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# Lesson 12: Linear Approximation and Differentials

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The tangent line to the graph of a function at a point can be thought of as a function itself. As such, it is the best linear function which agrees with the given function at the point. The function and its linear approximation will probably diverge away from the point at which they agree, but this "error" can be measured using the differential notation.

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### Lesson 12: Linear Approximation and Differentials

1. 1. Section 2.8 Linear Approximation and Differentials V63.0121.034, Calculus I October 12, 2009 Announcements Midterm Wednesday on Sections 1.1–2.4 . . . . . .
2. 2. Outline The linear approximation of a function near a point Examples Differentials The not-so-big idea Using differentials to estimate error Midterm Review Advanced Examples . . . . . .
3. 3. The Big Idea Question Let f be differentiable at a. What linear function best approximates f near a? . . . . . .
4. 4. The Big Idea Question Let f be differentiable at a. What linear function best approximates f near a? Answer The tangent line, of course! . . . . . .
5. 5. The Big Idea Question Let f be differentiable at a. What linear function best approximates f near a? Answer The tangent line, of course! Question What is the equation for the line tangent to y = f(x) at (a, f(a))? . . . . . .
6. 6. The Big Idea Question Let f be differentiable at a. What linear function best approximates f near a? Answer The tangent line, of course! Question What is the equation for the line tangent to y = f(x) at (a, f(a))? Answer L(x) = f(a) + f′ (a)(x − a) . . . . . .
7. 7. Example Example Estimate sin(61◦ ) by using a linear approximation (i) about a = 0 (ii) about a = 60◦ = π/3. . . . . . .
8. 8. Example Example Estimate sin(61◦ ) by using a linear approximation (i) about a = 0 (ii) about a = 60◦ = π/3. Solution (i) If f(x) = sin x, then f(0) = 0 and f′ (0) = 1. So the linear approximation near 0 is L(x) = 0 + 1 · x = x. Thus ( ) 61π 61π sin ≈ ≈ 1.06465 180 180 . . . . . .
9. 9. Example Example Estimate sin(61◦ ) by using a linear approximation (i) about a = 0 (ii) about a = 60◦ = π/3. Solution (i) Solution (ii) ( ) If f(x) = sin x, then f(0) = 0 We have f π = and and f′ (0) = 1. ( ) 3 f′ π = . 3 So the linear approximation near 0 is L(x) = 0 + 1 · x = x. Thus ( ) 61π 61π sin ≈ ≈ 1.06465 180 180 . . . . . .
10. 10. Example Example Estimate sin(61◦ ) by using a linear approximation (i) about a = 0 (ii) about a = 60◦ = π/3. Solution (i) Solution (ii) ( ) √ If f(x) = sin x, then f(0) = 0 We have f π = 3 and and f′ (0) = 1. ( ) 3 2 f′ π = . 3 So the linear approximation near 0 is L(x) = 0 + 1 · x = x. Thus ( ) 61π 61π sin ≈ ≈ 1.06465 180 180 . . . . . .
11. 11. Example Example Estimate sin(61◦ ) by using a linear approximation (i) about a = 0 (ii) about a = 60◦ = π/3. Solution (i) Solution (ii) ( ) √ If f(x) = sin x, then f(0) = 0 We have f π = 3 and and f′ (0) = 1. ( ) 3 2 f′ π = 1 . 3 2 So the linear approximation near 0 is L(x) = 0 + 1 · x = x. Thus ( ) 61π 61π sin ≈ ≈ 1.06465 180 180 . . . . . .
12. 12. Example Example Estimate sin(61◦ ) by using a linear approximation (i) about a = 0 (ii) about a = 60◦ = π/3. Solution (i) Solution (ii) ( ) √ If f(x) = sin x, then f(0) = 0 We have f π = 3 and and f′ (0) = 1. ( ) 3 2 f′ π = 1 . 3 2 So the linear approximation near 0 is So L(x) = L(x) = 0 + 1 · x = x. Thus ( ) 61π 61π sin ≈ ≈ 1.06465 180 180 . . . . . .
13. 13. Example Example Estimate sin(61◦ ) by using a linear approximation (i) about a = 0 (ii) about a = 60◦ = π/3. Solution (i) Solution (ii) ( ) √ If f(x) = sin x, then f(0) = 0 We have f π = 23 and and f′ (0) = 1. ( ) 3 f′ π = 1 . 3 2 √ So the linear approximation 3 1( π) near 0 is So L(x) = + x− 2 2 3 L(x) = 0 + 1 · x = x. Thus ( ) 61π 61π sin ≈ ≈ 1.06465 180 180 . . . . . .
14. 14. Example Example Estimate sin(61◦ ) by using a linear approximation (i) about a = 0 (ii) about a = 60◦ = π/3. Solution (i) Solution (ii) ( ) √ If f(x) = sin x, then f(0) = 0 We have f π = 23 and and f′ (0) = 1. ( ) 3 f′ π = 1 . 3 2 √ So the linear approximation 3 1( π) near 0 is So L(x) = + x− 2 2 3 L(x) = 0 + 1 · x = x. Thus Thus ( ) ( ) 61π 61π 61π sin ≈ sin ≈ ≈ 1.06465 180 180 180 . . . . . .
15. 15. Example Example Estimate sin(61◦ ) by using a linear approximation (i) about a = 0 (ii) about a = 60◦ = π/3. Solution (i) Solution (ii) ( ) √ If f(x) = sin x, then f(0) = 0 We have f π = 23 and and f′ (0) = 1. ( ) 3 f′ π = 1 . 3 2 √ So the linear approximation 3 1( π) near 0 is So L(x) = + x− 2 2 3 L(x) = 0 + 1 · x = x. Thus Thus ( ) ( ) 61π 61π 61π sin ≈ 0.87475 sin ≈ ≈ 1.06465 180 180 180 . . . . . .
16. 16. Example Example Estimate sin(61◦ ) by using a linear approximation (i) about a = 0 (ii) about a = 60◦ = π/3. Solution (i) Solution (ii) ( ) √ If f(x) = sin x, then f(0) = 0 We have f π = 23 and and f′ (0) = 1. ( ) 3 f′ π = 1 . 3 2 √ So the linear approximation 3 1( π) near 0 is So L(x) = + x− 2 2 3 L(x) = 0 + 1 · x = x. Thus Thus ( ) ( ) 61π 61π 61π sin ≈ 0.87475 sin ≈ ≈ 1.06465 180 180 180 Calculator check: sin(61◦ ) ≈ . . . . . .
17. 17. Example Example Estimate sin(61◦ ) by using a linear approximation (i) about a = 0 (ii) about a = 60◦ = π/3. Solution (i) Solution (ii) ( ) √ If f(x) = sin x, then f(0) = 0 We have f π = 23 and and f′ (0) = 1. ( ) 3 f′ π = 1 . 3 2 √ So the linear approximation 3 1( π) near 0 is So L(x) = + x− 2 2 3 L(x) = 0 + 1 · x = x. Thus Thus ( ) ( ) 61π 61π 61π sin ≈ 0.87475 sin ≈ ≈ 1.06465 180 180 180 Calculator check: sin(61◦ ) ≈ 0.87462. . . . . . .
18. 18. Illustration y . y . = sin x . x . . 1◦ 6 . . . . . .
19. 19. Illustration y . y . = L1 (x) = x y . = sin x . x . 0 . . 1◦ 6 . . . . . .
20. 20. Illustration y . y . = L1 (x) = x b . ig difference! y . = sin x . x . 0 . . 1◦ 6 . . . . . .
21. 21. Illustration y . y . = L1 (x) = x √ 3 1 ( ) y . = L2 (x) = 2 + 2 x− π 3 y . = sin x . . . x . 0 . . π/3 . 1◦ 6 . . . . . .
22. 22. Illustration y . y . = L1 (x) = x √ 3 1 ( ) y . = L2 (x) = 2 + 2 x− π 3 y . = sin x . . ery little difference! v . . x . 0 . . π/3 . 1◦ 6 . . . . . .
23. 23. Another Example Example √ Estimate 10 using the fact that 10 = 9 + 1. . . . . . .
24. 24. Another Example Example √ Estimate 10 using the fact that 10 = 9 + 1. Solution √ The key step is to use a linear approximation to f(x) = √ x near a = 9 to estimate f(10) = 10. . . . . . .
25. 25. Another Example Example √ Estimate 10 using the fact that 10 = 9 + 1. Solution √ The key step is to use a linear approximation to f(x) = √ x near a = 9 to estimate f(10) = 10. √ √d√ 10 ≈ 9+x (1) dx x=9 1 19 =3+ (1 ) = ≈ 3.167 2·3 6 . . . . . .
26. 26. Another Example Example √ Estimate 10 using the fact that 10 = 9 + 1. Solution √ The key step is to use a linear approximation to f(x) = √ x near a = 9 to estimate f(10) = 10. √ √ d√ 10 ≈ 9+ x (1) dx x=9 1 19 =3+ (1 ) = ≈ 3.167 2·3 6 ( )2 19 Check: = 6 . . . . . .
27. 27. Another Example Example √ Estimate 10 using the fact that 10 = 9 + 1. Solution √ The key step is to use a linear approximation to f(x) = √ x near a = 9 to estimate f(10) = 10. √ √ d√ 10 ≈ 9+ x (1) dx x=9 1 19 =3+ (1 ) = ≈ 3.167 2·3 6 ( )2 19 361 Check: = . 6 36 . . . . . .
28. 28. Dividing without dividing? Example Suppose I have an irrational fear of division and need to estimate 577 ÷ 408. I write 577 1 1 1 = 1 + 169 = 1 + 169 × × . 408 408 4 102 1 But still I have to ﬁnd . 102 . . . . . .
29. 29. Dividing without dividing? Example Suppose I have an irrational fear of division and need to estimate 577 ÷ 408. I write 577 1 1 1 = 1 + 169 = 1 + 169 × × . 408 408 4 102 1 But still I have to ﬁnd . 102 Solution 1 Let f(x) = . We know f(100) and we want to estimate 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. . . . . . .
30. 30. Outline The linear approximation of a function near a point Examples Differentials The not-so-big idea Using differentials to estimate error Midterm Review Advanced Examples . . . . . .
31. 31. 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? . . . . . .
32. 32. Answers 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? . . . . . .
33. 33. Answers 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?) . . . . . .
34. 34. 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? 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? . . . . . .
35. 35. Answers 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 (?) . . . . . .
36. 36. 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? 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? 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 vertical movement? . . . . . .
37. 37. Answers 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 vertical movement? . . . . . .
38. 38. Answers 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 vertical 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. . . . . . .
39. 39. Differentials are another way to express derivatives f(x + ∆x) − f(x) ≈ f′ (x) ∆x y . ∆y dy Rename ∆x = dx, so we can write this as . ∆y ≈ dy = f′ (x)dx. . dy . ∆y And this looks a lot like the . . dx = ∆ x Leibniz-Newton identity dy . x . = f ′ (x ) dx x x . . + ∆x . . . . . .
40. 40. Differentials are another way to express derivatives f(x + ∆x) − f(x) ≈ f′ (x) ∆x y . ∆y dy Rename ∆x = dx, so we can write this as . ∆y ≈ dy = f′ (x)dx. . dy . ∆y And this looks a lot like the . . dx = ∆ x Leibniz-Newton identity dy . x . = f ′ (x ) dx x x . . + ∆x Linear approximation means ∆y ≈ dy = f′ (x0 ) dx near x0 . . . . . . .
41. 41. Using differentials to estimate error y . If y = f(x), x0 and ∆x is known, and an estimate of ∆y is desired: Approximate: ∆y ≈ dy . Differentiate: . dy dy = f′ (x) dx . ∆y . Evaluate at x = x0 and . dx = ∆ x dx = ∆x. . x . x x . . + ∆x . . . . . .
42. 42. Example A sheet of plywood measures 8 ft × 4 ft. Suppose our plywood-cutting 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? . . . . . .
43. 43. Example A sheet of plywood measures 8 ft × 4 ft. Suppose our plywood-cutting 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? Solution 1 Write A(ℓ) = ℓ2 . We want to know ∆A when ℓ = 8 ft and 2 ∆ℓ = 1 in. . . . . . .
44. 44. Example A sheet of plywood measures 8 ft × 4 ft. Suppose our plywood-cutting 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? Solution 1 Write A(ℓ) = ℓ2 . We want to know ∆A when ℓ = 8 ft and 2 ∆ℓ = 1 in. ( ) 97 9409 (I) A(ℓ + ∆ℓ) = A = So 12 288 9409 ∆A = − 32 ≈ 0.6701. 288 . . . . . .
45. 45. Example A sheet of plywood measures 8 ft × 4 ft. Suppose our plywood-cutting 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? Solution 1 Write A(ℓ) = ℓ2 . We want to know ∆A when ℓ = 8 ft and 2 ∆ℓ = 1 in. ( ) 97 9409 (I) A(ℓ + ∆ℓ) = A = So 12 288 9409 ∆A = − 32 ≈ 0.6701. 288 dA (II) = ℓ, so dA = ℓ dℓ, which should be a good estimate for dℓ 1 ∆ℓ. When ℓ = 8 and dℓ = 12 , we have 8 2 dA = 12 = 3 ≈ 0.667. So we get estimates close to the hundredth of a square foot. . . . . . .
46. 46. Why? Why use linear approximations dy when the actual difference ∆y is known? Linear approximation is quick and reliable. Finding ∆y exactly depends on the function. These examples are overly simple. See the “Advanced Examples” later. In real life, sometimes only f(a) and f′ (a) are known, and not the general f(x). . . . . . .
47. 47. Outline The linear approximation of a function near a point Examples Differentials The not-so-big idea Using differentials to estimate error Midterm Review Advanced Examples . . . . . .
48. 48. Midterm Facts Covers sections 1.1–2.4 (Limits, Derivatives, Differentiation up to Quotient Rule) Calculator free Has about 7 problems each could have multiple parts Some ﬁxed-response, some free-response To study: outline do problems metacognition ask questions! . . . . . .
49. 49. Outline The linear approximation of a function near a point Examples Differentials The not-so-big idea Using differentials to estimate error Midterm Review Advanced Examples . . . . . .
50. 50. Gravitation Pencils down! 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. . . . . . .
51. 51. Gravitation Pencils down! 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. In fact, the force felt is GMm F (r ) = − , r2 where M is the mass of the earth and r is the distance from the center of the earth to the object. G is a constant. GMm At r = re the force really is F(re ) = = −mg. r2 e 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 relative error? The percentage error? . . . . . .
52. 52. Solution We wonder if ∆F = F(re + ∆r) − F(re ) is small. Using a linear approximation, dF GMm ∆F ≈ dF = dr = 2 3 dr dr r re ( e ) GMm dr ∆r = 2 = 2mg re re re ∆F ∆r The relative error is ≈ −2 F re re = 6378.1 km. If ∆r = 50 m, ∆F ∆r 50 ≈ −2 = −2 = −1.56 × 10−5 = −0.00156% F re 6378100 . . . . . .
53. 53. Systematic linear approximation √ √ 2 is irrational, but 9/4 is rational and 9/4 is close to 2. . . . . . .
54. 54. Systematic linear approximation √ √ 2 is irrational, but 9/4 is rational and 9/4 is close to 2. So √ √ √ 1 17 2= 9/4 − 1/4 ≈ 9/4 + (−1/4) = 2(3/2) 12 . . . . . .
55. 55. Systematic linear approximation √ √ 2 is irrational, but 9/4 is rational 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 better approximation since (17/12)2 = 289/144 . . . . . .
56. 56. Systematic linear approximation √ √ 2 is irrational, but 9/4 is rational 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 better approximation since (17/12)2 = 289/144 Do it again! √ √ √ 1 2= 289/144 − 1/144 ≈ 289/144+ (−1/144) = 577/408 2(17/12) ( )2 577 332, 929 1 Now = which is away from 2. 408 166, 464 166, 464 . . . . . .
57. 57. Illustration of the previous example . . . . . . .
58. 58. Illustration of the previous example . . . . . . .
59. 59. Illustration of the previous example . 2 . . . . . . .
60. 60. Illustration of the previous example . . 2 . . . . . . .
61. 61. Illustration of the previous example . . 2 . . . . . . .
62. 62. Illustration of the previous example . 2, 17 ) ( 12 . . . 2 . . . . . . .
63. 63. Illustration of the previous example . 2, 17 ) ( 12 . . . 2 . . . . . . .
64. 64. Illustration of the previous example . . 2, 17/12) ( . (9 2 . 4, 3) . . . . . .
65. 65. Illustration of the previous example . . 2, 17/12) ( .. . 9, 3) ( ( 289 17 ) 4 2 . 144 , 12 . . . . . .
66. 66. Illustration of the previous example . . 2, 17/12) ( .. . 9, 3) ( ( 577 ) ( 289 17 ) 4 2 . 2, 408 . 144 , 12 . . . . . .