Information about Lesson 13: Linear Approximation

New perspectives on an old idea: the derivative measures the slope of the tangent line, which is the line which best fits the graph near a point.

Outline The linear approximation of a function near a point Examples Differentials The not-so-big idea . . . . . .

The Big Idea Question Let f be differentiable at a. What linear function best approximates f near a? . . . . . .

The Big Idea Question Let f be differentiable at a. What linear function best approximates f near a? Answer The tangent line, of course! . . . . . .

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))? . . . . . .

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) . . . . . .

Example Example Estimate sin(61◦ ) by using a linear approximation (ii) about a = 60◦ = π/3. (i) about a = 0 . . . . . .

Example Example Estimate sin(61◦ ) by using a linear approximation (ii) about a = 60◦ = π/3. (i) about a = 0 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 ≈ ≈ 1.06465 sin 180 180 . . . . . .

Example Example Estimate sin(61◦ ) by using a linear approximation (ii) about a = 60◦ = π/3. (i) about a = 0 Solution (ii) Solution (i) () We have f π = and () If f(x) = sin x, then f(0) = 0 3 f′ π = . and f′ (0) = 1. 3 So the linear approximation near 0 is L(x) = 0 + 1 · x = x. Thus ( ) 61π 61π ≈ ≈ 1.06465 sin 180 180 . . . . . .

Example Example Estimate sin(61◦ ) by using a linear approximation (ii) about a = 60◦ = π/3. (i) about a = 0 Solution (ii) Solution (i) () √ We have f π = 3 and () If f(x) = sin x, then f(0) = 0 3 2 f′ π = . and f′ (0) = 1. 3 So the linear approximation near 0 is L(x) = 0 + 1 · x = x. Thus ( ) 61π 61π ≈ ≈ 1.06465 sin 180 180 . . . . . .

Example Example Estimate sin(61◦ ) by using a linear approximation (ii) about a = 60◦ = π/3. (i) about a = 0 Solution (ii) Solution (i) () √ We have f π = 3 and () If f(x) = sin x, then f(0) = 0 3 2 f′ π = 1 . and f′ (0) = 1. 3 2 So the linear approximation near 0 is L(x) = 0 + 1 · x = x. Thus ( ) 61π 61π ≈ ≈ 1.06465 sin 180 180 . . . . . .

Example Example Estimate sin(61◦ ) by using a linear approximation (ii) about a = 60◦ = π/3. (i) about a = 0 Solution (ii) Solution (i) () √ We have f π = 3 and () If f(x) = sin x, then f(0) = 0 3 2 f′ π = 1 . and f′ (0) = 1. 3 2 So the linear approximation So L(x) = near 0 is L(x) = 0 + 1 · x = x. Thus ( ) 61π 61π ≈ ≈ 1.06465 sin 180 180 . . . . . .

Example Example Estimate sin(61◦ ) by using a linear approximation (ii) about a = 60◦ = π/3. (i) about a = 0 Solution (ii) Solution (i) () √ We have f π = 23 and () If f(x) = sin x, then f(0) = 0 3 f′ π = 1 . and f′ (0) = 1. 3 2 √ 3 1( π) So the linear approximation x− So L(x) = + near 0 is L(x) = 0 + 1 · x = x. 2 2 3 Thus ( ) 61π 61π ≈ ≈ 1.06465 sin 180 180 . . . . . .

Example Example Estimate sin(61◦ ) by using a linear approximation (ii) about a = 60◦ = π/3. (i) about a = 0 Solution (ii) Solution (i) () √ We have f π = 23 and () If f(x) = sin x, then f(0) = 0 3 f′ π = 1 . and f′ (0) = 1. 3 2 √ 3 1( π) So the linear approximation x− So L(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 . . . . . .

Example Example Estimate sin(61◦ ) by using a linear approximation (ii) about a = 60◦ = π/3. (i) about a = 0 Solution (ii) Solution (i) () √ We have f π = 23 and () If f(x) = sin x, then f(0) = 0 3 f′ π = 1 . and f′ (0) = 1. 3 2 √ 3 1( π) So the linear approximation x− So L(x) = + near 0 is L(x) = 0 + 1 · x = x. 2 2 3 Thus Thus ( ) ( ) 61π ≈ 0.87475 61π 61π sin ≈ ≈ 1.06465 sin 180 180 180 . . . . . .

Example Example Estimate sin(61◦ ) by using a linear approximation (ii) about a = 60◦ = π/3. (i) about a = 0 Solution (ii) Solution (i) () √ We have f π = 23 and () If f(x) = sin x, then f(0) = 0 3 f′ π = 1 . and f′ (0) = 1. 3 2 √ 3 1( π) So the linear approximation x− So L(x) = + near 0 is L(x) = 0 + 1 · x = x. 2 2 3 Thus Thus ( ) ( ) 61π ≈ 0.87475 61π 61π sin ≈ ≈ 1.06465 sin 180 180 180 Calculator check: sin(61◦ ) ≈ . . . . . .

Example Example Estimate sin(61◦ ) by using a linear approximation (ii) about a = 60◦ = π/3. (i) about a = 0 Solution (ii) Solution (i) () √ We have f π = 23 and () If f(x) = sin x, then f(0) = 0 3 f′ π = 1 . and f′ (0) = 1. 3 2 √ 3 1( π) So the linear approximation x− So L(x) = + near 0 is L(x) = 0 + 1 · x = x. 2 2 3 Thus Thus ( ) ( ) 61π ≈ 0.87475 61π 61π sin ≈ ≈ 1.06465 sin 180 180 180 Calculator check: sin(61◦ ) ≈ 0.87462. . . . . . .

Illustration y . . = sin x y . x . . 1◦ 6 . . . . . .

Illustration y . . = L1 (x) = x y . = sin x y . x . . 1◦ 6 0 . . . . . . .

Illustration y . . = L1 (x) = x y . = sin x y b . ig difference! . x . . 1◦ 6 0 . . . . . . .

Illustration y . . = L1 (x) = x y ( ) √ x− π 3 1 . = L2 (x) = + y 2 2 3 . = sin x y . . . x . . 1◦ 6 0 . π/3 . . . . . . .

Illustration y . . = L1 (x) = x y ( ) √ x− π 3 1 . = L2 (x) = + y 2 2 3 . = sin x y . . ery little difference! v . . x . . 1◦ 6 0 . π/3 . . . . . . .

Another Example Example √ 10 using the fact that 10 = 9 + 1. Estimate . . . . . .

Another Example Example √ 10 using the fact that 10 = 9 + 1. Estimate Solution √ The key step is to √ a linear approximation to f(x) = x near a = 9 to use estimate f(10) = 10. . . . . . .

Another Example Example √ 10 using the fact that 10 = 9 + 1. Estimate Solution √ The key step is to √ a linear approximation to f(x) = x near a = 9 to use estimate f(10) = 10. √ √ d√ 10 ≈ 9+ (1) x dx x=9 1 19 ≈ 3.167 =3+ (1) = 2·3 6 . . . . . .

Another Example Example √ 10 using the fact that 10 = 9 + 1. Estimate Solution √ The key step is to √ a linear approximation to f(x) = x near a = 9 to use estimate f(10) = 10. √ √ d√ 10 ≈ 9+ (1) x dx x=9 1 19 ≈ 3.167 =3+ (1) = 2·3 6 ( )2 19 = Check: 6 . . . . . .

Another Example Example √ 10 using the fact that 10 = 9 + 1. Estimate Solution √ The key step is to √ a linear approximation to f(x) = x near a = 9 to use estimate f(10) = 10. √ √ d√ 10 ≈ 9+ (1) x dx x=9 1 19 ≈ 3.167 =3+ (1) = 2·3 6 ( )2 19 361 = Check: . 6 36 . . . . . .

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 . . . . . .

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 ≈ 1.41422. Calculator check: . . . . . .

Outline The linear approximation of a function near a point Examples Differentials The not-so-big idea . . . . . .

Differentials are another way to express derivatives f(x + ∆x) − f(x) ≈ f′ (x) ∆x y . dy ∆y 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 = f′ (x) . x . . . + ∆x xx dx . . . . . .

Differentials are another way to express derivatives f(x + ∆x) − f(x) ≈ f′ (x) ∆x y . dy ∆y 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 = f′ (x) . x . . . + ∆x xx dx Linear approximation means ∆y ≈ dy = f′ (x0 ) dx near x0 . . . . . . .

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? . . . . . .

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 12 Write A(ℓ) = ℓ . We want to know ∆A when ℓ = 8 ft and ∆ℓ = 1 in. 2 . . . . . .

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 12 Write A(ℓ) = ℓ . We want to know ∆A when ℓ = 8 ft and ∆ℓ = 1 in. 2 () 97 9409 9409 − 32 ≈ 0.6701. (I) A(ℓ + ∆ℓ) = A = So ∆A = 12 288 288 . . . . . .

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 12 Write A(ℓ) = ℓ . We want to know ∆A when ℓ = 8 ft and ∆ℓ = 1 in. 2 () 97 9409 9409 − 32 ≈ 0.6701. (I) A(ℓ + ∆ℓ) = A = So ∆A = 12 288 288 dA = ℓ, so dA = ℓ dℓ, which should be a good estimate for ∆ℓ. (II) 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. . . . . . .

Gravitation Pencils down! Example Drop a 1 kg ball off the roof of the Science Center (50m high). We usually say that a falling object feels a force F = −mg from gravity. . . . . . .

Gravitation Pencils down! Example Drop a 1 kg ball off the roof of the Science 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 ) = 2 = −mg. 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 relative error? The percentage error? . . . . . .

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) ∆r GMm dr = = 2mg 2 re re re ∆F ∆r ≈ −2 The relative error is F re re = 6378.1 km. If ∆r = 50 m, ∆F ∆r 50 = −1.56 × 10−5 = −0.00156% ≈ −2 = −2 F re 6378100 . . . . . .

Systematic linear approximation √ √ 9/4 is rational and 9/4 is close to 2. 2 is irrational, but . . . . . .

Systematic linear approximation √ √ 9/4 is rational and 9/4 is close to 2. So 2 is irrational, but √ √ √ 1 17 − 1/4 ≈ 2= 9/4 9/4 + (−1/4) = 2(3/2) 12 . . . . . .

Systematic linear approximation √ √ 9/4 is rational and 9/4 is close to 2. So 2 is irrational, but √ √ √ 1 17 − 1/4 ≈ 2= 9/4 9/4 + (−1/4) = 2(3/2) 12 This is a better approximation since (17/12)2 = 289/144 . . . . . .

Systematic linear approximation √ √ 9/4 is rational and 9/4 is close to 2. So 2 is irrational, but √ √ √ 1 17 − 1/4 ≈ 2= 9/4 9/4 + (−1/4) = 2(3/2) 12 This is a better approximation since (17/12)2 = 289/144 Do it again! √ √ √ 1 − 1/144 ≈ 2= 289/144 289/144 + (−1/144) = 577/408 2(17/12) ( )2 577 332, 929 1 = Now which is away from 2. 408 166, 464 166, 464 . . . . . .

Illustration of the previous example . . . . . . .

Illustration of the previous example . . . . . . .

Illustration of the previous example . 2 . . . . . . .

Illustration of the previous example . . 2 . . . . . . .

Illustration of the previous example . . 2 . . . . . . .

Illustration of the previous example . 2, 17 ) ( 12 .. . 2 . . . . . . .

Illustration of the previous example . 2, 17 ) ( 12 .. . 2 . . . . . . .

Illustration of the previous example . ( . 2, 17/12) (9 2 . 4, 3) . . . . . . .

Illustration of the previous example . ( . 2, 17/12) (9 2 . 4, 3) .. ( 289 17 ) . 144 , 12 . . . . . .

Illustration of the previous example . ( . 2, 17/12) . 9, 3) ( .. ( 577 ) ( 289 17 ) 4 2 . 144 , 12 . 2, 408 . . . . . .

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