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### Lesson16 -inverse_trigonometric_functions_041_slides

1. 1. Section 3.5 Inverse Trigonometric Functions V63.0121.041, Calculus I New York University November 1, 2010 Announcements Midterm grades have been submitted Quiz 3 this week in recitation on Section 2.6, 2.8, 3.1, 3.2 Thank you for the evaluations . . . . . .
2. 2. . . . . . . Announcements Midterm grades have been submitted Quiz 3 this week in recitation on Section 2.6, 2.8, 3.1, 3.2 Thank you for the evaluations V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 2 / 32
3. 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. V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 3 / 32
4. 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. V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 4 / 32
5. 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 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 4 / 32
6. 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. V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 5 / 32
7. 7. . . . . . . Outline Inverse Trigonometric Functions Derivatives of Inverse Trigonometric Functions Arcsine Arccosine Arctangent Arcsecant Applications V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 6 / 32
8. 8. . . . . . . arcsin Arcsin is the inverse of the sine function after restriction to [−π/2, π/2]. . .x .y .sin . .− π 2 . . π 2 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 7 / 32
9. 9. . . . . . . arcsin Arcsin is the inverse of the sine function after restriction to [−π/2, π/2]. . .x .y .sin . . . .− π 2 . . π 2 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 7 / 32
10. 10. . . . . . . arcsin Arcsin is the inverse of the sine function after restriction to [−π/2, π/2]. . .x .y .sin . . . .− π 2 . . π 2 .y = x V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 7 / 32
11. 11. . . . . . . arcsin Arcsin is the inverse of the sine function after restriction to [−π/2, π/2]. . .x .y .sin . . . .− π 2 . . π 2 . ..arcsin The domain of arcsin is [−1, 1] The range of arcsin is [ − π 2 , π 2 ] V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 7 / 32
12. 12. . . . . . . arccos Arccos is the inverse of the cosine function after restriction to [0, π] . .x .y .cos . .0 . .π V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 8 / 32
13. 13. . . . . . . arccos Arccos is the inverse of the cosine function after restriction to [0, π] . .x .y .cos . . . .0 . .π V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 8 / 32
14. 14. . . . . . . arccos Arccos is the inverse of the cosine function after restriction to [0, π] . .x .y .cos . . . .0 . .π .y = x V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 8 / 32
15. 15. . . . . . . arccos Arccos is the inverse of the cosine function after restriction to [0, π] . .x .y .cos . . . .0 . .π . ..arccos The domain of arccos is [−1, 1] The range of arccos is [0, π] V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 8 / 32
16. 16. . . . . . . arctan Arctan is the inverse of the tangent function after restriction to [−π/2, π/2]. . .x .y .tan .− 3π 2 .− π 2 . π 2 . 3π 2 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 9 / 32
17. 17. . . . . . . arctan Arctan is the inverse of the tangent function after restriction to [−π/2, π/2]. . .x .y .tan .− 3π 2 .− π 2 . π 2 . 3π 2 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 9 / 32
18. 18. . . . . . . arctan Arctan is the inverse of the tangent function after restriction to [−π/2, π/2]. . .x .y .tan .− 3π 2 .− π 2 . π 2 . 3π 2 .y = x V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 9 / 32
19. 19. . . . . . . arctan Arctan is the inverse of the tangent function after restriction to [−π/2, π/2]. . .x .y .arctan .− π 2 . π 2 The domain of arctan is (−∞, ∞) The range of arctan is ( − π 2 , π 2 ) lim x→∞ arctan x = π 2 , lim x→−∞ arctan x = − π 2 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 9 / 32
20. 20. . . . . . . arcsec Arcsecant is the inverse of secant after restriction to [0, π/2) ∪ (π, 3π/2]. . .x .y .sec .− 3π 2 .− π 2 . π 2 . 3π 2 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 10 / 32
21. 21. . . . . . . arcsec Arcsecant is the inverse of secant after restriction to [0, π/2) ∪ (π, 3π/2]. . .x .y .sec .− 3π 2 .− π 2 . π 2 . 3π 2 . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 10 / 32
22. 22. . . . . . . arcsec Arcsecant is the inverse of secant after restriction to [0, π/2) ∪ (π, 3π/2]. . .x .y .sec .− 3π 2 .− π 2 . π 2 . 3π 2 . . .y = x V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 10 / 32
23. 23. . . . . . . arcsec Arcsecant is the inverse of secant after restriction to [0, π/2) ∪ (π, 3π/2]. . .x .y . . . . . π 2 . 3π 2 The domain of arcsec is (−∞, −1] ∪ [1, ∞) The range of arcsec is [ 0, π 2 ) ∪ (π 2 , π ] lim x→∞ arcsec x = π 2 , lim x→−∞ arcsec x = 3π 2 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 10 / 32
24. 24. . . . . . . Values of Trigonometric Functions x 0 π 6 π 4 π 3 π 2 sin x 0 1 2 √ 2 2 √ 3 2 1 cos x 1 √ 3 2 √ 2 2 1 2 0 tan x 0 1 √ 3 1 √ 3 undef cot x undef √ 3 1 1 √ 3 0 sec x 1 2 √ 3 2 √ 2 2 undef csc x undef 2 2 √ 2 2 √ 3 1 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 11 / 32
25. 25. . . . . . . Check: Values of inverse trigonometric functions Example Find arcsin(1/2) arctan(−1) arccos ( − √ 2 2 ) V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 12 / 32
26. 26. . . . . . . Check: Values of inverse trigonometric functions Example Find arcsin(1/2) arctan(−1) arccos ( − √ 2 2 ) Solution π 6 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 12 / 32
27. 27. . . . . . . What is arctan(−1)? . . . . .3π/4 . .−π/4 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 13 / 32
28. 28. . . . . . . What is arctan(−1)? . . . . .3π/4 . .−π/4 .sin(3π/4) = √ 2 2 .cos(3π/4) = − √ 2 2 Yes, tan ( 3π 4 ) = −1 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 13 / 32
29. 29. . . . . . . What is arctan(−1)? . . . . .3π/4 . .−π/4 .sin(3π/4) = √ 2 2 .cos(3π/4) = − √ 2 2 Yes, tan ( 3π 4 ) = −1 But, the range of arctan is( − π 2 , π 2 ) V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 13 / 32
30. 30. . . . . . . What is arctan(−1)? . . . . .3π/4 . .−π/4 .sin(π/4) = − √ 2 2 .cos(π/4) = √ 2 2 Yes, tan ( 3π 4 ) = −1 But, the range of arctan is( − π 2 , π 2 ) Another angle whose tangent is −1 is − π 4 , and this is in the right range. V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 13 / 32
31. 31. . . . . . . What is arctan(−1)? . . . . .3π/4 . .−π/4 .sin(π/4) = − √ 2 2 .cos(π/4) = √ 2 2 Yes, tan ( 3π 4 ) = −1 But, the range of arctan is( − π 2 , π 2 ) Another angle whose tangent is −1 is − π 4 , and this is in the right range. So arctan(−1) = − π 4 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 13 / 32
32. 32. . . . . . . Check: Values of inverse trigonometric functions Example Find arcsin(1/2) arctan(−1) arccos ( − √ 2 2 ) Solution π 6 − π 4 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 14 / 32
33. 33. . . . . . . Check: Values of inverse trigonometric functions Example Find arcsin(1/2) arctan(−1) arccos ( − √ 2 2 ) Solution π 6 − π 4 3π 4 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 14 / 32
34. 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 . I use csc x for 1 sin x and arcsin x for the inverse of sin x. V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 15 / 32
35. 35. . . . . . . Outline Inverse Trigonometric Functions Derivatives of Inverse Trigonometric Functions Arcsine Arccosine Arctangent Arcsecant Applications V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 16 / 32
36. 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 (f−1 )′ (b) = 1 f′ (f−1 (b)) In Leibniz notation we have dx dy = 1 dy/dx V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 17 / 32
37. 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 (f−1 )′ (b) = 1 f′ (f−1 (b)) In Leibniz notation we have dx dy = 1 dy/dx Upshot: Many times the derivative of f−1 (x) can be found by implicit differentiation and the derivative of f: V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 17 / 32
38. 38. . . . . . . Illustrating the Inverse Function Theorem . . Example Use the inverse function theorem to find the derivative of the square root function. V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 18 / 32
39. 39. . . . . . . Illustrating the Inverse Function Theorem . . Example Use the inverse function theorem to find the derivative of the square root function. Solution (Newtonian notation) Let f(x) = x2 so that f−1 (y) = √ y. Then f′ (u) = 2u so for any b > 0 we have (f−1 )′ (b) = 1 2 √ b V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 18 / 32
40. 40. . . . . . . Illustrating the Inverse Function Theorem . . Example Use the inverse function theorem to find the derivative of the square root function. Solution (Newtonian notation) Let f(x) = x2 so that f−1 (y) = √ y. Then f′ (u) = 2u so for any b > 0 we have (f−1 )′ (b) = 1 2 √ b Solution (Leibniz notation) If the original function is y = x2 , then the inverse function is defined by x = y2 . Differentiate implicitly: 1 = 2y dy dx =⇒ dy dx = 1 2 √ x V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 18 / 32
41. 41. . . . . . . Derivation: The derivative of arcsin Let y = arcsin x, so x = sin y. Then cos y dy dx = 1 =⇒ dy dx = 1 cos y = 1 cos(arcsin x) V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 19 / 32
42. 42. . . . . . . Derivation: The derivative of arcsin Let y = arcsin x, so x = sin y. Then cos y dy dx = 1 =⇒ dy dx = 1 cos y = 1 cos(arcsin x) To simplify, look at a right triangle: . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 19 / 32
43. 43. . . . . . . Derivation: The derivative of arcsin Let y = arcsin x, so x = sin y. Then cos y dy dx = 1 =⇒ dy dx = 1 cos y = 1 cos(arcsin x) To simplify, look at a right triangle: . .1 .x V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 19 / 32
44. 44. . . . . . . Derivation: The derivative of arcsin Let y = arcsin x, so x = sin y. Then cos y dy dx = 1 =⇒ dy dx = 1 cos y = 1 cos(arcsin x) To simplify, look at a right triangle: . .1 .x . .y = arcsin x V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 19 / 32
45. 45. . . . . . . Derivation: The derivative of arcsin Let y = arcsin x, so x = sin y. Then cos y dy dx = 1 =⇒ dy dx = 1 cos y = 1 cos(arcsin x) To simplify, look at a right triangle: . .1 .x . .y = arcsin x . √ 1 − x2 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 19 / 32
46. 46. . . . . . . Derivation: The derivative of arcsin Let y = arcsin x, so x = sin y. Then cos y dy dx = 1 =⇒ dy dx = 1 cos y = 1 cos(arcsin x) To simplify, look at a right triangle: cos(arcsin x) = √ 1 − x2 . .1 .x . .y = arcsin x . √ 1 − x2 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 19 / 32
47. 47. . . . . . . Derivation: The derivative of arcsin Let y = arcsin x, so x = sin y. Then cos y dy dx = 1 =⇒ dy dx = 1 cos y = 1 cos(arcsin x) To simplify, look at a right triangle: cos(arcsin x) = √ 1 − x2 So d dx arcsin(x) = 1 √ 1 − x2 . .1 .x . .y = arcsin x . √ 1 − x2 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 19 / 32
48. 48. . . . . . . Graphing arcsin and its derivative The domain of f is [−1, 1], but the domain of f′ is (−1, 1) lim x→1− f′ (x) = +∞ lim x→−1+ f′ (x) = +∞ ..| .−1 .| .1 . ..arcsin . 1 √ 1 − x2 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 20 / 32
49. 49. . . . . . . Composing with arcsin Example Let f(x) = arcsin(x3 + 1). Find f′ (x). V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 21 / 32
50. 50. . . . . . . Composing with arcsin Example Let f(x) = arcsin(x3 + 1). Find f′ (x). Solution We have d dx arcsin(x3 + 1) = 1 √ 1 − (x3 + 1)2 d dx (x3 + 1) = 3x2 √ −x6 − 2x3 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 21 / 32
51. 51. . . . . . . Derivation: The derivative of arccos Let y = arccos x, so x = cos y. Then − sin y dy dx = 1 =⇒ dy dx = 1 − sin y = 1 − sin(arccos x) V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 22 / 32
52. 52. . . . . . . Derivation: The derivative of arccos Let y = arccos x, so x = cos y. Then − sin y dy dx = 1 =⇒ dy dx = 1 − sin y = 1 − sin(arccos x) To simplify, look at a right triangle: sin(arccos x) = √ 1 − x2 So d dx arccos(x) = − 1 √ 1 − x2 . .1 . √ 1 − x2 .x . .y = arccos x V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 22 / 32
53. 53. . . . . . . Graphing arcsin and arccos ..| .−1 .| .1 . ..arcsin . ..arccos V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 23 / 32
54. 54. . . . . . . Graphing arcsin and arccos ..| .−1 .| .1 . ..arcsin . ..arccos Note cos θ = sin (π 2 − θ ) =⇒ arccos x = π 2 − arcsin x So it’s not a surprise that their derivatives are opposites. V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 23 / 32
55. 55. . . . . . . Derivation: The derivative of arctan Let y = arctan x, so x = tan y. Then sec2 y dy dx = 1 =⇒ dy dx = 1 sec2 y = cos2 (arctan x) V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 24 / 32
56. 56. . . . . . . Derivation: The derivative of arctan Let y = arctan x, so x = tan y. Then sec2 y dy dx = 1 =⇒ dy dx = 1 sec2 y = cos2 (arctan x) To simplify, look at a right triangle: . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 24 / 32
57. 57. . . . . . . Derivation: The derivative of arctan Let y = arctan x, so x = tan y. Then sec2 y dy dx = 1 =⇒ dy dx = 1 sec2 y = cos2 (arctan x) To simplify, look at a right triangle: . .x .1 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 24 / 32
58. 58. . . . . . . Derivation: The derivative of arctan Let y = arctan x, so x = tan y. Then sec2 y dy dx = 1 =⇒ dy dx = 1 sec2 y = cos2 (arctan x) To simplify, look at a right triangle: . .x .1 . .y = arctan x V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 24 / 32
59. 59. . . . . . . Derivation: The derivative of arctan Let y = arctan x, so x = tan y. Then sec2 y dy dx = 1 =⇒ dy dx = 1 sec2 y = cos2 (arctan x) To simplify, look at a right triangle: . .x .1 . .y = arctan x . √ 1 + x2 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 24 / 32
60. 60. . . . . . . Derivation: The derivative of arctan Let y = arctan x, so x = tan y. Then sec2 y dy dx = 1 =⇒ dy dx = 1 sec2 y = cos2 (arctan x) To simplify, look at a right triangle: cos(arctan x) = 1 √ 1 + x2 . .x .1 . .y = arctan x . √ 1 + x2 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 24 / 32
61. 61. . . . . . . Derivation: The derivative of arctan Let y = arctan x, so x = tan y. Then sec2 y dy dx = 1 =⇒ dy dx = 1 sec2 y = cos2 (arctan x) To simplify, look at a right triangle: cos(arctan x) = 1 √ 1 + x2 So d dx arctan(x) = 1 1 + x2 . .x .1 . .y = arctan x . √ 1 + x2 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 24 / 32
62. 62. . . . . . . Graphing arctan and its derivative . .x .y .arctan . 1 1 + x2 .π/2 .−π/2 The domain of f and f′ are both (−∞, ∞) Because of the horizontal asymptotes, lim x→±∞ f′ (x) = 0 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 25 / 32
63. 63. . . . . . . Composing with arctan Example Let f(x) = arctan √ x. Find f′ (x). V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 26 / 32
64. 64. . . . . . . Composing with arctan Example Let f(x) = arctan √ x. Find f′ (x). Solution d dx arctan √ x = 1 1 + (√ x )2 d dx √ x = 1 1 + x · 1 2 √ x = 1 2 √ x + 2x √ x V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 26 / 32
65. 65. . . . . . . Derivation: The derivative of arcsec Try this first. V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 27 / 32
66. 66. . . . . . . Derivation: The derivative of arcsec Try this first. Let y = arcsec x, so x = sec y. Then sec y tan y dy dx = 1 =⇒ dy dx = 1 sec y tan y = 1 x tan(arcsec(x)) V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 27 / 32
67. 67. . . . . . . Derivation: The derivative of arcsec Try this first. Let y = arcsec x, so x = sec y. Then sec y tan y dy dx = 1 =⇒ dy dx = 1 sec y tan y = 1 x tan(arcsec(x)) To simplify, look at a right triangle: . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 27 / 32
68. 68. . . . . . . Derivation: The derivative of arcsec Try this first. Let y = arcsec x, so x = sec y. Then sec y tan y dy dx = 1 =⇒ dy dx = 1 sec y tan y = 1 x tan(arcsec(x)) To simplify, look at a right triangle: . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 27 / 32
69. 69. . . . . . . Derivation: The derivative of arcsec Try this first. Let y = arcsec x, so x = sec y. Then sec y tan y dy dx = 1 =⇒ dy dx = 1 sec y tan y = 1 x tan(arcsec(x)) To simplify, look at a right triangle: . .x .1 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 27 / 32
70. 70. . . . . . . Derivation: The derivative of arcsec Try this first. Let y = arcsec x, so x = sec y. Then sec y tan y dy dx = 1 =⇒ dy dx = 1 sec y tan y = 1 x tan(arcsec(x)) To simplify, look at a right triangle: . .x .1 . .y = arcsec x V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 27 / 32
71. 71. . . . . . . Derivation: The derivative of arcsec Try this first. Let y = arcsec x, so x = sec y. Then sec y tan y dy dx = 1 =⇒ dy dx = 1 sec y tan y = 1 x tan(arcsec(x)) To simplify, look at a right triangle: tan(arcsec x) = √ x2 − 1 1 . .x .1 . .y = arcsec x . √ x2 − 1 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 27 / 32
72. 72. . . . . . . Derivation: The derivative of arcsec Try this first. Let y = arcsec x, so x = sec y. Then sec y tan y dy dx = 1 =⇒ dy dx = 1 sec y tan y = 1 x tan(arcsec(x)) To simplify, look at a right triangle: tan(arcsec x) = √ x2 − 1 1 So d dx arcsec(x) = 1 x √ x2 − 1 . .x .1 . .y = arcsec x . √ x2 − 1 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 27 / 32
73. 73. . . . . . . Another Example Example Let f(x) = earcsec 3x . Find f′ (x). V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 28 / 32
74. 74. . . . . . . Another Example Example Let f(x) = earcsec 3x . Find f′ (x). Solution f′ (x) = earcsec 3x · 1 3x √ (3x)2 − 1 · 3 = 3earcsec 3x 3x √ 9x2 − 1 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 28 / 32
75. 75. . . . . . . Outline Inverse Trigonometric Functions Derivatives of Inverse Trigonometric Functions Arcsine Arccosine Arctangent Arcsecant Applications V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 29 / 32
76. 76. . . . . . . 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? V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 30 / 32
77. 77. . . . . . . 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? 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. V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 30 / 32
78. 78. . . . . . . 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. . .2 ft .y .130 ft/sec . .θ V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 31 / 32
79. 79. . . . . . . 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 dt = 1 1 + (y/2)2 · 1 2 dy dt . .2 ft .y .130 ft/sec . .θ V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 31 / 32
80. 80. . . . . . . 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 dt = 1 1 + (y/2)2 · 1 2 dy dt When y = 0 and y′ = −130, then dθ dt y=0 = 1 1 + 0 · 1 2 (−130) = −65 rad/sec . .2 ft .y .130 ft/sec . .θ V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 31 / 32
81. 81. . . . . . . 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 dt = 1 1 + (y/2)2 · 1 2 dy dt When y = 0 and y′ = −130, then dθ dt y=0 = 1 1 + 0 · 1 2 (−130) = −65 rad/sec The human eye can only track at 3 rad/sec! . .2 ft .y .130 ft/sec . .θ V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 31 / 32
82. 82. . . . . . . Summary y y′ arcsin x 1 √ 1 − x2 arccos x − 1 √ 1 − x2 arctan x 1 1 + x2 arccot x − 1 1 + x2 arcsec x 1 x √ x2 − 1 arccsc x − 1 x √ x2 − 1 Remarkable that the derivatives of these transcendental functions are algebraic (or even rational!) V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 32 / 32