Lesson 16: Inverse Trigonometric Functions

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  • 1. Section 3.5 Inverse Trigonometric Functions V63.0121.002.2010Su, Calculus I New York University June 7, 2010 Announcements Exams not graded yet Written HW due tomorrow Quiz 3 on Thursday covering 3.3, 3.4, 3.5, 3.7
  • 2. Announcements Exams not graded yet Written HW due tomorrow Quiz 3 on Thursday covering 3.3, 3.4, 3.5, 3.7 V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 2 / 30
  • 3. Objectives Know the definitions, domains, ranges, and other properties of the inverse trignometric functions: arcsin, arccos, arctan, arcsec, arccsc, arccot. Know the derivatives of the inverse trignometric functions. V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 3 / 30
  • 4. What is an inverse function? Definition Let f be a function with domain D and range E . The inverse of f is the function f −1 defined by: f −1 (b) = a, where a is chosen so that f (a) = b. V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 4 / 30
  • 5. What is an inverse function? Definition Let f be a function with domain D and range E . The inverse of f is the function f −1 defined by: f −1 (b) = a, where a is chosen so that f (a) = b. So f −1 (f (x)) = x, f (f −1 (x)) = x V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 4 / 30
  • 6. What functions are invertible? In order for f −1 to be a function, there must be only one a in D corresponding to each b in E . Such a function is called one-to-one The graph of such a function passes the horizontal line test: any horizontal line intersects the graph in exactly one point if at all. If f is continuous, then f −1 is continuous. V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 5 / 30
  • 7. Outline Inverse Trigonometric Functions Derivatives of Inverse Trigonometric Functions Arcsine Arccosine Arctangent Arcsecant Applications V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 6 / 30
  • 8. arcsin Arcsin is the inverse of the sine function after restriction to [−π/2, π/2]. y x π π sin − 2 2 V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 7 / 30
  • 9. arcsin Arcsin is the inverse of the sine function after restriction to [−π/2, π/2]. y x π π sin − 2 2 V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 7 / 30
  • 10. arcsin Arcsin is the inverse of the sine function after restriction to [−π/2, π/2]. y y =x x π π sin − 2 2 V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 7 / 30
  • 11. arcsin Arcsin is the inverse of the sine function after restriction to [−π/2, π/2]. y arcsin x π π sin − 2 2 The domain of arcsin is [−1, 1] π π The range of arcsin is − , 2 2 V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 7 / 30
  • 12. arccos Arccos is the inverse of the cosine function after restriction to [0, π] y cos x 0 π V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 8 / 30
  • 13. arccos Arccos is the inverse of the cosine function after restriction to [0, π] y cos x 0 π V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 8 / 30
  • 14. arccos Arccos is the inverse of the cosine function after restriction to [0, π] y y =x cos x 0 π V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 8 / 30
  • 15. arccos Arccos is the inverse of the cosine function after restriction to [0, π] arccos y cos x 0 π The domain of arccos is [−1, 1] The range of arccos is [0, π] V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 8 / 30
  • 16. arctan Arctan is the inverse of the tangent yfunction after restriction to [−π/2, π/2]. x 3π π π 3π − − 2 2 2 2 tan V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 9 / 30
  • 17. arctan Arctan is the inverse of the tangent yfunction after restriction to [−π/2, π/2]. x 3π π π 3π − − 2 2 2 2 tan V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 9 / 30
  • 18. arctan y =x Arctan is the inverse of the tangent yfunction after restriction to [−π/2, π/2]. x 3π π π 3π − − 2 2 2 2 tan V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 9 / 30
  • 19. arctan Arctan is the inverse of the tangent yfunction after restriction to [−π/2, π/2]. π 2 arctan x π − 2 The domain of arctan is (−∞, ∞) π π The range of arctan is − , 2 2 π π lim arctan x = , lim arctan x = − x→∞ 2 x→−∞ 2 V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 9 / 30
  • 20. arcsec y Arcsecant is the inverse of secant after restriction to [0, π/2) ∪ (π, 3π/2]. x 3π π π 3π − − 2 2 2 2 sec V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 10 / 30
  • 21. arcsec y Arcsecant is the inverse of secant after restriction to [0, π/2) ∪ (π, 3π/2]. x 3π π π 3π − − 2 2 2 2 sec V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 10 / 30
  • 22. arcsec y =x y restriction to [0, π/2) ∪ (π, 3π/2]. Arcsecant is the inverse of secant after x 3π π π 3π − − 2 2 2 2 sec V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 10 / 30
  • 23. 3π arcsec 2 y Arcsecant is the inverse of secant after restriction to [0, π/2) ∪ (π, 3π/2]. π 2 x The domain of arcsec is (−∞, −1] ∪ [1, ∞) π π The range of arcsec is 0, ∪ ,π 2 2 π 3π lim arcsec x = , lim arcsec x = x→∞ 2 x→−∞ 2 V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 10 / 30
  • 24. Values of Trigonometric Functions π π π π x 0 6 4 3 2 √ √ 1 2 3 sin x 0 1 2 2 2 √ √ 3 2 1 cos x 1 0 2 2 2 1 √ tan x 0 √ 1 3 undef 3 √ 1 cot x undef 3 1 √ 0 3 2 2 sec x 1 √ √ 2 undef 3 2 2 2 csc x undef 2 √ √ 1 2 3 V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 11 / 30
  • 25. Check: Values of inverse trigonometric functions Example Find arcsin(1/2) arctan(−1) √ 2 arccos − 2 V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 12 / 30
  • 26. Check: Values of inverse trigonometric functions Example Find arcsin(1/2) arctan(−1) √ 2 arccos − 2 Solution π 6 V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 12 / 30
  • 27. What is arctan(−1)? 3π/4 −π/4 V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 13 / 30
  • 28. What is arctan(−1)? 3π/4 3π Yes, tan = −1 4 √ 2 sin(3π/4) = 2 √ 2 cos(3π/4) = − 2 −π/4 V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 13 / 30
  • 29. What is arctan(−1)? 3π/4 3π Yes, tan = −1 4 √ But, the range of arctan is 2 π π sin(3π/4) = − , 2 2 2 √ 2 cos(3π/4) = − 2 −π/4 V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 13 / 30
  • 30. What is arctan(−1)? 3π/4 3π Yes, tan = −1 4 But, the range of arctan is √ π π 2 − , cos(π/4) = 2 2 2 Another angle whose π tangent is −1 is − , and √ 4 2 this is in the right range. sin(π/4) = − 2 −π/4 V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 13 / 30
  • 31. What is arctan(−1)? 3π/4 3π Yes, tan = −1 4 But, the range of arctan is √ π π 2 − , cos(π/4) = 2 2 2 Another angle whose π tangent is −1 is − , and √ 4 2 this is in the right range. sin(π/4) = − π 2 So arctan(−1) = − 4 −π/4 V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 13 / 30
  • 32. Check: Values of inverse trigonometric functions Example Find arcsin(1/2) arctan(−1) √ 2 arccos − 2 Solution π 6 π − 4 V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 14 / 30
  • 33. Check: Values of inverse trigonometric functions Example Find arcsin(1/2) arctan(−1) √ 2 arccos − 2 Solution π 6 π − 4 3π 4 V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 14 / 30
  • 34. Caution: Notational ambiguity sin2 x = (sin x)2 sin−1 x = (sin x)−1 sinn x means the nth power of sin x, except when n = −1! The book uses sin−1 x for the inverse of sin x, and never for (sin x)−1 . 1 I use csc x for and arcsin x for the inverse of sin x. sin x V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 15 / 30
  • 35. Outline Inverse Trigonometric Functions Derivatives of Inverse Trigonometric Functions Arcsine Arccosine Arctangent Arcsecant Applications V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 16 / 30
  • 36. The Inverse Function Theorem Theorem (The Inverse Function Theorem) Let f be differentiable at a, and f (a) = 0. Then f −1 is defined in an open interval containing b = f (a), and 1 (f −1 ) (b) = f (f −1 (b)) V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 17 / 30
  • 37. The Inverse Function Theorem Theorem (The Inverse Function Theorem) Let f be differentiable at a, and f (a) = 0. Then f −1 is defined in an open interval containing b = f (a), and 1 (f −1 ) (b) = f (f −1 (b)) Upshot: Many times the derivative of f −1 (x) can be found by implicit differentiation and the derivative of f : V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 17 / 30
  • 38. Derivation: The derivative of arcsin Let y = arcsin x, so x = sin y . Then dy dy 1 1 cos y = 1 =⇒ = = dx dx cos y cos(arcsin x) V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 18 / 30
  • 39. Derivation: The derivative of arcsin Let y = arcsin x, so x = sin y . Then dy dy 1 1 cos y = 1 =⇒ = = dx dx cos y cos(arcsin x) To simplify, look at a right triangle: V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 18 / 30
  • 40. Derivation: The derivative of arcsin Let y = arcsin x, so x = sin y . Then dy dy 1 1 cos y = 1 =⇒ = = dx dx cos y cos(arcsin x) To simplify, look at a right triangle: 1 x V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 18 / 30
  • 41. Derivation: The derivative of arcsin Let y = arcsin x, so x = sin y . Then dy dy 1 1 cos y = 1 =⇒ = = dx dx cos y cos(arcsin x) To simplify, look at a right triangle: 1 x y = arcsin x V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 18 / 30
  • 42. Derivation: The derivative of arcsin Let y = arcsin x, so x = sin y . Then dy dy 1 1 cos y = 1 =⇒ = = dx dx cos y cos(arcsin x) To simplify, look at a right triangle: 1 x y = arcsin x 1 − x2 V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 18 / 30
  • 43. Derivation: The derivative of arcsin Let y = arcsin x, so x = sin y . Then dy dy 1 1 cos y = 1 =⇒ = = dx dx cos y cos(arcsin x) To simplify, look at a right triangle: cos(arcsin x) = 1 − x2 1 x y = arcsin x 1 − x2 V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 18 / 30
  • 44. Derivation: The derivative of arcsin Let y = arcsin x, so x = sin y . Then dy dy 1 1 cos y = 1 =⇒ = = dx dx cos y cos(arcsin x) To simplify, look at a right triangle: cos(arcsin x) = 1 − x2 1 x So d 1 arcsin(x) = √ y = arcsin x dx 1 − x2 1 − x2 V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 18 / 30
  • 45. Graphing arcsin and its derivative 1 √ 1 − x2 The domain of f is [−1, 1], but the domain of f is arcsin (−1, 1) lim f (x) = +∞ x→1− | | lim f (x) = +∞ −1 1 x→−1+ V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 19 / 30
  • 46. Derivation: The derivative of arccos Let y = arccos x, so x = cos y . Then dy dy 1 1 − sin y = 1 =⇒ = = dx dx − sin y − sin(arccos x) V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 20 / 30
  • 47. Derivation: The derivative of arccos Let y = arccos x, so x = cos y . Then dy dy 1 1 − sin y = 1 =⇒ = = dx dx − sin y − sin(arccos x) To simplify, look at a right triangle: sin(arccos x) = 1 − x2 1 1 − x2 So d 1 y = arccos x arccos(x) = − √ dx 1 − x2 x V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 20 / 30
  • 48. Graphing arcsin and arccos arccos arcsin | | −1 1 V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 21 / 30
  • 49. Graphing arcsin and arccos arccos Note π arcsin cos θ = sin −θ 2 π =⇒ arccos x = − arcsin x 2 | | −1 1 So it’s not a surprise that their derivatives are opposites. V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 21 / 30
  • 50. Derivation: The derivative of arctan Let y = arctan x, so x = tan y . Then dy dy 1 sec2 y = 1 =⇒ = = cos2 (arctan x) dx dx sec2 y V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 22 / 30
  • 51. Derivation: The derivative of arctan Let y = arctan x, so x = tan y . Then dy dy 1 sec2 y = 1 =⇒ = = cos2 (arctan x) dx dx sec2 y To simplify, look at a right triangle: V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 22 / 30
  • 52. Derivation: The derivative of arctan Let y = arctan x, so x = tan y . Then dy dy 1 sec2 y = 1 =⇒ = = cos2 (arctan x) dx dx sec2 y To simplify, look at a right triangle: x 1 V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 22 / 30
  • 53. Derivation: The derivative of arctan Let y = arctan x, so x = tan y . Then dy dy 1 sec2 y = 1 =⇒ = = cos2 (arctan x) dx dx sec2 y To simplify, look at a right triangle: x y = arctan x 1 V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 22 / 30
  • 54. Derivation: The derivative of arctan Let y = arctan x, so x = tan y . Then dy dy 1 sec2 y = 1 =⇒ = = cos2 (arctan x) dx dx sec2 y To simplify, look at a right triangle: 1 + x2 x y = arctan x 1 V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 22 / 30
  • 55. Derivation: The derivative of arctan Let y = arctan x, so x = tan y . Then dy dy 1 sec2 y = 1 =⇒ = = cos2 (arctan x) dx dx sec2 y To simplify, look at a right triangle: 1 cos(arctan x) = √ 1 + x2 1 + x2 x y = arctan x 1 V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 22 / 30
  • 56. Derivation: The derivative of arctan Let y = arctan x, so x = tan y . Then dy dy 1 sec2 y = 1 =⇒ = = cos2 (arctan x) dx dx sec2 y To simplify, look at a right triangle: 1 cos(arctan x) = √ 1 + x2 1 + x2 x So d 1 arctan(x) = y = arctan x dx 1 + x2 1 V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 22 / 30
  • 57. Graphing arctan and its derivative y π/2 arctan 1 1 + x2 x −π/2 The domain of f and f are both (−∞, ∞) Because of the horizontal asymptotes, lim f (x) = 0 x→±∞ V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 23 / 30
  • 58. Example √ Let f (x) = arctan x. Find f (x). V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 24 / 30
  • 59. Example √ Let f (x) = arctan x. Find f (x). Solution d √ 1 d √ 1 1 dx arctan x = √ 2 dx x = 1 + x · 2√x 1+ x 1 = √ √ 2 x + 2x x V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 24 / 30
  • 60. Derivation: The derivative of arcsec Try this first. V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 25 / 30
  • 61. Derivation: The derivative of arcsec Try this first. Let y = arcsec x, so x = sec y . Then dy dy 1 1 sec y tan y = 1 =⇒ = = dx dx sec y tan y x tan(arcsec(x)) V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 25 / 30
  • 62. Derivation: The derivative of arcsec Try this first. Let y = arcsec x, so x = sec y . Then dy dy 1 1 sec y tan y = 1 =⇒ = = dx dx sec y tan y x tan(arcsec(x)) To simplify, look at a right triangle: V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 25 / 30
  • 63. Derivation: The derivative of arcsec Try this first. Let y = arcsec x, so x = sec y . Then dy dy 1 1 sec y tan y = 1 =⇒ = = dx dx sec y tan y x tan(arcsec(x)) To simplify, look at a right triangle: V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 25 / 30
  • 64. Derivation: The derivative of arcsec Try this first. Let y = arcsec x, so x = sec y . Then dy dy 1 1 sec y tan y = 1 =⇒ = = dx dx sec y tan y x tan(arcsec(x)) To simplify, look at a right triangle: x 1 V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 25 / 30
  • 65. Derivation: The derivative of arcsec Try this first. Let y = arcsec x, so x = sec y . Then dy dy 1 1 sec y tan y = 1 =⇒ = = dx dx sec y tan y x tan(arcsec(x)) To simplify, look at a right triangle: x y = arcsec x 1 V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 25 / 30
  • 66. Derivation: The derivative of arcsec Try this first. Let y = arcsec x, so x = sec y . Then dy dy 1 1 sec y tan y = 1 =⇒ = = dx dx sec y tan y x tan(arcsec(x)) To simplify, look at a right triangle: √ x2 − 1 tan(arcsec x) = 1 x x2 − 1 y = arcsec x 1 V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 25 / 30
  • 67. Derivation: The derivative of arcsec Try this first. Let y = arcsec x, so x = sec y . Then dy dy 1 1 sec y tan y = 1 =⇒ = = dx dx sec y tan y x tan(arcsec(x)) To simplify, look at a right triangle: √ x2 − 1 tan(arcsec x) = 1 x x2 − 1 So d 1 arcsec(x) = √ y = arcsec x dx x x2 − 1 1 V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 25 / 30
  • 68. Another Example Example Let f (x) = e arcsec x . Find f (x). V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 26 / 30
  • 69. Another Example Example Let f (x) = e arcsec x . Find f (x). Solution 1 f (x) = e arcsec x · √ x x2 − 1 V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 26 / 30
  • 70. Outline Inverse Trigonometric Functions Derivatives of Inverse Trigonometric Functions Arcsine Arccosine Arctangent Arcsecant Applications V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 27 / 30
  • 71. Application Example One of the guiding principles of most sports is to “keep your eye on the ball.” In baseball, a batter stands 2 ft away from home plate as a pitch is thrown with a velocity of 130 ft/sec (about 90 mph). At what rate does the batter’s angle of gaze need to change to follow the ball as it crosses home plate? V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 28 / 30
  • 72. Solution Let y (t) be the distance from the ball to home plate, and θ the angle the batter’s eyes make with home plate while following the ball. We know y = −130 and we want θ at the moment that y = 0. y 130 ft/sec θ 2 ft V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 29 / 30
  • 73. Solution Let y (t) be the distance from the ball to home plate, and θ the angle the batter’s eyes make with home plate while following the ball. We know y = −130 and we want θ at the moment that y = 0. We have θ = arctan(y /2). Thus dθ 1 1 dy = · dt 1 + (y /2)2 2 dt y 130 ft/sec θ 2 ft V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 29 / 30
  • 74. Solution Let y (t) be the distance from the ball to home plate, and θ the angle the batter’s eyes make with home plate while following the ball. We know y = −130 and we want θ at the moment that y = 0. We have θ = arctan(y /2). Thus dθ 1 1 dy = · dt 1 + (y /2)2 2 dt When y = 0 and y = −130, then y dθ 1 1 = · (−130) = −65 rad/sec 130 ft/sec dt y =0 1+0 2 θ 2 ft V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 29 / 30
  • 75. Solution Let y (t) be the distance from the ball to home plate, and θ the angle the batter’s eyes make with home plate while following the ball. We know y = −130 and we want θ at the moment that y = 0. We have θ = arctan(y /2). Thus dθ 1 1 dy = · dt 1 + (y /2)2 2 dt When y = 0 and y = −130, then y dθ 1 1 = · (−130) = −65 rad/sec 130 ft/sec dt y =0 1+0 2 The human eye can only track θ at 3 rad/sec! 2 ft V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 29 / 30
  • 76. Summary y y 1 arcsin x √ 1 − x2 1 arccos x −√ 1 − x2 Remarkable that the 1 derivatives of these arctan x transcendental functions are 1 + x2 1 algebraic (or even rational!) arccot x − 1 + x2 1 arcsec x √ x x2 − 1 1 arccsc x − √ x x2 − 1 V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 30 / 30