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# Lesson 12: Linear Approximation and Differentials (Section 41 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 Approximation and Differentials (Section 41 slides)

1. 1. Section 2.8 Linear Approximation and Differentials V63.0121.041, Calculus I New York University October 13, 2010Announcements Quiz 2 in recitation this week on §§1.5, 1.6, 2.1, 2.2 Midterm on §§1.1–2.5 . . . . . .
2. 2. Announcements Quiz 2 in recitation this week on §§1.5, 1.6, 2.1, 2.2 Midterm on §§1.1–2.5 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 2 / 27
3. 3. Objectives Use tangent lines to make linear approximations to a function. Given a function and a point in the domain, compute the linearization of the function at that point. Use linearization to approximate values of functions Given a function, compute the differential of that function Use the differential notation to estimate error in linear approximations. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 3 / 27
4. 4. OutlineThe linear approximation of a function near a point Examples QuestionsDifferentials Using differentials to estimate errorAdvanced Examples . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 4 / 27
5. 5. The Big IdeaQuestionLet f be differentiable at a. What linear function best approximates fnear a? . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 5 / 27
6. 6. The Big IdeaQuestionLet f be differentiable at a. What linear function best approximates fnear a?AnswerThe tangent line, of course! . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 5 / 27
7. 7. The Big IdeaQuestionLet f be differentiable at a. What linear function best approximates fnear a?AnswerThe tangent line, of course!QuestionWhat is the equation for the line tangent to y = f(x) at (a, f(a))? . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 5 / 27
8. 8. The Big IdeaQuestionLet f be differentiable at a. What linear function best approximates fnear a?AnswerThe tangent line, of course!QuestionWhat is the equation for the line tangent to y = f(x) at (a, f(a))?Answer L(x) = f(a) + f′ (a)(x − a) . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 5 / 27
9. 9. The tangent line is a linear approximation y . L(x) = f(a) + f′ (a)(x − a)is a decent approximation to f L . (x) .near a. f .(x) . f .(a) . . x−a . x . a . x . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 6 / 27
10. 10. The tangent line is a linear approximation y . L(x) = f(a) + f′ (a)(x − a)is a decent approximation to f L . (x) .near a. f .(x) .How decent? The closer x is toa, the better the approxmation f .(a) . . x−aL(x) is to f(x) . x . a . x . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 6 / 27
11. 11. Example.ExampleEstimate sin(61◦ ) = sin(61π/180) by using a linear approximation(i) about a = 0 (ii) about a = 60◦ = π/3.. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 7 / 27
12. 12. Example.ExampleEstimate sin(61◦ ) = sin(61π/180) 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. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 7 / 27
13. 13. Example.ExampleEstimate sin(61◦ ) = sin(61π/180) by using a linear approximation(i) about a = 0 (ii) about a = 60◦ = π/3.Solution (i) Solution (ii) (π) We have f = and If f(x) = sin x, then f(0) = 0 ( ) 3 and f′ (0) = 1. f′ π = . 3 So the linear approximation near 0 is L(x) = 0 + 1 · x = x. Thus ( ) 61π 61π sin ≈ ≈ 1.06465 180 180. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 7 / 27
14. 14. Example.ExampleEstimate sin(61◦ ) = sin(61π/180) by using a linear approximation(i) about a = 0 (ii) about a = 60◦ = π/3.Solution (i) Solution (ii) (π) √ 3 We have f = and If f(x) = sin x, then f(0) = 0 ( ) 3 2 and f′ (0) = 1. f′ π = . 3 So the linear approximation near 0 is L(x) = 0 + 1 · x = x. Thus ( ) 61π 61π sin ≈ ≈ 1.06465 180 180. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 7 / 27
15. 15. Example.ExampleEstimate sin(61◦ ) = sin(61π/180) by using a linear approximation(i) about a = 0 (ii) about a = 60◦ = π/3.Solution (i) Solution (ii) (π) √ 3 We have f = and If f(x) = sin x, then f(0) = 0 ( ) 3 2 and f′ (0) = 1. 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. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 7 / 27
16. 16. Example.ExampleEstimate sin(61◦ ) = sin(61π/180) by using a linear approximation(i) about a = 0 (ii) about a = 60◦ = π/3.Solution (i) Solution (ii) (π) √ 3 We have f = and If f(x) = sin x, then f(0) = 0 ( ) 3 2 and f′ (0) = 1. f′ π = 1 . 3 2 So the linear approximation So L(x) = near 0 is L(x) = 0 + 1 · x = x. Thus ( ) 61π 61π sin ≈ ≈ 1.06465 180 180. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 7 / 27
17. 17. Example.ExampleEstimate sin(61◦ ) = sin(61π/180) by using a linear approximation(i) about a = 0 (ii) about a = 60◦ = π/3.Solution (i) Solution (ii) (π) √ 3 We have f = and If f(x) = sin x, then f(0) = 0 ( ) 3 2 and f′ (0) = 1. f′ π = 1 . 3 2 √ So the linear approximation 3 1( π) So L(x) = + x− near 0 is L(x) = 0 + 1 · x = x. 2 2 3 Thus ( ) 61π 61π sin ≈ ≈ 1.06465 180 180. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 7 / 27
18. 18. Example.ExampleEstimate sin(61◦ ) = sin(61π/180) by using a linear approximation(i) about a = 0 (ii) about a = 60◦ = π/3.Solution (i) Solution (ii) (π) √ 3 We have f = and If f(x) = sin x, then f(0) = 0 ( ) 3 2 and f′ (0) = 1. f′ π = 1 . 3 2 √ So the linear approximation 3 1( π) So L(x) = + x− near 0 is L(x) = 0 + 1 · x = x. 2 2 3 Thus Thus ( ) ( ) 61π 61π 61π sin ≈ ≈ 1.06465 sin ≈ 180 180 180. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 7 / 27
19. 19. Example.ExampleEstimate sin(61◦ ) = sin(61π/180) by using a linear approximation(i) about a = 0 (ii) about a = 60◦ = π/3.Solution (i) Solution (ii) (π) √ 3 We have f = and If f(x) = sin x, then f(0) = 0 ( ) 3 2 and f′ (0) = 1. f′ π = 1 . 3 2 √ So the linear approximation 3 1( π) So L(x) = + x− near 0 is L(x) = 0 + 1 · x = x. 2 2 3 Thus Thus ( ) ( ) 61π 61π 61π sin ≈ ≈ 1.06465 sin ≈ 0.87475 180 180 180. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 7 / 27
20. 20. Example.ExampleEstimate sin(61◦ ) = sin(61π/180) by using a linear approximation(i) about a = 0 (ii) about a = 60◦ = π/3.Solution (i) Solution (ii) (π) √ 3 We have f = and If f(x) = sin x, then f(0) = 0 ( ) 3 2 and f′ (0) = 1. f′ π = 1 . 3 2 √ So the linear approximation 3 1( π) So L(x) = + x− near 0 is L(x) = 0 + 1 · x = x. 2 2 3 Thus Thus ( ) ( ) 61π 61π 61π sin ≈ ≈ 1.06465 sin ≈ 0.87475 180 180 180Calculator check: sin(61◦ ) ≈. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 7 / 27
21. 21. Example.ExampleEstimate sin(61◦ ) = sin(61π/180) by using a linear approximation(i) about a = 0 (ii) about a = 60◦ = π/3.Solution (i) Solution (ii) (π) √ 3 We have f = and If f(x) = sin x, then f(0) = 0 ( ) 3 2 and f′ (0) = 1. f′ π = 1 . 3 2 √ So the linear approximation 3 1( π) So L(x) = + x− near 0 is L(x) = 0 + 1 · x = x. 2 2 3 Thus Thus ( ) ( ) 61π 61π 61π sin ≈ ≈ 1.06465 sin ≈ 0.87475 180 180 180Calculator check: sin(61◦ ) ≈ 0.87462.. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 7 / 27
22. 22. Illustration y . y . = sin x . x . . 1◦ 6 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 8 / 27
23. 23. Illustration y . y . = L1 (x) = x y . = sin x . x . 0 . . 1◦ 6 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 8 / 27
24. 24. Illustration y . y . = L1 (x) = x b . ig difference! y . = sin x . x . 0 . . 1◦ 6 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 8 / 27
25. 25. Illustration y . y . = L1 (x) = x √ ( ) y . = L2 (x) = 2 3 + 1 2 x− π 3 y . = sin x . . . x . 0 . . π/3 . 1◦ 6 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 8 / 27
26. 26. Illustration y . y . = L1 (x) = x √ ( ) y . = L2 (x) = 2 3 + 1 2 x− π 3 y . = sin x . . ery little difference! v . . x . 0 . . π/3 . 1◦ 6 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 8 / 27
27. 27. Another ExampleExample √Estimate 10 using the fact that 10 = 9 + 1. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 9 / 27
28. 28. Another ExampleExample √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 = 9to estimate f(10) = 10. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 9 / 27
29. 29. Another ExampleExample √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 = 9to estimate f(10) = 10. √ √ d√ 10 ≈ 9 + x (1) dx x=9 1 19 =3+ (1) = ≈ 3.167 2·3 6 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 9 / 27
30. 30. Another ExampleExample √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 = 9to estimate f(10) = 10. √ √ d√ 10 ≈ 9 + x (1) dx x=9 1 19 =3+ (1) = ≈ 3.167 2·3 6 ( )2 19Check: = 6 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 9 / 27
31. 31. Another ExampleExample √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 = 9to estimate f(10) = 10. √ √ d√ 10 ≈ 9 + x (1) dx x=9 1 19 =3+ (1) = ≈ 3.167 2·3 6 ( )2 19 361Check: = . 6 36 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 9 / 27
32. 32. Dividing without dividing?ExampleSuppose I have an irrational fear of division and need to estimate577 ÷ 408. I write 577 1 1 1 = 1 + 169 = 1 + 169 × × . 408 408 4 102 1But still I have to find . 102 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 10 / 27
33. 33. Dividing without dividing?ExampleSuppose I have an irrational fear of division and need to estimate577 ÷ 408. I write 577 1 1 1 = 1 + 169 = 1 + 169 × × . 408 408 4 102 1But still I have to find . 102Solution 1Let 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 10 / 27
34. 34. QuestionsExampleSuppose 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? . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 11 / 27
35. 35. AnswersExampleSuppose 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? . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 12 / 27
36. 36. AnswersExampleSuppose 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?) . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 12 / 27
37. 37. QuestionsExampleSuppose 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?ExampleSuppose our factory makes MP3 players and the marginal cost iscurrently \$50/lot. How much will it cost to make 2 more lots? 3 morelots? 12 more lots? . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 13 / 27
38. 38. AnswersExampleSuppose our factory makes MP3 players and the marginal cost iscurrently \$50/lot. How much will it cost to make 2 more lots? 3 morelots? 12 more lots?Answer \$100 \$150 \$600 (?) . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 14 / 27
39. 39. QuestionsExampleSuppose 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?ExampleSuppose our factory makes MP3 players and the marginal cost iscurrently \$50/lot. How much will it cost to make 2 more lots? 3 morelots? 12 more lots?ExampleSuppose a line goes through the point (x0 , y0 ) and has slope m. If thepoint is moved horizontally by dx, while staying on the line, what is thecorresponding vertical movement? . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 15 / 27
40. 40. AnswersExampleSuppose a line goes through the point (x0 , y0 ) and has slope m. If thepoint is moved horizontally by dx, while staying on the line, what is thecorresponding vertical movement? . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 16 / 27
41. 41. AnswersExampleSuppose a line goes through the point (x0 , y0 ) and has slope m. If thepoint is moved horizontally by dx, while staying on the line, what is thecorresponding vertical movement?AnswerThe slope of the line is rise m= runWe are given a “run” of dx, so the corresponding “rise” is m dx. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 16 / 27
42. 42. OutlineThe linear approximation of a function near a point Examples QuestionsDifferentials Using differentials to estimate errorAdvanced Examples . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 17 / 27
43. 43. Differentials are another way to express derivatives f(x + ∆x) − f(x) ≈ f′ (x) ∆x y . ∆y dyRename ∆x = dx, so we canwrite this as . ∆y ≈ dy = f′ (x)dx. . dy . ∆yAnd this looks a lot like the . . dx = ∆xLeibniz-Newton identity dy . = f′ (x) x . dx x x . . + ∆x . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 18 / 27
44. 44. Differentials are another way to express derivatives f(x + ∆x) − f(x) ≈ f′ (x) ∆x y . ∆y dyRename ∆x = dx, so we canwrite this as . ∆y ≈ dy = f′ (x)dx. . dy . ∆yAnd this looks a lot like the . . dx = ∆xLeibniz-Newton identity dy . = f′ (x) x . dx x x . . + ∆xLinear approximation means ∆y ≈ dy = f′ (x0 ) dx near x0 . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 18 / 27
45. 45. Using differentials to estimate error y .If y = f(x), x0 and ∆x is known,and an estimate of ∆y isdesired: Approximate: ∆y ≈ dy . Differentiate: dy = f′ (x) dx . ∆y . dy Evaluate at x = x0 and . . dx = ∆x dx = ∆x. . x . x x . . + ∆x . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 19 / 27
46. 46. ExampleA sheet of plywood measures 8 ft × 4 ft. Suppose our plywood-cuttingmachine will cut a rectangle whose width is exactly half its length, butthe length is prone to errors. If the length is off by 1 in, how bad can thearea of the sheet be off by? . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 20 / 27
47. 47. ExampleA sheet of plywood measures 8 ft × 4 ft. Suppose our plywood-cuttingmachine will cut a rectangle whose width is exactly half its length, butthe length is prone to errors. If the length is off by 1 in, how bad can thearea of the sheet be off by?Solution 1 2Write A(ℓ) = ℓ . We want to know ∆A when ℓ = 8 ft and ∆ℓ = 1 in. 2 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 20 / 27
48. 48. ExampleA sheet of plywood measures 8 ft × 4 ft. Suppose our plywood-cuttingmachine will cut a rectangle whose width is exactly half its length, butthe length is prone to errors. If the length is off by 1 in, how bad can thearea of the sheet be off by?Solution 1 2Write A(ℓ) = ℓ . We want to know ∆A when ℓ = 8 ft and ∆ℓ = 1 in. 2 ( ) 97 9409 9409 (I) A(ℓ + ∆ℓ) = A = So ∆A = − 32 ≈ 0.6701. 12 288 288 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 20 / 27
49. 49. ExampleA sheet of plywood measures 8 ft × 4 ft. Suppose our plywood-cuttingmachine will cut a rectangle whose width is exactly half its length, butthe length is prone to errors. If the length is off by 1 in, how bad can thearea of the sheet be off by?Solution 1 2Write A(ℓ) = ℓ . We want to know ∆A when ℓ = 8 ft and ∆ℓ = 1 in. 2 ( ) 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 estimate for ∆ℓ. dℓ When ℓ = 8 and dℓ = 12 , we have dA = 12 = 2 ≈ 0.667. So we 1 8 3 get estimates close to the hundredth of a square foot. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 20 / 27
50. 50. Why?Why use linear approximations dy when the actual difference ∆y isknown? 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). . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 21 / 27
51. 51. OutlineThe linear approximation of a function near a point Examples QuestionsDifferentials Using differentials to estimate errorAdvanced Examples . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 22 / 27
52. 52. GravitationPencils 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. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 23 / 27
53. 53. GravitationPencils 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? . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 23 / 27
54. 54. Gravitation SolutionSolutionWe wonder if ∆F = F(re + ∆r) − F(re ) is small. Using a linear approximation, dF GMm ∆F ≈ dF = dr = 2 3 dr dr re re ( ) 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 24 / 27
55. 55. Systematic linear approximation √ √ 2 is irrational, but 9/4 is rational and 9/4 is close to 2. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 25 / 27
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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 25 / 27
57. 57. 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 25 / 27
58. 58. 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 25 / 27
59. 59. Illustration of the previous example . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 26 / 27
60. 60. Illustration of the previous example . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 26 / 27
61. 61. Illustration of the previous example . 2 . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 26 / 27
62. 62. Illustration of the previous example . . 2 . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 26 / 27
63. 63. Illustration of the previous example . . 2 . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 26 / 27
64. 64. Illustration of the previous example . 2, 17 ) ( 12 . . . 2 . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 26 / 27
65. 65. Illustration of the previous example . 2, 17 ) ( 12 . . . 2 . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 26 / 27
66. 66. Illustration of the previous example . . 2, 17/12) ( . . 4, 3) (9 2 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 26 / 27
67. 67. Illustration of the previous example . . 2, 17/12) ( .. ( . 9, 3) ( )4 2 289 17 . 144 , 12 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 26 / 27
68. 68. Illustration of the previous example . . 2, 17/12) ( .. ( . 9, 3) ( ( 577 ) )4 2 . 2, 408 289 17 . 144 , 12 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 26 / 27
69. 69. Summary Linear approximation: If f is differentiable at a, the best linear approximation to f near a is given by Lf,a (x) = f(a) + f′ (a)(x − a) Differentials: If f is differentiable at x, a good approximation to ∆y = f(x + ∆x) − f(x) is dy dy ∆y ≈ dy = · dx = · ∆x dx dx Don’t buy plywood from me. . . . . . .V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 27 / 27