Lesson 6: The Derivative

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  • 1. Section 2.1 The Derivative and Rates of Change V63.0121.002.2010Su, Calculus I New York University May 20, 2010 Announcements Written Assignment 1 is on the website
  • 2. Announcements Written Assignment 1 is on the website V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 2 / 28
  • 3. Format of written work Please: Use scratch paper and copy your final work onto fresh paper. Use loose-leaf paper (not torn from a notebook). Write your name, assignment number, and date at the top. Staple your homework together. See the website for more information. V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 3 / 28
  • 4. Objectives Understand and state the definition of the derivative of a function at a point. Given a function and a point in its domain, decide if the function is differentiable at the point and find the value of the derivative at that point. Understand and give several examples of derivatives modeling rates of change in science. V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 4 / 28
  • 5. Outline Rates of Change Tangent Lines Velocity Population growth Marginal costs The derivative, defined Derivatives of (some) power functions V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 5 / 28
  • 6. The tangent problem Problem Given a curve and a point on the curve, find the slope of the line tangent to the curve at that point. V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 6 / 28
  • 7. The tangent problem Problem Given a curve and a point on the curve, find the slope of the line tangent to the curve at that point. Example Find the slope of the line tangent to the curve y = x 2 at the point (2, 4). V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 6 / 28
  • 8. Graphically and numerically y x 2 − 22 x m= x −2 4 x 2 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
  • 9. Graphically and numerically y x 2 − 22 x m= x −2 3 9 4 x 2 3 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
  • 10. Graphically and numerically y x 2 − 22 x m= x −2 3 5 9 4 x 2 3 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
  • 11. Graphically and numerically y x 2 − 22 x m= x −2 3 5 2.5 6.25 4 x 2 2.5 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
  • 12. Graphically and numerically y x 2 − 22 x m= x −2 3 5 2.5 4.5 6.25 4 x 2 2.5 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
  • 13. Graphically and numerically y x 2 − 22 x m= x −2 3 5 2.5 4.5 2.1 4.41 4 x 2.1 2 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
  • 14. Graphically and numerically y x 2 − 22 x m= x −2 3 5 2.5 4.5 2.1 4.1 4.41 4 x 2.1 2 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
  • 15. Graphically and numerically y x 2 − 22 x m= x −2 3 5 2.5 4.5 2.1 4.1 2.01 4.0401 4 x 2.01 2 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
  • 16. Graphically and numerically y x 2 − 22 x m= x −2 3 5 2.5 4.5 2.1 4.1 2.01 4.01 4.0401 4 x 2.01 2 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
  • 17. Graphically and numerically y x 2 − 22 x m= x −2 3 5 2.5 4.5 2.1 4.1 2.01 4.01 4 1 1 x 1 2 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
  • 18. Graphically and numerically y x 2 − 22 x m= x −2 3 5 2.5 4.5 2.1 4.1 2.01 4.01 4 1 3 1 x 1 2 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
  • 19. Graphically and numerically y x 2 − 22 x m= x −2 3 5 2.5 4.5 2.1 4.1 2.01 4.01 4 1.5 2.25 1 3 x 1.5 2 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
  • 20. Graphically and numerically y x 2 − 22 x m= x −2 3 5 2.5 4.5 2.1 4.1 2.01 4.01 4 1.5 3.5 2.25 1 3 x 1.5 2 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
  • 21. Graphically and numerically y x 2 − 22 x m= x −2 3 5 2.5 4.5 2.1 4.1 2.01 4.01 4 1.9 3.61 1.5 3.5 1 3 x 1.9 2 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
  • 22. Graphically and numerically y x 2 − 22 x m= x −2 3 5 2.5 4.5 2.1 4.1 2.01 4.01 4 3.9601 1.9 3.9 3.61 1.5 3.5 1 3 x 1.99 1.9 2 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
  • 23. Graphically and numerically y x 2 − 22 x m= x −2 3 5 2.5 4.5 2.1 4.1 2.01 4.01 1.99 4 3.9601 1.9 3.9 1.5 3.5 1 3 x 1.99 2 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
  • 24. Graphically and numerically y x 2 − 22 x m= x −2 3 5 2.5 4.5 2.1 4.1 2.01 4.01 1.99 3.99 4 1.9 3.9 1.5 3.5 1 3 x 2 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
  • 25. Graphically and numerically y x 2 − 22 x m= x −2 3 5 9 2.5 4.5 2.1 4.1 2.01 4.01 6.25 limit 4 4.41 1.99 3.99 4.0401 4 3.9601 1.9 3.9 3.61 1.5 3.5 2.25 1 3 1 x 1 1.5 2.1 3 1.99 1.9 2.5 2.01 2 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
  • 26. The tangent problem Problem Given a curve and a point on the curve, find the slope of the line tangent to the curve at that point. Example Find the slope of the line tangent to the curve y = x 2 at the point (2, 4). Upshot If the curve is given by y = f (x), and the point on the curve is (a, f (a)), then the slope of the tangent line is given by f (x) − f (a) mtangent = lim x→a x −a V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 8 / 28
  • 27. Velocity Problem Given the position function of a moving object, find the velocity of the object at a certain instant in time. Example Drop a ball off the roof of the Silver Center so that its height can be described by h(t) = 50 − 5t 2 where t is seconds after dropping it and h is meters above the ground. How fast is it falling one second after we drop it? V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 9 / 28
  • 28. Numerical evidence h(t) − h(1) t vave = t −1 2 − 15 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 10 / 28
  • 29. Numerical evidence h(t) − h(1) t vave = t −1 2 − 15 1.5 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 10 / 28
  • 30. Numerical evidence h(t) − h(1) t vave = t −1 2 − 15 1.5 − 12.5 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 10 / 28
  • 31. Numerical evidence h(t) − h(1) t vave = t −1 2 − 15 1.5 − 12.5 1.1 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 10 / 28
  • 32. Numerical evidence h(t) − h(1) t vave = t −1 2 − 15 1.5 − 12.5 1.1 − 10.5 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 10 / 28
  • 33. Numerical evidence h(t) − h(1) t vave = t −1 2 − 15 1.5 − 12.5 1.1 − 10.5 1.01 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 10 / 28
  • 34. Numerical evidence h(t) − h(1) t vave = t −1 2 − 15 1.5 − 12.5 1.1 − 10.5 1.01 − 10.05 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 10 / 28
  • 35. Numerical evidence h(t) − h(1) t vave = t −1 2 − 15 1.5 − 12.5 1.1 − 10.5 1.01 − 10.05 1.001 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 10 / 28
  • 36. Numerical evidence h(t) − h(1) t vave = t −1 2 − 15 1.5 − 12.5 1.1 − 10.5 1.01 − 10.05 1.001 − 10.005 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 10 / 28
  • 37. Velocity Problem Given the position function of a moving object, find the velocity of the object at a certain instant in time. Example Drop a ball off the roof of the Silver Center so that its height can be described by h(t) = 50 − 5t 2 where t is seconds after dropping it and h is meters above the ground. How fast is it falling one second after we drop it? Solution The answer is (50 − 5t 2 ) − 45 5 − 5t 2 5(1 − t)(1 + t) v = lim = lim = lim t→1 t −1 t→1 t − 1 t→1 t −1 V63.0121.002.2010Su, Calculus I (NYU) t) = Section 2.1 = −10 = (−5) lim (1 + −5 · 2 The Derivative May 20, 2010 11 / 28
  • 38. y = h(t) Upshot If the height function is given by h(t), the instantaneous velocity at time t0 is given by h(t) − h(t0 ) v = lim t→t0 t − t0 h(t0 + ∆t) − h(t0 ) = lim ∆t→0 ∆t ∆t t t0 t V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 12 / 28
  • 39. Population growth Problem Given the population function of a group of organisms, find the rate of growth of the population at a particular instant. V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 13 / 28
  • 40. Population growth Problem Given the population function of a group of organisms, find the rate of growth of the population at a particular instant. Example Suppose the population of fish in the East River is given by the function 3e t P(t) = 1 + et where t is in years since 2000 and P is in millions of fish. Is the fish population growing fastest in 1990, 2000, or 2010? (Estimate numerically) V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 13 / 28
  • 41. Derivation Let ∆t be an increment in time and ∆P the corresponding change in population: ∆P = P(t + ∆t) − P(t) This depends on ∆t, so we want ∆P 1 3e t+∆t 3e t lim = lim − ∆t→0 ∆t ∆t→0 ∆t 1 + e t+∆t 1 + et V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 14 / 28
  • 42. Derivation Let ∆t be an increment in time and ∆P the corresponding change in population: ∆P = P(t + ∆t) − P(t) This depends on ∆t, so we want ∆P 1 3e t+∆t 3e t lim = lim − ∆t→0 ∆t ∆t→0 ∆t 1 + e t+∆t 1 + et Too hard! Try a small ∆t to approximate. V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 14 / 28
  • 43. Numerical evidence ∆P ∆P Use ∆t = 0.1, and use to approximate lim . ∆t ∆t→0 ∆t P(−10 + 0.1) − P(−10) r1990 ≈ = 0.1 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 15 / 28
  • 44. Numerical evidence ∆P ∆P Use ∆t = 0.1, and use to approximate lim . ∆t ∆t→0 ∆t P(−10 + 0.1) − P(−10) r1990 ≈ = 0.000143229 0.1 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 15 / 28
  • 45. Numerical evidence ∆P ∆P Use ∆t = 0.1, and use to approximate lim . ∆t ∆t→0 ∆t P(−10 + 0.1) − P(−10) r1990 ≈ = 0.000143229 0.1 P(0.1) − P(0) r2000 ≈ = 0.1 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 15 / 28
  • 46. Numerical evidence ∆P ∆P Use ∆t = 0.1, and use to approximate lim . ∆t ∆t→0 ∆t P(−10 + 0.1) − P(−10) r1990 ≈ = 0.000143229 0.1 P(0.1) − P(0) r2000 ≈ = 0.749376 0.1 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 15 / 28
  • 47. Numerical evidence ∆P ∆P Use ∆t = 0.1, and use to approximate lim . ∆t ∆t→0 ∆t P(−10 + 0.1) − P(−10) r1990 ≈ = 0.000143229 0.1 P(0.1) − P(0) r2000 ≈ = 0.749376 0.1 P(10 + 0.1) − P(10) r2010 ≈ = 0.1 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 15 / 28
  • 48. Numerical evidence ∆P ∆P Use ∆t = 0.1, and use to approximate lim . ∆t ∆t→0 ∆t P(−10 + 0.1) − P(−10) r1990 ≈ = 0.000143229 0.1 P(0.1) − P(0) r2000 ≈ = 0.749376 0.1 P(10 + 0.1) − P(10) r2010 ≈ = 0.0001296 0.1 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 15 / 28
  • 49. Population growth Problem Given the population function of a group of organisms, find the rate of growth of the population at a particular instant. Example Suppose the population of fish in the East River is given by the function 3e t P(t) = 1 + et where t is in years since 2000 and P is in millions of fish. Is the fish population growing fastest in 1990, 2000, or 2010? (Estimate numerically) Solution We estimates the rates of growth to be 0.000143229, 0.749376, and 0.0001296. So the population is growing fastest in 2000. V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 16 / 28
  • 50. Upshot The instantaneous population growth is given by P(t + ∆t) − P(t) lim ∆t→0 ∆t V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 17 / 28
  • 51. Marginal costs Problem Given the production cost of a good, find the marginal cost of production after having produced a certain quantity. V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 18 / 28
  • 52. Marginal costs Problem Given the production cost of a good, find the marginal cost of production after having produced a certain quantity. Example Suppose the cost of producing q tons of rice on our paddy in a year is C (q) = q 3 − 12q 2 + 60q We are currently producing 5 tons a year. Should we change that? V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 18 / 28
  • 53. Comparisons q C (q) 4 5 6 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 19 / 28
  • 54. Comparisons q C (q) 4 112 5 6 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 19 / 28
  • 55. Comparisons q C (q) 4 112 5 125 6 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 19 / 28
  • 56. Comparisons q C (q) 4 112 5 125 6 144 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 19 / 28
  • 57. Comparisons q C (q) AC (q) = C (q)/q 4 112 5 125 6 144 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 19 / 28
  • 58. Comparisons q C (q) AC (q) = C (q)/q 4 112 28 5 125 6 144 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 19 / 28
  • 59. Comparisons q C (q) AC (q) = C (q)/q 4 112 28 5 125 25 6 144 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 19 / 28
  • 60. Comparisons q C (q) AC (q) = C (q)/q 4 112 28 5 125 25 6 144 24 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 19 / 28
  • 61. Comparisons q C (q) AC (q) = C (q)/q ∆C = C (q + 1) − C (q) 4 112 28 5 125 25 6 144 24 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 19 / 28
  • 62. Comparisons q C (q) AC (q) = C (q)/q ∆C = C (q + 1) − C (q) 4 112 28 13 5 125 25 6 144 24 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 19 / 28
  • 63. Comparisons q C (q) AC (q) = C (q)/q ∆C = C (q + 1) − C (q) 4 112 28 13 5 125 25 19 6 144 24 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 19 / 28
  • 64. Comparisons q C (q) AC (q) = C (q)/q ∆C = C (q + 1) − C (q) 4 112 28 13 5 125 25 19 6 144 24 31 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 19 / 28
  • 65. Marginal costs Problem Given the production cost of a good, find the marginal cost of production after having produced a certain quantity. Example Suppose the cost of producing q tons of rice on our paddy in a year is C (q) = q 3 − 12q 2 + 60q We are currently producing 5 tons a year. Should we change that? Example If q = 5, then C = 125, ∆C = 19, while AC = 25. So we should produce more to lower average costs. V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 20 / 28
  • 66. Upshot The incremental cost ∆C = C (q + 1) − C (q) is useful, but depends on units. V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 21 / 28
  • 67. Upshot The incremental cost ∆C = C (q + 1) − C (q) is useful, but depends on units. The marginal cost after producing q given by C (q + ∆q) − C (q) MC = lim ∆q→0 ∆q is more useful since it’s unit-independent. V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 21 / 28
  • 68. Outline Rates of Change Tangent Lines Velocity Population growth Marginal costs The derivative, defined Derivatives of (some) power functions V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 22 / 28
  • 69. The definition All of these rates of change are found the same way! V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 23 / 28
  • 70. The definition All of these rates of change are found the same way! Definition Let f be a function and a a point in the domain of f . If the limit f (a + h) − f (a) f (x) − f (a) f (a) = lim = lim h→0 h x→a x −a exists, the function is said to be differentiable at a and f (a) is the derivative of f at a. V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 23 / 28
  • 71. Derivative of the squaring function Example Suppose f (x) = x 2 . Use the definition of derivative to find f (a). V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 24 / 28
  • 72. Derivative of the squaring function Example Suppose f (x) = x 2 . Use the definition of derivative to find f (a). Solution f (a + h) − f (a) (a + h)2 − a2 f (a) = lim = lim h→0 h h→0 h (a2 + 2ah + h2 ) − a2 2ah + h2 = lim = lim h→0 h h→0 h = lim (2a + h) = 2a. h→0 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 24 / 28
  • 73. Derivative of the reciprocal function Example 1 Suppose f (x) = . Use the x definition of the derivative to find f (2). V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 25 / 28
  • 74. Derivative of the reciprocal function Example 1 Suppose f (x) = . Use the x definition of the derivative to find x f (2). Solution 1/x − 1/2 2−x f (2) = lim = lim x x→2 x −2 x→2 2x(x − 2) −1 1 = lim =− x→2 2x 4 V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 25 / 28
  • 75. The Sure-Fire Sally Rule (SFSR) for adding Fractions In anticipation of the question, “How did you get that?” a c ad ± bc ± = b d bd So 1 1 2−x − x 2 = 2x x −2 x −2 2−x = 2x(x − 2) V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 26 / 28
  • 76. The Sure-Fire Sally Rule (SFSR) for adding Fractions In anticipation of the question, “How did you get that?” a c ad ± bc ± = b d bd So 1 1 2−x − x 2 = 2x x −2 x −2 2−x = 2x(x − 2) V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 26 / 28
  • 77. Worksheet V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 27 / 28
  • 78. Summary The derivative measures instantaneous rate of change The derivative has many interpretations: slope of the tangent line, velocity, marginal quantities, etc. The derivative is calculated with a limit. V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 28 / 28