6.1: Example projections. 6.1.2 Interpreting Points on the Graph We can visually detect where a function has positive or negative values: Important Fact 6.1.2. The function values f(x) control the height of the point P(x) = (x,f(x)) on the graph above the x-axis; if the function value f(x) is negative, the point P(x) is below the x-axis. P (x) = (x,f (x)) f (x) units above x-axis f(x3) units above the x-axis x2 P (x3) = (x3,f (x3)) x x3 x-axis \|f(x)\| units below the x-axis P(x2) = (x2,f(x2)) y-axis Figure 6.2: Interpreting points on a graph. In Figure 6.2 we can now divide the domain (in this case the whole number line) into segments where the function is above, below or crossing the axis. Keeping track of this information on a number line is called a sign plot for the function. We include a “shadow” of the graph in Figure 6.3 6.1. VISUAL ANALYSIS OF A GRAPH 75 positive positive negative negative x-axis Figure 6.3: Sign plot. to emphasize how we arrived at our “positive” and “negative” labeling of the sign plot; in practice we would only provide a labeled number line. By moving through a sequence of x values we can investigate how the corresponding points on the graph move “up and down”; this then gives us a dynamic visual sense of how the function values are changing. For example, in Figure 6.4, suppose we let x move from 1 to 5, left to right; we have indicated how the corresponding points on the curve will move and how the function values will change. #3: f(x) values move like this y-axis f(1) P f(2) f(3) f(4) 1 f(5) #2: points on graph move from P to Q graph of y = f(x) Q 5 #1: x values move from 1 to 5 x-axis Figure 6.4: Dynamic interpretation of a graph. 76 CHAPTER 6. GRAPHICAL ANALYSIS 6.1.3 Interpreting Intercepts of a Graph y-axis y-intercept = (0,f(0)) (x3,0) (x2,0) x-axis (x1,0) x-intercepts have the form “(x,0)” Figure 6.5: Intercepts of a graph. The places where a graph crosses the axes are often signiﬁcant. We isolate each as an important feature to look for when doing graphical analysis. The graph of the function y = f(x) crosses the y-axis at the point (0,f(0)); so, the y-intercept of the graph is just f(0). The graph of the function y = f(x) crosses the x-axis at points of the form (x0,f(x0)), where f(x0) = 0. The values x0 are called roots or zeros of the function f(x). There can be at most one y-intercept, but there can be several x-intercepts or no x-intercept: See Figure 6.5 The graph of a function y = f(x) crosses the vertical line x = h at the point (h,f(h)). To ﬁnd where the graph of a function y = f(x) crosses the horizontal line y = k, ﬁrst solve the equation k = f(x) for x. If the equation k = f(x) has solutions x1, x2, x3, x4, then the points of intersection would have the coordinates given. (h,(f(h)) y-axis graph of f(x) (x1,k) (x2,k) (x3,k) (x4,k) y = k x1 x2 x3 x4 x-axis x = h y-axis upper semicircle radius r = 2 centered at (2,1-axis (a) General curve. (b) Semicircle. Figure 6.6: Crossing horizontal and vertical lines. As another example, the graph in Figure 6.6(b) above will cross the horizontal line y = k twice if and only if 1 k < 3; the graph will cross the ≤ horizontal line y = k once if and only if k = 3. The graph will not intersect the line y = 1 2 and the graph will cross the vertical line x = h if and only if 0 4. h ≤ ≤ 6.1.4 Interpreting Increasing and Decreasing y-axis uphill x-axis = local extrema downhill Figure 6.7: Graphically interpreting increasing and decreasing. We use certain terms to describe how the function values are changing over some domain of x values. Typically, we want to study what is happening to the values f(x) as x moves from “left to right” in some interval. This can be linked graphically with the study of “uphill” and “downhill” portions of the function graph: If you were “walking to the right” along the graph, the function values are increasing if you are walking uphill. Likewise, if you were 6.2. CIRCLES AND SEMICIRCLES 77 “walking to the right” along the graph, the function values are decreasing if you are walking downhill. Once we understand where the graph is moving uphill and downhill, we can isolate the places where we change from moving uphill to downhill, or vice versa; these “peaks” and “valleys” are called local maxima and local minima. Some folks refer to either case as a local extrema. People have invested a lot of time (centuries!) and energy (lifetimes!) into the study of how to ﬁnd local extrema for particular function graphs. We will see some basic examples in this course and others will surface in future courses once you have the tools of calculus at your disposal. Examples range from business applications that involve optimizing proﬁt to understanding the three-dimensional shape a of biological molecule. Example 6.1.3. A hang glider launches from a gliderport in La Jolla. The launch point is located at the edge of a 500 ft. high cliff over the Paciﬁc Ocean. The elevation of the pilot above the gliderport after t minutes is given by the graph in Figure 6.8: 1. When is the pilot climbing and descending? 2. When is the pilot at the glider port elevation? 3. How much time does the pilot spend ﬂying level? ft above gliderport 5 minutes 10 600 400 200 −200 −400 −600 Figure 6.8: Hanglider elevations. Solution. 1. Graphically, we need to determine the portions of the graph that are increasing or decreasing. In this example, it is increasing when 5 and 0 9 9. And, it is decreasing when 3 2 and 7 10. Graphically, this question amounts to asking when the elevation is 0, which is the same as ﬁnding when the graph crosses the horizontal axis. We can read off there are four such times: t = 0, 4, 8, 10. 3. Graphically, we need to determine the portions of the graph that are 3 made up of horizontal line segments. This happens when 2 7. So, our pilot ﬂies level for a total of 3 minutes. and 5 ≤ ≤ t t ≤ ≤ 6.2 Circles and Semicircles Back in Chapter 3, we discussed equations whose graphs were circles: We found that the graph of the equation (x − h)2 + (y − k)2 = r2 (6.1) 78 CHAPTER 6. GRAPHICAL ANALYSIS is a circle of radius r centered at the point (h, k). It is possible to manipulate this equation and become confused. We could rewrite this as (y − k)2 = r2 − (x − h)2, then take the square root of each side. However, the resulting equivalent equation would be y = k r2 − (x − h)2 ± and the presence of that ± r2 − (x − h)2 or r2 − (x − h)2. sign is tricky; it means we have two equations: Each of these two equations deﬁnes a function: p f(x) = k + g(x) = k − p r2 − (x − h)2 or r2 − (x − h)2. (6.2) (6.3) p So, even though the Equation 6.1 is not a function, we were able to obtain two different functions f(x) and g(x) from the original equation. The relationship between the graph of the original equation and the graphs of the two functions in (6.2) and (6.3) is as follows: The upper semicircle is the graph of the function f(x) and the lower semicircle is the graph of the function g(x). y-axis upper semicircle y-axis (h,k) x-axis (h,k) lower semicircle x-axis Graph of y = f(x) Graph of y = g(x) Figure 6.9: Upper and lower semicircles. Example 6.2.1. A tunnel connecting two portions of a space station has a circular cross-section of radius 15 feet. Two walkway decks are constructed in the tunnel. Deck A is along a horizontal diameter and another parallel Deck B is 2 feet below Deck A. Because the space station is in a weightless environment, you can walk vertically upright along Deck A, or vertically upside down along Deck B. You have been assigned to paint “safety stripes” on each deck level, so that a 6 foot person can safely walk upright along either deck. Determine the width of the “safe walk zone” on each deck. 6.3. MULTIPART FUNCTIONS 79 Deck A x-axis r = 15ft Deck B (a) Cross-section of tunnel. Deck A Deck B = Safe walk zone (b) Walk zones. Figure 6.10: Space station tunnels. Solution. Impose a coordinate system so that the origin is at the center of the circular cross section of the tunnel; by symmetry the walkway is centered about the
origin. With this coordinate system, the graph of the equation x2 + y2 = 152 = 225 will be the circular cross-section of the tunnel. In the case of Deck A, we basically need to determine how close to each edge of the tunnel a 6 foot high person can stand without hitting his or her head on the tunnel; a similar remark applies to Deck B. This means we are really trying to ﬁt two six-foot-high rectangular safe walk zones into the picture: Our job is to ﬁnd the coordinates of the four points P, Q, R, and S. Let’s denote by x1, x2, x3,and x4 the xcoordinates of these four points, then P = (x1, 6), Q = (x2, 6), R = (x3, −8), and S = (x4, −8). To ﬁnd x1, x2, x3, and x4, we need to ﬁnd the intersection of the circle in Figure 6.10(b) with two horizontal lines: • • Intersecting the the upper semicircle with the horizontal line having equation y = 6 will determine x1 and x2; the upper semicircle is the graph of f(x) = √225 − x2. Intersecting the lower semicircle with the horizontal line having equation y = −8 will determine x3 and x4; the lower semicircle is the graph of g(x) = −√225 − x2. For Deck A, we simultaneously solve the system of equations y = √225 − x2 y = 6 . Plugging in y = 6 into the ﬁrst equation of the system gives x2 = 225 − 62 = 189; i.e., x = 13.75. This tells us that P = (−13.75, 6) and Q = (13.75, 6). In a similar way, for Deck B, we ﬁnd R = (−12.69, −8) and S = (12.69, −8). √189 = ± ± In the case of Deck A, we would paint a safety stripe 13.75 feet to the right and left of the centerline. In the case of Deck B, we would paint a safety stripe 12.69 feet to the right and left of the centerline. 6.3 Multipart Functions So far, in all of our examples we have been able to write f(x) as a nice compact expression in the variable x. Sometimes we have to work harder. 80 CHAPTER 6. GRAPHICAL ANALYSIS As an example of what we have in mind, consider the graph in Figure 6.11(a): y-axis 1 2 3 4 x-axis 1 −1 (a) Graphing a multipart function. −1 if 0 1 if 1 −1 if 2 1 if 1 if x = 4 f(x) =    (b) Writing a multipart function. Figure 6.11: A multipart function. x ≤ The curve we are trying to describe in this picture is made up of ﬁve pieces; four little line segments and a single point. The ﬁrst thing to 4, this curve will deﬁne the graph of notice is that on the domain 0 ≤ some function f(x). To see why this is true, imagine a vertical line moving from left to right within the domain 0 4 on the x-axis; any one of these vertical lines will intersect the curve exactly once, so by the vertical line test, the curve must be the graph of a function. Mathematicians use the shorthand notation above to describe this function. Notice how the rule for f(x) involves ﬁve cases; each of these cases corresponds to one of the ﬁve pieces that make up the curve. Finally, notice the care with the “open” and “closed” circles is really needed if we want to make sure the curve deﬁnes a function; in terms of the rule, these open and closed circles translate into strict inequalities like < or weak inequalities like . ≤ This is an example of what we call a multipart function. ≤ ≤ x The symbolic appearance of multipart functions can be somewhat frightening. The key point is that the graph (and rule) of the function will be broken up into a number of separate cases. To study the graph or rule, we simply “home in” on the appropriate case. For example, in the above illustration, suppose we wanted to compute f(3.56). First, we would ﬁnd which of the ﬁve cases covers x = 3.56, then apply that part of the rule to compute f(3.56) = 1. Our ﬁrst multipart function example illustrated how to go from a graph in the plane to a rule for f(x); we can reverse this process and go from the rule to the graph. Example 6.3.1. Sketch the graph of the multipart function 1 1 + √1 − x2 1 g(x) =    if x ≤ if −1 if x ≥ −1 x ≤ 1 1 ≤ 6.3. MULTIPART FUNCTIONS 81 y-axis ≤ Solution. The graph of g(x) will consist of three pieces. The ﬁrst case consists of the graph of the function y = −1, this consists of all points g(x) = 1 on the domain x on the horizontal line y = 1 to the left of and including the point (−1,1). We have “lassoed” this portion of the graph in Figure 6.12. Likewise, the third case in the deﬁnition yields the graph of the function y = g(x) = 1 on the domain 1; this is just all points on the horizontal line y = 1 x to the right of and including the point (1, 1). Finally, we need to analyze the middle case, which means we need to look at the graph of 1 + √1 − x2 on the domain −1 1. This is just the upper semicircle of the circle of radius 1 centered at (0,1). If we paste these three pieces together, we arrive at the graph of g(x). ﬁrst part of graph ≤ ≤ ≥ x = graph of g(x) second part of graph x-axis third part of graph Figure 6.12: Multipart function g(x). Example 6.3.2. You are dribbling a basketball and the function s = h(t) keeps track of the height of the ball’s center above the ﬂoor after t seconds. Sketch a reasonable graph of s = h(t). seconds 2 Solution. If we take the domain to be 0 2 (the ﬁrst 2 seconds), a reasonable graph might look like Figure 6.13. This is a multipart function. Three portions of the graph are decreasing and two portions are increasing. Why doesn’t the graph touch the t axis? ≤ ≤ t Figure 6.13: Dribbling. 82 CHAPTER 6. GRAPHICAL ANALYSIS 6.4 Exercises Problem 6.1. The absolute value function is deﬁned by the multipart rule: Problem 6.4. (a) Let f(x) = x + \|2x − 1\|. Find all solutions to the equation x if 0 x −x if x < 0 ≤ \|x\| = f(x) = 8. The graph of the absolute value function is pictured below: (b) Let g(x) = 3x − 3 + \|x + 5\|. Find all values of a which satisfy the equation y-axis g(a) = 2a + 8. y = \|x\| x-axis (c) Let h(x) = \|x\| − 3x + 4. Find all solutions to the equation h(x − 1) = x − 2. Problem 6.5. Express the area of the shaded region below as a function of x. The dimensions in the ﬁgure are centimeters. (a) Calculate: \|0\|, \|2\|, \| − 3\|. (b) Solve for x: \|x\| = 4; \|x\| = 0, \|x\| = −1. (c) Sketch the graph of y = 1 2 x + 2 and y = \|x\| in the same coordinate system. Find where the two graphs intersect, label the coordinates of these point(s), then ﬁnd the area of the region bounded by the two graphs. Problem 6.2. For each of the following functions, graph f(x) and g(x) = \|f(x)\|, and give the multipart rule for g(x). 3 (a) f(x) = −0.5x − 1 (b) f(x) = 2x − 5 (c) f(x) = x + 3 x 6 5 Problem 6.3. Solve each of the following equations for x. (a) g(x) = 17, where g(x) = \|3x + 5\| (b) f(x) = 1.5 where f(x) = (c) h(x) = −1 where 2x if x < 3, 3. 4 − x if x ≥ h(x) = −8 − 4x if x 1 + 1 −2, 3 x if x > −2. ≤ Problem 6.6. Pizzeria Buonapetito makes a triangular-shaped pizza with base width of 30 inches and height 20 inches as shown. Alice wants only a portion of the pizza and does so by making a vertical cut through the pizza and taking the shaded portion. Letting x be the bottom length of Alice’s portion and y be the length of the cut as shown, answer the following questions: 6.4. EXERCISES 83 y x 10 x 20 y 20 20 Problem 6.8. Arthur is going for a run. From his starting point, he runs due east at 10 feet per second for 250 feet. He then turns, and runs north at 12 feet per second for 400 feet. He then turns, and runs west at 9 feet per second for 90 feet. Express the (straight-line) distance from Arthur to his starting point as a function of t, the number of seconds since he started. Problem 6.9. A baseball diamond is a square with sides of length 90 ft. Assume Edgar hits a home run and races around the bases (counterclockwise) at a speed of 18 ft/sec. Express the distance between Edgar and home plate as a function of time t. (Hint: This will be a multipart function.) Try to sketch a graph of this function. (a) Find a formula for y as a multipart function of x, for 0 30. Sketch the graph of this function and calculate the range. ≤ ≤ x (b) Find a formula for the area of Alice’s portion as a multipart function of x, for 0 30. x ≤ ≤ (c) If Alice wants her portion to have half the area of the pizza, where should she make the cut? Problem 6.7. This problem deals with cars traveling between Bellevue and Spokane, which are 280 miles apart. Let t be the time in hours, measured from 12:00 noon; for example, t = −1 is 11:00 am. (a) Joan drives from Bellevue to Spokane at a constant speed, departing from Bellevue at 11:00 am and arriving in Spokane at 3:30 pm. Find a function j(t) that computes her distance from Bellevue at time t. Sketch the graph, specify the domain and determine the range. (b) Steve drives from Spokane to Bellevue at 70 mph, departing from Spokane at 12:00 noon. Find a function s(t) for his distance from Bellevue at time t. Sketch the graph, specify the domain and determine the range. (c) Find a function d(t) that computes the distance between Joan and Steve at time t. 90 ft Edgar d(t) home plate Problem 6.10. Pagliacci Pizza has designed a cardboard delivery box from a single piece of cardboard, as pictured. (a) Find a polynomial function v(x) that computes the volume of the box in terms of x. What is the degree of v? (b) Find a polynomial function a(x) that computes the exposed surface area of the closed box in terms of x. What is the degree of a? What are the explicit dimensions if the exposed surface area of the closed box is 600 sq. inches? 84 x x x x 50 in remove shaded squares and fold to get: CHAPTER 6. GRAPHICAL ANALYSIS (a) When will the ditch be completely full? (b) Find a multipart function that models the vertical cross-section of the ditch. 20 in (c) What is the width of the ﬁlled portion of the ditch after 1 hour and 18 minutes? (d) When will the ﬁlled portion of the ditch be 42 feet wide? 50 feet wide? 73 feet wide? Problem 6.12. The graph of a function y = g(x) on the domain −6 6 consistes of line ≤ segments and semicircles of radius 2 connecting the points (−6,0),(−4,4), (0,4), (4,4), (6,0). ≤ x y-axis x-axis Problem 6.11. The vertical cross-section of a drainage ditch is pictured below: (a) What i
s the range of g? (b) Where is the function increasing? Where is the function decreasing? (c) Find the multipart formula for y = g(x). (d) If we restrict the function to the smaller domain −5 x ≤ ≤ 0, what is the range? (e) If we restrict the function to the smaller domain 0 x ≤ ≤ 4, what is the range? 3D−view of ditch 20 ft 20 ft R R Problem 6.13. (a) Simply as far as possible R R vertical cross-section Here, R indicates a circle of radius 10 feet and all of the indicated circle centers lie along the common horizontal line 10 feet above and parallel to the ditch bottom. Assume that water is ﬂowing into the ditch so that the level above the bottom is rising 2 inches per minuteb) Find a, b, c that simultaneously satisfy these three equations: a + b − c = 5 2a − 3b + 1 Chapter 7 Quadratic Modeling If you kick a ball through the air enough times, you will ﬁnd its path tends to be parabolic. Before we can answer any detailed questions about this situation, we need to get our hands on a precise mathematical model for a parabolic shaped curve. This means we seek a function y = f(x) whose graph reproduces the path of the ball. ground level Figure 7.1: Possible paths for a kicked ball are parabolic. 7.1 Parabolas and Vertex Form OK, suppose we sit down with an xy-coordinate system and draw four random parabolas; let’s label them I, II, III, and IV: See Figure 7.2. The relationship between these parabolas and the ﬁxed coordinate system can vary quite a bit: The key distinction between these four curves is that only I and IV are the graphs of functions; this follows from the vertical line test. A parabola that is the graph of a function is called a standard parabola. We can see that any standard parabola has three basic features: 85 86 CHAPTER 7. QUADRATIC MODELING y-axis II I III not graphs of functions IV x-axis Figure 7.2: Relationship between a ﬁxed coordinate system and various parabolas. • • • the parabola will either open “upward” or “downward”; the graph will have either a “highest point” or “lowest point,” called the vertex; the parabola will be symmetric about some vertical line called the axis of symmetry. Our ﬁrst task is to describe the mathematical model for any standard parabola. In other words, what kind of function equations y = f(x) give us standard parabolas as their graphs? Our approach is geometric and visual: • Begin with one speciﬁc example, then show every other standard parabola can be obtained from it via some speciﬁc geometric maneuvers. • As we perform these geometric maneuvers, we keep track of how the function equation for the curve is changing. This discussion will amount to a concrete application of a more general set of tools developed in the following section of this chapter. y-axis 35 30 25 20 15 10 5 x-axis Using a graphing device, it is an easy matter to plot the graph of y = x2 and see we are getting the parabola pictured in Figure 7.3. The basic idea is to describe how we can manipulate this graph and obtain any standard parabola. In the end, we will see that standard parabolas are obtained as the graphs of functions having the form −6 −4 −2 2 4 6 Figure 7.3: Graph of y = x2. y = ax2 + bx + c, 7.1. PARABOLAS AND VERTEX FORM 87 = 0. A function of this type is for various constants a, b, and c, with a called a quadratic function and these play a central role throughout the course. We will divide our task into two steps. First we show every standard parabola arises as the graph of a func- tion having the form y = a(x − h)2 + k, = 0. This is called the vertex for some constants a, h, and k, with a form of a quadratic function. Notice, if we were to algebraically expand out this equation, we could rewrite it in the y = ax2 + bx + c form. For example, suppose we start with the vertex form y = 2(x − 1)2 + 3, so that a = 2, h = 1, k = 3. Then we can rewrite the equation in the form y = ax2 + bx + c as follows: 2(x − 1)2 + 3 = 2(x2 − 2x + 1) + 3 = 2x2 − 4x + 5, so a = 2, b = −4, c = 5. The second step is to show any quadratic function can be written in vertex form; the underlying algebraic technique used here is called completing the square. This is a bit more involved. For example, if you are simply handed the quadratic function y = −3x2+6x−1, it not at all obvious why the vertex form is obtained by this equality: −3x2 + 6x − 1 = −3(x − 1)2 + 2. The reason behind this equality is the technique of completing the square. In the end, we will almost always be interested in the vertex form of a quadratic. This is because a great deal of qualitative information about the parabolic graph can simply be “read off” from this form. 7.1.1 First Maneuver: Shifting Suppose we start with the graph in Figure 7.3 and horizontally shift it h units to the right. To be speciﬁc, consider the two cases h = 2 and h = 4. To visualize this, imagine making a wire model of the graph, set in on top of the curve, then slide the wire model h units to the right. What you will obtain are the two “dashed curves” in Figure 7.4. We will call the process just described a horizontal shift. Since the “dashed curves” are no longer the original parabola in Figure 7.3, the corresponding function equations must have changed. y-axis 40 30 20 10 x-axis −6 −4 −2 2 4 6 Figure 7.4: Shift to the right. Using a graphing device, you can check that the corresponding equa- tions for the dashed graphs would be y = (x − 2)2 = x2 − 4x + 4, 6 6 88 CHAPTER 7. QUADRATIC MODELING which is the plot with lowest point (2, 0) and y = (x − 4)2 = x2 − 8x + 16, which is the plot with lowest point (4, 0). In general, if h is positive, the graph of the function y = (x − h)2 is the parabola obtained by shifting the graph of y = x2 by h units to the right. y-axis 40 30 20 10 x-axis −6 −4 −2 2 4 6 Figure 7.5: Shift to the left. Next, if h is negative, shifting h units to the right is the same as shifting \|h\| units left! On the domain −6 6, Figure 7.5 indicates this for the cases h = −2, − 4, using “dashed curves” for the shifted graphs and a solid line for the graph of y = x2. Using a graphing device, we can check that the corresponding equations for the dashed graphs would be ≤ ≤ x y = (x − (−2))2 = (x + 2)2 = x2 + 4x + 4, which is the plot with lowest point (−2, 0) and y = (x − (−4))2 = (x + 4)2 = x2 + 8x + 16, which is the plot with lowest point (−4, 0). In general, if h is negative, the graph of the function y = (x − h)2 gives the parabola obtained by shifting the graph of y = x2 by \|h\| units to the LEFT. The conclusion thus far is this: Begin with the graph of y = x2 in Figure 7.3. Horizontally shifting this graph h units to the right gives a new (standard) parabola whose equation is y = (x − h)2. y-axis 40 30 20 10 x-axis −6 −4 −2 2 4 6 −10 Figure 7.6: Vertical shifts. We can also imagine vertically shifting the graph in Figure 7.3. This amounts to moving the graph k units vertically upward. It turns out that this vertically shifted graph corresponds to the graph of the function y = x2 + k. We can work out a few special cases and use a graphing device to illustrate what all this really means. Figure 7.6 illustrates the graphs of y = x2 + k in the cases when k = 4, 10 and k = −4, −10, leading to vertically shifted graphs. Positive values of k lead to the upper two “dashed curves” and negative values of k lead to the lower two “dashed curves”; the plot of y = x2 is again the solid line. The equations giving these graphs would be y = x2 − 10, y = x2 − 4, y = x2 + 4 and y = x2 + 10, from bottom to top dashed plot. If we combine horizontal and vertical shifting, we end up with the graphs of functions of the form y = (x − h)2 + k. Figure 7.7(a) illustrates 7.1. PARABOLAS AND VERTEX FORM 89 the four cases with corresponding equations y = (x identify which equation goes with each curve. ± 2)2 ± 4; as an exercise, 7.1.2 Second Maneuver: Reﬂection Next, we can reﬂect any of the curves y = p(x) obtained by horizontal or vertical shifting across the x-axis. This procedure will produce a new curve which is the graph of the new function y = −p(x). For example, begin with the four dashed curves in the previous ﬁgure. Here are the reﬂected parabolas and their equations are y = −(x 4: See Figure 7.7(b). 2)2 ± ± 7.1.3 Third Maneuver: Vertical Dilation y-axis 40 30 20 10 x-axis −4 −6 −2 2 4 6 (a) Combined shifts. y-axis If a is a positive number, the graph of y = ax2 is usually called a vertical dilation of the graph of y = x2. There are two cases to distinguish here: x-axis −6 −4 −2 30 20 10 −10 −20 −30 2 4 6 If a > 1, we have a vertically expanded graph. If 0 < a < 1, we have a vertically compressed graph. • • This is illustrated for a = 2 (upper dashed plot) and a = 1/2 (lower dashed plot): See Figure 7.7(c). 7.1.4 Conclusion Starting with y = x2 in Figure 7.3, we can combine together all three of the operations: shifting, reﬂection and dilation. This will lead to the graphs of functions that have the form: y = a(x − h)2 + k, (b) Reﬂections. y-axis 30 25 20 15 10 5 x-axis −6 −4 −2 2 4 6 (c) Vertical dilations. Figure 7.7: Shifts, reﬂections, and dilations. = 0. If you think about it for awhile, for some a, h and k, a it seems pretty easy to believe that any standard parabola arises from the one in Figure 7.3 using our three geometric maneuvers. In other words, what we have shown is that any standard parabola is the graph of a quadratic equation in vertex form. Let’s summarize. Important Fact 7.1.1. A standard parabola is the graph of a function y = f(x) = a(x − h)2 + k, for some constants a, h, and k and a = 0. The vertex of the parabola is (h, k) and the axis of symmetry is the line x = h. If a > 0, then the parabola opens upward; if a < 0, then the parabola opens downward. 6 6 90 CHAPTER 7. QUADRATIC MODELING Example 7.1.2. Describe a sequence of geometric operations leading from the graph of y = x2 to the graph of y = f(x) = −3(x − 1)2 + 2. reﬂect across x-axis horizontal shift by h = 1 −3(x − 1)2 + 2 vertical shift by k = 2 vertical dilate by 3 (a) What do the sym
bols of an equation mean? y-axis 20 15 10 5 x-axis −6 −4 −2 2 4 6 −5 −10 −15 (b) What does the equation look like? Figure 7.8: Interpreting an equation. Solution. To begin with, we can make some initial conclusions about the speciﬁc shifts, reﬂections and dilations involved, based on looking at the vertex form of the equation. In addition, by Fact 7.1.1, we know that the vertex of the graph of y = f(x) is (1, 2), the line x = 1 is a vertical axis of symmetry and the parabola opens downward. We need to be a little careful about the order in which we apply the four operations highlighted. We will illustrate a procedure that works. The full explanation for the success of our procedure involves function compositions and we will return to that at the end of Chapter 8. The order in which we will apply our geometric maneuvers is as follows: horizontal shift vertical dilate reﬂect vertical shift ⇒ ⇒ ⇒ Figure 7.8(b) illustrates the four curves obtained by applying these successive steps, in this order. As a reference, we include the graph of y = x2 as a “dashed curve”: • A horizontal shift by h = 1 yields the graph of y = (x − 1)2; this is the fat parabola opening upward with vertex (1, 0). • • • A dilation by a = 3 yields the graph of y = 3(x − 1)2; this is the skinny parabola opening upward with vertex (1, 0). A reﬂection yields the graph of y = −3(x − 1)2; this is the downward opening parabola with vertex (1, 0). A vertical shift by k = 2 yields the graph of y = −3(x − 1)2 + 2; this is the downward opening parabola with vertex (1, 2). 7.2 Completing the Square By now it is pretty clear we can say a lot about the graph of a quadratic function which is in vertex form. We need a procedure for rewriting a given quadratic function in vertex form. Let’s ﬁrst look at an example. 7.2. COMPLETING THE SQUARE 91 Example 7.2.1. Find the vertex form of the quadratic function y = −3x2 + 6x − 1. Solution. Since our goal is to put the function in vertex form, we can write down what this means, then try to solve for the unknown constants. Our ﬁrst step would be to write −3x2 + 6x − 1 = a(x − h)2 + k, for some constants a, h, k. Now, expand the right hand side of this equation and factor out coefﬁcients of x and x2: −3x2 + 6x − 1 = a(x − h)2 + k −3x2 + 6x − 1 = a(x2 − 2xh + h2) + k −3x2 + 6x − 1 = ax2 − 2xah + ah2 + k (−3)x2 + (6)x + (−1) = (a)x2 + (−2ah)x + (ah2 + k). If this is an equation, then it must be the case that the coefﬁcients of like powers of x match up on the two sides of the equation in Figure 7.9. Now we have three equations and three unknowns (the a, h, k) and we Equal (−3) x2 + (6) x + (−1) = (a) x2 + (−2ah) x + (ah2 + k) z}\|{ \| {z } \| {z } \|{z} z }\| { Equal \| {z } Equal Figure 7.9: Balancing the coefﬁcients. can proceed to solve for these: −3 = a 6 = −2ah −1 = ah2 + k The ﬁrst equation just hands us the value of a = −3. Next, we can plug this value of a into the second equation, giving us 6 = −2ah = −2(−3)h = 6h, 92 CHAPTER 7. QUADRATIC MODELING so h = 1. Finally, plug the now known values of a and h into the third equation: −1 = ah2 + k = −3(12) + k = −3 + k, so k = 2. Our conclusion is then −3x2 + 6x − 1 = −3(x − 1)2 + 2. Notice, this is the quadratic we studied in Example 7.1.2 on page 90. The procedure used in the preceding example will always work to rewrite a quadratic function in vertex form. We refer to this as completing the square. Example 7.2.2. Describe the relationship between the graphs of y = x2 and y = f(x) = −4x2 + 5x + 2. y-axis 20 10 −6 −4 −2 2 4 6 x-axis −10 −20 Figure 7.10: Maneuvering y = x2. Solution. We will go through the algebra to complete the square, then interpret what this all means in terms of graphical maneuvers. We have −4x2 + 5x + 2 = a(x − h)2 + k (−4)x2 + 5x + 2 = ax2 + (−2ah)x + (ah2 + k). This gives us three equations: −4 = a 5 = −2ah 2 = ah2 + k. We conclude that a = −4, h = 5 16 = 3.562. So, this tells us that we can obtain the graph of y = f(x) from that of y = x2 by these steps: 8 = 0.625 and k = 57 Horizontally shifting by h = 0.625 units gives y = (x − 0.625)2. Vertically dilate by the factor a = 4 gives y = 4(x − 0.625)2. Reﬂecting across the x-axis gives y = −4(x − 0.625)2. Vertically shifting by k = 3.562 gives y = f(x) = −4(x − 0.625)2 + 3.562. • • • • Example 7.2.3. A drainage canal has a cross-section in the shape of a parabola. Suppose that the canal is 10 feet deep and 20 feet wide at the top. If the water depth in the ditch is 5 feet, how wide is the surface of the water in the ditch? 7.3. INTERPRETING THE VERTEX 93 Solution. Impose an xy-coordinate system so that the parabolic cross-section of the canal is symmetric about the y-axis and its vertex is the origin. The vertex form of any such parabola is y = f(x) = ax2, for some a > 0; this is because (h, k) = (0, 0) is the vertex and the parabola opens upward! The dimension information given tells us that the points (10, 10) and (−10, 10) are on the graph of f(x). Plugging into the expression for f, we conclude that 10 = 100a, so a = 0.1 and f(x) = (0.1)x2. Finally, if the water is 5 feet deep, we must solve the equation: 5 = (0.1)x2, leading to x = 7.07. Conclude the surface of the water is 14.14 feet wide when the water is 5 feet deep. √50 = ± ± Figure 7.11: 20 feet centerline 10 feet A drainage canal. 7.3 Interpreting the Vertex minimum value f −b 2a maximum value f −b 2a vertex −b 2a vertex −b 2a Figure 7.12: The vertex as the extremum of the quadratic function. If we begin with a quadratic function y = f(x) = ax2+bx+c, we know the graph will be a parabola. Graphically, the vertex will correspond to either If a > 0, the vertex the “highest point” or “lowest point” on the graph. is the lowest point on the graph; if a < 0, the vertex is the highest point on the graph. The maximum or minimum value of the function is the second coordinate of the vertex and the value of the variable x for which this extreme value is achieved is the ﬁrst coordinate of the vertex. As we know, it is easy to read off the vertex coordinates when a quadratic function is written in vertex form. If instead we are given a quadratic function y = ax2 + bx + c, we can use the technique of completing the square and arrive at a formula for the coordinates of the vertex in terms of a, b, and c. We summarize this below and label the two situations (upward or downward opening parabola) in the Figure 7.12. Keep in 94 CHAPTER 7. QUADRATIC MODELING mind, it is always possible to obtain this formula by simply completing the square. Important Fact 7.3.1. In applications involving a quadratic function f(x) = ax2 + bx + c, the vertex has coordinates P = . The second coordinate of the vertex will detect the maximum or minimum value of f(x); this is often a key step in problem solving. −b 2a , f −b 2a Example 7.3.2. Discuss the graph of the quadratic function y = f(x) = −2x2 + 11x − 4. −4 −2 2 4 6 8 −20 −40 −60 −80 Figure 7.13: Sketching y = f(x). Solution. We need to place the equation y = f(x) in vertex form. We can simply compute a = −2, h = −b 4 and k = f( 11 8 , using Fact 7.3.1: 2a = 11 4 ) = 89 f(x) = −2x2 + 11x − 4 = −2 x − 11 4 2 + 89 8 . This means that the graph of f(x) is a parabola opening 4 ; see Figdownward with vertex ure 7.13. and axis x = 11 4 , 89 11 8 7.4 Quadratic Modeling Problems The real importance of quadratic functions stems from the connection with motion problems. Imagine one of the three kicked ball scenarios in Figure 7.1 and impose a coordinate system with the kicker located at the origin. We can study the motion of the ball in two ways: • • Regard time t as the important variable and try to ﬁnd a function y(t) which describes the height of the ball t seconds after the ball is kicked; this would just be the y-coordinate of the ball at time t. The function y(t) is a quadratic function. If we had this function in hand, we could determine when the ball hits the ground by solving the equation 0 = y(t), but we would not be able to determine where the ball hits the ground. A second approach is to forget about the time variable and simply try to ﬁnd a function y = f(x) whose graph models the exact path of the ball. In particular, we could ﬁnd where the ball hits the ground by solving 0 = f(x), but we would not be able to determine when the ball hits the ground. 7.4. QUADRATIC MODELING PROBLEMS 95 Example 7.4.1. Figure 7.14(a) shows a ball is located on the edge of a cliff. The ball is kicked and its height (in feet) above the level ground is given by the function s = y(t) = −16t2 + 48t + 50, where t represents seconds elapsed after kicking the ball. What is the maximum height of the ball and when is this height achieved? When does the ball hit the ground? How high is the cliff? Solution. The function y(t) is a quadratic function with a negative leading coefﬁcient, so its graph in the tscoordinate system will be a downward opening parabola. We use a graphing device to get the picture in Figure 7.14(b). The vertex is the highest point on the graph, which can be found by writing y(t) in standard form using Fact 7.3.1: y(t) = −16t2 + 48t + 50 = −16 t − 3 2 2 + 86. path of kicked ball cliff ground level (a) What it looks like physically. 50 −2 −1 1 2 3 4 5 −50 −100 (b) What it looks like graphically. Figure 7.14: Different views of the ball’s trajectory. The vertex of the graph of y(t) is , so the maximum height of the ball above the level ground is 86 feet, occuring at time t = 3 2. 3 2, 86 The ball hits the ground when its height above the ground is zero; using the quadratic formula: y(t) = −16t2 + 48t + 50 = 0 t = −48 ± (48)2 − 4 (−16) (50) = 3.818 sec or − 0.818 sec p 2 16 · Conclude the ball hits the ground after 3.818 seconds. Finally, the height of the cliff is the height of the ball zero seconds after release; i.e., y(0) = 50 feet is the height of the cliff. Here are two items to consider carefully: 1. The graph of y(t) is NOT the path followed by the ball! Finding the actual path of the ball is not possible unless additional inf
ormation is given. Can you see why? !!! CAUTION !!! 2. The function y(t) is deﬁned for all t; however, in the context of the problem, there is no physical meaning when t < 0. 96 CHAPTER 7. QUADRATIC MODELING The next example illustrates how we must be very careful to link the question being asked with an appropriate function. Example 7.4.2. A hot air balloon takes off from the edge of a mountain lake. Impose a coordinate system as pictured in Figure 7.15 and assume that the path of the balloon follows the graph of y = f(x) = − 2 5x. The land rises at a constant incline from the lake at the rate of 2 vertical feet for each 20 horizontal feet. What is the maximum height of the balloon above lake level? What is the maximum height of the balloon above ground level? Where does the balloon land on the ground? Where is the balloon 50 feet above the ground? 2500 x2 + 4 height above lake (ft) 200 100 lake 500 balloon balloon path Solution. In the coordinate system indicated, the origin is the takeoff point and the graph of y = f(x) is the path of the balloon. Since f(x) is a quadratic function with a negative leading coefﬁcient, its graph will be a parabola which opens downward. The difﬁculty with this problem is that at any instant during the balloon’s ﬂight, the “height of the balloon above the ground” and the “height of the balloon above the lake level” are different! The picture in Figure 7.16 highlights this difference; consequently, two different functions will be needed to study these two different quantities. 1000 (ft) Figure 7.15: Visualizing. 200 100 lake A B 500 1000 height of balloon above lake level A B height of balloon above ground level height of ground above lake level Figure 7.16: The height of the balloon y as a function of x. The function y = f(x) keeps track of the height of the balloon above lake level at a given x location on the horizontal axis. The line ℓ with slope 20 = 1 m = 2 10 passing through the origin models the ground level. This says that the function y = 1 10 x keeps track of the height of the ground above lake level at a given x location on the horizontal axis. We can determine the maximum height of the balloon above lake level by analyzing the parabolic graph of y = f(x). Putting f(x) in vertex form, via Fact 7.3.1, f(x) = − 2 2500 (x − 500)2 + 200. 7.4. QUADRATIC MODELING PROBLEMS 97 The vertex of the graph of y = f(x) is (500, 200). This just tells us that the maximum height of the balloon above lake level is 200 feet. To ﬁnd the landing point, we need to solve the system of equations 2500 x2 + 4 5x y = − 2 y = 1 10x . As usual, plugging the second equation into the ﬁrst and solving for x, we get 1 10 x = − 2 2500 x2 + 4 5 x x2 = 875x x2 − 875x = 0 x(x − 875) = 0 height above lake level (feet) vertex (high point above lake) 200 100 landing point horizontal distance from launch (feet) takeoff point 500 1000 Figure 7.17: Locating the takeoff and landing points. From the algebra, we see there are two solutions: x = 0 or x = 875; these correspond to the takeoff and landing points of the balloon, which are the two places the ﬂight path and ground coincide. (Notice, if we had divided out x from the last equation, we would only get one solution; the tricky point is that we can’t divide by zero!) The balloon lands at the position where x = 875 and to ﬁnd the y coordinate of this landing point we plug x = 875 into our function for the balloon height above lake level: y = f(875) = 87.5 feet. So, the landing point has coordinates (875,87.5). Next, we want to study the height of the balloon above the ground. Let y = g(x) be the function which represents the height of the balloon above 98 CHAPTER 7. QUADRATIC MODELING the ground when the horizontal coordinate is x. We ﬁnd g(x) = height of the balloon above lake level with horizontal coordinate x − elevation of ground above lake level with horizontal coordinate x = f(x) − g(x) = − 2 2500 x2 + 4 5 x (balloon above lake level) − minus 1 10 x (ground above lake level) \|{z} {z } (x − 437.5)2 + 153.12, \| = − 2 2500 \| {z } Notice that g(x) itself is a NEW quadratic function with a negative leading coefﬁcient, so the graph of y = g(x) will be a downward opening parabola. The vertex of this parabola will be (437.5, 153.12), so the highest elevation of the balloon above the ground is 153.12 feet. We can now sketch the graph of g(x) and the horizontal line determined by y = 50 in a common coordinate system, as below. Finding where the balloon is 50 feet above the ground amounts to ﬁnding where these two graphs intersect. We need to now solve the system of equations y = − 2 y = 50 2500 x2 + 7 10x . Plug the second equation into the ﬁrst and apply the quadratic formula to get x = 796.54 or 78.46. This tells us the two possible x coordinates when the balloon is 50 feet above the ground. In terms of the original coordinate system imposed, the two places where the balloon is 50 feet above the ground are (78.46, 57.85) and (796.54, 129.6). 7.4.1 How many points determine a parabola? We all recall from elementary geometry that two distinct points in the plane will uniquely determine a line; in fact, we used this to derive equations for lines in the plane. We could then ask if there is a similar characterization of parabolas. Important Fact 7.4.3. Let P = (x1, y1), Q = (x2, y2) and R = (x3, y3) be three distinct non-collinear points in the plane such that the x-coordinates are all different. Then there exists a unique standard parabola passing through these three points. This parabola is the graph of a quadratic function y = f(x) = ax2 + bx + c and we can ﬁnd these coefﬁcients by simultaneously 7.4. QUADRATIC MODELING PROBLEMS 99 vertex (high point above ground) 100 graph of height above ground function line y = 50 500 1000 places where 50 feet above ground Figure 7.18: Finding heights above the ground. solving the system of three equations and three unknowns obtained by assuming P, Q and R are points on the graph of y = f(x): ax2 1 + bx1 + c = y1 ax2 2 + bx2 + c = y2 ax2 3 + bx3 + c = y3.    .    Example 7.4.4. Assume the value of a particular house in Seattle has increased in value according to a quadratic function y = v(x), where the units of y are in dollars and x represents the number of years the property has been owned. Suppose the house was purchased on January 1, 1970 and valued at $50,000. In 1980, the value of the house on January 1 was $80,000. Finally, on January 1, 1990 the value was $200,000. Find the value function v(x), determine the value on January 1, 1996 and ﬁnd when the house will be valued at $1,000,000. Solution. The goal is to explicitly ﬁnd the value of the function y = v(x). We are going to work in a xy-coordinate system in which the ﬁrst coordinate of any point represents time and the second coordinate represents value. We need to decide what kind of units will be used. The x-variable, which represents time, will denote the number of years the house is owned. For the y-variable, which represents value, we could use dollars. But, instead, we will follow a typical practice in real estate and use the units of K, where K = $1,000. For example, a house valued 100 CHAPTER 7. QUADRATIC MODELING at $235,600 would be worth 235.6 K. These will be the units we use, which essentially saves us from drowning in a sea of zeros! We are given three pieces of information about the value of a particular house. This leads to three points in our coordinate system: P = (0, 50), Q = (10, 80) and R = (20, 200). If we plot these points, they do not lie on a common line, so we know there is a unique quadratic function v(x) = ax2 + bx + c whose graph (which will be a parabola) passes through these three points. In order to ﬁnd the coefﬁcients a, b, and c, we need to solve the system of equations: a02 + b0 + c = 50 a(10)2 + b(10) + c = 80 a(20)2 + b(20) + c = 200   which is equivalent to the system     , = 50 c 100a + 10b + c = 80 400a + 20b + c = 200 .     Plugging c = 50 into the second two equations gives the system   100a + 10b = 30 400a + 20b = 150 Solve the ﬁrst equation for a, obtaining a = 30−10b 100 the second equation to get: (7.1) , then plug this into 400 30 − 10b 100 + 20b = 150 120 − 40b + 20b = 150 3 2 b = − . 2 into the ﬁrst equation of Equation 7.1 to get 100a + Now, plug b = − 3 10 = 30; i.e., a = 9 − 3 2 y = v(x) = 9 20 x2 − 20. We conclude that 3 2 x + 50, keeping in mind the units here are K. To ﬁnd the value of the house on January 1, 1996, we simply note this is after x = 26 years of ownership. Plugging in, we get y = v(26) = 9 20(26)2 − 3 226 + 50 = 315.2; i.e., the value of the house is $315,200. To ﬁnd when the house will be worth $1,000,000, we note that $1,000,000 = 1,000 K and need to solve the equation 1000 = v(x) = 0 = 9 20 9 20 x2 − x2 − 3 2 3 2 x + 50 x − 950. 7.5. WHAT’S NEEDED TO BUILD A QUADRATIC MODEL? 101 By the quadratic formula 20 √1712.25 0.9 = 47.64 or − 44.31. 1.5 ± = 9 20 (−950) Because x represents time, we can ignore the negative solution and so the value of the house will be $1,000,000 after approximately 47.64 years of ownership. 7.5 What’s Needed to Build a Quadratic Model? Back in Fact 4.7.1 on page 44, we highlighted the information required to determine a linear model. We now describe the quadratic model analog. Important Facts 7.5.1. A quadratic model is completely determined by: 1. Three distinct non-collinear points, or 2. The vertex and one other point on the graph. The ﬁrst approach is just Fact 7.4.3. The second approach is based on the vertex form of a quadratic function. The idea is that we know any quadratic function f(x) has the form f(x) = a(x − h)2 + k, where (h, k) is the vertex. If we are given h and k, together with another point (x0, y0) on the graph, then plugging in gives this equation: y0 = a(x0 − h)2 + k. The only unknown in this equation is a, which we can solve for using algebra. A couple of the exercises will depend upon these observations. 7.6 Summary A
quadratic function is one of the form • f(x) = ax2 + bx + c. where a = 0. 6 102 CHAPTER 7. QUADRATIC MODELING • The graph of a quadratic function is a parabola which is symmetric about the vertical line through the highest (or lowest) point on the graph. This highest (or lowest) point is known as the vertex of the graph; its location is given by (h,k) where h = − b 2a and k = f(h). If a > 0, then the vertex is the lowest (or minimum) point on the graph, and the parabola ”opens upward”. If a < 0, then the vertex is the highest (or maximum) point on the graph, and the parabola ”opens downward”. Every quadratic function can be expressed in the form • • f(x) = a(x − h)2 + k where (h,k) is the vertex of the function’s graph. 7.7. EXERCISES 7.7 Exercises Problem 7.1. Write the following quadratic functions in vertex form, ﬁnd the vertex, the axis of symmetry and sketch a rough graph. (a) f(x) = 2x2 − 16x + 41. (b) f(x) = 3x2 − 15x − 77. (c) f(x) = x2 − 3 7 x + 13. (d) f(x) = 2x2. (e) f(x) = 1 100 x2. Problem 7.2. In each case, ﬁnd a quadratic function whose graph passes through the given points: (a) (0,0), (1,1) and (3, − 1). (b) (−1,1), (1, − 2) and (3,4). (c) (2,1), (3,2) and (5,1). (d) (0,1), (1,1) and (1,3). Problem 7.3. (a) Sketch the graph of the function f(x) = x2 − 3x + 4 on the interval −3 5. What is the maximum value of f(x) on that interval? What is the minimum value of f(x) on that interval? ≤ ≤ x (b) Sketch the graph of the function f(x) = x2 − 3x + 4 on the interval 2 7. What is the maximum value of f(x) on that interval? What is the minimum value of f(x) on that interval? ≤ ≤ x (c) Sketch the graph of the function g(x) = −(x + 3)2 + 3 on the interval 0 4. What is the maximum value of g(x) on that interval? What is the minimum value of g(x) on that interval? ≤ ≤ x Problem 7.4. If the graph of the quadratic function f(x) = x2 + dx + 3d has its vertex on the x-axis, what are the possible values of d? What if f(x) = x2 + 3dx − d2 + 1 ? 103 (a) Find the multipart function s(t) giving the stock price after t days. If you buy 1000 shares after 30 days, what is the cost? (b) To maximize proﬁt, when should you sell shares? How much will the proﬁt be on your 1000 shares purchased in (a)? the graph Problem 7.6. Sketch of y = x2 − 2x − 3. Label the coordinates of the x and y intercepts of the graph. In the same coordinate system, sketch the graph of y = \|x2 − 2x − 3\|, give the multipart rule and label the x and y intercepts of the graph. Problem 7.7. A hot air balloon takes off from the edge of a plateau. Impose a coordinate system as pictured below and assume that the path the balloon follows is the graph of the quadratic function y = f(x) = − 4 5 x. The land drops at a constant incline from the plateau at the rate of 1 vertical foot for each 5 horizontal feet. Answer the following questions: 2500 x2 + 4 height above plateau (feet) balloon takeoff horizontal distance from launch (feet) ground incline (a) What is the maximum height of the bal- loon above plateau level? (b) What is the maximum height of the bal- loon above ground level? (c) Where does the balloon land on the ground? (d) Where is the balloon 50 feet above the ground? Problem 7.5. The initial price of buzz.com stock is $10 per share. After 20 days the stock price is $20 per share and after 40 days the price is $25 per share. Assume that while the price of the stock is not zero it can be modeled by a quadratic function. Problem 7.8. (a) Suppose f(x) = 3x2 − 2. Does the point (1,2) lie on the graph of y = f(x)? Why or why not? (b) If b is a constant, where does the line y = 1 + 2b intersect the graph of y = x2 + bx + b? 104 CHAPTER 7. QUADRATIC MODELING (c) If a is a constant, where does the line y = 1 − a2 intersect the graph of y = x2 − 2ax + 1? (d) Where does the graph of y = −2x2 + 3x + 10 intersect the graph of y = x2 + x − 10? Problem 7.9. Sylvia has an apple orchard. One season, her 100 trees yielded 140 apples per tree. She wants to increase her production by adding more trees to the orchard. However, she knows that for every 10 additional trees she plants, she will lose 4 apples per tree (i.e., the yield per tree will decrease by 4 apples). How many trees should she have in the orchard to maximize her production of apples? Problem 7.10. Rosalie is organizing a circus performance to raise money for a charity. She is trying to decide how much to charge for tickets. From past experience, she knows that the number of people who will attend is a linear function of the price per ticket. If she charges 5 dollars, 1200 people will attend. If she charges 7 dollars, 970 people will attend. How much should she charge per ticket to make the most money? Problem 7.11. A Norman window is a rectangle with a semicircle on top. Suppose that the perimeter of a particular Norman window is to be 24 feet. What should its dimensions be in order to maximize the area of the window and, therefore, allow in as much light as possible? Problem 7.12. Jun has 300 meters of fencing to make a rectangular enclosure. She also wants to use some fencing to split the enclosure into two parts with a fence parallel to two of the sides. What dimensions should the enclosure have to have the maximum possible area? Problem 7.13. You have $6000 with which to build a rectangular enclosure with fencing. The fencing material costs $20 per meter. You also want to have two parititions across the width of the enclosure, so that there will be three separated spaces in the enclosure. The material for the partitions costs $15 per meter. What is the maximum area you can achieve for the enclosure? a piece of wire 60 inches long and cuts it into two pieces. Steve takes the ﬁrst piece of wire and bends it into the shape of a perfect circle. He then proceeds to bend the second piece of wire into the shape of a perfect square. Where should Steve cut the wire so that the total area of the circle and square combined is as small as possible? What is this minimal area? What should Steve do if he wants the combined area to be as large as possible? Problem 7.15. Two particles are moving in the xy-plane. The move along straight lines at constant speed. At time t, particle A’s position is given by x = t + 2, y = 1 2 t − 3 and particle B’s position is given by x = 12 − 2t, y = 6 − 1 3 t. (a) Find the equation of the line along which particle A moves. Sketch this line, and label A’s starting point and direction of motion. (b) Find the equation of the line along which particle B moves. Sketch this line on the same axes, and label B’s starting point and direction of motion. (c) Find the time (i.e., the value of t) at which the distance between A and B is minimal. Find the locations of particles A and B at this time, and label them on your graph. Problem 7.16. Sven starts walking due south at 5 feet per second from a point 120 feet north of an intersection. At the same time Rudyard starts walking due east at 4 feet per second from a point 150 feet west of the intersection. (a) Write an expression for the distance between Sven and Rudyard t seconds after they start walking. (b) When are Sven and Rudyard closest? What is the minimum distance between them? Problem 7.14. Steve likes to entertain friends at parties with “wire tricks.” Suppose he takes Problem 7.17. After a vigorous soccer match, Tina and Michael decide to have a glass of 7.7. EXERCISES 105 their favorite refreshment. They each run in a straight line along the indicated paths at a speed of 10 ft/sec. Parametrize the motion of Tina and Michael individually. Find when and where Tina and Michael are closest to one another; also compute this minimum distance. Find values of α that make this equation true (your answer will involve x). Problem 7.19. For each of the following equations, ﬁnd the value(s) of the constant α so that the equation has exactly one solution, and determine the solution for each value. (−50,275) beet juice (200,300) soy milk (400,50) Tina Michael Problem 7.18. Consider the equation: αx2 + 2α2x + 1 = 0. Find the values of x that make this equation true (your answer will involve α). (a) αx2 + x + 1 = 0 (b) x2 + αx + 1 = 0 (c) x2 + x + α = 0 (d) x2 + αx + 4α + 1 = 0 Problem 7.20. (a) Solve for t s = 2(t − 1)2 + 1 (b) Solve for x y = x2 + 2x + 3 106 CHAPTER 7. QUADRATIC MODELING Chapter 8 Composition A new home takes its shape from basic building materials and the skillful use of construction tools. Likewise, we can build new functions from known functions through the application of analogous mathematical tools. There are ﬁve tools we want to develop: composition, reﬂection, shifting, dilation, arithmetic. We will handle composition in this section, then discuss the others in the following two sections. To set the stage, let’s look at a simple botany experiment. Imagine a plant growing under a particular steady light source. Plants continually give off oxygen gas to the environment at some rate; common units would be liters/hour. If we leave this plant unbothered, we measure that the plant puts out 1 liter/hour; so, the oxygen output is a steady constant rate. However, if we apply a ﬂash of high intensity green light at the time t = 1 and measure the oxygen output of the plant, we obtain the plot in Figure 8.1(a). Using what we know from the previous section on quadratic functions, we can check that a reasonable model for the graph is this multipart function f(t) (on the domain 0 10): t ≤ ≤ f(t) =   1 2 3t2 − 8 1 if t 3t + 3 if 1 if .8 0.6 0.4 0.2 1 0.8 0.6 0.4 0.2 oxygen rate 1 hr hours 2 4 6 8 10 (a) Flash at t = 1. oxygen rate 1 hr 2 4 6 hours 8 10 (b) Flash at t = 5. Figure 8.1: Light ﬂashes.  Suppose we want to model the oxygen consumption when a green light pulse occurs at time t = 5 (instead of time t = 1), what is the mathematical model? For starters, it is pretty easy to believe that the graph for this new situation will look like the new graph in Figure 8.1(b). But, can we somehow use the model f(t) in hand (the known functio
n) to build the model we want (the new function)? We will return in Exam- 107 108 CHAPTER 8. COMPOSITION ple 8.2.4 to see the answer is yes; ﬁrst, we need to develop the tool of function composition. 8.1 The Formula for a Composition The basic idea is to start with two functions f and g and produce a new function called their composition. There are two basic steps in this process and we are going to focus on each separately. The ﬁrst step is fairly mechanical, though perhaps somewhat unnatural. It involves combining the formulas for the functions f and g together to get a new formula; we will focus on that step in this subsection. The next step is of varying complexity and involves analyzing how the domains and ranges of f and g affect those of the composition; we defer that to the next subsection once we have the mechanics down. in out in out u x y the “g” function x = g(u) the “f” function y = f(x) in out “composed” function y = f(g(u)) Figure 8.2: Visualizing a composite function y = h(u) = f(g(u)). Here is a very common occurrence: We are handed a function y = f(x), which means given an x value, the rule f(x) produces a new y value. In addition, it may happen that the variable x is itself related to a third variable u by some different function equation x = g(u). Given u, the rule g(u) will produce a value of x; from this x we can use the rule f(x) In other words, we can regard y as a function to produce a y value. depending on the new independent variable u. It is important to know the mechanics of working with this kind of setup. Abstractly, we have just described a situation where we take two functions and build a new 8.1. THE FORMULA FOR A COMPOSITION 109 function which “composes” the original ones together; schematically the situation looks like this: Example 8.1.1. A pebble is tossed into a pond. The radius of the ﬁrst circular ripple is measured to increase at the constant rate of 2.3 ft/sec. What is the area enclosed by the leading ripple after 6 seconds have elapsed? How much time must elapse so that the area enclosed by the leading ripple is 300 square feet? Solution. We know that an object tossed into a pond will generate a series of concentric ripples, which grow steadily larger. We are asked questions that relate the area of the circular region bounded by the leading ripple and time elapsed. leading ripple after t seconds leading ripple after 2 seconds r = r(t) Let r denote the radius of the leading ripple after t seconds; units of feet. The area A of a disc bounded by a leading ripple will be A = A(r) = πr2. This exhibits A as a function in the variable r. However, the radius is changing with respect to time: leading ripple after 1 second Figure 8.3: Concentric ripples. r = r(t) = radius after t seconds = 2.3 feet sec t seconds = 2.3t feet. So, r = r(t) is a function of t. In the expression A = A(r), replace “r” by “r(t),” then A = π(2.3t)2 = 5.29πt2. The new function a(t) = 5.29πt2 gives a precise relationship between area and time. To answer our ﬁrst question, a(6) = 598.3 feet2 is the area of the region bounded by the leading ripple after 6 seconds. On the other hand, if a(t) = 300 ft2;, 300 = 5.29πt2, so t = 4.25. Since t represents time, only the positive solution t = 4.25 seconds makes sense. p We can formalize the key idea used in solving this problem, which is 300/(5.29π) = ± ± familiar from previous courses. Suppose that y = f(x) and that additionally the independent variable x is itself a function of a different independent variable t; i.e., x = g(t). Then we can replace every occurrence of “x” in f(x) by the expression “g(t),” thereby obtaining y as a function in the independent variable t. We usually denote this new function of t: y = f(g(t)). 110 CHAPTER 8. COMPOSITION We refer to f(g(t)) as the composition of f and g or the compositefunction. The process of forming the composition of two functions is a mechanical procedure. If you are handed the actual formulas for y = f(x) and x = g(t), then Procedure 8.1.2 is what you need. Important Procedure 8.1.2. To obtain the formula for f(g(t)), replace every occurrence of “x” in f(x) by the expression “g(t).” Here are some examples of how to do this: Examples 8.1.3. Use the composition procedure in each of these cases. (i) If y = f(x) = 2 and x = g(t) = 2t, then f(g(t)) = f(2t) = 2. (ii) If y = 3x − 7 and x = g(t) = 4, then f(g(t)) = f(4) = 3 4 − 7 = 5. · (iii) If y = f(x) = x2 + 1 and x = g(t) = 2t − 1, then f(g(t)) = f(2t − 1) = (2t − 1)2 + 1 = 4t2 − 4t + 2. (iv) If y = f(x) = 2 + 1 + (x − 3)2 and x = g(t) = 2t2 − 1, then p f(g(t)) = f(2t2 − 1) = 2 + 1 + (2t2 − 1 − 3)2 = 2 + 4t4 − 16t2 + 17. p (v) If y = f(x) = x2 and x = g(t) = t + p , then ♥ ) f(g(t)) = f(t + ♥ )2 = (t + ♥ = t2 + 2t ♥ + 2. ♥ It is natural to ask: What good is this whole business about compositions? One way to think of it is that we can use composite functions to break complicated functions into simpler parts. For example, y = h(x) = x2 + 1 p can be written as the composition f(g(x)), where y = f(z) = √z and z = g(x) = x2+1. Each of the functions f and g is “simpler” than the original h, which can help when studying h. Examples 8.1.4. Here we use composite functions to “simplify” a given function. 8.1. THE FORMULA FOR A COMPOSITION 111 (i) The function y = where y = f(z) = 1 1 (x−3)2+2 can be written as a composition y = f(g(x)), z2+2 and z = g(x) = x − 3. (ii) The upper semicircle of radius 2 centered at (1,2) is the graph of the 4 − (x − 1)2. This function can be written as a comfunction y = 2 + position y = f(g(x)), where y = f(z) = 2 + √4 − z2 and z = g(x) = x − 1. p 8.1.1 Some notational confusion In our discussion above, we have used different letters to represent the domain variables of two functions we are composing. Typically, we have been writing: If y = f(x) and x = g(t), then y = f(g(t)) is the composition. This illustrates that the three variables t, x, and y can all be of different types. For example, t might represent time, x could be speed and y could be distance. If we are given two functions that involve the same independent variable, like f(x) = x2 and g(x) = 2x + 1, then we can still form a new function f(g(x)) by following the same prescription as in Procedure 8.1.2: Important Procedure 8.1.5. To obtain the formula for f(g(x)), replace every occurrence of “x” in f(x) by the expression “g(x).” For our example, this gives us: f(g(x)) = f(2x + 1) = (2x + 1)2. Here are three other examples: If f(x) = √x, g(x) = 2x2 + 1, then f(g(x)) = √2x2 + 1. If f(x) = 1 x, g(x) = 2x + 1, then f(g(x)) = 1 2x+1 . If f(x) = x2, g(x) = △ − x, then f(g(x)) = 2 − 2x △ △ + x2. • • • 112 y-axis 10 8 6 4 2 x-axis −3 −2 −1 1 2 3 Figure 8.4: Sketching composite functions. CHAPTER 8. COMPOSITION Example 8.1.6. Let f(x) = x2, g(x) = x + 1 and h(x) = x − 1. Find the formulas for f(g(x)), g(f(x)), f(h(x)) and h(f(x)). Discuss the relationship between the graphs of these four functions. Solution. If we apply Procedure 8.1.5, we obtain the composition formulas. The four graphs are given on the domain −3 3, together with the graph of f(x) = x2. x ≤ ≤ f(g(x)) = f(x + 1) = (x + 1)2 g(f(x)) = g(x2) = x2 + 1 f(h(x)) = f(x − 1) = (x − 1)2 h(f(x)) = h(x2) = x2 − 1. We can identify each graph by looking at its vertex: f(x) has vertex (0,0) f(g(x)) has vertex (-1,0) g(f(x)) has vertex (0,1) f(h(x)) has vertex (1,0) h(f(x)) has vertex (0,-1) • • • • • Horizontal or vertical shifting of the graph of f(x) = x2 gives the other four graphs: See Figure 8.4. 8.2 Domain, Range, etc. for a Composition A function is a “package” consisting of a rule, a domain of allowed input values, and a range of output values. When we start to compose functions, we sometimes need to worry about how the domains and ranges of the composing functions affect the composed function. First off, when you form the composition f(g(x)) of f(x) and g(x), the range values for g(x) must lie within the domain values for f(x). This may require that you modify the range values of g(x) by changing its domain. The domain values for f(g(x)) will be the domain values for g(x). 8.2. DOMAIN, RANGE, ETC. FOR A COMPOSITION 113 in out in out domain g the “g” function the “f” function range of f(g(x)) range of g domain of f Figure 8.5: What is the domain and range of a composite function? In practical terms, here is how one deals with the domain issues for a composition. This is a reﬁnement of Procedure 8.1.5 on page 111. Important Procedure 8.2.1. To obtain the formula for f(g(x)), replace every occurrence of “x” in f(x) by the expression “g(x).” In addition, if there is a condition on the domain of f that involves x, then replace every occurrence of “x” in that condition by the expression “g(x).” The next example illustrates how to use this principle. Example 8.2.2. Start with the function y = f(x) = x2 on the domain −1 1. Find the rule and domain of y = f(g(x)), where g(x) = x − 1. x ≤ ≤ Solution. We can apply the ﬁrst statement in Procedure 8.2.1 to ﬁnd the rule for y = f(g(x)): y = f(g(x)) = f(x − 1) = (x − 1)2 = x2 − 2x + 1. To ﬁnd the domain of y = f(g(x)), we apply the second statement in Procedure 8.2.1; this will require that we solve an inequality equation: −1 −1 0 ≤ ≤ ≤ g(x) x − 1 x 1 1 2 ≤ ≤ ≤ The conclusion is that y = f(g(x)) = x2 −2x+1 on the domain 0 x ≤ ≤ 2. Example 8.2.3. Let y = f(z) = √z, z = g(x) = x + 1. What is the largest possible domain so that the composition f(g(x)) makes sense? 114 CHAPTER 8. COMPOSITION y range y = f(z) z domain z z = g(x) = x + 1 −1 desired range x required domain (a) y = √z. (b) z = x = 1. Figure 8.6: Finding the largest domain for f(g(x)). Solution. The largest possible domain for y = f(z) will consist of all nonnegative real numbers; this is also the range of the function f(z): See Figure 8.6(a). To ﬁnd the largest domain for the composition, we try to ﬁnd a domain of x-values so that the range of z = g(x) is the domain of y = f(z). So, in this case, we want the range of g(x) to be all non
-negative real numbers, z. We graph z = g(x) in the xz-plane, mark the desired range denoted 0 0 z on the vertical z-axis, then determine which x-values would lead to points on the graph with second coordinates in this zone. We ﬁnd that the domain of all x-values greater or equal to −1 (denoted −1 x) leads to the desired range. In summary, the composition y = f(g(x)) = √x + 1 is deﬁned on the domain of x-values −1 x. ≤ ≤ ≤ ≤ Let’s return to the botany experiment that opened this section and see how composition of functions can be applied to the situation. Recall that plants continually give off oxygen gas to the environment at some rate; common units would be liters/hour. Example 8.2.4. A plant is growing under a particular steady light source. If we apply a ﬂash of high intensity green light at the time t = 1 and measure the oxygen output of the plant, we obtain the plot below and the mathematical model f(t). f(t) =   1 2 3t2 − 8 1 if t 3t + 3 if 1 if 3 1 t t ≤ ≤ ≤ 3 ≤  Now, suppose instead we apply the ﬂash of high intensity green light at the time t = 5. Verify that the mathematical model for this experiment is given by f(g(t)), where g(t) = t − 4. 8.2. DOMAIN, RANGE, ETC. FOR A COMPOSITION 115 Solution. Our expectation is that the plot for this new experiment will have the “parabolic dip” shifted over to occur starting at time t = 5 instead of at time t = 1. In other words, we expect the graph in Figure 8.7(b). Our job is to verify that this graph is obtained from the function f(g(t)), where g(t) = t − 4. This is a new terrain for us, since we need to look at a composition involving a multipart function. Here is how to proceed: When we are calculating a composition involving a multipart function, we need to look at each of the parts separately, so there will be three cases to consider: First part: f(t) = 1 when t 1. To get the formula for f(g(t)), we now appeal to Procedure 8.2.1 and just replace every occurrence of t in f(t) by g(t). That gives us this NEW domain condition and function equation: ≤ 1 0.8 0.6 0.4 0.2 1 0.8 0.6 0.4 0.2 oxygen rate 1 hr hours 2 4 6 8 10 (a) Flash at t = 1. oxygen rate 1 hr hours 2 4 6 8 10 (b) Flash at t = 5. f(g(t)) = f(t − 4) = 1 when t − 4 1 ≤ = 1 when t Second part: f(t) = 2 5. ≤ 3t2 − 8 3. We now appeal to Procedure 8.2.1 and just replace every occurrence of t in this function by g(t). That gives us this NEW domain condition and function equation: 3t + 3 when 1 ≤ ≤ t Figure 8.7: Applying light at time t. f(g(t)) = f(t − 4) = (t − 4)2 − 8 3 (t − 4) + 3 when 1 t − 4 3 ≤ ≤ 2 3 Third part: f(t) = 1 when 3 t2 − 8t + = when 5 t ≤ ≤ 7. t. We now appeal to Procedure 8.2.1 and just replace every occurrence of t in this function by g(t). That gives us this NEW domain condition and function equation: ≤ 2 3 73 3 f(g(t)) = f(t − 4) = 1 when 3 = 1 when 7 t. ≤ t − 4 ≤ The multipart rule for this composition can now be written down and using a graphing device you can verify its graph is the model for our experiment. 1 3t2 − 8t + 73 2 1 3 if t if 5 if (g(t)) =    116 CHAPTER 8. COMPOSITION 8.3 Exercises Problem 8.1. For this problem, f(t) = t − 1, g(t) = −t − 1 and h(t) = \|t\|. (a) Compute the multipart rules for h(f(t)) and h(g(t)) and sketch their graphs. (b) Compute the multipart rules for f(h(t)) and g(h(t)) and sketch their graphs. (c) Compute the multipart rule for h(h(t)−1) and sketch the graph. Problem 8.2. Write each of the following functions as a composition of two simpler functions: (There is more than one correct answer.) (a) y = (x − 11)5. (b) y = 3√1 + x2. (c) y = 2(x − 3)5 − 5(x − 3)2 + 1 2 (x − 3) + 11. (d) y = 1 x2+3 . (e) y = √x + 1. p (f) y = 2 − 5 − (3x − 1)2. p Problem 8.3. (a) Let f(x) be a linear function, f(x) = ax + b for constants a and b. Show that f(f(x)) is a linear function. (b) Find a function g(x) such that g(g(x)) = 6x − 8. Problem 8.4. Let f(x) = 1 2 x + 3. (a) Sketch the graphs of f(x),f(f(x)),f(f(f(x))) on the interval −2 x 10. ≤ ≤ (b) Your graphs should all intersect at the point (6,6). The value x = 6 is called a ﬁxed point of the function f(x) since f(6) = 6; that is, 6 is ﬁxed - it doesn’t move when f is applied to it. Give an explanation for why 6 is a ﬁxed point for any function f(f(f(...f(x)...))). (c) Linear functions (with the exception of f(x) = x) can have at most one ﬁxed point. Quadratic functions can have at most two. Find the ﬁxed points of the function g(x) = x2 − 2. Problem 8.5. A car leaves Seattle heading east. The speed of the car in mph after m minutes is given by the function C(m) = 70m2 10 + m2 . (a) Find a function m = f(s) that converts seconds s into minutes m. Write out the formula for the new function C(f(s)); what does this function calculate? (b) Find a function m = g(h) that converts hours h into minutes m. Write out the formula for the new function C(g(h)); what does this function calculate? (c) Find a function z = v(s) that converts mph s into ft/sec z. Write out the formula for the new function v(C(m); what does this function calculate? Problem 8.6. Compute the compositions f(g(x)), f(f(x)) and g(f(x)) in each case: (a) f(x) = x2, g(x) = x + 3. (b) f(x) = 1/x, g(x) = √x. (c) f(x) = 9x + 2, g(x) = 1 (d) f(x) = 6x2 + 5, g(x) = x − 4. (e) f(x) = 4x3 − 3, g(x) = 3√2x + 6 (f) f(x) = 2x + 1, g(x) = x3. 9 (x − 2). (g) f(x) = 3, g(x) = 4x2 + 2x + 1. (h) f(x) = −4, g(x) = 0. Problem 8.7. Let y = f(z) = √4 − z2 and z = g(x) = 2x + 3. Compute the composition y = f(g(x)). Find the largest possible domain of x-values so that the composition y = f(g(x)) is deﬁned. Problem 8.8. Suppose you have a function y = f(x) such that the domain of f(x) is 1 6 and the range of f(x) is −3 5. y x ≤ ≤ (a) What is the domain of f(2(x − 3)) ? ≤ ≤ (b) What is the range of f(2(x − 3)) ? (c) What is the domain of 2f(x) − 3 ? (d) Give a quadratic function whose ﬁxed (d) What is the range of 2f(x) − 3 ? points are x = −2 and x = 3. (e) Can you ﬁnd constants B and C so that the domain of f(B(x − C)) is 8 x ≤ ≤ 9? 8.3. EXERCISES 117 (f) Can you ﬁnd constants A and D so that this simpliﬁed expression: the range of Af(x) + D is 0 y ≤ ≤ 1? Problem 8.9. For each of the given functions y = f(x), simplify the following expression so that h is no longer a factor in the denominator, then calculate the result of setting h = 0 in f(x + h) − f(x) h . (a) f(x) = 1 x−1 . (b) f(x) = (2x + 1)2. (c) f(x) = √25 − x2. 118 CHAPTER 8. COMPOSITION Chapter 9 Inverse Functions The experimental sciences are loaded with examples of functions relating time and some measured quantity. In this case, time represents our “input” and the quantity we are measuring is the “output.” For example, maybe you have just mixed together some chemical reactants in a vessel. As time goes by, you measure the fraction of reactants remaining, tabulate your results, then sketch a graph as indicated in Figure 9.1. 1 fraction Viewing the input value as “time” and the output value as “fraction of product,” we could ﬁnd a function y = f(t) modeling this data. Using this function, you can easily compute the fraction of reactants remaining at any time in the future. However, it is probably just as interesting to know how to predict the time when a given fraction of reactants exists. In other words, we would like a new function that allows us to input a “fraction of reactants” and get out the “time” when this occurs. This “reverses” the input/output roles in the original function. Is there a systematic way to ﬁnd the new function if we know y = f(t)? The answer is yes and depends upon the general theory of inverse functions. 0 time Figure 9.1: Fraction of reactants as a function of time. 9.1 Concept of an Inverse Function Suppose you are asked to solve the following three equations for x. How do you proceed? (x + 2) = 64 (x + 2)2 = 64 (x + 2)3 = 64. In the ﬁrst equation, you add “−2” to each side, then obtain x = 62. In the third equation, you take the cube root of both sides of the equation, giving you x + 2 = 4, then subtract 2 getting x = 2. In second equation, you take a square root of both sides, BUT you need to remember both the 119 120 CHAPTER 9. INVERSE FUNCTIONS positive and negative results when doing this. So, you are reduced down to x + 2 = 8 or that x = −10 or 6. Why is it that in two of these cases you obtain a single solution, while in the remaining case there are two different answers? We need to sort this out, since the underlying ideas will surface when we address the inverse circular functions in Chapter 20. ± In x Out f(x) The function Figure 9.2: A function as a process. Let’s recall the conceptual idea of a function: A function is a process which takes a number x and outputs a new number f(x). So far, we’ve only worked with this process from “left to right;” i.e., given x, we simply put it into a symbolic rule and out pops a new number f(x). This is all pretty mechanical and straightforward. 9.1.1 An Example Let’s schematically interpret what happens for the speciﬁc concrete example y = f(x) = 3x − 1, when x = −1, − 1 2, 1, 2: See Figure 9.3. 2 , 0, 1 −1 − 1 2 in in in out 3x − 1 −4 3x − 1 − 5 2 out out 0 3x − 1 −1 in in in 1 2 1 2 out 3x − 1 out 3x − 1 out 3x − 1 1 2 2 5 Figure 9.3: Function process y = 3x − 1. We could try to understand the function process in this example in “reverse order,” going “right to left;” namely, you might ask what x value can be run through the process so you end up with the number 11? This is somewhat like the “Jeopardy” game show: You know what the answer is, you want to ﬁnd the question. For our example, if we start out with some given y values, then we can deﬁne a “reverse process” x = 1 3(y + 1), which returns the x value required so that f(x) = y: See Figure 9.4. 9.1. CONCEPT OF AN INVERSE FUNCTION 121 eplacements −1 − 1 2 0 (y+1) 3 (y+1) 3 (y+1) 3 −4 − 5 2 1 2 1 −1 2 (y+1) 3 (y+1) 3 (y+1) 3 1 2 2 5 Figure 9.4: The reverse process x = 1 3 (y + 1). 9.1.2 A Second Example If we begin with a linear function y = f(x) = mx + b, where m = 0, then we c
an always ﬁnd a “reverse process” for the function. To ﬁnd it, you must solve the equation y = f(x) for x in terms of y: y = mx + b y − b = mx 1 m (y − b) = x So, if m = 3 and b = −1, we just have the ﬁrst example above. For another example, suppose y = −0.8x + 2; then m = −0.8 and b = 2. In this case, the reverse process is −1.25(y − 2) = x. If we are given the value y = 11, we simply compute that x = −11.25; i.e., f(−11.25) = 11. 9.1.3 A Third Example The previous examples hide a subtle point that can arise when we try to understand the “reverse process” for a given function. Suppose we begin with the function y = f(x) = (x − 1)2 + 1. Figure 9.5 is a schematic of how the function works when we plug in x = −1, − 1 2, 1, 2; what is being illustrated is a “forward process”, in that each input generates a unique output. 2, 0, 1 For this example, if we start out with some given y values, then we can try to deﬁne a “reverse process” x = ??? which returns an x value required so that f(x) = y. Unfortunately, there is no way to obtain a single formula for this reverse process; Figure 9.6(a) shows what happens if you are given y = 3 and you try to solve for x. 6 122 CHAPTER 9. INVERSE FUNCTIONS in in in −1 − 1 2 0 (x − 1)2 + 1 (x − 1)2 + 1 (x − 1)2 + 1 out out out 5 13 4 2 in in in 1 2 1 2 (x − 1)2 + 1 (x − 1)2 + 1 (x − 1)2 + 1 out out out 5 4 1 2 Figure 9.5: Function process y = f(x) = (x − 1)2 + 1. The conclusion is that the “reverse process” has two outputs. This violates the rules required for a function, so this is NOT a function. The solution is to create two new “reverse processes.” 1 + √2 reverse process x = − +√y − 1 3 1 − √2 (a) Reverse process but not a function. reverse process 1 + √2 x = +√y − 1 reverse process 1 − √2 x = −√y − 1 3 3 (b) Two new reverse processes that are functions. Figure 9.6: What to do if a reverse process is not a function. Each of these “reverse processes” has a unique output; in other words, each of these “reverse processes” deﬁnes a function. So, given y = 3, there are TWO possible x values, √2, so that f(1 + √2) = 3 and f(1 − √2) = 3. namely x = 1 In other words, the reverse process is not given by a single equation; there are TWO POSSIBLE reverse processes. ± 9.2 Graphical Idea of an Inverse We have seen that ﬁnding inverses is related to solving equations. However so far, the discussion has been symbolic; we have pushed around a few equations and in the end generated some confusion. Let’s use the tools of Chapter 6 to visualize what is going on here. Suppose we are given the graph of a function f(x) as in Figure 9.7(a). What input x values result in an output value of 3? This involves ﬁnding all x such that f(x) = 3. Graphically, this means we are trying to ﬁnd points on the graph of f(x) so that their y-coordinates are 3. The easiest way to to do this is to draw the line y = 3 and ﬁnd where it intersects the graph. In Figure 9.7(b) we can see the points of intersection are (−5, 3), (−1, 3), and (9, 3). That means that x = −5, −1, 9 produce the output value 3; i.e., f(−5) = f(−1) = f(9) = 3. 9.2. GRAPHICAL IDEA OF AN INVERSE 123 y = f(x) 5 5 y = f(x) y = 3 −5 5 10 −5 5 10 −5 −5 (a) Given the function y = f(x). (b) What values of x give f(x) = 3? Figure 9.7: Using the horizontal line y = 3 to ﬁnd values on the x-axis. This leads to our ﬁrst important fact about the “reverse process” for a function: Important Fact 9.2.1. Given a number c, the x values such that f(x) = c can be found by ﬁnding the x-coordinates of the intersection points of the graphs of y = f(x) and y = c. Example 9.2.2. Graph y = f(x) = x2 and discuss the meaning of Fact 9.2.1 when c = 3, 1, 6. Solution. We graph y = x2 and the lines y = 1, y = 3 and y = 6. Let’s use c = 6 as an example. We need to simultaneously solve the equations y = x2 and y = 6. Putting these together, we get x2 = 6 or x = 2.449; i.e., 1.732) = 3. 2.449) = 6. If c = 3, we get x = f( Finally, if c = 1, we get x = √6 1.732; i.e., f( ± 1; i.e., f( ≈ ± ± 1) = 1. ± ± ± ± y = x2 y = 6 y = 3 y = 1 The pictures so far indicate another very important piece of information. For any number c, we can tell exactly “how many” input x values lead to the same output value c, just by counting the number of times the graphs of y = f(x) and y = c intersect. Figure 9.8: Graph of y = f(x) = x2. Important Fact 9.2.3. For any function f(x) and any number c, the number of x values so that f(x) = c is the number of times the graphs of y = c and y = f(x) intersect. 124 CHAPTER 9. INVERSE FUNCTIONS Examples 9.2.4. (i) If f(x) is a linear function f(x) = mx + b, m = 0, then the graph of f(x) intersects a given horizontal line y = c EXACTLY once; i.e., the equation c = f(x) always has a unique solution. (ii) If f(x) = d is a constant function and c = d, then every input x value = d, in the domain leads to the output value c. On the other hand, if c then no input x value will lead to the output value c. For example, if f(x) = 1 and c = 1, then every real number can be input to produce an output of 1; if c = 2, then no input value of x will lead to an output of 2. y-axis f(x) = mx + b y-axis y = c x-axis the only input which leads to an output of c f(x) = 1 x-axis any of these inputs leads to an output of 1 linear functions constant functions Figure 9.9: Does a horizontal line y = c intersect a curve once or more than once? 9.2.1 One-to-one Functions For a speciﬁed domain, one-to-onefunctions are functions with the property: Given any number c, there is at most one input x value in the domain so that f(x) = c. Among our examples thus far, linear functions (degree 1 polynomials) are always one-to-one. However, f(x) = x2 is not one-to-one; we’ve already seen that it can have two values for some of its inverses. By Fact 9.2.3, we can quickly come up with what’s called the horizontal line test. Important Fact 9.2.5 (Horizontal Line Test). On a given domain of x-values, if the graph of some function f(x) has the property that every horizontal line crosses the graph at most only once, then the function is one-to-one on this domain. 6 6 9.3. INVERSE FUNCTIONS Example 9.2.6. By the horizontal line test, it is easy to see that f(x) = x3 is one-to-one on the domain of all real numbers. Although it isn’t common, it’s quite nice when a function is one-to-one because we don’t need to worry as much about the number of input x values producing the same output y value. In effect, this is saying that we can deﬁne a “reverse process” for the function y = f(x) which will also be a function; this is the key theme of the next section. 125 y-axis x-axis horizontal lines Figure 9.10: A one-to-one function f(x) = x3. 9.3 Inverse Functions Let’s now come face to face with the problem of ﬁnding the “reverse process” for a given function y = f(x). It is important to keep in mind that the domain and range of the function will both play an important role in this whole development. For example, Figure 9.11 shows the function f(x) = x2 with three different domains speciﬁed and the corresponding range values. range range domain domain range domain Figure 9.11: Possible domains for a given range. These comments set the stage for a third important fact. Since the domain and range of the function and its inverse rule are going to be intimately related, we want to use notation that will highlight this fact. We have been using the letters x and y for the domain (input) and range (output) variables of f(x) and the “reverse process” is going to reverse these roles. It then seems natural to simply write y (instead of c) for the input values of the “reverse process” and x for its output values. 126 CHAPTER 9. INVERSE FUNCTIONS Important Fact 9.3.1. Suppose a function f(x) is one-to-one on a domain of x values. Then deﬁne a NEW FUNCTION by the rule f−1(y) = the x value so that f(x) = y. The domain of y values for the function f−1(y) is equal to the range of the function f(x). The rule deﬁned here is the “reverse process” for the given function. It is referred to as the inverse function and we read f−1(y) as “...eff inverse of y...”. Both the “domain” of f(x) and the “rule” f(x) have equal inﬂuence on whether the inverse rule is a function. Keep in mind, you do NOT get an inverse function automatically from functions that are not one-to-one! !!! CAUTION !!! 9.3.1 Schematic Idea of an Inverse Function Suppose that f(x) is one-to-one, so that f−1(y) is a function. As a result, we can model f−1(y) as a black box. What does it do? If we put in y in the input side, we should get out the x such that f(x) = y. in y f−1(y) out x such that f(x) = y Figure 9.12: A new function x = f−1(y). Now, let’s try to unravel something very special that is happening on a symbolic level. What would happen if we plugged f(a) into the inverse function for some number a? Then the inverse rule f−1(f(a)) tells us that we want to ﬁnd some x so that f(x) = f(a). But, we already know x = a works and since f−1(y) is a function (hence gives us unique answers), the output of f−1(f(a)) is just a. Symbolically, this means we have Fact 9.3.2. Important Fact 9.3.2. For every a value in the domain of f(x), we have f−1(f(a)) = a. (9.1) This is better shown in the black box picture of Figure 9.13. in a out in f(x) f(a) f−1(x) out a Figure 9.13: Visualizing f−1(f(a)) = a. A good way to get an idea of what an inverse function is doing is to remember that f−1(y) reverses the process of f(x). We can think of f−1(y) as a “black box” running f(x) backwards. 9.4. TRYING TO INVERT A NON ONE-TO-ONE FUNCTION 127 9.3.2 Graphing Inverse Functions How can we get the graph of an inverse function? The idea is to manipulate the graph of our original one-to-one function in some prescribed way, ending up with the graph of f−1(y). This isn’t as hard as it sounds, but some confusion with the variables enters into play. Remember that a typical point on the graph of a function y = f(x) looks like (x, f(x)). Now let’s take a look at the inverse function x = f−1(y). Given a num
ber y in the domain of f−1(y), y = f(x) for some x in the domain of f(x); i.e., we are using the fact that the domain of f−1 equals the range of f. The function f−1(y) takes the number f(x) and sends it to x, by Fact 9.3.2. So when f(x) is the input value, x becomes the output value. Conclude a point on the graph of f−1(y) looks like (f(x), x). It’s similar to the graph of y = f(x), only the x and y coordinates have reversed! What does that do to the graph? Essentially, you reorient the picture so that the positive x-axis and positive y-axis are interchanged. Figure 9.14 shows the process for the function y = f(x) = x3 and its inverse function x = f−1(y) = 3√y. We symbols on the graph to help keep track of what is happenplace some ing. ∗ +y−axis * +x−axis * * rotate 900 clockwise * * * +x−axis +y−axis ﬂip across horiz axis +x−axis * * * +y−axis Figure 9.14: Graphically ﬁnding x = f−1(y) = 3√y. 9.4 Trying to Invert a Non one-to-one Func- tion Suppose we blindly try to show that √y is the inverse function for y = x2, without worrying about all of this one-to-one stuff. We’ll start out with the number −7. If f−1(y) = √y, then we know that f−1(f(−7)) = f−1(49) = 7. On the other hand, the formula in Fact 9.3.2 tells us that we must have f−1(f(−7)) = −7, so we have just shown 7 = −7! So clearly f−1(y) = √y. Even if we try f−1(y) = −√y, we produce a contradiction. It seems that if you didn’t have 6 128 CHAPTER 9. INVERSE FUNCTIONS to worry about negative numbers, things would be all right. Then you could say that f−1(y) = √y. Let’s try to see what this means graphically. Let’s set f(x) = x2, but only for non-negative x-values. That means that we want to erase the graph to the left of the y-axis (so remember - no negative x-values allowed). The graph would then look like Figure 9.15. +y y = x2 +x inverse function √y +x +y domain non-negative x domain non-negative y Figure 9.15: Restricting the domain: No negative x-values. This is a now a one-to-one function! And now, one can see that its inverse function is √y. Similarly, we could have taken f(x) = x2 but only for the non-positive x-values. In that case, f−1(y) = −√y. In effect, we have split the graph of y = x2 into two parts, each of which is the graph of a one-to-one function; Figure 9.16. +y y = x2 +x domain non-negative y +x +y domain non-positive x inverse function −√y Figure 9.16: Restricting the domain: No positive x-values. It is precisely this splitting into two cases that leads us to multiple solutions of an equation like x2 = 5. We obtain x = √5 and x = −√5; one 9.5. SUMMARY 129 solution comes from the side of the graph to the left of the y-axis, and the other from the right of the y-axis. This is because we have separate inverse functions for the left and right side of the graph of y = x2. 9.5 Summary • • • • • • Two functions f and g are inverses if f(g(x)) = x and g(f(x)) = x for all x in the domain of f and the domain of g. A function f is one-to-one if every equation f(x) = k If there is a value of k such that the has at most one solution. equation f(x) = k has more than one solution, then f is not one-toone. A function is one-to-one if every horizontal line intersects the function’s graph at most once. A function has an inverse if the function is one-to-one, and every one-to-one function has an inverse. The domain of a function is the range of its inverse, and the range of a function is the domain of its inverse. The graph of a function and its inverse are mirror images of each other across the line y = x. 130 CHAPTER 9. INVERSE FUNCTIONS 9.6 Exercises Problem 9.1. Let f(x) = 2 domain for which the formula makes sense. 3x−4 on the largest A (a) Find the domain and range of f(x), then sketch the graph. (b) Find the domain, range and rule for the inverse function f−1, then sketch its graph. C Problem 9.2. Find the inverse function of each of the following functions. Specify the domains of the inverse functions. B D (a) f(x) = (b) h(xc) g(x) = 4√3 − x − 7 (d) j(x) = √x + √x − 1 (e) k(x) = 16 − x2, 0 p 4 x ≤ ≤ Problem 9.3. For this problem, y = f(x) = 2x2 − 3x − 1 on the domain of all real numbers. (a) Sketch the function graph and ﬁnd the coordinates of the vertex P = (a,b). (b) Explain why y = f(x) does not have an inverse function on the domain of all real numbers. (c) Restrict y = f(x) to the domain {a ≤ x} and ﬁnd the formula for the inverse function f−1(y). What are the domain and range of the inverse function? (d) Restrict y = f(x) to the domain {x ≤ a} and ﬁnd the formula for the inverse function f−1(y). What are the domain and range of the inverse function? Problem 9.5. Show that, for every value of a, the function f(x) = a + 1 x − a is its own inverse. Problem 9.6. Clovis is standing at the edge of a cliff, which slopes 4 feet downward from him for every 1 horizontal foot. He launches a small model rocket from where he is standing. With the origin of the coordinate system located where he is standing, and the x-axis extending horizontally, the path of the rocket is described by the formula y = −2x2 + 120x. (a) Give a function h = f(x) relating the height h of the rocket above the sloping ground to its x-coordinate. (b) Find the maximum height of the rocket above the sloping ground. What is its x-coordinate when it is at its maximum height? (c) Clovis measures its height h of the rocket above the sloping ground while it is going up. Give a function x = g(h) relating the x-coordinate of the rocket to h. (d) Does this function still work when the rocket is going down? Explain. Problem 9.4. Which of the following graphs are one-to-one? If they are not one-to-one, section the graph up into parts that are one-toone. Problem 9.7. For each of the following functions: (1) sketch the function, (2) ﬁnd the inverse function, and (3) sketch the inverse function. In each case, indicate the correct domains and ranges. (4) Finally, make sure you 9.6. EXERCISES 131 test each of the functions you propose as an inverse with the following compositions: (b) After how many hours will the surface of the water have width of 6 feet? and f(f−1(x)) ?= x f−1(f(x)) ?= x. (c) Give a function t = f−1(w) relating the time to the width of the surface of the water. Make sure to specify the domain and compute the range too. (a) f(x) = 3x − 2 (b) f(x) = 1 2 x + 5 (c) f(x) = −x2 + 3, x 0 ≥ (d) f(x) = x2 + 2x + 5, x (e) f(x) = √4 − x2, 0 −1 2 ≤ x ≤ ≤ Problem 9.8. A trough has a semicircular cross section with a radius of 5 feet. Water starts ﬂowing into the trough in such a way that the depth of the water is increasing at a rate of 2 inches per hour. water 5 ft cross-section of trough (a) Give a function w = f(t) relating the width w of the surface of the water to the time t, in hours. Make sure to specify the domain and compute the range too. Problem 9.9. A biochemical experiment involves combining together two protein extracts. Suppose a function φ(t) monitors the amount (nanograms) of extract A remaining at time t (nanoseconds). Assume you know these facts: 1. The function φ is invertible; i.e., it has an inverse function. 2. φ(0) = 6, φ(1) = 5, φ(2) = 3, φ(3) = 1, φ(4) = 0.5, φ(10) = 0. (a) At what time do you know there will be 3 nanograms of extract A remaining? (b) What is φ−1(0.5) and what does it tell you? (c) (True or False) There is exactly one time when the amount of extract A remaining is 4 nanograms. (d) Calculate φ(φ−1(1)) = (e) Calculate φ−1(φ(6)) = (f) What is the domain and range of φ? 132 CHAPTER 9. INVERSE FUNCTIONS Chapter 10 Exponential Functions If we start with a single yeast cell under favorable growth conditions, then it will divide in one hour to form two identical “daughter cells”. In turn, after another hour, each of these daughter cells will divide to produce two identical cells; we now have four identical “granddaughter cells” of the original parent cell. Under ideal conditions, we can imagine how this “doubling effect” will continue: cells Figure 10.1: Observing cell growth. The question is this: Can we ﬁnd a function of t that will predict (i.e. model) the number of yeast cells after t hours? If we tabulate some data (as at right), the conclusion is that the formula N(t) = 2t predicts the number of yeast cells after t hours. Now, let’s make a very slight change. Suppose that instead of starting with a single cell, we begin with a population of 3 If we assume 106 cells; a more realistic situation. × 133 TIME t=0 hours t=1 hours t=2 hours t=3 hours Total hours 0 1 2 3 4 5 6 Number of yeast cells 1=20 2=21 4=22 8=23 16=24 32=25 64=26 Table 10.1: Cell growth data. 134 CHAPTER 10. EXPONENTIAL FUNCTIONS that the population of cells will double every hour, then reasoning as above will lead us to conclude that the formula N(t) = (3 106)2t × gives the population of cells after t hours. Now, as long as t represents a non-negative integer, we know how to calculate N(t). For example, if t = 6, then N(t) = (3 = (3 × × 106)26 106)(2 106)64 106. = (3 × = 192 × ) The key point is that computing N(t) only involves simple arithmetic. But what happens if we want to know the population of cells after 6.37 hours? That would require that we work with the formula N(t) = (3 106)26.37 × and the rules of arithmetic do not sufﬁce to calculate N(t). We are stuck, since we must understand the meaning of an expression like 26.37. In order to proceed, we will need to review the algebra required to make sense of raising a number (such as 2) to a non-integer power. We need to understand the precise meaning of expressions like: 26.37, 2√5, 2−π, etc. 10.1 Functions of Exponential Type this is a variable x y = b this is a fixed positive integer b y = x this is a fixed number this is a variable Exponential Picture Monomial Picture Figure 10.2: Viewing the difference between exponential and monomial functions. On a symbolic level, the class of functions we are trying to motivate is easily introduced. We have already studied the monomials y = xb, where x was our input variable and b wa
s a ﬁxed positive integer exponent. What happens if we turn this around, interchanging x and b, deﬁning a new rule: y = f(x) = bx. (10.1) 10.1. FUNCTIONS OF EXPONENTIAL TYPE 135 We refer to x as the power and b the base. An expression of this sort is called a function of exponential type. Actually, if your algebra is a bit rusty, it is easy to initially confuse functions of exponential type and monomials (see Figure 10.2). 10.1.1 Reviewing the Rules of Exponents To be completely honest, making sense of the expression y = bx for all numbers x requires the tools of Calculus, but it is possible to establish a reasonable comfort level 0 by handling the case when x is a rational number. If b and n is a positive integer (i.e. n = 1, 2, 3, 4, . . . ), then we can try to solve the equation ≥ tn = b. (10.2) A solution t to this equation is called an nth root of b. This leads to complications, depending on whether n is even or odd. In the odd case, for any real number b, notice that the graph of y = b will always cross the graph of y = tn exactly once, leading to one solution of (10.2). On the other hand, if n is even and b < 0, then the graph of y = tn will miss the graph of y = b, implying there are no solutions to the equation in (10.2). (There will be complex solutions to equations such as t2 = −1, involv√−1, but we ing the imaginary complex numbers are only working with real numbers in this course.) Also, again in the case when n is even, it can happen that there are two solutions to (10.2). We do not want to constantly worry about this even/odd distinction, so we will henceforth assume b > 0. To eliminate possible ambiguity, we will single out a particular nth-root; we deﬁne the symbols1.5 -1 -0.5 0 0.5 1 1.5 -1 -2 -3 3 2 1 0 n=2 n=4 n=6 n=3 n=5 n=7 -1.5 -1 -0.5 0 0.5 1 1.5 -1 -2 -3 Figure 10.3: Even and odd monomials. n√b = b 1 n = the largest real nth root of b. (10.3) ± Thus, whereas 1 are both 4th-roots of 1, we have deﬁned 4√1 = 1. In order to manipulate y = bx for rational x, we need to recall some basic facts from algebra. Important Facts 10.1.1 (Working with rational exponents). For all positive integers p and q, and any real number base b > 0, we have p q = b q√b p = q√bp. For any rational numbers r and s, and for all positive bases a and b: 136 CHAPTER 10. EXPONENTIAL FUNCTIONS y-axis n y=t n even = solution y-axis n y=t n odd y=b t-axis y=b* no solution or two solutions y=b t-axis exactly one solution Figure 10.4: How many solutions to tn = b?. 1. Product of power rule: brbs = br+s 2. Power of power rule: (br)s = brs 3. Power of product rule: (ab)r = arbr 4. Zero exponent rule: b0 = 1 5. Negative power rule: b−r = 1 br These rules have two important consequences, one theoretical and the other more practical. On the ﬁrst count, recall that any rational number r can be written in the form r = p q , where p and q are integers. Consequently, using these rules, we see that the expression y = bx deﬁnes a function of x, whenever x is a rational number. On the more practical side of things, using the rules we can calculate and manipulate certain expressions. For example, 27 2 3 = 3√27 2 = 32 = 9; 8− 5 3 = 3√8 −5 = 2−5 = 1 25 = 1 32 . The sticky point which remains is knowing that f(x) = bx actually deﬁnes a function for all real values of x. This is not easy to verify and we are simply going to accept it as a fact. The difﬁculty is that we need the fundamentally new concept of a limit, which is the starting point of a Calculus course. Once we know the expression does deﬁne a function, we can also verify that the rules of Fact 10.1.1 carry through for all real 10.2. THE FUNCTIONS Y = A0BX 137 exponent powers. Your calculator should have a “y to the x key”, allowing you to calculate expressions such as π√2 involving non-rational powers. Here are the key modeling functions we will work with in this Chapter. Deﬁnition 10.1.2. A function of exponential type has the form A(x) = A0bx, for some b > 0, b = 1, and A0 6 = 0. We will refer to the formula in Deﬁnition 10.1.2 as the standard exponential form. Just as with standard forms for quadratic functions, we sometimes need to do a little calculation to put an equation in standard form. The constant A0 is called the initial value of the exponential function; this is because if x represents time, then A(0) = A0b0 = A0 is the value of the function at time x = 0; i.e. the initial value of the function. Example 10.1.3. Write the equations y = 83x and y = 7 exponential form. 2x−1 in standard 1 2 Solution. In both cases, we just use the rules of exponents to maneuver the given equation into standard form: and y = 83x = (83)x = 512x 2x−1 2x 1 2 1 2 x 2 −1 2 ! x 1 4 = 14 10.2 The Functions y = A0bx We know f(x) = 2x deﬁnes a function of x, so we can study basic qualitative features of its graph. The data assembled in the solution of the “Doubling Effect” beginning this Chapter, plus the rules of exponents, produce a number of points on the graph. This graph exhibits four key qualitative features that deserve mention: 6 138 CHAPTER 10. EXPONENTIAL FUNCTIONS x 2x ... ... -2 1/4 -1 1/2 0 1 2 3 ... 1 2 4 8 ... Point on the graph of y = 2x ... (-2, 1/4) (-1, 1/2) (0, 1) (1, 2) (2, 4) (3, 8) ... x y = 2 (3,8) (−1,1/2) (−2,1/4) (2,4) (1,2) (0,1) −1 1 (a) Data points from y = 2x. (b) Graph of y = 2x. Figure 10.5: Visualizing y = 2x. • • • • The graph is always above the horizontal axis; i.e. values are always positive. the function The graph has y-intercept 1 and is increasing. The graph becomes closer and closer to the horizontal axis as we the x-axis is a horizontal asymptote for the leftmove left; i.e. hand portion of the graph. The graph becomes higher and higher above the horizontal axis as we move to the right; i.e., the graph is unbounded as we move to the right. The special case of y = 2x is representative of the function y = bx, but there are a few subtle points that need to be addressed. First, recall we are always assuming that our base b > 0. We will consider three separate cases: b = 1, b > 1, and 0 < b < 1. 10.2.1 The case b = 1 In the case b = 1, we are working with the function y = 1x = 1; this is not too exciting, since the graph is just a horizontal line. We will ignore this case. 10.2.2 The case b > 1 If b > 1, the graph of the function y = bx is qualitatively similar to the situation for b = 2, which we just considered. The only difference is the exact amount of “concavity” in the graph, but the four features highlighted above are still valid. Figure 10.6(a) indicates how these graphs 10.2. THE FUNCTIONS Y = A0BX 139 compare for three different values of b. Functions of this type exhibit what is typically referred to as exponential growth; this codiﬁes the fact that the function values grow rapidly as we move to the right along the x-axis. y-axis y-axis all graphs pass through (0,1) x-axis all graphs pass through (0,1) x-axis (a) Graph of y = bx, b > 1. (b) Graph of y = bx, 0 < b < 1. Figure 10.6: Visualizing cases for b. 10.2.3 The case 0 < b < 1 We can understand the remaining case 0 < b < 1, by using the remarks above and our work in Chapter 13. First, with this condition on b, notice that 1 b > 1, so the graph of y = is of the type in Figure 10.6(a). Now, using the rules of exponents: 1 b x y = −x = 1 b 1 b x −1 ! = bx. −x 1 b By the reﬂection principle, the graph of y = is obtained by reﬂecting the graph of y = ( 1 b)x about the y-axis. Putting these remarks to- gether, if 0 < b < 1, we conclude that the graph of y = bx will look like Figure 10.6(b). Notice, the graphs in Figure 10.6(b) share qualitative features, mirroring the features outlined previously, with the “asymptote” and “unbounded” portions of the graph interchanged. Graphs of this sort are often said to exhibit exponential decay, in the sense that the function values rapidly approach zero as we move to the right along the x-axis. Important Facts 10.2.1 (Features of Exponential Type Functions). Let b be a positive real number, not equal to 1. The graph of y = bx has these four properties: 1. The graph is always above the horizontal axis. 2. The graph has y-intercept 1. 3. If b > 1 (resp. 0 < b < 1), the graph becomes closer and closer to the horizontal axis as we move to the left (resp. move to the right); this says the x-axis is a horizontal asymptote for the left-hand portion of the graph (resp. right-hand portion of the graph). 140 CHAPTER 10. EXPONENTIAL FUNCTIONS 4. If b > 1 (resp. 0 < b < 1), the graph becomes higher and higher above the horizontal axis as we move to the right (resp. move to the left); this says that the graph is unbounded as we move to the right (resp. move to the left). If A0 > 0, the graph of the function y = A0bx is a vertically expanded or compressed version of the graph of y = bx. If A0 < 0, we additionally reﬂect about the x-axis. 10.3 Piano Frequency Range A sound wave will cause your eardrum to move back and forth. In the case of a so-called pure tone, this motion is modeled by a function of the form d(t) = A sin(2πft), where f is called the frequency, in units of “periods/unit time”, called “Hertz” and abbreviated “Hz”. The coefﬁcient A is related to the actual displacement of the eardrum, which is, in turn, related to the loudness of the sound. A person can typically perceive sounds ranging from 20 Hz to 20,000 Hz. A# C# D# F# G# A# C# D# F# G#A# C#D# F# G# A# C# D# F# G#A# C# D# F# G# A# C# D# F# G# A# C# D# F# G#A# A B DC 220 Hz middle C Figure 10.7: A piano keyboard. A piano keyboard layout is shown in Figure 10.7. The white keys are labelled A, B, C, D, E, F, and G, with the sequence running from left to right and repeating for the length of the keyboard. The black keys ﬁt into this sequence as “sharps”, so that the black key between A and B is “A sharp”, denoted A#. Thus, starting at any A key, the 12 keys to the right are A, A#, B, C, C#, D, D#, E, F, F#, G, and G#. The sequence then repeats. Notice that between some adjacent pairs of white keys there is no b
lack key. A piano keyboard is commonly tuned according to a rule requiring that each key (white and black) has a frequency 21/12 times the frequency of the key to its immediate left. This makes the ratio of adjacent keys always the same (21/12), and it means that keys 12 keys apart have a ratio of frequencies exactly equal to 2 (since (21/12)12 = 2). Two such keys are 10.3. PIANO FREQUENCY RANGE 141 said to be an octave apart. Assuming that the key A below middle C has a frequency of 220 Hz, we can determine the frequency of every key on the keyboard. For instance, the A# to the right of this key has frequency 233.08188Hz. The B to the right of this 220 key has frequency 233.08188 1.059463094... 246.94165Hz. 21/12 = 220 ≈ 21/12 × × × ≈ 142 CHAPTER 10. EXPONENTIAL FUNCTIONS 10.4 Exercises Problem 10.1. Let’s brush up on the required calculator skills. Use a calculator to approximate: (a) 3π (b) 42+√5 (c) ππ (d) 5−√3 2 (e) 3π (f) √11π−7 Problem 10.2. Put each equation in standard exponential form: (a) y = 3(2−x) (b) y = 4−x/2 (c) y = ππx (d) y = 1 1 3 3+ x 2 (e) y = 5 0.3452x−7 (f) y = 4(0.0003467)−0.4x+2 (d) Anja, a third member of your lab working with the same yeast cells, took these 106 cells after two measurements: 7.246 106 cells after 6 hours. 4 hours; 16.504 Should you be worried by Anja’s results? If Anja’s measurements are correct, does your model over estimate or under estimate the number of yeast cells at time t? × × Problem 10.4. middle C. (a) Find the frequency of (b) Find the frequency of A above middle C. (c) What is the frequency of the lowest note on the keyboard? Is there a way to solve this without simply computing the frequency of every key below A220? (d) The Bosendorfer piano is famous, due in part, to the fact it includes additional keys at the left hand end of the keyboard, extending to the C below the bottom A on a standard keyboard. What is the lowest frequency produced by a Bosendorfer? Problem 10.3. A colony of yeast cells is estimated to contain 106 cells at time t = 0. After collecting experimental data in the lab, you decide that the total population of cells at time t hours is given by the function y = 106e0.495105t. Problem 10.5. You have a chess board as pictured, with squares numbered 1 through 64. You also have a huge change jar with an unlimited number of dimes. On the ﬁrst square you place one dime. On the second square you stack 2 dimes. Then you continue, always doubling the number from the previous square. (a) How many cells are present after one hour? (a) How many dimes will you have stacked on the 10th square? (b) (True or False) The population of yeast cells will double every 1.4 hours. (b) How many dimes will you have stacked on the nth square? (c) Cherie, another member of your lab, looks at your notebook and says : ...that formula is wrong, my calculations predict the formula for the number of yeast cells is given by the function (c) How many dimes will you have stacked on the 64th square? (d) Assuming a dime is 1 mm thick, how high will this last pile be? y = 106(2.042727)0.693147t . Should you be worried by Cherie’s remark? (e) The distance from the earth to the sun is approximately 150 million km. Relate the height of the last pile of dimes to this distance. 10.4. EXERCISES 143 63 64 calculates the fraction of hemoglobin saturated with oxygen at a given pressure p. (a) The graphs of M(p) and H(p) are given 100; which p below on the domain 0 is which? ≤ ≤ 1 2 3 10 9 8 Problem 10.6. Myoglobin and hemoglobin are oxygen carrying molecules in the human body. Hemoglobin is found inside red blood cells, which ﬂow from the lungs to the muscles through the bloodstream. Myoglobin is found in muscle cells. The function p 1 + p Y = M(p) = calculates the fraction of myoglobin saturated with oxygen at a given pressure p torrs. For example, at a pressure of 1 torr, M(1) = 0.5, which means half of the myoglobin (i.e. 50%) is oxygen saturated. (Note: More precisely, you need to use something called the “partial pressure”, but the distinction is not important for this problem.) Likewise, the function Y = H(p) = p2.8 262.8 + p2.8 fraction 1 0.8 0.6 0.4 0.2 20 40 60 80 p 100 (b) If the pressure in the lungs is 100 torrs, what is the level of oxygen saturation of the hemoglobin in the lungs? (c) The pressure in an active muscle is 20 torrs. What is the level of oxygen saturation of myoglobin in an active muscle? What is the level of hemoglobin in an active muscle? (d) Deﬁne the efﬁciency of oxygen transport at a given pressure p to be M(p) − H(p). What is the oxygen transport efﬁciency at 20 torrs? At 40 torrs? At 60 torrs? Sketch the graph of M(p) − H(p); are there conditions under which transport efﬁciency is maximized (explain)? 144 CHAPTER 10. EXPONENTIAL FUNCTIONS Chapter 11 Exponential Modeling Example 11.0.1. A computer industry spokesperson has predicted that the number of subscribers to geton.com, an internet provider, will grow exponentially for the ﬁrst 5 years. Assume this person is correct. If geton.com has 100,000 subscribers after 6 months and 750,000 subscribers after 12 months, how many subscribers will there be after 5 years? Solution. The solution to this problem offers a template for many exponential modeling applications. Since, we are assuming that the number of subscribers N(x), where x represents years, is a function of exponential type, N(x) = N bx, ◦ for some N values of N(x): ◦ and b > 1. We are given two pieces of information about the N(0.5) = 100,000; i.e., N0b0.5 = 100,000, and N(1) = 750,000; i.e., N0b = 750,000. We can use these two equations to solve for the two unknowns N as follows: If we divide the second equation by the ﬁrst, we get ◦ and b = 7.5 b1 b0.5 b0.5 = b1/2 = √b = 7.5 ∴ b = 56.25. Plugging this value of b into either equation (say the ﬁrst one), we can = 100,000 (56.25)0.5 = 13,333. We conclude that the number of solve for N geton.com subscribers will be predicted by : N ◦ ◦ N(x) = 13,333(56.25)x. In ﬁve years, we obtain N(5) = 7,508,300,000,000 subscribers, which exceeds the population of the Earth (which is between 5 and 6 billion)! 145 146 Q CHAPTER 11. EXPONENTIAL MODELING There are two important conclusions we can draw from this problem. First, the given information provides us with two points on the graph of the function N(x): P 0.2 0.4 0.6 0.8 1 P = (0.5, 100,000) Q = (1, 750,000). 800000 600000 400000 200000 Figure 11.1: Finding the equation for N(x) = N0bx. More importantly, this example illustrates a very important principal we can use when modeling with functions of exponential type. Important Fact 11.0.2. A function of exponential type can be determined if we are given two data points on its graph. !!! CAUTION !!! When you use the above strategy to ﬁnd the base b of the exponential model, make sure to write down a lengthy decimal approximation. As a rule of thumb, go for twice as many signiﬁcant digits as you are otherwise using in the problem. 11.1 The Method of Compound Interest You walk into a Bank with P0 dollars (usually called principal), wishing to invest the money in a savings account. You expect to be rewarded by the Bank and paid interest, so how do you compute the total value of the account after t years? The future value of the account is really a function of the number of years t elapsed, so we can write this as a function P(t). Our goal is to see that P(t) is a function of exponential type. In order to compute the future value of the account, the Bank provides any savings account investor with two important pieces of information: r = annual (decimal) interest rate n = the number of compounding periods per year The number n tells us how many times each year the Bank will compute the total value P(t) of the account. For example, if n = 1, the calculation is done at one-year intervals; if n = 12, the calculation is done each month, etc. The bank will compute the value of your account after a typical compounding period by using the periodic rate of return r n. For example, if the interest rate percentage is 12% and the compounding period is monthly (i.e., n = 12), then the annual (decimal) interest rate is 0.12 and the periodic rate is 0.12 12 = 0.01. !!! CAUTION !!! The number r always represents the decimal interest rate, which is a decimal between 0 and 1. If you are given the interest rate percentage (which is a positive number between 0 and 100), you need to convert to a decimal by dividing by 100. 11.1. THE METHOD OF COMPOUND INTEREST 147 11.1.1 Two Examples Let’s consider an example: P0 = $1,000 invested at the annual interest percentage of 8% compounded yearly, so n = 1 and r = 0.08. To compute the value P(1) after one year, we will have P(1) = P0 + (periodic rate)P0 = P0 + rP0 = P0(1 + r) = $1,000(1 + 0.08) = $1,080. To compute the value after two years, we need to apply the periodic rate to the value of the account after one year: P(2) = P(1) + (periodic rate)P(1) = P0(1 + r) + rP0(1 + r) = P0(1 + r)2 = $1,000(1 + 0.08)2 = $1,166.40. Notice, the amount the Bank has paid after two years is $166.40, which is slightly bigger than twice the $80 paid after one year. To compute the value after three years, we need to apply the periodic rate to the value of the account after two years: P(3) = P(2) + (periodic rate)P(2) = P0(1 + r)2 + rP0(1 + r)2 = P0(1 + r)3 = $1,000(1 + 0.08)3 = $1,259.71. Again, notice the amount the Bank is paying after three years is $259.71, which is slightly larger than three times the $80 paid after one year. Continuing on in this way, to ﬁnd the value after t years, we arrive at the formula P(t) = P0(1 + r)t = $1,000(1.08)t. In particular, after 5 and 10 years, the value of the account (to the nearest dollar) will be $1,469 and $2,159, respectively. As a second example, suppose we begin with the same $1,000 and the same annual interest percentage 8%, but now compound monthly, so n = 12 and r = 0.08. The value of the account after one compounding period is P(1/12), since a month is one-t
welfth of a year. Arguing as before, paying special attention that the periodic rate is now r 12 , we have n = 0.08 P(1/12) = P0 + (periodic rate)P0 = P0 1 + .08 12 = $1,000(1 + 0.006667) = $1,006.67. 148 CHAPTER 11. EXPONENTIAL MODELING After two compounding periods, the value is P(2/12), P(2/12) = P(1/12) + (periodic rate)P(1/12) = P0 1 + = P0 1 + .08 12 .08 12 0.08 12 P0 1 + .08 12 + 2 = $1,000(1 + 0.006667)2 = $1,013.38. Continuing on in this way, after k compounding periods have elapsed, the value will be P , which is computed as k 12 P(k/12) = P0 1 + .08 12 k . It is possible to rewrite this formula to give us the value after t years, noting that t years will lead to 12t compounding periods; i.e., set k = 12t in the previous formula: P(t) = P0 1 + .08 12 12t For example, after 1, 5 and 10 years, the value of the account, to the nearest dollar, would be $1,083, $1,490, and $2,220. 11.1.2 Discrete Compounding The two examples above highlight a general formula for computing the future value of an account. Important Fact 11.1.1 (Discrete compounding). Suppose an account is opened with P0 principal. If the decimal interest rate is r and the number of compounding periods per year is n, then the value P(t) of the account after t years will be P(t) = P0 1 + nt . r n Notice, the future value P(t) is a function of exponential type; the base , which will be greater than one. Since P0 > 0, the is the number graph will be qualitatively similar to the ones pictured in Figure 10.6(a). 1 + r n Example 11.1.2. At birth, your Uncle Hans secretly purchased a $5,000 U.S. Savings Bond for $2,500. The conditions of the bond state that the U.S. Government will pay a minimum annual interest rate of r = 8.75%, compounded quarterly. Your Uncle has given you the bond as a gift, subject to the condition that you cash the bond at age 35 and buy a red Porsche. 11.2. THE NUMBER E AND THE EXPONENTIAL FUNCTION 149 On your way to the Dealer, you receive a call from your tax accountant informing you of a 28% tax on the capital gain you realize through cashing in the bond; the capital gain is the selling price of the bond minus the purchase price. Before stepping onto the showroom ﬂoor, compute how much cash will you have on hand, after the U.S. Government shares in your proﬁts. Solution. The value of your bond after 35 years is computed by the formula in Fact 11.1.1, using P0 = $2,500, r = 0.0875, n = 4, and t = 35. Plugging this all in, we ﬁnd that the selling price of the bond is P(35) = $2,500 1 + 0.0875 4 4(35) = $51,716.42. The capital gain will be $51,716.42 - $2,500 = $49,216.42 and the tax due is $(49,216.42)(0.28) = $13,780.60. You are left with $51,716.42 $13,780.60 = $37,935.82. Better make that a used Porsche! 11.2 The Number e and the Exponential Function What happens to the future value of an investment of P0 dollars as the number of compounding periods is increased? For example, return to our earlier example: P0 = $1000 and an annual interest percentage of 8%. After 1 year, the table below indicates the value of the investment for various compounding periods: yearly, quarterly, monthly, weekly, daily, and hourly. n Compounding Period Value after 1 year (to nearest dollar) 1 yearly $1,000(1 + 0.08)1 = $1,080.00 4 quarterly $1,000 1 + 0.08 4 4 = $1,082.43 12 monthly $1,000 52 weekly $1,000 365 daily $1,000 1 + 0.08 12 12 1 + 0.08 52 52 1 + 0.08 365 365 = $1,083.00 = $1,083.22 = $1,083.28 8,760 hourly $1,000 1 + 0.08 8,760 8,760 =$1,083.29 150 CHAPTER 11. EXPONENTIAL MODELING We could continue on, considering “minute” and “second” compounding and what we will ﬁnd is that the value will be at most $1,083.29. This illustrates a general principal: Important Fact 11.2.1. Initially increasing the number of compounding periods makes a signiﬁcant difference in the future value; however, eventually there appears to be a limiting value. Let’s see if we can understand mathematically why this is happening. The ﬁrst step is to recall the discrete compounding formula: P(t) = P0 1 + nt . r n If our desire is to study the effect of increasing the number of compounding periods, this means we want to see what happens to this formula as n gets BIG. To analyze this, it is best to rewrite the expression using a substitution trick: Set z = n z . Plugging in, we have r , so that n = rz and r n = 1 P(t) = P0 1 + nt rzt r n 1 z 1 z 1 + 1 + = P0 = P0 z rt . (11.1) z 1 + 1 z So, since r is a ﬁxed number and z = n r , letting n get BIG is the same as letting z become BIG in (11.1). This all means we need to answer as z becomes this new question: What happens to the expression large? On the one hand, the power in the expression is getting large; at the same time, the base is getting close to 1. This makes it very tricky to make quick predictions about the outcome. It is best to ﬁrst tabulate 1 + 1 some numerical data for the values of y = g(z) = and look at a z 100: See Figure 11.2. z plot of this function graph on the domain 0.01 ≤ ≤ z approaches the 1 + 1 You can see from this plot, the graph of y = z “dashed” horizontal asymptote, as z becomes BIG. We will let the letter “e” represent the spot where this horizontal line crosses the vertical axis and 2.7182818. This number is only an approximation, since e is known to e be an irrational number. What sets this irrational number apart from the ones you are familiar with (e.g. √2, π, etc.) is that deﬁning the number e requires a “limiting” process. This will be studied a lot more in your Calculus course. The new number e is a positive number greater than 1, so we can study the function: ≈ z y = ex. (11.2) Since e > 1, the graph will share the properties in Figure 10.6(a). This function is usually referred to as THE exponential function. Scientiﬁc calculators will have a key of the form “ exp(x) ” or “ ex ”. 11.2. THE NUMBER E AND THE EXPONENTIAL FUNCTION 151 z z 1 + 1 z 2 1 2.25 2 2.37037 3 2.4414 4 2.65329 20 100 2.70481 1000 2.71692 109 2.71828 (a) Data points for 1 + 1 z . z 2.5 2 1.5 1 0.5 20 40 60 80 100 (b) The graph of 1 + 1 z z . Figure 11.2: What happens when z get very large? 11.2.1 Calculator drill Plugging in x = 1, you can compute an approximation to e on your calculator; you should get e = 2.7183 to four decimal places. Make sure you can compute expressions like e3, eπ, and √e = e1/2; to four decimal places, you should get 20.0855, 23.1407, and 1.6487. 11.2.2 Back to the original problem... We can now return to our future value formula (11.1) and conclude that as the number of compounding periods increases, the future value is approaching a limiting value: P(t) = P0 1 + 1 z z rt = P0ert. The right hand limiting formula Q(t) = P0ert computes the future value using what is usually referred to as continuous compounding. From the investors viewpoint, this is the best possible scheme for computing future value. ⇒ Important Fact 11.2.2 (Continuous compounding). The future value of P0 dollars principal invested at an annual decimal interest rate of r under continuous compounding after t years is Q(t) = P0ert; this value is alnt, for any discrete compounding 1 + r ways greater than the value of P0 n scheme. In fact, P0ert is the limiting value. 152 CHAPTER 11. EXPONENTIAL MODELING 11.3 Exercises Problem 11.1. In 1968, the U.S. minimum wage was $1.60 per hour. In 1976, the minimum wage was $2.30 per hour. Assume the minimum wage grows according to an exponential model w(t), where t represents the time in years after 1960. (a) Find a formula for w(t). (b) What does the model predict for the min- imum wage in 1960? (c) If the minimum wage was $5.15 in 1996, is this above, below or equal to what the model predicts. Problem 11.2. The Pinedale, Wyoming, is experiencing a population boom. In 1990, the population was 860 and ﬁve years later it was 1210. town of (a) Find a linear model l(x) and an exponential model p(x) for the population of Pinedale in the year 1990+x. (b) What do these models estimate the population of Pinedale to be in the year 2000? Problem 11.3. In 1989, research scientists published a model for predicting the cumulative number of AIDS cases reported in the United States: a(t) = 155 t − 1980 10 3 , (thousands) where t is the year. This paper was considered a “relief”, since there was a fear the correct model would be of exponential type. Pick two data points predicted by the research model a(t) to construct a new exponential model b(t) for the number of cumulative AIDS cases. Discuss how the two models differ and explain the use of the word “relief”. Problem 11.4. Deﬁne two new functions: and y = cosh(x) = y = sinh(x) = ex + e−x 2 ex − e−x 2 . These are called the basic hyperbolic trigonometric functions. (a) Sketch rough graphs of these two func- tions. (b) The graph of the equation x2 − y2 = 1 is shown below; this is called the unit hyperbola. For any value a, show that the point (x,y) = (cosh(a), sinh(a)) is on the unit hyperbola. (Hint: Verify that [cosh(x)]2 − [sinh(x)]2 = 1, for all x.) y (−1,0) (1,0) x (c) A hanging cable is modeled by a portion of the graph of the function y = a cosh( x − h a ) + C, for appropriate constants a, h and C. The constant h depends on how the coordinate system is imposed. A cable for a suspension bridge hangs from two 100 ft. high towers located 400 ft. apart. Impose a coordinate system so that the picture is symmetric about the y-axis and the roadway coincides with the x-axis. The hanging cable constant is a = 500 and h = 0. Find the minimum distance from the cable to the road. towers cable d 400 ft 100 ft roadway Chapter 12 Logarithmic Functions If we invest P0 = $1,000 at an annual rate of r = 8% compounded continuously, how long will it take for the account to have a value of $5000? The formula P(t) = 1,000e0.08t gives the value after t years, so we need to solve the equation: 5,000 = 1,000e0.08t 5 = e0.08t. Unfortunately, algebraic manipulation will not lead to a further simpliﬁcation of this equation; we are stuck! The required technique involv
es the theory of inverse functions. Assuming we can ﬁnd the inverse function of f(t) = et, we can apply f−1(t) to each side of the equation and solve for t: f−1(5) = f−1(e0.08t) = 0.08t (12.5)f−1(5) = t The goal in this section is to describe the function f−1, which is usually denoted by the symbol f−1(t) = ln(t) and called the natural logarithm function. On your calculator, you will ﬁnd a button dedicated to this function and we can now compute ln(5) = 1.60944. Conclude that the solution is t = 20.12 years. 12.1 The Inverse Function of y = ex If we sketch a picture of the exponential function on the domain of all real numbers and keep in mind the properties in Fact 10.2.1, then every horizontal line above the x-axis intersects the graph of y = ex exactly once: See Figure 12.1(a). The range of the exponential function will consist of all possible y-coordinates of points on the graph. Using the graphical techniques of Chapter 6, we can see that the range of will be all POSITIVE real numbers: See Figure 12.1(b). 153 154 CHAPTER 12. LOGARITHMIC FUNCTIONS PSfrag x y = e range = all positive numbers y = ex graphs of y=c, c>1 cross exponential graph exactly once. these horizontal lines miss graph of exponential function. −1 1 −1 1 domain = all real numbers (a) Horizontal line test for y = ex. (b) The domain and range for y = ex. Figure 12.1: Properties needed to ﬁnd the inverse of f(x) = ex. By the horizontal line test, this means the exponential function is one- to-one and the inverse rule f−1(c) will deﬁne a function f−1(c) =    reflecting line y = x x y = e y=ln (x) −1 1 Figure 12.2: Visualizing the y = ln(x). soluthe unique tion of the equation c = ex ! , if c > 0 (undeﬁned), if c 0. ≤ (12.1) This inverse function is called the natural logarithm function, denoted ln(c). We can sketch the graph of the the natural logarithm as follows: First, by Fact 9.2.1, the domain of the function ln(y) = x is just the range of the exponential function, which we noted is all positive numbers. Likewise, the range of the function ln(y) = x is the domain of the exponential function, which we noted is all Interchanging x and y, the graph of the real numbers. natural logarithm function y = ln(x) can be obtained by ﬂipping the graph of y = ex across the line y = x: Important Facts 12.1.1 (Graphical features of natural log). The function y = ln(x) has these features: • • • The largest domain is the set of positive numbers; e.g. ln(−1) makes no sense. The graph has x-intercept 1 and is increasing. The graph becomes closer and closer to the vertical axis as we approach x = 0; i.e. the y-axis is a vertical asymptote for the graph. The graph is unbounded as we move to the right. • Any time we are working with an inverse function, symbolic properties are useful. Here are the important ones related to the natural logarithm. Important Facts 12.1.2 (Natural log properties). We have the following properties: 12.1. THE INVERSE FUNCTION OF Y = EX 155 (a) For any real number x, ln(ex) = x. (b) For any positive number x, eln(x) = x. (c) ln(bt) = t ln(b), for b > 0 and t any real number; (d) ln(ba) = ln(a) + ln(b), for all a, b > 0; (e) ln b a = ln(b) − ln(a), for all a, b > 0. The properties (c)-(e) are related to three of the rules of exponents in Facts 10.1.1. Here are the kinds of basic symbolic maneuvers you can pull off using these properties: Examples 12.1.3. (i) ln(83) = 3 ln(8) = 6.2383; ln(6π) = ln(6) + ln(π) = 2.9365; ln ln(3) − ln(5) = −0.5108. 3 5 = (ii) ln √x = ln = 1 2 ln(x); ln x5 = ln x1/2 x5 x2+1 ln(x + 1); ln Examples 12.1.4. Given the equation 3x+1 = 12, we can solve for x: − ln x2 + 1 = 5 ln(x) − ln x2 + 1 . x2 − 1 = ln((x − 1)(x + 1)) = ln(x − 1) + 3x+1 = 12 3x+1 ln = ln(12) (x + 1) ln(3) = ln(12) ln(12) ln(3) x = − 1 = 1.2619. Example 12.1.5. If $2,000 is invested in a continuously compounding savings account and we want the value after 12 years to be $130,000, what is the required annual interest rate? If, instead, the same $2,000 is invested in a continuously compounding savings account with r = 6.4% annual interest, when will the exact account value be be $130,000? Solution. In the ﬁrst scenario, 130,000 = 2,000e12r 65 = e12r ln(65) = ln ln(65) = 12r ln(65) 12 e12r r = = 0.3479. 156 CHAPTER 12. LOGARITHMIC FUNCTIONS This gives an annual interest rate of 34.79%. In the second scenario, we study the equation 130,000 = 2,000e(0.064)t 65 = e(0.064)t ln(65) 0.064 t = = 65.22. So, it takes over 65 years to accumulate $130,000 under the second scheme. 12.2 Alternate form for functions of exponential type The standard model for an exponential function is A(t) = A0bt, for some = 0. Using the properties of the natural logarithm b > 0, b function, = 1, and A0 6 bt = eln(b) t = et ln(b). This means that every function as in Deﬁnition 10.1.2 can be re-written using the exponential function et. Another way of saying this is that you really only need the function keys “ et ” and “ ln(t) ” on your calculator. Important Fact 12.2.1 (Observation). A function of exponential type can be written in the form A(t) = A0eat, for some constants A0 6 = 0 and a = 0. By studying the sign of the constant a, we can determine whether the function exhibits exponential growth or decay. For example, given the function A(t) = eat, if a > 0 (resp. a < 0), then the function exhibits exponential growth (resp. decay). Examples 12.2.2. (a) The function A(t) = 200 (2t) exhibits exponential growth and can be re-written as: A(t) = 200 et ln(2) = 200e0.69315t (b) The function A(t) = 4e−0.2t exhibits exponential decay and can be re- written as: A(t) = 4e−0.2t = 4 e−0.2 t = 4 0.81873t . 6 6 12.3. THE INVERSE FUNCTION OF Y = BX 157 12.3 The Inverse Function of y = bx For some topics in Chemistry and Physics (e.g. acid base equilibria and acoustics) it is useful to have on hand an inverse function for y = bx, = 1. Just as above, we would show that f(x) = bx is where b > 0 and b one-to-one, the range is all positive numbers and obtain the graph using ideas in Figure 12.2. We will refer to the inverse rule as the logarithm function base b, denoted logb(x), deﬁned by the rule: the unique solution of the equation c = bx ! , if c > 0 (undeﬁned), if c 0. ≤ logb(c) =    We will need to consider two cases, depending on the magnitude of b: The important qualitative features of the logarithm function y = logb(x) mirror Fact 12.1.1: x y = b reflecting line y = x x y = b reflecting line y = x y=log (x)b −1 1 −1 1 y=log (x) b The case b > 1 The case 0 < b < 1 Figure 12.3: Cases to consider for b. Important Facts 12.3.1 (Graphical features of general logs). The function y = logb(x) has these features: • • • The largest domain is the set of positive numbers; e.g. logb(−1) is not deﬁned. The graph has x-intercept 1 and is increasing if b > 1 (resp. decreasing if 0 < b < 1). The graph becomes closer and closer to the vertical axis as we approach x = 0; this says the y-axis is a vertical asymptote for the graph. 6 158 • CHAPTER 12. LOGARITHMIC FUNCTIONS The graph is unbounded as we move to the right. Important Facts 12.3.2 (Log properties). Fix a positive base b, b = 1. (a) For any real number x, logb(bx) = x. (b) For any positive number x, blogb(x) = x. (c) logb(rt) = t logb(r), for r > 0 and t any real number; s) = logb(r) − logb(s), for all r, s > 0. (d) logb(rs) = logb(r) + logb(s), for all r, s > 0; (e) logb( r It is common to simplify terminology and refer to the function logb(x) as the log base b function, dropping the longer phrase “logarithm”. Some scientiﬁc calculators will have a key devoted to this function. Other calculators may have a key labeled “log(x)”, which is usually understood to mean the log base 10. However, many calculators only have the key “ln(x)”. This is not cause for alarm, since it is always possible to express logb(x) in terms of the natural log function. Let’s see how to do this, since it is a great application of the Log Properties listed in Fact 12.3.2. Suppose we start with y = logb(x). We will rewrite this in terms of the natural log by carrying out a sequence of algebraic steps below; make sure you see why each step is justiﬁed. y = logb(x) by = x ln(by) = ln(x) y ln(b) = ln(x) ln(x) ln(b) y = . We have just veriﬁed a useful conversion formula: Important Fact 12.3.3 (Log conversion formula). For x a positive number and b > 0, b = 1 a base, logb(x) = ln(x) ln(b) . For example, log10(5) = log0.02(11) = log20 1 2 = ln(5) ln(10) ln(11) ln(0.02) ln( 1 2) ln(20) = 0.699 = −0.613 = −0.2314 6 6 12.4. MEASURING THE LOUDNESS OF SOUND 159 The conversion formula allows one to proceed slightly differently when solving equations involving functions of exponential type. This is illustrated in the next example. Example 12.3.4. Ten years ago, you purchased a house valued at $80,000. Your plan is to sell the house at some point in the future, when the value is at least $1,000,000. Assume that the future value of the house can be computed using quarterly compounding and an annual interest rate of 4.8%. How soon can you sell the house? Solution. We can use the future value formula to obtain the equation 1,000,000 = 80,000 1 + 12.5 = (1.012)4t 0.048 4 4t Using the log base b = 1.012, log1.012(12.5) = log1.012 log1.012(12.5) = 4t (1.012)4t t = ln(12.5) 4 ln(1.012) = 52.934. Since you have already owned the house for 10 years, you would need to wait nearly 43 years to sell at the desired price. 12.4 Measuring the Loudness of Sound As we noted earlier, the reception of a sound wave by the ear gives rise to a vibration of the eardrum with a deﬁnite frequency and a deﬁnite amplitude. This vibration may also be described in terms of the variation of air pressure at the same point, which causes the eardrum to move. The perception that rustling leaves and a jet aircraft sound different involves two concepts: (1) the fact that the frequencies involved may differ; (2) the intuitive notion of “loudness”. This loudness is directly related to the force being exer
ted on the eardrum, which we refer to as the intensity of the sound. We can try to measure the intensity using some sort of scale. This becomes challenging, since the human ear is an amazing instrument, capable of hearing a large range of sound intensities. For that reason, a logarithmic scale becomes most useful. The sound pressure level β of a sound is deﬁned by the equation β = 10 log10 , I I0 (12.2) where I0 is an arbitrary reference intensity which is taken to correspond with the average faintest sound which can be heard and I is the intensity 160 CHAPTER 12. LOGARITHMIC FUNCTIONS of the sound being measured. The units used for β are called decibels, (Historically, the units of loudness were called bels, abbreviated “db”. in honor of Alexander Graham Bell, referring to the quantity log10 .) Notice, in the case of sound of intensity I = I0, we have a sound pressure level of I I0 β = 10 log10 I0 I0 = 10 log10(1) = 10(0) = 0. We refer to any sound of intensity I0 as having a sound pressure level at the threshold of hearing. At the other end of the scale, a sound of intensity the maximum the eardrum can tolerate has an average sound pressure level of about 120 db. The Table 12.4(a) gives a hint of the sound pressure levels associated to some common sounds. Source of Noise Threshold of pain Riveter Busy Street Trafﬁc Ordinary Conversation Quiet Auto Background Radio Whisper Rustle of Leaves Threshold of Hearing Sound Pressure Level in db 120 95 70 65 50 40 20 10 0 pain threshold Zone of Hearing 120 100 80 60 40 20 0 db 20 100 1000 10,000 20,000 hearing threshold Hz (a) Sources of noise levels. (b) Graphing noise levels. Figure 12.4: Considering noise levels. It turns out that the above comments on the threshold of hearing and pain are really only averages and depend upon the frequency of the given sound. In fact, while the threshold of pain is on average close to 120 db across all frequencies between 20 Hz and 20,000 Hz, the threshold of hearing is much more sensitive to frequency. For example, for a tone of 20 Hz (something like the ground-shaking rumble of a passing freight train), the sound pressure level needs to be relatively high to be heard; 100 db on average. As the frequency increases, the required sound pressure level for hearing tends to drop down to 0 db around 2000 Hz. An examination by a hearing specialist can determine the precise sensitivities of your ear across the frequency range, leading to a plot of your “envelope of hearing”; a sample plot is given in Figure 12.4(b). Such a plot would differ from person to person and is helpful in isolating hearing problems. Example 12.4.1. A loudspeaker manufacturer advertises that their model no. 801 speaker produces a sound pressure level of 87 db when a reference test tone is applied. A competing speaker company advertises that 12.4. MEASURING THE LOUDNESS OF SOUND 161 their model X-1 speaker produces a sound pressure level of 93 db when fed the same test signal. What is the ratio of the two sound intensities produced by these speakers? If you wanted to ﬁnd a speaker which produces a sound of intensity twice that of the no. 801 when fed the test signal, what is its sound pressure level? Solution. If we let I1 and I2 refer to the sound intensities of the two speakers reproducing the test signal, then we have two equations: 87 = 10 log10 93 = 10 log10 I1 I0 I2 I0 Using log properties, we can solve the ﬁrst equation for I1: 87 = 10 log10 I1 I0 log10(I1) = 8.7 + log10(I0) 10log10(I1) = 108.7+log10(I0) = 10 log10(I1) − 10 log10(I0) I1 = 108.710log10(I0) = 108.7I0. Similarly, we ﬁnd that I2 = 109.3I0. This means that the ratio of the intensities will be I2 I1 = 109.3I0 108.7I0 = 100.6 = 3.98. This means that the test signal on the X − 1 speaker produces a sound pressure level nearly 4 times that of the same test signal on the no. 801 speaker. To ﬁnish the problem, imagine a third speaker which produces a sound pressure level β, which is twice that of the ﬁrst speaker. If I3 is the corresponding intensity of the sound, then as above, I3 = 10(β/10)I0. We are assuming that I3 = 2I1, so this gives us the equation I1 = 108.7I0 = 1 2 1 2 I3 10(β/10)I0 log10 108.7 = log10 8.7 = log10 1 2 1 2 10(β/10) + log10 8.7 = −0.30103 + 90 = β β 10 10(β/10) 162 CHAPTER 12. LOGARITHMIC FUNCTIONS So, the test signal on the third speaker must produce a sound pressure level of 90 db. 12.5. EXERCISES 12.5 Exercises Problem 12.1. These problems will help you develop your skills with logarithms. (a) Compute: log5 3, loge 11, log√2 π, log2 10, log10 2. (b) Solve for x: 35 = ex, log3 x = e, log3 5 = xe3. (c) Solve each of these equations for x in terms of y: y = 10x, 3y = 10x, y = 103x. Problem 12.2. As light from the surface penetrates water, its intensity is diminished. In the clear waters of the Caribbean, the intensity is decreased by 15 percent for every 3 meters of depth. Thus, the intensity will have the form of a general exponential function. (a) If the intensity of light at the water’s sur, ﬁnd a formula for I(d), the inface is I tensity of light at a depth of d meters. Your formula should depend on I and d. ◦ ◦ (b) At what depth will the light intensity be decreased to 1% of its surface intensity? Problem 12.3. Rewrite each function in the eat, for appropriate constants A form y = A and a. ◦ ◦ (a) y = 13(3t) (b) y = 2( 1 8 )t (c) y = −7(1.567)t−3 (d) y = −17(2.005)−t (e) y = 3(14.24)4t Problem 12.4. (a) If you invest Po dollars at 7% annual interest and the future value is computed by continuous compounding, how long will it take for your money to double? 163 (c) A rule of thumb used by many people to determine the length of time to double an investment is the rule of 70. The rule says it takes about t = 70 r years to double the investment. Graphically compare this rule to the one isolated in part b. of this problem. Problem 12.5. The length of some ﬁsh are modeled by a von Bertalanffy growth function. For Paciﬁc halibut, this function has the form L(t) = 200 (1 − 0.956 e−0.18t) where L(t) is the length (in centimeters) of a ﬁsh t years old. (a) What is the length of a new-born halibut at birth? (b) Use the formula to estimate the length of a 6–year–old halibut. (c) At what age would you expect the hal- ibut to be 120 cm long? (d) What is the practical (physical) signiﬁcance of the number 200 in the formula for L(t)? Problem 12.6. A cancerous cell lacks normal biological growth regulation and can divide continuously. Suppose a single mouse skin cell is cancerous and its mitotic cell cycle (the time for the cell to divide once) is 20 hours. The number of cells at time t grows according to an exponential model. (a) Find a formula C(t) for the number of cancerous skin cells after t hours. (b) Assume a typical mouse skin cell is 10−4 cm. Find the spherical of radius 50 combined volume of all cancerous skin cells after t hours. When will the volume of cancerous cells be 1 cm3? × (b) Suppose you invest Po dollars at r% aninterest and the future value is nual computed by continuous compounding. If you want the value of the account to double in 2 years, what is the required interest rate? Problem 12.7. Your Grandfather purchased a house for $55,000 in 1952 and it has increased in value according to a function y = v(x), where x is the number of years owned. These questions probe the future value of the house under various mathematical models. 164 CHAPTER 12. LOGARITHMIC FUNCTIONS (a) Suppose the value of the house is $75,000 in 1962. Assume v(x) is a linear function. Find a formula for v(x). What is the value of the house in 1995? When will the house be valued at $200,000? (b) Suppose the value of the house is $75,000 in 1962 and $120,000 in 1967. Assume v(x) is a quadratic function. Find a formula for v(x). What is the value of the house in 1995? When will the house be valued at $200,000? (c) Suppose the value of the house is $75,000 in 1962. Assume v(x) is a function of exponential type. Find a formula for v(x). What is the value of the house in 1995? When will the house be valued at $200,000? Problem 12.8. Solve the following equations for x: (a) log3(5) = log2(x) (b) 10log2(x) = 3 x (c) 35 = 7 (d) log2(ln(x)) = 3 (e) ex = 105 (f) 23x+5 = 32 (d) There also was an exponentially-growing population of anteaters on board. At the start of the voyage there were 17 anteaters, and the population of anteaters doubled every 2.8 weeks. How long into the voyage were there 200 ants per anteater? Problem 12.10. The populations of termites and spiders in a certain house are growing exponentially. The house contains 100 termites the day you move in. After 4 days, the house contains 200 termites. Three days after moving in, there are two times as many termites as spiders. Eight days after moving in, there were four times as many termites as spiders. How long (in days) does it take the popula- tion of spiders to triple? Problem 12.11. In 1987, the population of Mexico was estimated at 82 million people, with an annual growth rate of 2.5%. The 1987 population of the United States was estimated at 244 million with an annual growth rate of 0.7 %. Assume that both populations are growing exponentially. (a) When will Mexico double its 1987 popu- lation? (b) When will the United States and Mexico have the same population? Problem 12.9. A ship embarked on a long voyage. At the start of the voyage, there were 500 ants in the cargo hold of the ship. One week into the voyage, there were 800 ants. Suppose the population of ants is an exponential function of time. (a) How long did it take the population to double? (b) How long did it take the population to triple? (c) When were there be 10,000 ants on board? Problem 12.12. The cities of Abnarca and Bonipto have populations that are growing exponentially. In 1980, Abnarca had a population of 25,000 people. In 1990, its population was 29,000. Bonipto had a population of 34,000 in 1980. The population of Bonipto doubles every 55 years. (a) How long does it take t
he population of Abnarca to double? (b) When will Abnarca’s population equal that of Bonipto? Chapter 13 Three Construction Tools Sometimes the composition of two functions can be understood by graphical manipulation. When we discussed quadratic functions and parabolas in the previous section, certain key graphical maneuvers were laid out. In this section, we extend those graphical techniques to general function graphs. 13.1 A Low-tech Exercise This section is all about building new functions from ones we already have in hand. This can be approached symbolically or graphically. Let’s begin with a simple hands-on exercise involving the curve in Figure 13.1. y = f(x) By the vertical line test, we know this represents the graph of a function y = f(x). With this picture and a piece of bendable wire we can build an INFINITE number of new functions from the original function. Begin by making a “model” of this graph by bending a piece of wire to the exact shape of the graph and place it right on top of the curve. The wire model can be manipulated in a variety of ways: slide the model back and forth horizontally, up and down vertically, expand or compress the model horizontally or vertically. Figure 13.1: Start with some curve. Another way to build new curves from old ones is to exploit the built in symmetry of the xy-coordinate system. For example, imagine reﬂecting the graph of y = f(x) across the x-axis or the y-axis. In all of the above cases, we moved from the original wire model of our function graph to a new curve that (by the vertical line test) is the graph of a new function. The big caution in all this is that we are NOT ALLOWED to rotate or twist the curve; this kind of maneuver does lead to a new curve, but it may not be the graph of a function: See Figure 13.2. rotate NOT a function graph Figure 13.2: Rotating a curve. The pictures in Figure 13.3 highlight most of what we have to say in 165 166 CHAPTER 13. THREE CONSTRUCTION TOOLS this section; the hard work remaining is a symbolic reinterpretation of these graphical operations. left right up Horizontal shift. Vertical shift. down pull pull push push expand horizontally compress horizontally Horizontal expansion. Horizontal compression. pull pull expand vertically push push compress vertically Vertical expansion. Vertical compression. reﬂect across x-axis reﬂect across y-axis Vertical reﬂection. Horizontal reﬂection. Figure 13.3: Shifting, dilating, and reﬂecting y = f(x). 13.2 Reﬂection In order to illustrate the technique of reﬂection, we will use a concrete example: Function: y = p(x) = 2x + 2 2 Domain: 6 Range: −2 −2 x y ≤ ≤ ≤ ≤ As we know, the graph of y = p(x) on the domain −2 2 is a line of slope 2 with y-intercept 2, as pictured in Figure 13.4(a). Now, start with ≤ ≤ x 13.2. REFLECTION 167 the function equation y = p(x) = 2x + 2 and replace every occurrence of “y” by “−y.” This produces the new equation −y = 2x + 2; or, equivalently, y = q(x) = −2x − 2. x ≤ −y The domain is still −2 2, but the range will ≤ change; we obtain the new range by replacing “y” by “−y” −6. The in the original range: −2 graph of this function is a DIFFERENT line; this one has slope −2 and y-intercept −2. We contrast these two curves in Figure 13.4(b), where q(x) is graphed as the “dashed line” in the same picture with the original p(x). Once we do this, it is easy to see how the graph of q(x) is really just the original line reﬂected across the x-axis. 6; so 2 ≥ ≤ ≥ ≤ y Next, take the original function equation y = p(x) = 2x + 2 and replace every occurrence of “x” by “−x.” This produces a new equation y = 2(−x) + 2; or, equivalently, −2 −1 y = r(x) = −2x + 2. −2 −1 y-axis x-axis 1 2 8 6 4 2 −2 −4 −6 −8 (a) Graph of p(x) = 2x + 2. y-axis x-axis 1 2 8 6 4 2 −2 −4 −6 −8 x ≥ ≥ The domain must also be checked by replacing “x” by “−x” in the original domain condition: −2 2, so −2. It just so happens in this case, the domain 2 is unchanged. This is yet another DIFFERENT line; this one has slope −2 and y-intercept 2. We contrast these two curves in Figure 13.4(c), where r(x) is graphed as the “dashed line” in the same picture with the original p(x). Once we do this, it is easy to see how the graph of r(x) is really just the original curve reﬂected across the y-axis. −x ≤ ≤ This example illustrates a general principle referred to as the reﬂection principle. Important Facts 13.2.1 (Reﬂection). Let y = f(x) be a function equation. (b) Graph of q(x) = −2x − 2. y-axis x-axis 1 2 8 6 4 2 −2 −4 −6 −8 −2 −1 (c) Graph of r(x) = −2x + 2. Figure 13.4: Reﬂecting y = p(x). (i) We can reﬂect the graph across the x-axis and the resulting curve is the graph of the new function obtained by replacing “y” by “−y” in the original equation. The domain is the same as the domain for y = f(x). If the range for y = f(x) d. In other words, is c the reﬂection across the x-axis is the graph of y = −f(x). d, then the range of −y = f(x) is c −y ≤ ≤ ≤ ≤ y (ii) We can reﬂect the graph across the y-axis and the resulting curve is the graph of the new function obtained by replacing “x” by “−x” in the original equation. The range is the same as the range for y = f(x). If the domain for y = f(x) is a b, then the domain x ≤ ≤ 168 CHAPTER 13. THREE CONSTRUCTION TOOLS of y = f(−x) is a b. Using composition no≤ tation, the reﬂection across the y-axis is the graph of y = f(−x). −x ≤ Example 13.2.2. Consider the parallelogram-shaped region with vertices (0, 2), (0, −2), (1, 0), and (−1, 0). Use the reﬂection principle to ﬁnd functions whose graphs bound 0,2) l 2 Q = (1,0) x-axis l 1 Region R y-axis Figure 13.5: The region . R y-axis y = f(−x) y = f(x) −4 −2 2 4 x-axis −y = f(x) Figure 13.6: Reﬂecting the semicircle. R Solution. Here is a picture of the region : First off, using the two point formula for the equation of a line, we ﬁnd that the line ℓ1 passing through the points P = (0, 2) and Q = (1, 0) is the graph of the function y = f1(x) = −2x + 2. By Fact 13.2.1 (i), ℓ2 is the graph of the equation −y = −2x + 2, which we can write as the function y = f2(x) = 2x − 2. By Fact 13.2.1 (ii) applied to ℓ2, the line ℓ3 is the graph of the function y = f3(x) = −2x − 2. Finally, by Fact 13.2.1 (i) applied to ℓ3, the line ℓ4 is the graph of the equation −y = −2x − 2, which we can write as the function y = f4(x) = 2x + 2. Figure 13.6 illustrates the fact that we need to be careful about the domain of the original function when using the reﬂection principle. For example, consider 1 − (x − 3)2. The largest possible domain of y = f(x) = 1 + x-values is 2 4 and the graph is an upper semicircle x p ≤ of radius 1 centered at the point (3, 1). ≤ Reﬂection across the x-axis gives the graph of reﬂection 1 − (x + 3)2 y = −1 − across the y-axis gives the graph of y = 1 + on the new domain −4 1 − (x − 3)2 on the same domain; −2. p x ≤ ≤ p 13.3 Shifting 2.5 2 1.5 1 0.5 −2 −1 0 y-axis x x-axis Let’s start out with the function y = f(x) = √4 − x2, which has a largest possible domain −2 2. From Chapter 6, the graph of this equation is an upper semicircle of radius 2 centered at the origin (0, 0). Sliding the graph back and forth horizontally or vertically (or both), never rotating or twisting, we are led to the “dashed curves” below (contrasted with the original graph which is plotted with a solid curve). This describes some shifted curves on a pictorial level, but what are the underlying equations? For this example, we can use the fact that all of the shifted curves are still semicircles and Chapter 6 tells us how to ﬁnd their equations. ≤ ≤ 2 1 Figure 13.7: Graph of y = √4 − x2. 13.3. SHIFTING 169 The lower right-hand dashed semicircle is of radius 2 and is centered at (3, 0), so the corresponding equation must be y = 4 − (x − 3)2. The upper left-hand dashed semicircle is of radius 2 and centered at (0, 3), so the corresponding equation must be y = 3+√4 − x2. The upper right-hand dashed semicircle is of radius 2 and centered at (3, 3), so the corresponding equation must be y = 3 + 4 − (x − 3)2. p Keeping this same example, we can continue this kind p of shifting more generally by thinking about the effect of making the following three replacements in the equation y = √4 − x2x and y) (x − h and y − k). These substitutions lead to three new equations, each the equation of a semicircle: 5 4 3 2 1 −2 −1 1 2 3 4 5 Figure 13.8: Shifting the upper semicircle. y = 4 − (x − h)2 − x2; and, 4 − (x − h)2. ⇐ ⇐ Upper semicircle with radius 2 and center (h, 0). Upper semicircle with radius 2 and center (0, k). Upper semicircle with radius 2 and center (h, k). There are three potentially confusing points with this ⇐ y-axis original curve x-axis h negative h positive curve shifted h units right if h is positive curve shifted \|h\| units left if h is negative Figure 13.9: Potentially confusing points. example: • • ± ) of h and k. In FigBe careful with the sign (i.e., ure 13.9, if h = 1, we horizontally shift the semicircle 1 unit to the right; whereas, if h = −1, we horizontally shift the semicircle −1 units to the right. But, shifting −1 unit to the right is the same as shifting 1 unit to the left! In other words, if h is positive, then a horizontal shift by h will move the graph \|h\| units to the right; if h is negative, then a horizontal shift by h will move the graph \|h\| units to the left. If k is positive, then a vertical shift by k will move the graph \|k\| units up; if k is negative, then a vertical shift by k will move the graph \|k\| units down. These conventions insure that the “positivity” of h and k match up with “rightward” and “upward” movement of the graph. 170 CHAPTER 13. THREE CONSTRUCTION TOOLS • When shifting, the domain of allowed x-values may change. This example illustrates an important general principle referred to as the shifting principle. Important Facts 13.3.1 (Shifting). Let y = f(x) be a function equation. (i) If we replace “x” by “x − h” in the original function equation, then the graph of the resulting new function y = f(x − h) is obt
ained by horizontally shifting the graph of f(x) by h. If h is positive, the picture shifts to the right h units; if h is negative, the picture shifts to the left h units. If the domain of f(x) is an interval a b, then the domain ≤ of f(x − h) is a x − h b. The range remains unchanged under ≤ horizontal shifting. ≤ ≤ x (ii) If we replace “y” by “y − k” in the original function equation, then the graph of the resulting new function y = f(x) + k is obtained by vertically shifting the graph of f(x) by k. If k is positive, the picture shifts upward k units; if k is negative, the picture shifts downward k units. If the range of f(x) is an interval c d, then the range y of f(x) + k is c d. The domain remains unchanged under ≤ vertical shifting. y − k ≤ ≤ ≤ 13.4 Dilation 1 0.75 0.5 0.25 y-axis 2 To introduce the next graphical principle we will look at the function y = f(x) = x x2 + 1 . −6 −4 −2 4 6 x-axis −0.25 −0.5 −0.75 −1 Figure 13.10: Graph of y = f(x) = x x2+1 . ≤ Using a graphing device, we have produced a plot of the graph on the domain −6 6. Figure 13.10 shows the x ≤ curve has a high point H (like a “mountain peak”) and a low point L (like a “valley”). Using a graphing device, we can determine that the high point is H = (1, 1 2) (it lies on −1, − 1 the line with equation y = 1 (it lies on 2 the line with equation line y = − 1 1 2 . Draw two new horizontal lines with equations y = 2 1. Grab the high point H on the curve and uniformly pull straight up, so that the high point now lies on the horizontal line y = 1 at (1, 1). Repeat this process by pulling L straight downward, so that the low point is now on the line y = −1 at (−1, −1). We end up with the ”stretched dashed curve” illustrated in Figure 13.11(a). In terms of the original function equation y = x x2+1, we are simply describing the graphical effect of multiplying the y-coordinate of every point on the curve by the positive number 2. In other words, the dashed curve is the graph of y = 2x x2+1 = 2 2 ) and the low point is L = 2 ), so the range is − 1 · ± x x2+1 1 2 = 2 ≤ ≤ ± y . 13.4. DILATION 171 1 2 = 4 at (1, 1 Next, draw the two horizontal lines with equations y = 1 1 4. Grab the high point H on the curve (in Figure 13.10) and uniformly push straight down, so that the high point now lies on the horizontal line y = 1 4). Repeat this process at the low point L by pushing the curve straight upward, so that the low point is now on the line y = − 1 4 at (−1, − 1 4). We end up with the new ”dashed curve” illustrated in Figure 13.11(b). In terms of the original function equation y = x x2+1 , we are simply describing the graphical effect of multiplying the y-coordinate of every point on the curve by the positive number 1 2. In other words, the dashed curve is the graph of y = x 2 · ± ± 2(x2+1) . We could repeat this process systematically: • • • Pick a positive number c. 1 2. If c > 1, then 2 is parallel and above the graph of 2. On the other hand, if 0 < c < 1, then this new Draw the two horizontal lines y = c the graph of y = c y = 1 line is parallel and below y = 1 2. · ± Uniformly deform the original graph (in Figure 13.10) so that the new curve has it’s high and c low points just touching y = 2 . This will involve vertically stretching or compressing, depending on whether 1 < c or 0 < c < 1, respectively. A number of possibilities are pictured in Figure 13.11(c). ± We refer to each new dashed curve as a verticaldilation of the original (solid) curve. This example illustrates an important principle. Important Facts 13.4.1 (Vertical dilation). Let c > 0 be a positive number and y = f(x) a function equation. (i) If we replace “ y ” by “ y c ” in the original equation, then the graph of the resulting new equation is obtained by vertical dilation of the graph of y = f(x). The domain of x-values is not affected. (ii) If c > 1, then the graph of y c = f(x) (i.e., y = cf(x)) is a vertically stretched version of the original graph. −6 −4 −2 y-axis 1 0.75 0.5 0.25 2 −0.25 −0.5 −0.75 −1 4 6 x-axis (a) Vertical expansion. y-axis 1 0.75 0.5 0.25 −6 −4 −2 2 −0.25 −0.5 −0.75 −1 4 6 x-axis (b) Vertical compression. y-axis 1 0.75 0.5 0.25 −6 −4 −2 4 6 x-axis 2 −0.25 −0.5 −0.75 −1 (c) Many possibilities. Figure 13.11: Dilating y = f(x). (iii) If 0 < c < 1, then the graph of y c = f(x) (i.e., y = cf(x)) is a vertically compressed version of the original graph. If we combine dilation with reﬂection across the x-axis, we can determine the graphical relationship between y = f(x) and y = cf(x), for any constant c. The key observation is that reﬂection across the x-axis corresponds to the case c = −1. 172 CHAPTER 13. THREE CONSTRUCTION TOOLS Example 13.4.2. Describe the relationship between the graphs of y = f(x) = 1 − (x + 1)2, p y = −f(x) = − 1 − (x + 1)2, and p y = −4f(x) = −4 1 − (x + 1)2. p step 1: start with upper semicircle x-axis step 2: reﬂect across step 3: stretch curve from step 2: to get y = −4f(x) Solution. The graph of y = f(x) is an upper semicircle of radius 1 centered at the point (−1,0). To obtain the picture of the graph of y = −4f(x), we ﬁrst reﬂect y = f(x) across the x-axis; this gives us the graph of y = −f(x). Then, we vertically dilate this picture by a factor of c = 4 to get the graph of y 4 = −f(x), which is the same as the graph of the equation y = −4f(x). See Figure 13.12. Figure 13.12: Reﬂecting and dilating a lower semicircle. Let’s return to the original example y = x x2+1 and investigate a different type of dilation where the action is taking place in the horizontal direction (whereas it was in the vertical direction before). Grab the right-hand end of the graph (in Figure 13.10) and pull to the right, while at the same time pulling the left-hand end to the left. We can quantify this by stipulating that the 2, 1 high point H = 2 and the low point L = of the original curve moves to the new location moves to the new location 1, 1 2 . −1, − 1 2 −2, − 1 2 y-axis 1 0.75 0.5 0.25 y-axis 1 0.75 0.5 0.25 −6 −4 −2 2 −0.25 −0.5 −0.75 −1 (a) Stretching. 4 6 x-axis −6 −4 −2 4 6 x-axis 2 −0.25 −0.5 −0.75 −1 Figure 13.13: Horizontally dilating y = x (b) Compressing. x2+1 . The result will be somewhat analogous to stretching a spring. By the same token, we could push the right-hand end to the left and push the left-hand end to the right, like compressing a spring. We can quantify this by stipulating that the high point H = of the original curve moves moves to the new to the new location location . These two situations are indicated in Figure 13.13. We refer to each of the dashed curves as a horizontal dilation of the original (solid) curve. and the low point L = −1, − 1 2 2, − 1 1, 1 2 2, 1 − 1 2 1 2 13.5. VERTEX FORM AND ORDER OF OPERATIONS 173 The tricky point is to understand what happens on the level of the original equation. In the case of the stretched graph in Figure 13.13(a), you can use a graphing device to verify that this looks like the graph of y = x/2 (x/2)2+1 ; in other words, we replaced “x” by “x/2” in the original equation. In the case of the compressed graph in Figure 13.13(b), you can use a graphing device to verify that this looks like the graph of y = 2x (2x)2+1; in other words, we replaced “x” by “ x 1/2 = 2x” in the original equation. The process just described leads to a general principle. Important Facts 13.4.3 (Horizontal dilation). Let c > 0 be a positive number and y = f(x) a function equation. (i) If we replace “ x” by “ x graph of the resulting new function y = f tal dilation of the graph of y = f(x). If the domain of f(x) is a then the domain of y = f is a c ” in the original function equation, then the is obtained by a horizonb, ≤ ≤ x x c x c b. x c ≤ is a horizontal stretch. ≤ x c (ii) If c > 1, then the graph of y = f (iii) If 0 < c < 1, then the graph of y = f is a horizontal compression. x c 13.5 Vertex Form and Order of Operations Using the language of function compositions we can clarify our discussion in Example 7.1.2. Let’s revisit that example: Example 13.5.1. The problem is to describe a sequence of geometric maneuvers that transform the graph of y = x2 into the graph of y = −3(x − 1)2 + 2. Solution. The idea is to rewrite y = −3(x−1)2 +2 as a composition of y = x2 with four other functions, each of which corresponds to a horizontal shift, vertical shift, reﬂection or dilation. Once we have done this, we can read off the order of geometric operations using the order of composition. Along the way, pay special attention to the exact order in which we will be composing our functions; this will make a big difference. To begin with, we can isolate four key numbers in the equation: Reﬂect − z}\|{ 3 (x − Dilate by 3 Horizontal shift by h = 1 1 z}\|{ )2 + 2 Vertical shift by k = 2 \|{z} \|{z} We want to use each number to deﬁne a new function, then compose these in the correct order. We will also give our starting function y = x2 a speciﬁc name to make things deﬁnite: f(x) = x2 v(x) = x + 2 d(x) = 3x. h(x) = x − 1 r(x) = −x 174 CHAPTER 13. THREE CONSTRUCTION TOOLS Now, verify that f h(x[x − 1] d{(x − 1)2}   r(3(x − 1)2) = v i h = v[−3(x − 1)2] = −3(x − 1)2 + 2 = −3x2 + 6x − 1. 13.6 Summary of Rules 2 −2 2 Figure 13.14: A multipart function. For quick reference, we summarize the consequence of shifting and expanding symbolically and pictorially. The running example for Tables 13.1, 13.2, and 13.3 will be a multipart function y = f(x) whose graph, seen in Figure 13.14, consists of a line segment and a quarter circle on the domain −2 2: x ≤ ≤ f(x) = x + 2 √4 − x2 0 x if −2 if 0 ≤ x ≤ ≤ ≤ 2 13.6. SUMMARY OF RULES 175 Reﬂection Symbolic Change New Equation Graphical Consequence Replace x with −x. y = f(−x) Replace with −f(x). f(x) y = −f(x) A reﬂection across the y-axis. A reﬂection across the x-axis. Picture 2 2 2 −2 −2 −2 Table 13.1: Reﬂecting y = f(x). 176 CHAPTER 13. THREE CONSTRUCTION TOOLS Shifting (Assume c > 0 ) Symbolic Change New Equation Graphical Consequence Picture Replace x with (x − c). y = f(x − c) A shift to th
e right c units. 2 Replace x with (x + c). y = f(x + c) A shift to the left c units. Replace with (f(x) + c). f(x) y = f(x) + c A shift up c units. 2 4 2 −4 −2 4 2 −2 2 Replace with (f(x) − c). f(x) y = f(x) − c A shift down c units. −2 2 2 Table 13.2: Shifting y = f(x). 13.6. SUMMARY OF RULES 177 Dilation Symbolic Change New Equation Graphical Consequence Picture If c > 1, replace x x with c . y = f x c A horizontal expansion. c = 2 4 3 2 1 −1−2−3−4 0 1 2 3 4 If 0 < c < 1, replace x with . x c y = f x c A horizontal compression. c = 1 2 4 3 2 1 If c > 1, replace f(x) with (cf(x)). y = cf(x) A vertical expansion. If 0 < c < 1, replace f(x) with (cf(x)). y = cf(x) A vertical compression. Table 13.3: Dilating y = f(x). −4 −1−2−1−2−3−1−2−3−4 0 1 2 3 4 178 CHAPTER 13. THREE CONSTRUCTION TOOLS 13.7 Exercises Problem 13.1. On a single set of axes, sketch a picture of the graphs of the following four equations: y = −x+√2, y = −x−√2, y = x+√2, and y = x − √2. These equations determine lines, which in turn bound a diamond shaped region in the plane. (a) Show that the unit circle sits inside this diamond tangentially; i.e. show that the unit circle intersects each of the four lines exactly once. (b) Find the intersection points between the unit circle and each of the four lines. (c) Construct a diamond shaped region in which the circle of radius 1 centered at (−2, − 1) sits tangentially. Use the techniques of this section to help. y π π/2 f(x) −1 1 x Problem 13.2. The graph of a function y = f(x) is pictured with domain −2.5 3.5. x ≤ ≤ Problem 13.4. (a) Each of the six functions y = f(x) below can be written in the “standard form” y = A\|B(x − C)\| + D, for some constants A,B,C,D. Find these constants, describe the precise order of graphical operations involved in going from the graph of y = \|x\| to the graph of y = f(x) (paying close attention to the order), write out the multipart rule, sketch the graph and calculate the coordinates of the “vertex” of the graph. Sketch the graph of each of the new func- tions listed below. (a) g(x) = 2f(x + 1) (b) h(x) = 1 2 f(2x − 1) (c) j(x) = 5f( 1 3 x + 2) − 2 (a1) f(x) = \|x − 2\| (a2) f(x) = 2\|x + 3\| (a3) f(x) = \|2x − 1\| (a4) f(x) = \|2(x − 1)\| (a5) f(x) = 3\|2x − 1\| + 5 (a6) f(x) = −2\|x + 3\| − 1 Problem 13.3. The graph of a function y = f(x) is pictured with domain −1 1. Sketch the graph of the new function ≤ ≤ x y = g(x) = 1 π f(3x) − 0.5. Find the largest possible domain of the function y = g(x). p (b) Solve the following inequalities using your work in the previous part of this problem: (b1) \|x − 2\| 3 (b2) 1 ≤ ≤ 2\|x + 3\| 5 ≤ (b3) y = 3\|2x − 1\| + 5 10 ≥ 13.7. EXERCISES 179 (c) The graphs of y = 3\|2x − 1\| + 5 and y = −\|x − 3\| + 10 intersect to form a bounded region of the plane. Find the vertices of this region and sketch a picture. (c) The graph of z = a(x) from part (a) is given below. Sketch the graph and ﬁnd the rule for the function z = 2a(3x+3)+1; make sure to specify the domain and range of this new function. Problem 13.5. Consider the function y = f(x) with multipart deﬁnition 0 2x + 2 −x + 2 0 f(x) =   0 if x ≤ if −1 if 0 if x ≤ ≥ −1 x ≤ 2 ≤ ≤ x 2 (a) Sketch the graph of y = f(x).  (b) Is y = f(x) an even function? Is y = f(x) an odd function? (A function y = f(x) is called even if f(x) = f(−x) for all x in the domain. A function y = f(x) is called odd if f(−x) = −f(x) for all x in the domain.) (c) Sketch the reﬂection of the graph across the x-axis and y-axis. Obtain the resulting multipart equations for these reﬂected curves. (d) Sketch the vertical dilations y = 2f(x) and y = 1 2 f(x). (e) Sketch the horizontal dilations y = f(2x) and y = f( 1 2 x). (f) Find a number c > 0 so that the highest point on the graph of the vertical dilation y = cf(x) has y-coordinate 11. (g) Using horizontal dilation, ﬁnd a number c > 0 so that the function values f( x c ) are non-zero for all − 5 2 < x < 5. (h) Using horizontal dilation, ﬁnd positive numbers c,d > 0 so that the function values f( 1 c (x − d)) are non-zero precisely when 0 < x < 1. Problem 13.6. An isosceles triangle has sides of length x, x and y. In addition, assume the triangle has perimeter 12. (a) Find the rule for a function that computes the area of the triangle as a function of x. Describe the largest possible domain of this function. (b) Assume that the maximum value of the function a(x) in (a) occurs when x = 4. Find the maximum value of z = a(x) and z = 2a(3x + 3) + 1. 10 y-axis 8 6 4 2 2 4 6 8 10 x-axis Problem 13.7. Describe how each graph differs from that of y = x2. (a) y = 2x2 (b) y = x2 − 5 (c) y = (x − 4)2 (d) y = (3x − 12)2 (e) y = 2(3x − 12)2 − 5 Problem 13.8. In each case, start with the function y = \|x\| and perform the operations described to the graph, in the order speciﬁed. Write out the resulting rule for the function and sketch the ﬁnal graph you obtain. (a) (1) horizontally compress by a factor of 2; (2) horizontally shift to the left by 2; (3) vertically stretch by a factor of 7; (4) vertically shift up 2 units. (b) (1) horizontally stretch by a factor of 2; (2) horizontally shift to the right by 2; (3) vertically compress by a factor of 7; (4) vertically shift down 2 units. (c) (1) horizontally shift to the right 2 units; (2) horizontally compress by a factor of 3. Problem 13.9. (a) Begin with the function y = f(x) = 2x. (a1) Rewrite each of the following functions in standard exponential form: f(2x), f(x − 1), f(2x − 1), f(2(x − 1)), 3f(x), 3f(2(x − 1)). 180 CHAPTER 13. THREE CONSTRUCTION TOOLS (a2) Is the function 3f(2(x−1))+1 a func- tion of exponential type? (a3) Sketch the graphs of f(x), f(2x), f(2(x− 1)), 3f(2(x − 1)) and 3f(2(x − 1)) + 1 in the same coordinate system and explain which graphical operation(s) (vertical shifting, vertical dilation, horizontal shifting, horizontal dilation) have been carried out. (b) In general, explain what happens when you apply the four construction tools of Chapter 13 (vertical shifting, verti- cal dilation, horizontal shifting, horizontal dilation) to the standard exponential model y = Aobx. For which of the four operations is the resulting function still a standard exponential model? Problem 13.10. Begin with a sketch of the graph of the function y = 2x on the domain of all real numbers. Describe how to use the “four tools” of Chapter 13 to obtain the graphs of these functions: y = −2x, y = 2−x, y = 3(2x), 3 (2x), y = 3 + 2x, y = 2x − 2, y = 2x−2, y = 1 y = 2x+2, y = 23x, y = 2x/3. Chapter 14 Rational Functions A rationalfunction is a function of the form f(x) = p(x) are polynomials. For example, the following are all rational functions. q(x) where p(x) and q(x) f(x) = x 3x + 4 g(x) = x2 + 1 3x − 5 h(x) = 4x5 − 4x2 − 8 x3 + x2 − x + 1 j(x) = x6 x8 + 5x − 1 2 There is a huge variety of rational functions. In this course, we will concentrate our efforts exclusively toward understanding the simplest type of rational functions: linear-to-linear rational functions. Linearto-linear rational functions are rational functions in which the numerator and the denominator are both linear polynomials. The following are linear-to-linear-rational functions. k(x) = x 3x + 4 m(x) = 5x − 6 2x + 1 n(x) = 0.34x − 0.113 x − 1 p(x) = 4x + 3 4 5 8x − 1.117 The simplest example of a linear-to-linear rational function is f(x) = 1 x whose graph is shown in Figure 14.1. This is an important example for the study of this class of functions, as we shall see. Let’s consider the graph of this function, f(x) = 1 x . We ﬁrst begin by considering the domain of f. Since the numerator of 1 x is a constant, and the denominator is just x, the only way we can run into difﬁculty when evaluating this function is if we try to divide by zero; that is, the only value of x not in the domain of this function is zero. The function is deﬁned for all x except x = 0. At x = 0, there must be a gap, or hole, in the graph. -10 -5 10 5 0 0 -5 -10 5 10 Figure 14.1: The graph of f(x) = 1 x . To get the graph started we might simply give ourselves a point on the graph. For instance, we see that f(1) = 1/1 = 1, so the point (1,1) is on the graph. Then, if we use larger values of x, we see that 1/x becomes smaller as x grows. For instance, f(2) = 1/2, f(10) = 1/10, and f(1000) = 1/1000. In addition, we see that, other than the fact that 1/x > 0 for positive x, there is nothing preventing us from 181 182 CHAPTER 14. RATIONAL FUNCTIONS making 1/x as small as we want simply by taking x big enough. Want 1/x to be less than 0.001? Just pick x bigger than 1000. Want 1/x to be less than 0.000001? Just use x bigger than 1,000,000. What this means graphically is that as x gets bigger (starting from x = 1), the curve y = 1/x gets closer and closer to the x-axis. As a result, we say that the x-axis is a horizontal asymptote for this function. We see the same behavior for negative values of x. If x is large, and negative (think -1000, or -1000000), then 1/x is very small (i.e., close to zero), and it gets smaller the larger x becomes. Graphically, this means that as x gets large in the negative direction, the curve y = 1/x gets closer and closer to the x-axis. We say that, in both the positive and negative directions, y = 1/x is asymptotic to the x-axis. A similar thing happens when we consider x near zero. If x is a small positive number (think 1/2, or 1/10, or 1/10000), then 1/x is a large positive number. What’s more, if we think of x as getting closer to zero, 1/x gets bigger and bigger. Plus, there is no bound on how big we can make 1/x simply by taking x as close to zero as we need to. For instance, can 1/x be as big as 10000? All you need to do is pick a positive x smaller than 1/10000. Graphically, what this means is that as x approaches zero from the positive side, the y value gets larger and larger. As a result, the curve approaches the y-axis as y gets larger. We say that the y-axis is a vertical asymptote for the curve y = 1/x. We see the same phenomenon as x approaches zero from the negative side: y = 1/x
gets larger in the negative direction (i.e., it gets more and more negative). The curve gets closer and closer to the negative y-axis as y becomes more and more negative. Again, we say that the y − axis is a vertical asymptote for the curve y = 1/x. It turns out that every linear-to-linear rational function has a graph that looks essentially the same as the graph of y = 1/x. Let’s see why. Consider the linear-to-linear rational function f(x) = ax+b cx+d . If we divide cx + d into ax + b, the result is f(x) = ax + b cx + d = a c + b − ad c cx + d which we can rewrite as f(x) = a c + b − ad c · 1 cx + d = a c + b − ad c · 1 c(x + d c ) = a c + bc − ad c2 1 x + d c . · If we now let A = , B = a c then we can write bc − ad c2 , and C = d c , f(x) = A + B 1 x + C . 183 If we let g(x) = 1/x then we have shown that f(x) = A + Bg(x + C), and so the graph of the function f is just a horizontally shifted, vertically shifted and vertically dilated version of the graph of g. Also, if B turned out to be negative the graph would be vertically ﬂipped, too. Why is that useful? It means that the graph of a linear-to-linear ra- tional function can only take one of two forms. Either it looks like this: Or like this: We can sketch an accurate graph of a linear-to-linear rational function by sketching the asymptotes and then sketching in just one point on the graph. That will be enough information to nail down a decent sketch. We can always plot more points to give us more precision, but one point is enough to capture the essence of the graph. Given a linear-to-linear function that we wish to graph, we must ﬁrst determine the asymptotes. There will be one horizontal asymptote and one vertical asymptote. The vertical asymptote will be a vertical line with equation x = k where k is the one x value which is not in the domain of the function. That is, ﬁnd the value of x which makes the denominator zero and that will tell you the vertical asymptote. The horizontal asymptote is a little more involved. However, we can quickly get to a shortcut. The essence of a horizontal asymptote is that it describes what value the function is approximately equal to for very large values of x. To study what a linear-to-linear function is like when x is very large, we can perform the following algebraic manipulation: f(x) = ax + b cx + d = ax + b cx + d · 1/x 1/ While dividing by x is troublesome if x equals zero, here we are assuming x is very large, so it is certainly not zero. Now, consider this last expression in the above equation. If x is very large, then b x ≈ 0 184 (where ≈ CHAPTER 14. RATIONAL FUNCTIONS means ”is approximately”) and likewise d x ≈ 0. Hence, when x is very large, f(x . ≈ We can interpret this by saying that when x is very large, the function f(x) is is close to a constant, and that constant is a c . Thus, the horizontal asymptote of f(x) = is the horizontal line y = a c . ax + b cx + d Example 14.0.1. Sketch the graph of the function f(x) = 3x−1 2x+7 . Solution. We begin by ﬁnding the asymptotes of f. The denominator is equal to zero when 2x + 7 = 0, i.e., when x = −7/2. As a result, the vertical asymptote for this function is the vertical line x = −7/2. By taking the ratio of the coefﬁcients of x in the numerator and denominator, we can ﬁnd that the horizontal asymptote is the horizontal line y = 3 2 . -10 -5 10 5 0 0 -5 -10 5 10 Figure 14.2: The graph of f(x) = 3x−1 2x+7 . We then sketch these two asymptotes. The last thing we need is a single point. For instance, we may evaluate f(0): f(0) = −1 7 14.1. MODELINGWITHLINEAR-TO-LINEARRATIONALFUNCTIONS185 and so the point (0, −1/7) is on the graph. With this information, we know that the curve lies below the horizontal asymptote to the right of the vertical asymptote, and consequently the curve lies above the horizontal asymptote to the left of the vertical asymptote. We graph the result in Figure 14.2. 14.1 Modeling with Linear-to-linear Rational Functions As we have done with other sorts of functions, such as linear and quadratic, we can also model using linear-to-linear rational functions. One reason for using this type of function is their asymptotic nature. Many changing quantities in the world continually increase or decrease, but with bounds on how large or small they can get. For instance, a population may steadily decrease, but a population can never be negative. Conversely, a population may steadily increase, but due to environmental and other factors we may hypothesize that the population will always stay below some upper bound. As a result, the population may ”level off”. This leveling off behavior is exempliﬁed by the asymptotic nature of the linear-to-linear rational functions, and so this type of function provides a way to model such behavior. Given any linear-to-linear rational function, we can always divide the numerator and the denominator by the coefﬁcient of x in the denominator. In this way, we can always assume that the coefﬁcient of x in the denominator of a function we seek is 1. This is illustrated in the next example. Example 14.1.1. Let f(x) = 2x+3 5x−7. Then f(x) = = 1 5 1 5 2x + 3 5x − 7 · 5x + 3 2 5 x − 7 5 . Thus, in general, when we seek a linear-to-linear rational function, we will be looking for a function of the form f(x) = ax + b x + c and thus there are three parameters we need to determine. Note that for a function of this form, the horizontal asymptote is y = a and the vertical asymptote is x = −c. Since these functions have three parameters (i.e., a, b and c), we will need three pieces of information to nail down the function. 186 CHAPTER 14. RATIONAL FUNCTIONS There are essentially three types of modeling problems that require the determination of a linear-to-linear function. The three types are based on the kind of information given about the function. The three types are: 1. You know three points the the graph of the function passes through; 2. You know one of the function’s asymptotes and two points the graph passes through; 3. You know both asymptotes and one point the graph passes through. Notice that in all cases you know three pieces of information. Since a linear-to-linear function is determined by three parameters, this is exactly the amount of information needed to determine the function. The worst case, in terms of the amount of algebra you need to do, is the ﬁrst case. Let’s look at an example of the algebra involved with this sort. Example 14.1.2. Find the linear-to-linear rational function f(x) such that f(10) = 20, f(20) = 32 and f(25) = 36. Solution. Since f(x) is a linear-to-linear rational function, we know f(x) = ax + b x + c for constants a, b, and c. We need to ﬁnd a, b and c. We know three things. First, f(10) = 20. So f(10) = 10a + b 10 + c = 20, which we can rewrite as 10a + b = 200 + 20c. Second, f(20) = 32. So f(20) = 20a + b 20 + c = 32, which we can rewrite as 20a + b = 640 + 32c. Third, f(25) = 36. So f(25) = 25a + b 25 + c = 36, (14.1) (14.2) 14.1. MODELINGWITHLINEAR-TO-LINEARRATIONALFUNCTIONS187 which we can rewrite as 25a + b = 900 + 36c. (14.3) These three numbered equations are enough algebraic material to solve for a, b, and c. Here is one way to do that. Subtract equation 14.1 from equation 14.2 to get 10a = 440 + 12c and subtract equation 14.2 from equation 14.3 to get 5a = 260 + 4c (14.4) (14.5) Note that we’ve eliminated b. Now multiply this last equation by 2 to get 10a = 520 + 8c Subtract equation 14.4 from this to get 0 = 80 − 4c which easily give us c = 20. Plugging this value into equation 14.4, we can ﬁnd a = 68, and then we can ﬁnd b = −80. Thus, f(x) = 68x − 80 x + 20 . We can check that we have done the algebra correctly by evaluating If we get f(10) = 20, f(20) = 32 and f(x) at x = 10, x = 20 and x = 25. f(25) = 36, then we’ll know our work is correct. Algebraically, this was the worst situation of the three, since it reIf, instead of knowing three points, we know quired the most algebra. one or both of the asymptotes, then we can easily ﬁnd a and/or c, and so cut down on the amount of algebra needed. However, the method is essentially identical. Let’s now apply these ideas to a real world problem. Example 14.1.3. Clyde makes extra money selling tickets in front of the Safeco Field. The amount he charges for a ticket depends on how many he has. If he only has one ticket, he charges $100 for it. If he has 10 tickets, he charges $80 a piece. But if he has a large number of tickets, he will sell them for $50 each. How much will he charge for a ticket if he holds 20 tickets? 188 CHAPTER 14. RATIONAL FUNCTIONS Solution. We want to give a linear-to-linear rational function relating the price of a ticket y to the number of tickets x that Clyde is holding. As we saw above, we can assume the function is of the form y = ax + b x + c where a, b and c are numbers. Note that y = a is the horizontal asymptote. When x is very large, y is close to 50. This means the line y = 50 is a horizontal asymptote. Thus a = 50 and y = 50x + b x + c . Next we plug in the point (1,100) to get a linear equation in b and c. 50 1 + b 100 = · 1 + c (1 + c) = 50 + b 50 = b − 100c 100 · Similarly, plugging in (10,80) and doing a little algebra (do it now!) gives another linear equation 300 = b − 80c. Solving these two linear equations simultaneously gives c = 12.5 and b = 1300. Thus our function is y = 50x + 1300 x + 12.5 and, if Clyde holds 20 tickets, he will charge y = 50 20 + 1300 · 20 + 12.5 = $70.77 per ticket. 14.2 Summary • Every linear-to-linear rational function has a graph which is a shifted, scaled version of the curve y = 1/x. As a result, they have one vertical asymptote, and one horizontal asymptote. • Every linear-to-linear rational function f can be expressed in the form f(x) = ax + b x + c . This function has horizontal asymptote y = a and vertical asymptote x = −c. 14.3. EXERCISES 14.3 Exercises Problem 14.1. Give the domain of each of the following functions. Find the x- and yintercepts of each function.
Sketch a graph and indicate any vertical or horizontal asymptotes. Give equations for the asymptotes. (a) f(x) = 2x x−1 (c) h(x) = x+1 x−2 (e) k(x) = 8x+16 5x− 1 2 (b) g(x) = 3x+2 2x−5 (d) j(x) = 4x−12 x+8 (f) m(x) = 9x+24 35x−100 Problem 14.2. Oscar is hunting magnetic ﬁelds with his gauss meter, a device for measuring the strength and polarity of magnetic ﬁelds. The reading on the meter will increase as Oscar gets closer to a magnet. Oscar is in a long hallway at the end of which is a room containing an extremely strong magnet. When he is far down the hallway from the room, the meter reads a level of 0.2. He then walks down the hallway and enters the room. When he has gone 6 feet into the room, the meter reads 2.3. Eight feet into the room, the meter reads 4.4. (a) Give a linear-to-linear rational model relating the meter reading y to how many feet x Oscar has gone into the room. (b) How far must he go for the meter to reach 10? 100? (c) Considering your function from part (a) and the results of part (b), how far into the room do you think the magnet is? Problem 14.3. In 1975 I bought an old MarIn 1995 a similar uke tin ukulele for $200. was selling for $900. In 1980 I bought a new Kamaka uke for $100. In 1990 I sold it for $400. (a) Give a linear model relating the price p of the Martin uke to the year t. Take t = 0 in 1975. (b) Give a linear model relating the price q of the Kamaka uke to the year t. Again take t = 0 in 1975. (c) When is the value of the Martin twice the value of the Kamaka? (d) Give a function f(t) which gives the ratio of the price of the Martin to the price of the Kamaka. 189 (e) In the long run, what will be the ratio of the prices of the ukuleles? Problem 14.4. Isobel is producing and selling casette tapes of her rock band. When she had sold 10 tapes, her net proﬁt was $6. When she had sold 20 tapes, however, her net proﬁt had shrunk to $4 due to increased production expenses. But when she had sold 30 tapes, her net proﬁt had rebounded to $8. (a) Give a quadratic model relating Isobel’s net proﬁt y to the number of tapes sold x. (b) Divide the proﬁt function in part (a) by the number of tapes sold x to get a model relating average proﬁt w per tape to the number of tapes sold. (c) How many tapes must she sell in order to make $1.20 per tape in net proﬁt? Problem 14.5. Find the linear-to-linear function whose graph passes through the points (1,1), (5,2) and (20,3). What is its horizontal asymptote? Problem 14.6. Find the linear-to-linear function whose graph has y = 6 as a horizontal asymptote and passes through (0,10) and (3,7). Problem 14.7. The more you study for a certain exam, the better your performance on it. If you study for 10 hours, your score will be 65%. If you study for 20 hours, your score will be 95%. You can get as close as you want to a perfect score just by studying long enough. Assume your percentage score is a linear-tolinear function of the number of hours that you study. If you want a score of 80%, how long do you need to study? Problem 14.8. A street light is 10 feet above a straight bike path. Olav is bicycling down the path at a rate of 15 MPH. At midnight, Olav is 33 feet from the point on the bike path directly (See the picture). The below the street light. relationship between the intensity C of light (in candlepower) and the distance d (in feet) from the light source is given by C = k d2 , where k is a constant depending on the light source. 190 CHAPTER 14. RATIONAL FUNCTIONS (a) From 20 feet away, the street light has an intensity of 1 candle. What is k? (b) Find a function which gives the intensity of the light shining on Olav as a function of time, in seconds. (c) When will the light on Olav have maxi- mum intensity? (d) When will the intensity of the light be 2 candles? Problem 14.10. The number of customers in a local dive shop depends on the amount of money spent on advertising. If the shop spends nothing on advertising, there will be the shop spends 100 customers/day. $100, there will be 200 customers/day. As the amount spent on advertising increases, the number of customers/day increases and approaches (but never exceeds) 400 customers/day. If olav path 33ft 10ft Problem 14.9. For each of the following ﬁnd the linear to linear function f(x) satisfying the given requirements: (a) f(0) = 0, f(10) = 10, f(20) = 15 (b) f(0) = 10, f(5) = 4, f(20) = 3 (c) f(10) = 20, f(30) = 25, and the graph of f(x) has y = 30 as its horizontal asymptote (a) Find a linear to linear rational function y = f(x) that calculates the number y of customers/day if $x is spent on advertising. (b) How much must the shop spend on advertising to have 300 customers/day. (c) Sketch the graph of the function y = f(x) x on the domain 0 5000. ≤ ≤ (d) Find the rule, domain and range for the inverse function from part (c). Explain in words what the inverse function calculates. Chapter 15 Measuring an Angle So far, the equations we have studied have an algebraic character, involving the variables x and y, arithmetic operations and maybe extraction of roots. Restricting our attention to such equations would limit our ability to describe certain natural phenomena. An important example involves understanding motion around a circle, and it can be motivated by analyzing a very simple scenario: Cosmo the dog, tied by a 20 foot long tether to a post, begins walking around a circle. A number of very natural questions arise: Q S Cosmo motors around the circle P 20 feet R Figure 15.1: Cosmo the dog walking a circular path. Natural Questions 15.0.1. How can we measure the angles ∠SPR, ∠QPR, and ∠QPS? How can we measure the arc lengths arc(RS), arc(SQ) and arc(RQ)? How can we measure the rate Cosmo is moving around the circle? If we know how to measure angles, can we compute the coordinates of R, S, and Q? Turning this around, if we know how to compute the coordinates of R, S, and Q, can we then measure the angles ∠SPR, ∠QPR, and ∠QPS ? Finally, how can we specify the direction Cosmo is traveling? We will answer all of these questions and see how the theory which evolves can be applied to a variety of problems. The deﬁnition and basic properties of the circular functions will emerge as a central theme in this Chapter. The full problem-solving power of these functions will become apparent in our discussion of sinusoidal functions in Chapter 19. The xy-coordinate system is well equipped to study straight line motion between two locations. For problems of this sort, the important tool is the distance formula. However, as Cosmo has illustrated, not all two-dimensional motion is along a straight line. In this section, we will describe how to calculate length along a circular arc, which requires a quick review of angle measurement. 191 192 CHAPTER 15. MEASURING AN ANGLE 15.1 Standard and Central Angles An angle is the union of two rays emanating from a common point called the vertex of the angle. A typical angle can be dynamically generated by rotating a single ray from one position to another, sweeping counterclockwise or clockwise: See Figure 15.2. We often insert a curved arrow to indicate the direction in which we are sweeping out the angle. The ray ℓ1 is called the initial side and ℓ2 the terminal side of the angle ∠AOB. (terminal side) l2 B SWEEP CLOCKWISE O A l1 vertex O l1 (initial side) A vertex A O (initial side) l1 START SWEEP COUNTERCLOCKWISE (terminal side) B l2 Figure 15.2: Angle ∠AOB. Working with angles, we need to agree on a standard frame of reference for viewing them. Within the usual xy-coordinate system, imagine a model of ∠AOB in Figure 15.2 constructed from two pieces of rigid wire, welded at the vertex. Sliding this model around inside the xy-plane will not distort its shape, only its position relative to the coordinate axis. So, we can slide the angle into position so that the initial side is coincident with the positive x-axis and the vertex is the origin. Whenever we do this, we say the angle is in standard position. Once an angle is in standard position, we can construct a circle centered at the origin and view our standard angle as cutting out a particular “pie shaped wedge” of the corresponding disc. Notice, the shaded regions in Figure 15.3 depend on whether we sweep the angle counterclockwise or clockwise from the initial side. The portion of the “pie wedge” along the circle edge, which is an arc, is called the arc subtended by the angle. We can keep track of this arc using the notation arc(AB). A central angle is any angle with vertex at the center of a circle, but its initial side may or may not be the positive x-axis. For example, ∠QPS in the Figure beginning this Chapter is a central angle which is not in standard position. 15.2. AN ANALOGY 193 y-axis l2 B l2 y-axis B arc subtended arc(AB) arc(AB) vertex O x-axis l1 A vertex O x-axis l1 A COUNTERCLOCKWISE CLOCKWISE Figure 15.3: Standard angles and arcs. 15.2 An Analogy To measure the dimensions of a box you would use a ruler. In other words, you use an instrument (the ruler) as a standard against which you measure the box. The ruler would most likely be divided up into either English units (inches) or metric units (centimeters), so we could express the dimensions in a couple of different ways, depending on the units desired. By analogy, to measure the size of an angle, we need a standard against which any angle can be compared. In this section, we will describe two standards commonly used: the degree method and the radian method of angle measurement. The key idea is this: Beginning with a circular region, describe how to construct a “basic” pie shaped wedge whose interior angle becomes the standard unit of angle measurement. 15.3 Degree Method Begin by drawing a circle of radius r, call it Cr, centered at the origin. Divide this circle into 360 equal sized pie shaped wedges, beginning with the the point (r,0) on the circle; i.e. the place where the circle crosses the x-axis. We will refer to the arcs along the outside edge
s of these wedges as one-degree arcs. Why 360 equal sized arcs? The reason for doing so is historically tied to the fact that the ancient Babylonians did so as they developed their study of astronomy. (There is actually an alternate system based on dividing the circular region into 400 equal sized wedges.) Any central angle which subtends one of these 360 equal sized arcs is 194 CHAPTER 15. MEASURING AN ANGLE circle Cr typical wedge (r,0) r r this angle is DEFINED to have measure 1 degree a total of 360 equal sized pie shaped wedges inside this disk etc. etc. *NOT TO SCALE* Figure 15.4: Wedges as 1◦ arcs. deﬁned to have a measure of one degree, denoted 1◦. We can now use one-degree arcs to measure any angle: Begin by sliding the angle ∠AOB into standard central position, as in Figure 15.3. Piece together consecutive one-degree arcs in a counterclockwise or clockwise direction, beginning from the initial side and working toward the terminal side, approximating the angle ∠AOB to the nearest degree. If we are allowed to divide a one-degree arc into a fractional portion, then we could precisely determine the number m of one-degree arcs which consecutively ﬁt together into the given arc. If we are sweeping counterclockwise from the initial side of the angle, m is deﬁned to be the degree measure of the angle. If we sweep in a clockwise direction, then −m is deﬁned to be the degree measure of the angle. So, in Figure 15.3, the left-hand angle has positive degree measure while the right-hand angle has negative degree measure. Simple examples would be angles like the ones in Figure 15.5. Notice, with our conventions, the rays determining an angle with measure −135◦ sit inside the circle in the same position as those for an angle of measure 225◦; the minus sign keeps track of sweeping the positive x-axis clockwise (rather than counterclockwise). We can further divide a one-degree arc into 60 equal arcs, each called a one minute arc. Each one-minute arc can be further divided into 60 equal arcs, each called a one second arc. This then leads to angle measures of one minute, denoted 1 ′ and one second, denoted 1 ′′: 1◦ = 60 minutes = 3600 seconds. 15.3. DEGREE METHOD 195 90◦ 180◦ 270◦ 45◦ 315◦ −135◦ Figure 15.5: Examples of common angles. For example, an angle of measure θ = 5 degrees 23 minutes 18 seconds is usually denoted 5◦ 23 ′ 18 ′′. We could express this as a decimal of degrees: 5◦ 23 ′ 18 ′′ = 5 + 23 60 = 5.3883◦. In degrees! + 18 3600 ◦ ← As another example, suppose we have an angle with measure 75.456◦ and we wish to convert this into degree/minute/second units. First, since 75.456◦ = 75◦+0.456◦, we need to write 0.456◦ in minutes by the calculation: 0.456 degree 60 × minutes degree = 27.36 ′. This tells us that 75.456◦ = 75◦27.36 ′ = 75◦27 ′ + 0.36 ′. Now we need to write 0.36 ′ in seconds via the calculation: 0.36 minutes 60 seconds/minute = 21.6 ′′. × In other words, 75.456◦ = 75◦ 27 ′ 21.6 ′′. Degree measurement of an angle is very closely tied to direction in the plane, explaining its use in map navigation. With some additional work, it is also possible to relate degree measure and lengths of circular arcs. To do this carefully, ﬁrst go back to Figure 15.3 and recall the situation 196 CHAPTER 15. MEASURING AN ANGLE where an arc arc(AB) is subtended by the central angle ∠AOB. In this situation, the arc length of arc(AB), commonly denoted by the letter s, is the distance from A to B computed along the circular arc; keep in mind, this is NOT the same as the straight line distance between the points A and B. For example, consider the six angles pictured above, of measures 90◦, 180◦, 270◦, 45◦, −135◦, and 315◦. If the circle is of radius r and we want to compute the lengths of the arcs subtended by these six angles, then this can be done using the formula for the circumference of a circle (on the back of this text) and the following general principle: Important Fact 15.3.1. (length of a part) = (fraction of the part) (length of the whole) × For example, the circumference of the entire circle of radius r is 2πr; this is the “length of the whole” in the general principle. The arc subtended by a 90◦ angle is 90 4 of the entire circumference; this is the “fraction of the part” in the general principle. The boxed formula implies: 360 = 1 s = arc length of the 90◦ arc = 1 4 2πr = πr 2 . s = distance along the arc r θ degrees r Cr Similarly, a 180◦ angle subtends an arc of length s = πr, 2πr = 7πr a 315◦ angle subtends an arc of length s = 4 , etc. In general, we arrive at this formula: 315 360 Important Fact 15.3.2 (Arc length in degrees). Start with a central angle of measure θ degrees inside a circle of radius r. Then this angle will subtend an arc of length Figure 15.6: The deﬁnition of arc length. s = 2π 360 rθ 15.4 Radian Method The key to understanding degree measurement was the description of a “basic wedge” which contained an interior angle of measure 1◦; this was straightforward and familiar to all of us. Once this was done, we could proceed to measure any angle in degrees and compute arc lengths as in Fact 15.3.2. However, the formula for the length of an arc subtended by an angle measured in degrees is sort of cumbersome, involving the curious factor 2π 360. Our next goal is to introduce an alternate angle measurement scheme called radian measure that begins with a different “basic wedge”. As will become apparent, a big selling point of radian measure is that arc length calculations become easy. 15.4. RADIAN METHOD 197 r r equilateral wedge r (r,0) circle Cr r r r this angle is DEFINED to have measure 1 radian Figure 15.7: Constructing an equilateral wedge. As before, begin with a circle Cr of radius r. Construct an equilateral wedge with all three sides of equal length r; see Figure 15.7. We deﬁne the measure of the interior angle of this wedge to be 1 radian. Once we have deﬁned an angle of measure 1 radian, we can deﬁne an angle of measure 2 radians by putting together two equilateral wedges. Likewise, an angle of measure 1 2 radian is obtained by symmetrically dividing an equilateral wedge in half, etc. Reasoning in this way, we can piece together equilateral wedges or fractions of such to compute the radian measure of any angle. It is important to notice an important relationship between the radian measure of an angle and arc length calculations. In the ﬁve angles pictured above, 1 radian, 2 radian, 3 radian, 1 2 radian and 1 4 radian, the length of the arcs 2 r, and 1 subtended by these angles θ are r, 2r, 3r, 1 4r. In other words, a pattern emerges that gives a very simple relationship between the length s of an arc and the radian measure of the subtended angle: Important Fact 15.4.1 (Arc length in radians). Start with a central angle of measure θ radians inside a circle of radius r. Then this angle will subtend an arc of length s = θr. These remarks allow us to summarize the deﬁnition of the radian measure θ of ∠AOB inside a circle of radius r by the formula: s = distance along the arc r θ radians r Cr θ = s r − s r if angle is swept counterclockwise if angle is swept clockwise Figure 15.9: Deﬁning arc length when angles are measured in radians. 198 CHAPTER 15. MEASURING AN ANGLE r r 1 radian = θ r 2r 3r r 2 radians = θ r r r 1 2 radian = θ r 2 3 radians = θ r r 1 4 radian = θ r r r 4 Figure 15.8: Measuring angles in radians. y-axis length of arc(AB) = s B θ radian O A x-axis The units of θ are sometimes abbreviated as rad. It is important to appreciate the role of the radius of the circle Cr when using radian measure of an angle: An angle of radian measure θ will subtend an arc of length \|θ\| on the unit circle. In other words, radian measure of angles is exactly the same as arc length on the unit circle; we couldn’t hope for a better connection! Figure 15.10: Arc length after imposing a coordinate system. circle radius r The difﬁculty with radian measure versus degree measure is really one of familiarity. Let’s view a few common angles in radian measure. It is easiest to start with the case of angles in central standard position within the unit circle. Examples of basic angles would be fractional parts of one complete revolution around the unit circle; for example, 1 12 revolution, 1 4 revolution. One revolution around the unit circle describes an arc of length 2π and so the subtended angle (1 revolution) is 2π radians. We can now easily ﬁnd the radian measure of these six angles. For example, 1 12 revolution would describe an angle of measure ( 1 6 rad. Similarly, the other ﬁve angles pictured below have measures π 2 rad, π rad and 3π 2 rad. 2 revolution and 3 8 revolution, 1 4 revolution, 1 6 revolution, 1 12)2π rad= π 3 rad, π 4 rad, π All of these examples have positive radian measure. For an angle with 2 radians, we would locate B 4 revolution clockwise, etc. From these calculations and our negative radian measure, such as θ = − π by rotating 1 15.5. AREAS OF WEDGES 199 eplacements π 6 π 4 π 3 1 12 revolution 1 8 revolution 1 6 revolution π 2 π 3π 2 1 4 revolution 1 2 revolution 3 4 revolution Figure 15.11: Common angles measured in radians. previous examples of degree measure we ﬁnd that 180 degrees = π radians. (15.1) Solving this equation for degrees or radians will provide conversion formulas relating the two types of angle measurement. The formula also helps explain the origin of the curious conversion factor π 180 = 2π in Fact 15.3.2. 360 15.5 Areas of Wedges The beauty of radian measure is that it is rigged so that we can easily compute lengths of arcs and areas of circular sectors (i.e. “pie-shaped regions”). This is a key reason why we will almost always prefer to work with radian measure. Example 15.5.1. If a 16 inch pizza is cut into 12 equal slices, what is the area of a single slice? This can be solved using a general principle: (Area of a part) = (area of the whole) (fraction of the part) × 200 CHAPTER 15. MEASURING AN ANGLE So, for our pizza: (area one slice) = (area whole p
ie) 1 12 = (82π) = 16π 3 . (fraction of pie) × B Rθ θ O A Cr Figure 15.12: Finding the area of a “pie shaped wedge”. Let’s apply the same reasoning to ﬁnd the area of a circular sector. We know the area of the circular disc bounded by a circle of radius r is πr2. Let Rθ be the “pie shaped wedge” cut out by an angle ∠AOB with positive measure θ radians. Using the above principle area(Rθ) = (area of disc bounded by Cr) (portion of disc accounted for by Rθ) × = (πr2) θ 2π = 1 2 r2θ. For example, if r = 3 in. and θ = π the pie shaped wedge is 9 8π sq. in. 4 rad, then the area of Important Fact 15.5.2 (Wedge area). Start with a central angle with positive measure θ radians inside a circle of radius r. The area of the “pie shaped region” bounded by the angle is 1 2r2θ. Example 15.5.3. A water drip irrigation arm 1200 feet long rotates around a pivot P once every 12 hours. How much area is covered by the arm in one hour? in 37 minutes? How much time is required to drip irrigate 1000 square feet? Solution. The irrigation arm will complete one revolution in 12 hours. The angle swept out by one complete revolution is 2π radians, so after t hours the arm sweeps out an angle θ(t) given by θ(t) = 2π radians 12 hours × t hours = π 6 t radians. Consequently, by Important Fact 15.5.2 the area A(t) of the irrigated region after t hours is A(t) = 1 2 (1200)2θ(t) = 1 2 (1200)2 π 6 t = 120,000πt square feet. After 1 hour, the irrigated area is A(1) = 120,000π = 376,991 sq. ft. Likewise, after 37 minutes, which is 37 60 hours, the area of the irrigated region is A( 37 60) = 232,500 square feet. To answer the ﬁnal question, we need to solve the equation A(t) = 1000; i.e., 120,000πt = 1000, so 60) = 120,000π( 37 t = 1 120π hours × 3600 seconds hour = 9.55 seconds. 15.5. AREAS OF WEDGES 201 15.5.1 Chord Approximation Our ability to compute arc lengths can be used as an estimating tool for distances between two points. Let’s return to the situation posed at the beginning of this section: Cosmo the dog, tied by a 20 foot long tether to a post in the ground, begins at location R and walks counterclockwise to location S. Furthermore, let’s suppose you are standing at the center of the circle determined by the tether and you measure the angle from R to S to be 5◦; see the left-hand ﬁgure. Because the angle is small, notice that the straight line distance d from R to S is approximately the same as the arc length s subtended by the angle ∠RPS; the right-hand picture in Figure 15.13 is a blow-up: 5◦ P 20 feet S R S s d 5◦ P 20 feet R Figure 15.13: Using the arc length s to approximate the chord d. Example 15.5.4. Estimate the distance from R to S. Solution. We ﬁrst convert the angle into radian measure via (15.1): 5◦ = 0.0873 radians. By Fact 15.3.2, the arc s has length 1.745 feet = 20.94 inches. This is approximately equal to the distance from R to S, since the angle is small. 202 CHAPTER 15. MEASURING AN ANGLE S chord s R O We call a line segment connecting two points on a circle a chord of the circle. The above example illustrates a general principal for approximating the length of any chord. A smaller angle will improve the accuracy of the arc length approximation. Important Fact 15.5.5 (Chord Approximation). In Figure 15.14, if the central angle is small, then s \|RS\|. ≈ Figure 15.14: Chord approximation. 15.6 Great Circle Navigation A basic problem is to ﬁnd the shortest route between any two locations on the earth. We will review how to coordinatize the surface of the earth and recall the fact that the shortest path between two points is measured along a great circle. View the earth as a sphere of radius r = 3,960 miles. We could slice P0 which is both perpendicular the earth with a two-dimensional plane to a line connecting the North and South poles and passes through the center of the earth. Of course, the resulting intersection will trace out a circle of radius r = 3,960 miles on the surface of the earth, which we call the equator. We call the plane P0 the equatorial plane. Slicing the earth with any other plane P0, we can consider the right triangle pictured below and the angle θ: parallel to P North Pole line of latitude North Pole r equatorial plane P0 equator b r θ r equator b θ r 90◦ − θ◦ South Pole center of earth center of earth South Pole Figure 15.15: Measuring latitude. Essentially two cases arise, depending on whether or not the plane P slices the surface of is above or below the equatorial plane. The plane the earth in a circle, which we call a line of latitude. This terminology is somewhat incorrect, since these lines of latitude are actually circles on the surface of the earth, but the terminology is by now standard. Depending on whether this line of latitude lies above or below the equatorial plane, we refer to it as the θ◦ North line of latitude (denoted θ◦ N) or the θ◦ South line of latitude (denoted θ◦ S). Notice, the radius b of a line of latitude can vary from a maximum of 3,960 miles (in the case of θ = 0◦), P 15.6. GREAT CIRCLE NAVIGATION 203 to a minimum of 0 miles, (when θ = 90◦). When b = 0, we are at the North or South poles on the earth. In a similar spirit, we could imagine slicing the earth with a plane Q which is perpendicular to the equatorial plane and passes through the center of the earth. The resulting intersection will trace out a circle of radius 3,960 miles on the surface of the earth, which is called a line of longitude. Half of a line of longitude from the North Pole to the South Pole is called a meridian. We distinguish one such meridian; the one which passes through Greenwich, England as the Greenwich meridian. Longitudes are measured using angles East or West of Greenwich. Pictured below, the longitude of A is θ. Because θ is east of Greenwich, θ measures longitude East, typically written θ◦ E; west longitudes would be denoted as θ◦ W. All longitudes are between 0◦ and 180◦. The meridian which is 180◦ West (and 180◦ E) is called the International Date Line. Introducing the grid of latitude and longitude lines on the earth amounts to imposing a coordinate system. In other words, any position on the earth can be determined by providing the longitude and latitude of the point. The usual convention is to list longitude ﬁrst. For example, Seattle has coordinates 122.0333◦ W, 47.6◦ N. Since the labels “N and S” are attached to latitudes and the labels “E and W” are attached to longitudes, there is no ambiguity here. This means that Seattle is on the line of longitude 122.0333◦ West of the Greenwich meridian and on the line of latitude 47.6◦ North of the equator. In the ﬁgure below, we indicate the key angles ψ = 47.6◦ and θ = 122.0333◦ by inserting the three indicated radial line segments. Greenwich meridian center of earth θ r A r Greenwich, England North Pole line of longitude International Date Line Equator South Pole Figure 15.16: The International Date Line. Seattle, WA N great circles not great circle Ψ r θ S Greenwich Meridian not great circle Figure 15.17: Distances along great circles. 204 CHAPTER 15. MEASURING AN ANGLE Now that we have imposed a coordinate system on the earth, it is natural to study the distance between two locations. A great circle of a sphere is deﬁned to be a circle lying on the sphere with the same center as the sphere. For example, the equator and any line of longitude are great circles. However, lines of latitude are not great circles (except the special case of the equator). Great circles are very important because they are used to ﬁnd the shortest distance between two points on the earth. The important fact from geometry is summarized below. Important Fact 15.6.1 (Great Circles). The shortest distance between two points on the earth is measured along a great circle connecting them. Example 15.6.2. What is the shortest distance from the North Pole to Seattle, WA ? W E N O S equator Greenwich Meridian Figure 15.18: Distance between the North Pole and Seattle, Washington. Solution. The line of longitude 122.0333◦ W is a great circle connecting the North Pole and Seattle. So, the shortest distance will be the arc length s subtended by the angle ∠NOW pictured in Figure 15.18. Since the latitude of Seattle is 47.6◦, the angle ∠EOW has measure 47.6◦. Since ∠EON is a right angle (i.e., 90◦), ∠NOW has measure 42.4◦. By Fact 15.4.1 and Equation 15.1, s = (3960 miles)(42.4◦)(0.01745 radians/degree) = 2943.7 miles, which is the shortest distance from the pole to Seattle. 15.7 Summary 360◦ = 2π radians A circular arc with radius r and angle θ has length s, with s = rθ when θ is measured in radians. A circular wedge with radius r and angle θ has area A, with • • • A = 1 2 r2θ when θ is measured in radians. 15.8. EXERCISES 15.8 Exercises Problem 15.1. Let ∠AOB be an angle of measure θ. (a) Convert θ = 13.4o into degrees/ min- utes/ seconds and into radians. (b) Convert θ = 1o4 ′44 ′′ into degrees and ra- dians. (c) Convert θ = 0.1 radian into degrees and degrees/ minutes/ seconds. Problem 15.2. A nautical mile is a unit of distance frequently used in ocean navigation. It is deﬁned as the length of an arc s along a great circle on the earth when the subtending angle has measure 1 ′ = “one minute” = 1/60 of one degree. Assume the radius of the earth is 3,960 miles. (a) Find the length of one nautical mile to the nearest 10 feet. (b) A vessel which travels one nautical mile in one hours time is said to have the speed of one knot; this is the usual navigational measure of speed. If a vessel is traveling 26 knots, what is the speed in mph (miles per hour)? (c) If a vessel is traveling 18 mph, what is the speed in knots? Problem 15.3. The rear window wiper blade on a station wagon has a length of 16 inches. The wiper blade is mounted on a 22 inch arm, 6 inches from the pivot point. 6" 16" 205 (c) Suppose bug A lands on the end of the blade farthest from the pivot. Assume the wiper turns through an angle of 110◦. In one cycle (back and forth) of the wiper blade, how far has the bug
traveled? (d) Suppose bug B lands on the end of the wiper blade closest to the pivot. Assume the wiper turns through an angle of 110◦. In one cycle of the wiper blade, how far has the bug traveled? (e) Suppose bug C lands on an intermediate location of the wiper blade. Assume the wiper turns through an angle of 110◦. If bug C travels 28 inches after one cycle of the wiper blade, determine the location of bug C on the wiper blade. Problem 15.4. A water treatment facility operates by dripping water from a 60 foot long arm whose end is mounted to a central pivot. The water then ﬁlters through a layer of charcoal. The arm rotates once every 8 minutes. (a) Find the area of charcoal covered with water after 1 minute. (b) Find the area of charcoal covered with water after 1 second. (c) How long would it take to cover 100 square feet of charcoal with water? (d) How long would it take to cover 3245 square feet of charcoal with water? (a) If the wiper turns through an angle of 110◦, how much area is swept clean? (b) Through how much of an angle would the wiper sweep if the area cleaned was 10 square inches? Problem 15.5. Astronomical measurements are often made by computing the small angle formed by the extremities of a distant object and using the estimating technique in 15.5.1. In the picture below, the full moon is shown to form an angle of 1 when the distance indi2 cated is 248,000 miles. Estimate the diameter of the moon. o 206 CHAPTER 15. MEASURING AN ANGLE moon (a) θ = 45◦ (b) θ = 80o (c) θ = 3 radians (d) θ = 2.46 radians (e) θ = 97o23 ′3 ′′ (f) θ = 35o24 ′2 ′′ 248,000 miles o 1/2 earth Problem 15.6. An aircraft is ﬂying at the speed of 500 mph at an elevation of 10 miles above the earth, beginning at the North pole and heading South along the Greenwich meridian. A spy satellite is orbiting the earth at an elevation of 4800 miles above the earth in a circular orbit in the same plane as the Greenwich meridian. Miraculously, the plane and satellite always lie on the same radial line from the center of the earth. Assume the radius of the earth is 3960 miles. satellite plane earth Problem 15.8. Matilda is planning a walk around the perimeter of Wedge Park, which is shaped like a circular wedge, as shown below. The walk around the park is 2.1 miles, and the park has an area of 0.25 square miles. If θ is less than 90 degrees, what is the value of the radius, r? θ r r Problem 15.9. Let C6 be the circle of radius 6 inches centered at the origin in the xycoordinate system. Compute the areas of the shaded regions in the picture below; the inner circle in the rightmost picture is the unit circle: y=x y=−x (a) When is the plane directly over a location with latitude 74◦30 ′18 ′′ N for the ﬁrst time? (b) How fast is the satellite moving? (c) When is the plane directly over the equa- tor and how far has it traveled? (d) How far has the satellite traveled when the plane is directly over the equator? Problem 15.7. Find the area of the sector of a circle of radius 11 inches if the measure θ of a central angle of this sector is: C 6 C 6 y=x y=−(1/4)x + 2 C 6 Chapter 16 Measuring Circular Motion IfCosmobeginsatlocation R andwalkscounterclockwise, always maintaining a tight tether, how can we measure Cosmo’s speed? This is a “dynamic question” and requires that we disIn contrast, if cuss ways of measuring circular motion. we take a snapshot and ask to measure the speciﬁc angle ∠RPS, this is a “static question”, which we answered in the previous section. Cosmo moves counterclockwise, maintaining S a tight tether. R P 20 feet Figure 16.1: How fast is Cosmo moving? 16.1 Different ways to measure Cosmo’s speed If Cosmo starts at location R and arrives at location S after some amount of time, we could study ω = measure ∠RPS time required to go from R to S . ”, “ degrees minute degrees second The funny Greek letter “ω” on the left of side of the equation is pronounced “oh-meg-a”. We will refer to this as an angular speed. Typical radians ”, etc. For example, if the angle units are “ minute swept out by Cosmo after 8 seconds is 40◦, then Cosmo’s angular speed 5◦ = is . Using (15.1), we can convert to radian units and get sec ω = π sec = 5 3π rad min . This is a new example of a rate and we can ask to 36 ﬁnd the total change, in the spirit of (1.2). If we are given ω in units of “ ”, we have 40◦ 8 seconds ” or “ ”, “ rad rad time deg time θ = ωt, which computes the measure of the angle θ swept out after time t (i.e. the total change in the angle). Angular speed places emphasis upon the 207 208 CHAPTER 16. MEASURING CIRCULAR MOTION “size of the angle being swept out per unit time” by the moving object, starting from some initial position. We need to somehow indicate the direction in which the angle is being swept out. This can be done by indicating “clockwise” our “counterclockwise”. Alternatively, we can adopt the convention that the positive rotational direction is counterclockwise, then insert a minus sign to indicate rotation clockwise. For example, saying that Cosmo is moving at an angular speed of ω = − π rad sec means he 2 is moving clockwise π 2 rad sec . Another way to study the rate of a circular motion is to count the number of complete circuits of the circle per unit time. This sort of rate has the form Number of Revolutions Unit of Time ; If we take “minutes” to be we will also view this as an angular speed. the preferred unit of time, we arrive at the common measurement called revolutionsper minute, usually denoted RPM or rev/min. For example, if Cosmo completes one trip around the circle every 2 minutes, then Cosmo is moving at a rate of 1 2 RPM. If instead, Cosmo completes one trip around the circle every 12 seconds, then we could ﬁrst express Cosmo’s speed in units of revolutions/second as 1 12 rev/second, then convert to RPM units: 1 12 rev sec 60 sec min = 5 RPM. As a variation, if we measure that Cosmo completed 3 2 minutes, then Cosmo’s angular speed is computed by 7 of a revolution in 3 7rev 2 min = 3 14 RPM. The only possible ambiguity involves the direction of revolution: the object can move clockwise or counterclockwise. The one shortcoming of using angular speed is that we are not directly keeping track of the distance the object is traveling. This is fairly easy to remedy. Returning to Figure 16.1, the circumference of the circle of motion is 2π(20) = 40π feet. This is the distance traveled per revolution, so we can now make conversions of angular speed into “distance traveled per unit time”; this is called the linear speed. 2 RPM, then he has a linear speed of If Cosmo is moving 1 v = 1 rev 2 min 40π ft rev = 20π ft min . Likewise, if Cosmo is moving π 7 rad sec , then v = π rad 7 sec 1 rev 2π rad 40π ft rev = 20π 7 ft sec . 16.2. DIFFERENT WAYS TO MEASURE CIRCULAR MOTION 209 Important Fact 16.1.1. This discussion is an example of what is usually called “units analysis”. The key idea we have illustrated is how to convert between two different types of units: rev min converts to − ft min 16.2 Different Ways to Measure → Circular Motion The discussion of Cosmo applies to circular motion of any object. As a matter of convention, we usually use the Greek letter ω to denote angular speed and v for linear speed. If an object is moving around a circle of radius r at a constant rate, then we can measure it’s speed in two ways: • • The angular speed ω = “revolutions” “unit time” or “degrees swept” “unit time” or “radians swept” “unit time” . The linear speed v = “distance traveled” “per unit time” . Important Facts 16.2.1 (Measuring and converting). We can convert between angular and linear speeds using these facts: 1 revolution = 360◦ = 2π radians; The circumference of a circle of radius r units is 2πr units. • • 16.2.1 Three Key Formulas If an object begins moving around a circle, there are a number of quantities we can try to relate. Some of these are “static quantities”: Take a visual “snapshot” of the situation after a certain amount of time has elapsed, then we can measure the radius, angle swept, arc length and time elapsed. Other quantities of interest are “dynamic quantities”: This means something is CHANGING with respect to time; in our case, the linear speed (which measures distance traveled per unit time) and angular speed (which measures angle swept per unit time) fall into this category. 210 CHAPTER 16. MEASURING CIRCULAR MOTION STATIC QUANTITIES DYNAMIC QUANTITIES S 1. arc length s S 5. angular speed ω 2. angle swept in time r 6. linear speed v θ P R 3. radius r 4. elapsed time t P R ...take a “snapshot” after time t... ...see what happens per unit time... Figure 16.2: Measuring linear and angular speed. We now know two general relationships for circular motion: (i) s = rθ, where s=arclength (a linear distance), r=radius of the circular path and θ=angle swept in RADIAN measure; this was the content of Fact 15.4.1 on page 197. (ii) θ = ωt, where θ is the measure of an angle swept, ω= angular speed and t represents time elapsed. This is really just a consequence of units manipulation. If r=20 feet and θ = 1.3 Notice how the units work in these formulas. radians, then the arc length s = 20(1.3) feet= 26 feet; this is the length If ω = 3 of the arc of radius 20 feet that is subtending the angle θ. rad/second and t = 5 seconds, then θ = 3 rad 5 seconds = 15 radians. If we replace “θ” in s = rθ of (i) with θ = ωt in (ii), then we get seconds × s = rωt. This gives us a relationship between arclength s (a distance) and time t. Plug in the fact that the linear speed is deﬁned to be v = “distance” and we get t v = s t = rωt t = rω. All of these observations are summarized below. Important Facts 16.2.2 (Three really useful formulas). If we measure angles θ in RADIANS and ω in units of radians per unit time, we have these three formulas: s = rθ θ = ωt v = rω (16.1) (16.2) (16.3) 16.2. DIFFERENT WAYS TO MEASURE CIRCULAR MOTION 211 Example 16.2.3. You are riding a stationary exercise bike and the speedometer reads a st
eady speed of 40 MPH (miles per hour). If the rear wheel is 28 inches in diameter, determine the angular speed of a location on the rear tire. A pebble becomes stuck to the tread of the rear tire. Describe the location of the pebble after 1 second and 0.1 second. * pebble sticks to tread here Figure 16.3: Where is the pebble after t seconds? Solution. The tires will be rotating in a counterclockwise direction and the radius r = 1 228 = 14 inches. The other given quantity, “40 MPH”, involves miles, so we need to decide which common units to work with. Either will work, but since the problem is focused on the wheel, we will utilize inches. If the speedometer reads 40 MPH, this is the linear speed of a speciﬁed location on the rear tire. We need to convert this into an angular speed, using unit conversion formulas. First, the linear speed of the wheel is v = 40 = 704 miles hr in sec . 5280 ft mile in ft 12 1 hr 60 min 1 min 60 sec Now, the angular speed ω of the wheel will be ω = 704 inches second 2(14)π inches revolution revolution second = 8 = 480 RPM It is then an easy matter to convert this to ω = 8 = 2,880 revolution second degrees second . 360 degrees revolution If the pebble begins at the “6 o’clock” position (the place the tire touches the ground on the wheel), then after 1 second the pebble will go through 8 revolutions, so will be in the “6 o’clock” position again. After 0.1 seconds, the pebble will go through rev sec 8 (0.1 sec) = 0.8 rev = (0.8 rev) 360 = 288◦. deg rev 212 CHAPTER 16. MEASURING CIRCULAR MOTION Keeping in mind that the rotation is counterclockwise, we can view the location of the pebble after 0.1 seconds as pictured below: 288◦ wise rotation counterclock- after 0.1 second * located here at time = 0.1 sec starts here * pebble sticks to tread in 6 o’clock position Figure 16.4: Computing the pebble’s position after t = 0.1 sec. We solved the previous problem using the “unit conversion method”. There is an alternate approach available, which uses one of the formulas in Fact 16.2.2. Here is how you could proceed: First, as above, we know the linear speed is v = 704 in/sec. Using the “v = ωr” formula, we have 704 in sec = ω(14 in) ω = 50.28 rad sec . Notice how the units worked out in the calculation: the “time” unit comes from v and the “angular” unit will always be radians. As a comparison with the solution above, we can convert ω into RPM units: rad sec 1 rev 2π rad ω = 50.28 = 8 rev sec . All of the problems in this section can be worked using either the “unit conversion method” or the “v = ωr method”. 16.3 Music Listening Technology The technology of reproducing music has gone through a revolution since the early 1980’s. The “old” stereo long playing record (the LP) and the 16.3. MUSIC LISTENING TECHNOLOGY 213 “new” digital compact disc (the CD) are two methods of storing musical data for later reproduction in a home stereo system. These two technologies adopt different perspectives as to which notion of circular speed is best to work with. Long playing stereo records are thin vinyl plastic discs of radius 6 inches onto which small spiral grooves are etched into the surface; we can approximately view this groove as a circle. The LP is placed on a ﬂat 12 inch diameter platter which turns at a constant angular speed of 33 1 3 RPM. An arm on a pivot (called the tone arm) has a needle mounted on the end (called the cartridge), which is placed in the groove on the outside edge of the record. Because the grooves wobble microscopically from side-to-side, the needle will mimic this motion. In turn, this sets a magnet (mounted on the opposite end of the needle) into motion. This moving magnet sits inside a coil of wire, causing a small varying voltage; the electric signal is then fed to your stereo, ampliﬁed and passed onto your speakers, reproducing music! LP turning at 33 1 3 RPM amp speakers tonearm needle Figure 16.5: Reproducing music using analogue technology. This is known as analogue technology and is based upon the idea of maintaining a constant angular speed of 33 1 3 RPM for the storage medium (our LP). (Older analogue technologies used 45 RPM and 78 RPM records. However, 33 1 3 RPM became the consumer standard for stereo music.) With an LP, the beginning of the record (the lead-in groove) would be on the outermost edge of the record and the end of the record (the exit groove) would be close to the center. Placing the needle in the lead-in groove, the needle gradually works its way to the exit groove. However, whereas the angular speed of the LP is a constant 33 1 3 RPM, the linear speed at the needle can vary quite a bit, depending on the needle location. 214 CHAPTER 16. MEASURING CIRCULAR MOTION 6 1 Figure 16.6: Lead-in and exit grooves. Example 16.3.1 (Analogue LP’s). The “lead-in groove” is 6 inches from the center of an LP, while the “exit groove” is 1 inch from the center. What is the linear speed (MPH) of the needle in the “lead-in groove”? What is the linear speed (MPH) of the needle in the “exit groove”? Find the location of the needle if the linear speed is 1 MPH. Solution. This is a straightforward application of Fact 16.2.1. Let v6 (resp. v1) be the linear speed at the lead in groove (resp. exit groove); the subscript keeps track of the needle radial location. Since the groove is approximately a circle, 2(6)π inches rev v6 = 33 1 3 = 1257 rev min in min 1257 in min 5280 ft mile = 1.19 MPH = 60 min hour 12 in ft Similarly, v1 = 0.2 MPH. To answer the remaining question, let r be the radial distance from the center of the LP to the needle location on the record. If vr = 1 MPH: 1 mile hour = vr = 33 1 3 rev min 2rπ in rev 60 min hr 1 ft 12 in 1 mile 5280 ft So, when the needle is r = 5.04 inches from the center, the linear speed is 1 MPH. laser laser support arm moves back and forth spinning CD Figure 16.7: Reproducing music using digital technology. In the early 1980’s, a new method of storing and reproducing music was introduced; this medium is called the digital compact disc, referred to as a CD for short. This is a thin plastic disc of diameter 4.5 inches, which appears to the naked eye to have a shiny silver coating on one side. Upon microscopic examination one would ﬁnd concentric circles of pits in the silver coating. This disc is placed in a CD player, which spins the disc. A laser located above the spinning disc will project onto the spinning disc. The pits in the silver coating will cause the reﬂected laser light to vary in intensity. A sensor detects this variation, converting it to a digital signal (the analogue to digital or AD conversion). This is fed into a digital to analogue or DA conversion device, which sends a signal to your stereo, again producing music. 16.4. BELT AND WHEEL PROBLEMS 215 The technology of CD ′s differs from that of LP ′s in two crucial ways. First, the circular motion of the spinning CD is controlled so that the target on the disc below the laser is always moving at a constant linear speed of 1.2 meters minute . Secondly, the beginning location of the laser will be on the inside portion of the disc, working its way outward to the end. In this context, it makes sense to study how the angular speed of the CD is changing, as the laser position changes. sec = 2,835 inches Example 16.3.2 (Digital CD’s). What is the angular speed (in RPM) of a CD if the laser is at the beginning, located 3 4 inches from the center of the disc ? What is the angular speed (in RPM) of a CD if the laser is at the end, located 2 inches from the center of the disc ? Find the location of the laser if the angular speed is 350 RPM. Solution. This is an application of Fact 16.2.1. Let ω3/4 be the angular speed at the start and ω2 the angular speed at the end of the CD; the subscript is keeping track of the laser distance from the CD center. ω2 = (2835 inches/min) (2(2)π inches/rev) = 225.6 RPM ω3/4 = 2835 inches/min) (2(0.75)π inches/rev) = 601.6 RPM ′′ 2 ′′ 3 4 Start of CD End of CD Figure 16.8: Computing the angular speed of a CD. To answer the remaining question, let r be the radial distance from the center of the CD to the laser location on the CD. If the angular speed ωr at this location is 350 RPM, we have the equation 350 RPM = ωr 2,835 inches minute 2rπ inches revolution 1.289 inches = r. = So, when the laser is 1.289 inches from the center, the CD is moving 350 RPM. 16.4 Belt and Wheel Problems The industrial revolution spawned a number of elaborate machines involving systems of belts and wheels. Computing the speed of various belts and wheels in such a system may seem complicated at ﬁrst glance. The situation can range from a simple system of two wheels with a belt connecting them, to more elaborate designs. We call problems of this sort belt and wheel problems, or more generally, connected wheel problems. Solving problems of this type always uses the same strategy, which we will ﬁrst highlight by way of an example. 216 CHAPTER 16. MEASURING CIRCULAR MOTION Figure 16.9: Two typical connected wheel scenarios. front sprocket radius = 5 inches rear sprocket radius = 2 inches (a) A stationary exercise bike. radius A = 14 inches A B radius B = 2 inches C radius C = 5 inches (b) A model of the bike’s connected wheels. Figure 16.10: Visualizing the connected wheels of an exercise bike. Example 16.4.1. You are riding a stationary exercise bike. Assume the rear wheel is 28 inches in diameter, the rear sprocket has radius 2 inches and the front sprocket has radius 5 inches. How many revolutions per minute of the front sprocket produces a forward speed of 40 MPH on the bike (miles per hour)? Solution. There are 3 wheels involved with a belt (the bicycle chain) connecting two of the wheels. In this problem, we are provided with the linear speed of wheel A (which is 40 MPH) and we need to ﬁnd the angular speed of wheel C=front sprocket. Denote by vA, vB, and vC the linear speeds of each of the wheels A, B, and C, respectively. Likewise, let ωA, ωB, and ωC denote the angula
r speeds of each of the wheels A, B, and C, respectively. In addition, the chain connecting the wheels B and C will have a linear speed, which we will denote by vchain. The strategy is broken into a sequence of steps which leads us from the known linear speed vA to the angular speed ωC of wheel C: • • • • Step 1: Given vA, ﬁnd ωA. Use the fact ωA = vA rA . Step 2: Observe ωA = ωB; this is because the wheel and rear sprocket are both rigidly mounted on a common axis of rotation. Step 3: Given ωB, ﬁnd vB. Use the fact vB = rBωB = rBωA = vA. rB rA Step 4: Observe vB = vchain = vC; this is because the chain is directly connecting the two sprockets and assumed not to slip. 16.4. BELT AND WHEEL PROBLEMS 217 • Step 5: Given vC, ﬁnd ωC. Use the fact ωC = vC rC vB rC rB rArC vA. = = Saying that the speedometer reads 40 MPH is the same as saying that the linear speed of a location on the rear wheel is vA = 40 MPH. Converting this into angular speed was carried out in our solution to Example 16.2.3 above; we found that ωA = 480 RPM. This completes Step 1 and so by Step 2, ωA = ωB = 480 RPM. For Step 3, we convert ωB = 480 RPM into linear speed following Fact 16.2.1: vB = 480 = 6,032 revolution minute inches minute . (2(2)π) inches revolution By Step 4, conclude that the linear speed of wheel C is vC = 6,032 inches/min. Finally, to carry out Step 5, we convert the linear speed into angular speed: ωC = 6,032 inches min 2(5)π inches rev = 192 RPM = 3.2 . rev sec In conclusion, the bike rider must pedal the front sprocket at the rate of 3.2 rev sec. This example indicates the basic strategy used in all belt/wheel prob- lems. Important Facts 16.4.2 (Belt and Wheel Strategy). Three basic facts are used in all such problems: • • Using “unit conversion” or Fact 16.2.2 allows us to go from linear speed v to angular speed ω, and vice versa. If two wheels are fastened rigidly to a common axle, then they have the same angular speed. (Caution: two wheels fastened to a common axle typically do not have the same linear speed!) • If two wheels are connected by a belt (or chain), the linear speed of the belt coincides with the linear speed of each wheel. 218 CHAPTER 16. MEASURING CIRCULAR MOTION 16.5 Exercises Problem 16.1. The restaurant in the Space Needle in Seattle rotates at the rate of one revolution per hour. (a) Through how many radians does it turn in 100 minutes? (b) How long does it take the restaurant to rotate through 4 radians? (c) How far does a person sitting by the window move in 100 minutes if the radius of the restaurant is 21 meters? Problem 16.2. You are riding a bicycle along a level road. Assume each wheel is 26 inches in diameter, the rear sprocket has a radius of 3 inches and the front sprocket has a radius of 7 inches. How fast do you need to pedal (in revolutions per minute) to achieve a speed of 35 mph? front wheel rear wheel rear sprocket front sprocket Problem 16.3. Answer the following angular speed questions. (a) A wheel of radius 22 ft. is rotating 11 RPM counterclockwise. Considering a point on the rim of the rotating wheel, what is the angular speed ω in rad/sec and the linear speed v in ft/sec? (b) A wheel of radius 8 in. is rotating 15o/sec. What is the linear speed v, the angular speed in RPM and the angular speed in rad/sec? (c) You are standing on the equator of the earth (radius 3960 miles). What is your linear and angular speed? Problem 16.4. Lee is running around the perimeter of a circular track at a rate of 10 ft/sec. The track has a radius of 100 yards. After 10 seconds, Lee turns and runs along a radial line to the center of the circle. Once he reaches the center, he turns and runs along a radial line to his starting point on the perimeter. Assume Lee does not slow down when he makes these two turns. (a) Sketch a picture of the situation. (b) How far has Lee traveled once he returns to his starting position? (c) How much time will elapse during Lee’s circuit? (d) Find the area of the pie shaped sector enclosed by Lee’s path. Problem 16.5. John has been hired to design an exciting carnival ride. Tiff, the carnival owner, has decided to create the worlds greatest ferris wheel. Tiff isn’t into math; she simply has a vision and has told John these constraints on her dream: (i) the wheel should rotate counterclockwise with an angular speed of 12 RPM; (ii) the linear speed of a rider should be 200 mph; (iii) the lowest point on the ride should be 4 feet above the level ground. 12 RPM θ P 4 feet (a) Find the radius of the ferris wheel. (b) Once the wheel is built, John suggests that Tiff should take the ﬁrst ride. The wheel starts turning when Tiff is at the location P, which makes an angle θ with the horizontal, as pictured. It takes her 1.3 seconds to reach the top of the ride. Find the angle θ. (d) An auto tire has radius 12 inches. If you are driving 65 mph, what is the angular speed in rad/sec and the angular speed in RPM? (c) Poor engineering causes Tiff’s seat to ﬂy off in 6 seconds. Describe where Tiff is located (an angle description) the instant she becomes a human missile. 16.5. EXERCISES 219 Problem 16.6. Michael and Aaron are on the “UL-Tossum” ride at Funworld. This is a merry-go-round of radius 20 feet which spins counterclockwise 60 RPM. The ride is driven by a belt connecting the outer edge of the ride to a drive wheel of radius 3 feet: Drive wheel radius 3 ft drive belt Aaron O P Michael main ride radius 20 ft (a) Assume Michael is seated on the edge of the ride, as pictured. What is Michael’s linear speed in mph and ft/sec? (b) What is the angular speed of the drive wheel in RPM? (c) Suppose Aaron is seated 16 feet from the center of the ride. What is the angular speed of Aaron in RPM? What is the linear speed of Aaron in ft/sec? (d) After 0.23 seconds Michael will be located at S as pictured. What is the angle ∠POS in degrees? What is the angle ∠POS in radians? How many feet has Michael traveled? S θ (e) Assume Michael has traveled 88 feet from the position P to a new position Q. How many seconds will this take? What will be the angle swept out by Michael? length of arc (PQ) is 88 ft O P Q Problem 16.7. You are riding a bicycle along a level road. Assume each wheel is 28 inches in diameter, the rear sprocket has radius 3 inches and the front sprocket has radius r inches. Suppose you are pedaling the front sprocket at the rate of 1.5 rev sec and your forward speed is 11 mph on the bike. What is the radius of the front sprocket? Problem 16.8. You are designing a system of wheels and belts as pictured below. You want wheel A to rotate 20 RPM while wheel B rotates 42 RPM. Wheel A has a radius of 6 inches, wheel B has a radius of 7 inches and wheel C has a radius of 1 inch. Assume wheels C and D are rigidly fastened to the same axle. What is the radius r of wheel D? O P A D C B 220 CHAPTER 16. MEASURING CIRCULAR MOTION Chapter 17 The Circular Functions SupposeCosmobeginsatlocation R andwalksinacounterclockwise direction, always maintaining a tight 20 ft long tether. As Cosmo moves around the circle, how can we describe his location at any given instant? In one sense, we have already answered this question: The measure of ∠RPS1 exactly pins down a location on the circle of radius 20 feet. But, we really might prefer a description of the horizontal and vertical coordinates of Cosmo; this would tie in better with the coordinate system we typically use. Solving this problem will require NEW functions, called the circular functions. S2 S3 S1 P 20 feet R S4 Figure 17.1: Cosmo moves counterclockwise maintaining a tight tether. Where’s Cosmo? 17.1 Sides and Angles of a Right Triangle Example 17.1.1. You are preparing to make your ﬁnal shot at the British Pocket Billiard World Championships. The position of your ball is as in Figure 17.2, and you must play the ball off the left cushion into the lower-right corner pocket, as indicated by the dotted path. For the big money, where should you aim to hit the cushion? Solution. This problem depends on two basic facts. First, the angles of entry and exit between the path the cushion will be equal. Secondly, the two obvious right triangles in this picture are similar triangles. Let x represent the distance from the bottom left corner to the impact point of the ball’s path: Properties of similar triangles tell us that the ratios of 5−x = 12 x . If we solve this equation 4 = 3.75 feet. common sides are equal: for x, we obtain x = 15 4 221 The billiard table layout. 4 ft 5 ft ﬁnd this location this pocket for the big money 6 ft 12 ft 4 θ θ 5 − x x Mathmatically modeling the bank shot. Figure 17.2: A pocket billiard banking problem. 222 CHAPTER 17. THE CIRCULAR FUNCTIONS This discussion is enough to win the tourney. But, of course, there are still other questions we can ask about this simple example: What is the angle θ? That is going to require substantially more work; indeed the bulk of this Chapter! It turns out, there is a lot of mathematical mileage in the idea of studying ratios of sides of right triangles. The ﬁrst step, which will get the ball rolling, is to introduce new functions whose very deﬁnition involves relating sides and angles of right triangles. 17.2 The Trigonometric Ratios B hypotenuse side opposite θ A θ side adjacent θ C Figure 17.3: Labeling the sides of a right triangle. From elementary geometry, the sum of the angles of any triangle will equal 180◦. Given a right triangle ABC, since one of the angles is 90◦, the remaining two angles must be acute angles; i.e., angles of measure between 0◦ and 90◦. If we specify one of the acute angles in a right triangle ABC, say angle θ, we can label the three sides using △ this terminology. We then consider the following three ratios of side lengths, referred to as trigonometric ratios: △ sin(θ) def= length of side opposite θ length of hypotenuse cos(θ) def= length of side adjacent θ length of hypotenuse tan(θ) def= length of side opposite θ length of side adjacent to θ . (17.1) (17.2) (17.3)
2, cos(θ) = √3 13, tan(θ) = 5 12. , cos(θ) = 1 √2 2 , tan(θ) = 1 √3 For example, we have three right triangles in Figure 17.4; you can verify that the Pythagorean Theorem holds in each of the cases. In the left-hand triangle, sin(θ) = 5 13, cos(θ) = 12 In the middle triangle, sin(θ) = 1 , tan(θ) = 1. In the right-hand triangle, √2 sin(θ) = 1 . The symbols “sin”, “cos”, and “tan” are abbreviations for the words sine, cosine and tangent, respectively. As we have deﬁned them, the trigonometric ratios depend on the dimensions of the triangle. However, the same ratios are obtained for any right triangle with acute angle θ. This follows from the properties of similar ADE are similar. If triangles. Consider Figure 17.5. Notice \|AB\| . On the other we use \|AD\|. Since the ratios of com= \|AE\| \|AD\|, △ ADE, we obtain cos(θ) = \|AE\| hand, if we use mon sides of similar triangles must agree, we have cos(θ) = \|AC\| \|AB\| ABC to compute cos(θ), then we ﬁnd cos(θ) = \|AC\| ABC and △ △ △ 17.2. THE TRIGONOMETRIC RATIOS 223 13 θ 12 5 √2 θ 1 1 2 θ √3 1 Figure 17.4: Computing trigonometric ratios for selected right triangles. which is what we wanted to be true. The same argument can be used to show that sin(θ) and tan(θ) can be computed using any right triangle with acute angle θ. Except for some “rigged” right triangles, it is not easy to calculate the trigonometric ratios. Before the 1970’s, approximate values of sin(θ), cos(θ), tan(θ) were listed in long tables or calculated using a slide rule. Today, a scientiﬁc calculator saves the day on these computations. Most scientiﬁc calculators will give an approximation for the values of the trigonometric ratios. However, it is good to keep in mind we can compute the EXACT values of the trigonometric ratios when θ = 0, π 4 , π 2 radians or, equivalently, when θ = 0◦, 30◦, 45◦, 60◦, 90◦. 17.5: Applying Figure trigonometric ratios to any right triangle. Angle θ Trigonometric Ratio Deg Rad sin(θ) cos(θ) tan(θ) 0◦ 30◦ 45◦ 60◦ 902 2 √3 2 1 1 √3 2 √2 2 1 2 0 0 1 √3 1 √3 Undeﬁned Table 17.1: Exact Trigonometric Ratios 224 CHAPTER 17. THE CIRCULAR FUNCTIONS !!! CAUTION !!! Some people make a big deal of “approximate” vs. “exact” answers; we won’t worry about it here, unless we are speciﬁcally asked for an exact answer. However, here is something we will make a big deal about: When computing values of cos(θ), sin(θ), and tan(θ) on your calculator, make sure you are using the correct “angle mode” when entering θ; i.e. “degrees” or “radians”. For example, if θ = 1◦, then cos(1◦) = 0.9998, sin(1◦) = 0.0175, and In contrast, if θ = 1 radians, then cos(1) = 0.5403, tan(1◦) = 0.0175. sin(1) = 0.8415, and tan(1) = 1.5574. 17.3 Applications h θ h cos(θ) h sin(θ) a tan(θ) θ a Figure 17.6: What do these ratios mean? When confronted with a situation involving a right triangle where the measure of one acute angle θ and one side are known, we can solve for the remaining sides using the appropriate trigonometric ratios. Here is the key picture to keep in mind: Important Facts 17.3.1 (Trigonometric ratios). Given a right triangle, the trigonometric ratios relate the lengths of the sides as shown in Figure 17.6. Example 17.3.2. To measure the distance across a river for a new bridge, surveyors placed poles at locations A, B and C. The length \|AB\| = 100 feet and the measure of the angle ∠ABC is 31◦18 ′. Find the distance to span the river. If the measurement of the angle ∠ABC is only accurate within 2 ′, ﬁnd the possible error in \|AC\|. ± C d A B 100 310 18 ′ Solution. The trigonometric ratio relating these two sides would be the tangent and we can convert θ into decimal form, arriving at: tan(31◦18 ′) = tan(31.3◦) = \|AC\| \|BA\| = d 100 Figure 17.7: The distance spanning a river. therefore d = 60.8 feet. This tells us that the bridge needs to span a gap of If the measurement of the angle was in error by +2 ′, then 60.8 feet. tan(31◦20 ′) = tan(31.3333◦) = 0.6088 and the span is 60.88 ft. On the other hand, if the measurement of the angle was in error by −2 ′, then tan(31◦16 ′) = tan(31.2667◦) = 0.6072 and the span is 60.72 ft. 17.3. APPLICATIONS 225 Example 17.3.3. A plane is ﬂying 2000 feet above sea level toward a mountain. The pilot observes the top of the mountain to be 18◦ above the horizontal, then immediately ﬂies the plane at an angle of 20◦ above horizontal. The airspeed of the plane is 100 mph. After 5 minutes, the plane is directly above the top of the mountain. How high is the plane above the top of the mountain (when it passes over)? What is the height of the mountain? T E L P 2000 ft S sealevel Figure 17.8: Flying toward a mountain. Solution. We can compute the hypotenuse of ing the speed and time information about the plane: △ LPT by us- \|PT \| = (100 mph)(5 minutes)(1 hour/60 minutes) = 25 3 miles. The deﬁnitions of the trigonometric ratios show: \|TL\| = \|PL\| = 25 3 25 3 sin(20◦) = 2.850 miles, and cos(20◦) = 7.831 miles. With this data, we can now ﬁnd \|EL\|: \|EL\| = \|PL\| tan(18◦) = 2.544 miles. The height of the plane above the peak is \|TE\| = \|TL\| − \|EL\| = 2.850 − 2.544 = 0.306 miles = 1,616 feet. The elevation of the peak above sea level is given by: Peak elevation = plane altitude + \|EL\| = \|SP\| + \|EL\| = 2,000 + (2.544)(5,280) = 15,432 feet. Example 17.3.4. A Forest Service helicopter needs to determine the width of a deep canyon. While hovering, they measure the angle γ = 48◦ at position B (see picture), then descend 400 feet to position A and make two measurements of α = 13◦ (the measure of ∠EAD), β = 53◦ (the measure of ∠CAD). Determine the width of the canyon to the nearest foot. 400 ft B γ A α β C E canyon D Figure 17.9: Finding the width of a canyon. Solution. We will need to exploit three right triangles in ACE. Our goal is to compute \|ED\| = ACD, and the picture: \|CD\| − \|CE\|, which suggests more than one right triangle will come into play. BCD, △ △ △ 226 CHAPTER 17. THE CIRCULAR FUNCTIONS The ﬁrst step is to use ACD to obtain a system of two equations and two unknowns involving some of the side lengths; we will then solve the system. From the deﬁnitions of the trigonometric ratios, BCD and △ △ \|CD\| = (400 + \|AC\|) tan(48◦) \|CD\| = \|AC\| tan(53◦). Plugging the second equation into the ﬁrst and rearranging we get \|AC\| = 400 tan(48◦) tan(53◦) − tan(48◦) = 2,053 feet. Plugging this back into the second equation of the system gives \|CD\| = (2053) tan(53◦) = 2724 feet. The next step is to relate ACE, which can now be done ACD and △ in an effective way using the calculations above. Notice that the measure of ∠CAE is β − α = 40◦. We have △ \|CE\| = \|AC\| tan(40◦) = (2053) tan(40◦) = 1,723 feet. As noted above, \|ED\| = \|CD\| − \|CE\| = 2,724 − 1,723 = 1,001 feet is the width of the canyon. 17.4 Circular Functions S = (x,y) y 20 θ x P R If Cosmo is located somewhere in the ﬁrst quadrant of Figure 17.1, represented by the location S, we can use the trigonometric ratios to describe his coordinates. Impose the indicated xy-coordinate system with origin at P and extract the pictured right triangle with vertices at P and S. The radius is 20 ft. and applying Fact 17.3.1 gives S = (x, y) = (20 cos(θ), 20 sin(θ)). Figure 17.10: Cosmo on a circular path. Unfortunately, we run into a snag if we allow Cosmo to wander into the second, third or fourth quadrant, since then the angle θ is no longer acute. 17.4. CIRCULAR FUNCTIONS 227 17.4.1 Are the trigonometric ratios functions? Recall that sin(θ), cos(θ), and tan(θ) are deﬁned for acute angles θ inside a right triangle. We would like to say that these three equations actually deﬁne functions where the variable is an angle θ. Having said this, it is natural to ask if these three equations can be extended to be deﬁned for ANY angle θ. For example, we need to explain how sin is deﬁned. 2π 3 To start, we begin with the unit circle pictured in the xy-coordinate system. Let θ = ∠ROP be the angle in standard central position shown in Figure 17.11. If θ is positive (resp. negative), we adopt the convention that θ is swept out by counterclockwise (resp. clockwise) rotation of the initial side OR. The objective is to ﬁnd the coordinates of the point P in this ﬁgure. Notice that each coordinate of P (the x-coordinate and the y-coordinate) will depend on the given angle θ. For this reason, we need to introduce two new functions involving the variable θ. P 1 A unit circle with radius = 1. θ O 1 R Figure 17.11: Coordinates of points on the unit circle. Deﬁnition 17.4.1. Let θ be an angle in standard central position inside the unit circle, as in Figure 17.11. This angle determines a point P on the unit circle. Deﬁne two new functions, cos(θ) and sin(θ), on the domain of all θ values as follows: cos(θ) sin(θ) def= horizontal x-coordinate of P on unit circle def= vertical y-coordinate of P on unit circle. r = 1 kilometer We refer to sin(θ) and cos(θ) as the basiccircularfunctions. Keep in mind that these functions have variables which are angles (either in degree or radian measure). These functions will be on your calculator. Again, BE CAREFUL to check the angle mode setting on your calculator (“degrees” or “radians”) before doing a calculation. Figure 17.12: driving track. 0.025 rad sec Michael starts here A circular Example 17.4.2. Michael is test driving a vehicle counterclockwise around a desert test track which is circular of radius 1 kilometer. He starts at the location pictured, traveling 0.025 rad Impose coordinates as pictured. Where is sec . Michael located (in xy-coordinates) after 18 seconds? Solution. Let M(t) be the point on the circle of motion representing Michael’s location after t seconds and θ(t) the angle swept out the by Michael after t seconds. Since we are given the angular speed, we get θ(t) = 0.025t radians. y-axis M(t) = (x(t),y(t)) θ(t) 0.025 rad sec x-axis Michael starts here Figure 17.13: Modeling Michael’s location. 228 CHAPTER 17. THE CIRCULAR FUNCTIONS Since the angle θ(t) is in central standard position, we get M(t) = (cos(θ(t)), sin(θ(t))) = (cos(0.025t), sin(0.0

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