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# Lesson 12: Linear Approximations and Differentials (slides)

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The line tangent to a curve is also the line which best "fits" the curve near that point. So derivatives can be used for approximating complicated functions with simple linear ones. Differentials are another set of notation for the same problem.

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### Lesson 12: Linear Approximations and Differentials (slides)

1. 1. Sec on 2.8 Linear Approxima ons and Diﬀeren als V63.0121.011: Calculus I Professor Ma hew Leingang New York University March 2, 2011.
2. 2. Announcements Quiz in recita on this week on 1.5, 1.6, 2.1, 2.2 Midterm March 7 in class on 1.1–2.5 No quiz in recita on next week
3. 3. Midterm FAQ Ques on What sec ons are covered on the midterm?
4. 4. Midterm FAQ Ques on What sec ons are covered on the midterm? Answer The midterm will cover Sec ons 1.1–2.5 (The Chain Rule).
5. 5. Midterm FAQ Ques on What sec ons are covered on the midterm? Answer The midterm will cover Sec ons 1.1–2.5 (The Chain Rule). Ques on Is Sec on 2.6 going to be on the midterm?
6. 6. Midterm FAQ Ques on What sec ons are covered on the midterm? Answer The midterm will cover Sec ons 1.1–2.5 (The Chain Rule). Ques on Is Sec on 2.6 going to be on the midterm? Answer The midterm will cover Sec ons 1.1–2.5 (The Chain Rule).
7. 7. Midterm FAQ, continued Ques on What format will the exam take?
8. 8. Midterm FAQ, continued Ques on What format will the exam take? Answer There will be both ﬁxed-response (e.g., mul ple choice) and free-response ques ons.
9. 9. Midterm FAQ, continued Ques on Will explana ons be necessary?
10. 10. Midterm FAQ, continued Ques on Will explana ons be necessary? Answer Yes, on free-response problems we will expect you to explain yourself. This is why it was required on wri en homework.
11. 11. Midterm FAQ, continued Ques on Is (topic X) going to be tested?
12. 12. Midterm FAQ, continued Ques on Is (topic X) going to be tested? Answer Everything covered in class or on homework is fair game for the exam. No topic that was not covered in class nor on homework will be on the exam. (This is not the same as saying all exam problems are similar to class examples or homework problems.)
13. 13. Midterm FAQ, continued Ques on Will there be a review session?
14. 14. Midterm FAQ, continued Ques on Will there be a review session? Answer The recita ons this week will review for the exam.
15. 15. Midterm FAQ, continued Ques on Will calculators be allowed?
16. 16. Midterm FAQ, continued Ques on Will calculators be allowed? Answer No. The exam is designed for pencil and brain.
17. 17. Midterm FAQ, continued Ques on How should I study?
18. 18. Midterm FAQ, continued Ques on How should I study? Answer The exam has problems; study by doing problems. If you get one right, think about how you got it right. If you got it wrong or didn’t get it at all, reread the textbook and do easier problems to build up your understanding. Break up the material into chunks. (related) Don’t put it all oﬀ un l the night before. Ask ques ons.
19. 19. Midterm FAQ, continued Ques on How many ques ons are there?
20. 20. Midterm FAQ, continued Ques on How many ques ons are there? Answer Does this ques on contribute to your understanding of the material?
21. 21. Midterm FAQ, continued Ques on Will there be a curve on the exam?
22. 22. Midterm FAQ, continued Ques on Will there be a curve on the exam? Answer Does this ques on contribute to your understanding of the material?
23. 23. Midterm FAQ, continued Ques on When will you grade my get-to-know-you and photo extra credit?
24. 24. Midterm FAQ, continued Ques on When will you grade my get-to-know-you and photo extra credit? Answer Does this ques on contribute to your understanding of the material?
25. 25. Objectives Use tangent lines to make linear approxima ons to a func on. Given a func on and a point in the domain, compute the lineariza on of the func on at that point. Use lineariza on to approximate values of func ons Given a func on, compute the diﬀeren al of that func on Use the diﬀeren al nota on to es mate error in linear approxima ons.
26. 26. Outline The linear approxima on of a func on near a point Examples Ques ons Diﬀeren als Using diﬀeren als to es mate error Advanced Examples
27. 27. The Big Idea Ques on What linear func on best approximates f near a?
28. 28. The Big Idea Ques on What linear func on best approximates f near a? Answer The tangent line, of course!
29. 29. The Big Idea Ques on What linear func on best approximates f near a? Answer The tangent line, of course! Ques on What is the equa on for the line tangent to y = f(x) at (a, f(a))?
30. 30. The Big Idea Ques on What linear func on best approximates f near a? Answer The tangent line, of course! Ques on What is the equa on for the line tangent to y = f(x) at (a, f(a))? Answer L(x) = f(a) + f′ (a)(x − a)
31. 31. tangent line = linear approximation y The func on L(x) L(x) = f(a) + f′ (a)(x − a) f(x) is a decent approxima on to f near a. f(a) x−a . x a x
32. 32. tangent line = linear approximation y The func on L(x) L(x) = f(a) + f′ (a)(x − a) f(x) is a decent approxima on to f near a. f(a) x−a How decent? The closer x is to a, the be er the approxima on L(x) is to f(x) . x a x
33. 33. Example Example Es mate sin(61◦ ) = sin(61π/180) by using a linear approxima on (i) about a = 0 (ii) about a = 60◦ = π/3.
34. 34. Example Example Es mate sin(61◦ ) = sin(61π/180) by using a linear approxima on (i) about a = 0 (ii) about a = 60◦ = π/3. Solu on (i) If f(x) = sin x, then f(0) = 0 and f′ (0) = 1. So the linear approxima on near 0 is L(x) = 0 + 1 · x = x. ( ) 61π 61π sin ≈ ≈ 1.06465 180 180
35. 35. Example Example Es mate sin(61◦ ) = sin(61π/180) by using a linear approxima on (i) about a = 0 (ii) about a = 60◦ = π/3. Solu on (i) Solu on (ii) ( ) We have f π = ( ) 3 and If f(x) = sin x, then f(0) = 0 f′ π = . and f′ (0) = 1. 3 So the linear approxima on near 0 is L(x) = 0 + 1 · x = x. ( ) 61π 61π sin ≈ ≈ 1.06465 180 180
36. 36. Example Example Es mate sin(61◦ ) = sin(61π/180) by using a linear approxima on (i) about a = 0 (ii) about a = 60◦ = π/3. Solu on (i) Solu on (ii) ( ) √ 3 We have f π = ( ) 3 2 and If f(x) = sin x, then f(0) = 0 f′ π = . and f′ (0) = 1. 3 So the linear approxima on near 0 is L(x) = 0 + 1 · x = x. ( ) 61π 61π sin ≈ ≈ 1.06465 180 180
37. 37. Example Example Es mate sin(61◦ ) = sin(61π/180) by using a linear approxima on (i) about a = 0 (ii) about a = 60◦ = π/3. Solu on (i) Solu on (ii) ( ) √ 3 We have f π = ( ) 3 2 and If f(x) = sin x, then f(0) = 0 f′ π = 1 . and f′ (0) = 1. 3 2 So the linear approxima on near 0 is L(x) = 0 + 1 · x = x. ( ) 61π 61π sin ≈ ≈ 1.06465 180 180
38. 38. Example Example Es mate sin(61◦ ) = sin(61π/180) by using a linear approxima on (i) about a = 0 (ii) about a = 60◦ = π/3. Solu on (i) Solu on (ii) ( ) √ 3 We have f π = ( ) 3 2 and If f(x) = sin x, then f(0) = 0 f′ π = 1 . and f′ (0) = 1. 3 2 So the linear approxima on So the linear approxima on is near 0 is L(x) = 0 + 1 · x = x. L(x) = ( ) 61π 61π sin ≈ ≈ 1.06465 180 180
39. 39. Example Example Es mate sin(61◦ ) = sin(61π/180) by using a linear approxima on (i) about a = 0 (ii) about a = 60◦ = π/3. Solu on (i) Solu on (ii) ( ) √ 3 We have f π = ( ) 3 2 and If f(x) = sin x, then f(0) = 0 f′ π = 1 . and f′ (0) = 1. 3 2 So the linear approxima on So the linear approxima on is √ 3 1( π) near 0 is L(x) = 0 + 1 · x = x. L(x) = + x− ( ) 2 2 3 61π 61π sin ≈ ≈ 1.06465 180 180
40. 40. Example Example Es mate sin(61◦ ) = sin(61π/180) by using a linear approxima on (i) about a = 0 (ii) about a = 60◦ = π/3. Solu on (i) Solu on (ii) ( ) √ 3 We have f π = ( ) 3 2 and If f(x) = sin x, then f(0) = 0 f′ π = 1 . and f′ (0) = 1. 3 2 So the linear approxima on So the linear approxima on is √ 3 1( π) near 0 is L(x) = 0 + 1 · x = x. L(x) = + x− ( ) 2 2 3 61π 61π ( ) sin ≈ ≈ 1.06465 61π 180 180 sin ≈ 180
41. 41. Example Example Es mate sin(61◦ ) = sin(61π/180) by using a linear approxima on (i) about a = 0 (ii) about a = 60◦ = π/3. Solu on (i) Solu on (ii) ( ) √ 3 We have f π = ( ) 3 2 and If f(x) = sin x, then f(0) = 0 f′ π = 1 . and f′ (0) = 1. 3 2 So the linear approxima on So the linear approxima on is √ 3 1( π) near 0 is L(x) = 0 + 1 · x = x. L(x) = + x− ( ) 2 2 3 61π 61π ( ) sin ≈ ≈ 1.06465 61π 180 180 sin ≈ 0.87475 180
42. 42. Example Example Es mate sin(61◦ ) = sin(61π/180) by using a linear approxima on (i) about a = 0 (ii) about a = 60◦ = π/3. Solu on (i) Solu on (ii) ( ) √ 3 We have f π = ( ) 3 2 and If f(x) = sin x, then f(0) = 0 f′ π = 1 . and f′ (0) = 1. 3 2 So the linear approxima on So the linear approxima on is √ 3 1( π) near 0 is L(x) = 0 + 1 · x = x. L(x) = + x− ( ) 2 2 3 61π 61π ( ) sin ≈ ≈ 1.06465 61π 180 180 sin ≈ 0.87475 180
43. 43. Example Example Es mate sin(61◦ ) = sin(61π/180) by using a linear approxima on (i) about a = 0 (ii) about a = 60◦ = π/3. Solu on (i) Solu on (ii) ( ) √ 3 We have f π = ( ) 3 2 and If f(x) = sin x, then f(0) = 0 f′ π = 1 . and f′ (0) = 1. 3 2 So the linear approxima on So the linear approxima on is √ 3 1( π) near 0 is L(x) = 0 + 1 · x = x. L(x) = + x− ( ) 2 2 3 61π 61π ( ) sin ≈ ≈ 1.06465 61π 180 180 sin ≈ 0.87475 180
44. 44. Illustration y y = sin x . x 61◦
45. 45. Illustration y y = L1 (x) = x y = sin x . x 0 61◦
46. 46. Illustration y y = L1 (x) = x big diﬀerence! y = sin x . x 0 61◦
47. 47. Illustration y y = L1 (x) = x √ ( ) y = L2 (x) = 2 3 + 1 2 x− π 3 y = sin x . x 0 π/3 61◦
48. 48. Illustration y y = L1 (x) = x √ ( ) y = L2 (x) = 2 3 + 1 2 x− π 3 y = sin x very li le diﬀerence! . x 0 π/3 61◦
49. 49. Another Example Example √ Es mate 10 using the fact that 10 = 9 + 1.
50. 50. Another Example Example √ Es mate 10 using the fact that 10 = 9 + 1. Solu on √ The key step is to use a linear approxima on to f(x) = x near a = 9 to es mate √ f(10) = 10.
51. 51. Another Example Example √ Es mate 10 using the fact that 10 = 9 + 1. Solu on √ The key step is to use a linear approxima on to f(x) = x near a = 9 to es mate √ f(10) = 10. f(10) ≈ L(10) = f(9) + f′ (9)(10 − 9) 1 19 =3+ (1) = ≈ 3.167 2·3 6
52. 52. Another Example Example √ Es mate 10 using the fact that 10 = 9 + 1. Solu on √ The key step is to use a linear approxima on to f(x) = x near a = 9 to es mate √ f(10) = 10. f(10) ≈ L(10) = f(9) + f′ (9)(10 − 9) 1 19 =3+ (1) = ≈ 3.167 2·3 6 ( )2 19 Check: = 6
53. 53. Another Example Example √ Es mate 10 using the fact that 10 = 9 + 1. Solu on √ The key step is to use a linear approxima on to f(x) = x near a = 9 to es mate √ f(10) = 10. f(10) ≈ L(10) = f(9) + f′ (9)(10 − 9) 1 19 =3+ (1) = ≈ 3.167 2·3 6 ( )2 19 361 Check: = . 6 36
54. 54. Dividing without dividing? Example A student has an irra onal fear of long division and needs to es mate 577 ÷ 408. He writes 577 1 1 1 = 1 + 169 = 1 + 169 × × . 408 408 4 102 1 Help the student es mate . 102
55. 55. Dividing without dividing? Solu on 1 Let f(x) = . We know f(100) and we want to es mate f(102). x 1 1 f(102) ≈ f(100) + f′ (100)(2) = − (2) = 0.0098 100 1002 577 =⇒ ≈ 1.41405 408 577 Calculator check: ≈ 1.41422. 408
56. 56. Questions Example Suppose we are traveling in a car and at noon our speed is 50 mi/hr. How far will we have traveled by 2:00pm? by 3:00pm? By midnight?
57. 57. Questions Example Suppose we are traveling in a car and at noon our speed is 50 mi/hr. How far will we have traveled by 2:00pm? by 3:00pm? By midnight? Answer 100 mi 150 mi 600 mi (?) (Is it reasonable to assume 12 hours at the same speed?)
58. 58. Questions Example Suppose our factory makes MP3 players and the marginal cost is currently \$50/lot. How much will it cost to make 2 more lots? 3 more lots? 12 more lots?
59. 59. Questions Example Suppose our factory makes MP3 players and the marginal cost is currently \$50/lot. How much will it cost to make 2 more lots? 3 more lots? 12 more lots? Answer \$100 \$150 \$600 (?)
60. 60. Questions Example Suppose a line goes through the point (x0 , y0 ) and has slope m. If the point is moved horizontally by dx, while staying on the line, what is the corresponding ver cal movement?
61. 61. Questions Example Suppose a line goes through the point (x0 , y0 ) and has slope m. If the point is moved horizontally by dx, while staying on the line, what is the corresponding ver cal movement? Answer The slope of the line is rise m= run We are given a “run” of dx, so the corresponding “rise” is m dx.
62. 62. Outline The linear approxima on of a func on near a point Examples Ques ons Diﬀeren als Using diﬀeren als to es mate error Advanced Examples
63. 63. Diﬀerentials are derivatives The fact that the the tangent line is an approxima on means that y f(x + ∆x) − f(x) ≈ f′ (x) ∆x ∆y dy Rename ∆x = dx, so we can write this as dy ∆y ≈ dy = f′ (x)dx. ∆y dx = ∆x Note this looks a lot like the Leibniz-Newton iden ty dy . x = f′ (x) x x + ∆x dx
64. 64. Using diﬀerentials to estimate error Es ma ng error with diﬀeren als y If y = f(x), x0 and ∆x is known, and an es mate of ∆y is desired: Approximate: ∆y ≈ dy ∆y dy Diﬀeren ate: dy = f′ (x) dx dx = ∆x Evaluate at x = x0 and . x dx = ∆x. x x + ∆x
65. 65. Using diﬀerentials to estimate error Example A regular sheet of plywood measures 8 ft × 4 ft. Suppose a defec ve plywood-cu ng machine will cut a rectangle whose width is exactly half its length, but the length is prone to errors. If the length is oﬀ by 1 in, how bad can the area of the sheet be oﬀ by?
66. 66. Solution Solu on 1 Write A(ℓ) = ℓ2 . We want to know ∆A when ℓ = 8 ft and 2 ∆ℓ = 1 in.
67. 67. Solution Solu on 1 Write A(ℓ) = ℓ2 . We want to know ∆A when ℓ = 8 ft and 2 ∆ℓ = 1 in. ( ) 97 9409 9409 (I) A(ℓ + ∆ℓ) = A = So ∆A = − 32 ≈ 0.6701. 12 288 288
68. 68. Solution Solu on 1 Write A(ℓ) = ℓ2 . We want to know ∆A when ℓ = 8 ft and 2 ∆ℓ = 1 in. ( ) 97 9409 9409 (I) A(ℓ + ∆ℓ) = A = So ∆A = − 32 ≈ 0.6701. 12 288 288 dA (II) = ℓ, so dA = ℓ dℓ, which should be a good es mate for ∆ℓ. dℓ
69. 69. Solution Solu on 1 Write A(ℓ) = ℓ2 . We want to know ∆A when ℓ = 8 ft and 2 ∆ℓ = 1 in. ( ) 97 9409 9409 (I) A(ℓ + ∆ℓ) = A = So ∆A = − 32 ≈ 0.6701. 12 288 288 dA (II) = ℓ, so dA = ℓ dℓ, which should be a good es mate for ∆ℓ. dℓ When ℓ = 8 and dℓ = 12 , we have dA = 12 = 2 ≈ 0.667. 1 8 3
70. 70. Solution Solu on 1 Write A(ℓ) = ℓ2 . We want to know ∆A when ℓ = 8 ft and 2 ∆ℓ = 1 in. ( ) 97 9409 9409 (I) A(ℓ + ∆ℓ) = A = So ∆A = − 32 ≈ 0.6701. 12 288 288 dA (II) = ℓ, so dA = ℓ dℓ, which should be a good es mate for ∆ℓ. dℓ When ℓ = 8 and dℓ = 12 , we have dA = 12 = 2 ≈ 0.667. 1 8 3 So we get es mates close to the hundredth of a square foot.
71. 71. Why should we care? Why use linear approxima ons dy when the actual diﬀerence ∆y is known? Linear approxima on is quick and reliable. Finding ∆y exactly depends on the func on. With more complicated func ons, linear approxima on much simpler. See the “Advanced Examples” later. In real life, some mes only f(a) and f′ (a) are known, and not the general f(x).
72. 72. Outline The linear approxima on of a func on near a point Examples Ques ons Diﬀeren als Using diﬀeren als to es mate error Advanced Examples
73. 73. Gravitation Example Drop a 1 kg ball oﬀ the roof of the Silver Center (50m high). We usually say that a falling object feels a force F = −mg from gravity.
74. 74. Gravitation Example Drop a 1 kg ball oﬀ the roof of the Silver Center (50m high). We usually say that a falling object feels a force F = −mg from gravity. GMm In fact, the force felt is F(r) = − 2 , where M is the mass of the earth and r r is the distance from the center of the earth to the object. G is a constant.
75. 75. Gravitation Example Drop a 1 kg ball oﬀ the roof of the Silver Center (50m high). We usually say that a falling object feels a force F = −mg from gravity. GMm In fact, the force felt is F(r) = − 2 , where M is the mass of the earth and r r is the distance from the center of the earth to the object. G is a constant. GMm GM At r = re the force really is F(re ) = 2 , and g is deﬁned to be 2 re re
76. 76. Gravitation Example Drop a 1 kg ball oﬀ the roof of the Silver Center (50m high). We usually say that a falling object feels a force F = −mg from gravity. GMm In fact, the force felt is F(r) = − 2 , where M is the mass of the earth and r r is the distance from the center of the earth to the object. G is a constant. GMm GM At r = re the force really is F(re ) = 2 , and g is deﬁned to be 2 re re What is the maximum error in replacing the actual force felt at the top of the building F(re + ∆r) by the force felt at ground level F(re )? The rela ve error? The percentage error?
77. 77. Gravitation Solution Solu on We wonder if ∆F = F(re + ∆r) − F(re ) is small. Using a linear approxima on, dF GMm ∆F ≈ dF = dr = 2 3 dr dr re ( re ) GMm dr ∆r = = 2mg r2 e re re ∆F ∆r The rela ve error is ≈ −2 F re
78. 78. Solution continued re = 6378.1 km. If ∆r = 50 m, ∆F ∆r 50 ≈ −2 = −2 = −1.56 × 10−5 = −0.00156% F re 6378100
79. 79. Systematic linear approximation √ √ 2 is irra onal, but 9/4 is ra onal and 9/4 is close to 2.
80. 80. Systematic linear approximation √ √ 2 is irra onal, but 9/4 is ra onal and 9/4 is close to 2. So √ √ √ 1 17 2= 9/4 − 1/4 ≈ 9/4 + (−1/4) = 2(3/2) 12
81. 81. Systematic linear approximation √ √ 2 is irra onal, but 9/4 is ra onal and 9/4 is close to 2. So √ √ √ 1 17 2= 9/4 − 1/4 ≈ 9/4 + (−1/4) = 2(3/2) 12 This is a be er approxima on since (17/12)2 = 289/144
82. 82. Systematic linear approximation √ √ 2 is irra onal, but 9/4 is ra onal and 9/4 is close to 2. So √ √ √ 1 17 2= 9/4 − 1/4 ≈ 9/4 + (−1/4) = 2(3/2) 12 This is a be er approxima on since (17/12)2 = 289/144 Do it again! √ √ √ 1 2 = 289/144 − 1/144 ≈ 289/144 + 17 (−1/144) = 577/408 2( /12) ( )2 577 332, 929 1 Now = which is away from 2. 408 166, 464 166, 464
83. 83. Illustration of the previousexample .
84. 84. Illustration of the previousexample .
85. 85. Illustration of the previousexample . 2
86. 86. Illustration of the previousexample . 2
87. 87. Illustration of the previousexample . 2
88. 88. Illustration of the previousexample (2, 17 ) 12 . 2
89. 89. Illustration of the previousexample (2, 17 ) 12 . 2
90. 90. Illustration of the previousexample (2, 17/12) (9, 3) 4 2
91. 91. Illustration of the previousexample (2, 17/12) ( 289 ()9 , 3 ) 17 4 2 144 , 12
92. 92. Illustration of the previousexample (2, 17/12) 9 3 ( 577 ) ( 289 17()4 , 2 ) 2, 408 144 , 12
93. 93. Summary Linear approxima on: If f is diﬀeren able at a, the best linear approxima on to f near a is given by Lf,a (x) = f(a) + f′ (a)(x − a) Diﬀeren als: If f is diﬀeren able at x, a good approxima on to ∆y = f(x + ∆x) − f(x) is dy dy ∆y ≈ dy = · dx = · ∆x dx dx Don’t buy plywood from me.