lesson 12: linear approximation

Section 2.8Linear Approximation and Differentials

V63.0121.002.2010Su, Calculus I

New York University

May 26, 2010

Announcements

I Quiz 2 Thursday on Sections 1.5–2.5I No class Monday, May 31I Assignment 2 due Tuesday, June 1

. . . . . .

. . . . . .

Announcements

I Quiz 2 Thursday onSections 1.5–2.5

I No class Monday, May 31I Assignment 2 due

Tuesday, June 1

V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 2 / 27

. . . . . .

Objectives

I Use tangent lines to makelinear approximations to afunction.

I Given a function and apoint in the domain,compute thelinearization of thefunction at that point.

I Use linearization toapproximate values offunctions

I Given a function, computethe differential of thatfunction

I Use the differentialnotation to estimate errorin linear approximations.

. . . . . .

Outline

The linear approximation of a function near a pointExamplesQuestions

DifferentialsUsing differentials to estimate error

Advanced Examples

. . . . . .

The Big Idea

QuestionLet 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)

. . . . . .

The Big Idea

Answer

L(x) = f(a) + f′(a)(x− a)

. . . . . .

The Big Idea

Answer

L(x) = f(a) + f′(a)(x− a)

. . . . . .

The Big Idea

Answer

L(x) = f(a) + f′(a)(x− a)

. . . . . .

The tangent line is a linear approximation

L(x) = f(a) + f′(a)(x− a)

is a decent approximation to fnear a.

How decent? The closer x is toa, the better the approxmationL(x) is to f(x)

. .x

.y

.

.f(a)

.f(x).L(x)

.a .x

.x − a

Page 10: Lesson 12: Linear Approximation

. . . . . .

The tangent line is a linear approximation

L(x) = f(a) + f′(a)(x− a)

is a decent approximation to fnear a.How decent? The closer x is toa, the better the approxmationL(x) is to f(x)

. .x

.y

.

.f(a)

.f(x).L(x)

.a .x

.x − a

Page 11: Lesson 12: Linear Approximation

. . . . . .

Example.

.

Example

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

Solution (i)

I If f(x) = sin x, then f(0) = 0and f′(0) = 1.

I So the linear approximationnear 0 is L(x) = 0+ 1 · x = x.

I Thus

sin(61π180

)≈ 61π

180≈ 1.06465

Solution (ii)

I We have f(π3)=

√32

andf′(π3)=

12

.

I So L(x) =

√32

+12

(x− π

3

)

I Thus

sin(61π180

)≈

0.87475

Calculator check: sin(61◦) ≈

0.87462.

Page 12: Lesson 12: Linear Approximation

. . . . . .

Example.

.

Example

Solution (i)

I Thus

sin(61π180

)≈ 61π

180≈ 1.06465

Solution (ii)

I We have f(π3)=

√32

andf′(π3)=

12

.

I So L(x) =

√32

+12

(x− π

3

)

I Thus

sin(61π180

)≈

0.87475

0.87462.

Page 13: Lesson 12: Linear Approximation

. . . . . .

Example.

.

Example

Solution (i)

I Thus

sin(61π180

)≈ 61π

180≈ 1.06465

Solution (ii)

I We have f(π3)=

√32

andf′(π3)=

12

.

I So L(x) =

√32

+12

(x− π

3

)

I Thus

sin(61π180

)≈

0.87475

0.87462.

Page 14: Lesson 12: Linear Approximation

. . . . . .

Example.

.

Example

Solution (i)

I Thus

sin(61π180

)≈ 61π

180≈ 1.06465

Solution (ii)

I We have f(π3)=

√32 and

f′(π3)=

12

.

I So L(x) =

√32

+12

(x− π

3

)

I Thus

sin(61π180

)≈

0.87475

0.87462.

Page 15: Lesson 12: Linear Approximation

. . . . . .

Example.

.

Example

Solution (i)

I Thus

sin(61π180

)≈ 61π

180≈ 1.06465

Solution (ii)

I We have f(π3)=

√32 and

f′(π3)= 1

2 .

I So L(x) =

√32

+12

(x− π

3

)

I Thus

sin(61π180

)≈

0.87475

0.87462.

Page 16: Lesson 12: Linear Approximation

. . . . . .

Example.

.

Example

Solution (i)

I Thus

sin(61π180

)≈ 61π

180≈ 1.06465

Solution (ii)

I We have f(π3)=

√32 and

f′(π3)= 1

2 .

I So L(x) =

√32

+12

(x− π

3

)I Thus

sin(61π180

)≈

0.87475

0.87462.

Page 17: Lesson 12: Linear Approximation

. . . . . .

Example.

.

Example

Solution (i)

I Thus

sin(61π180

)≈ 61π

180≈ 1.06465

Solution (ii)

I We have f(π3)=

√32 and

f′(π3)= 1

2 .

I So L(x) =√32

+12

(x− π

3

)

I Thus

sin(61π180

)≈

0.87475

0.87462.

Page 18: Lesson 12: Linear Approximation

. . . . . .

Example.

.

Example

Solution (i)

I Thus

sin(61π180

)≈ 61π

180≈ 1.06465

Solution (ii)

I We have f(π3)=

√32 and

f′(π3)= 1

2 .

I So L(x) =√32

+12

(x− π

3

)I Thus

sin(61π180

)≈

0.87475

0.87462.

Page 19: Lesson 12: Linear Approximation

. . . . . .

Example.

.

Example

Solution (i)

I Thus

sin(61π180

)≈ 61π

180≈ 1.06465

Solution (ii)

I We have f(π3)=

√32 and

f′(π3)= 1

2 .

I So L(x) =√32

+12

(x− π

3

)I Thus

sin(61π180

)≈ 0.87475

0.87462.

Page 20: Lesson 12: Linear Approximation

. . . . . .

Example.

.

Example

Solution (i)

I Thus

sin(61π180

)≈ 61π

180≈ 1.06465

Solution (ii)

I We have f(π3)=

√32 and

f′(π3)= 1

2 .

I So L(x) =√32

+12

(x− π

3

)I Thus

sin(61π180

)≈ 0.87475

0.87462.

Page 21: Lesson 12: Linear Approximation

. . . . . .

Example.

.

Example

Solution (i)

I Thus

sin(61π180

)≈ 61π

180≈ 1.06465

Solution (ii)

I We have f(π3)=

√32 and

f′(π3)= 1

2 .

I So L(x) =√32

+12

(x− π

3

)I Thus

sin(61π180

)≈ 0.87475

Calculator check: sin(61◦) ≈ 0.87462.

Page 22: Lesson 12: Linear Approximation

. . . . . .

Illustration

. .x

.y

.y = sin x

.61◦

.y = L1(x) = x

..0

.

.big difference!

.y = L2(x) =√32 + 1

2(x− π

3)

..π/3

.

.very little difference!

Page 23: Lesson 12: Linear Approximation

. . . . . .

Illustration

. .x

.y

.y = sin x

.61◦

.y = L1(x) = x

..0

.

.big difference!.y = L2(x) =

√32 + 1

2(x− π

3)

..π/3

.

Page 24: Lesson 12: Linear Approximation

. . . . . .

Illustration

. .x

.y

.y = sin x

.61◦

.y = L1(x) = x

..0

.

.big difference!

.y = L2(x) =√32 + 1

2(x− π

3)

..π/3

.

Page 25: Lesson 12: Linear Approximation

. . . . . .

Illustration

. .x

.y

.y = sin x

.61◦

.y = L1(x) = x

..0

.

.big difference!

.y = L2(x) =√32 + 1

2(x− π

3)

..π/3

.

Page 26: Lesson 12: Linear Approximation

. . . . . .

Illustration

. .x

.y

.y = sin x

.61◦

.y = L1(x) = x

..0

.

.big difference!

.y = L2(x) =√32 + 1

2(x− π

3)

..π/3

. .very little difference!

Page 27: Lesson 12: Linear Approximation

. . . . . .

Another Example

Example

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

SolutionThe key step is to use a linear approximation to f(x) =

√x near a = 9

to estimate f(10) =√10.

√10 ≈

√9+

ddx

√x∣∣∣∣x=9

(1)

= 3+1

2 · 3(1) =

196

≈ 3.167

Check:(196

)2=

36136

.

Page 28: Lesson 12: Linear Approximation

. . . . . .

Another Example

Example

√x near a = 9

√10 ≈

√9+

ddx

√x∣∣∣∣x=9

(1)

= 3+1

2 · 3(1) =

196

≈ 3.167

Check:(196

)2=

36136

.

Page 29: Lesson 12: Linear Approximation

. . . . . .

Another Example

Example

√x near a = 9

√10 ≈

√9+

ddx

√x∣∣∣∣x=9

(1)

= 3+1

2 · 3(1) =

196

≈ 3.167

Check:(196

)2=

36136

.

Page 30: Lesson 12: Linear Approximation

. . . . . .

Another Example

Example

√x near a = 9

√10 ≈

√9+

ddx

√x∣∣∣∣x=9

(1)

= 3+1

2 · 3(1) =

196

≈ 3.167

Check:(196

)2=

36136

.

Page 31: Lesson 12: Linear Approximation

. . . . . .

Another Example

Example

√x near a = 9

√10 ≈

√9+

ddx

√x∣∣∣∣x=9

(1)

= 3+1

2 · 3(1) =

196

≈ 3.167

Check:(196

)2=

36136

.

Page 32: Lesson 12: Linear Approximation

. . . . . .

Dividing without dividing?

Example

Suppose I have an irrational fear of division and need to estimate577÷ 408. I write

577408

= 1+ 1691

408= 1+ 169× 1

4× 1

102.

But still I have to find1

102.

Solution

Let f(x) =1x. We know f(100) and we want to estimate f(102).

f(102) ≈ f(100) + f′(100)(2) =1

100− 1

1002(2) = 0.0098

=⇒ 577408

≈ 1.41405

Calculator check:577408

≈ 1.41422.

Page 33: Lesson 12: Linear Approximation

. . . . . .

Dividing without dividing?

Example

Suppose I have an irrational fear of division and need to estimate577÷ 408. I write

577408

= 1+ 1691

408= 1+ 169× 1

4× 1

102.

But still I have to find1

102.

Solution

Let f(x) =1x. We know f(100) and we want to estimate f(102).

f(102) ≈ f(100) + f′(100)(2) =1

100− 1

1002(2) = 0.0098

=⇒ 577408

≈ 1.41405

Calculator check:577408

≈ 1.41422.

Page 34: Lesson 12: Linear Approximation

. . . . . .

Questions

Example

Suppose we are traveling in a car and at noon our speed is 50mi/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 iscurrently $50/lot. How much will it cost to make 2 more lots? 3 morelots? 12 more lots?

Example

Suppose 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?

Page 35: Lesson 12: Linear Approximation

. . . . . .

Answers

Example

Answer

I 100miI 150miI 600mi (?) (Is it reasonable to assume 12 hours at the same

speed?)

Page 36: Lesson 12: Linear Approximation

. . . . . .

Answers

Example

Answer

I 100miI 150miI 600mi (?) (Is it reasonable to assume 12 hours at the same

speed?)

Page 37: Lesson 12: Linear Approximation

. . . . . .

Questions

Example

Page 38: Lesson 12: Linear Approximation

. . . . . .

Answers

Example

Answer

I $100I $150I $600 (?)

Page 39: Lesson 12: Linear Approximation

. . . . . .

Questions

Example

Page 40: Lesson 12: Linear Approximation

. . . . . .

Answers

Example

AnswerThe slope of the line is

m =riserun

We are given a “run” of dx, so the corresponding “rise” is mdx.

Page 41: Lesson 12: Linear Approximation

. . . . . .

Answers

Example

AnswerThe slope of the line is

m =riserun

We are given a “run” of dx, so the corresponding “rise” is mdx.

Page 42: Lesson 12: Linear Approximation

. . . . . .

Outline

Advanced Examples

Page 43: Lesson 12: Linear Approximation

. . . . . .

Differentials are another way to express derivatives

f(x+∆x)− f(x)︸︷︷︸∆y

≈ f′(x)∆x︸︷︷︸dy

Rename ∆x = dx, so we canwrite this as

∆y ≈ dy = f′(x)dx.

And this looks a lot like theLeibniz-Newton identity

dydx

= f′(x) . .x

.y

.

.x .x+∆x

.dx = ∆x

.∆y.dy

Linear approximation means ∆y ≈ dy = f′(x0)dx near x0.

Page 44: Lesson 12: Linear Approximation

. . . . . .

Differentials are another way to express derivatives

f(x+∆x)− f(x)︸︷︷︸∆y

≈ f′(x)∆x︸︷︷︸dy

Rename ∆x = dx, so we canwrite this as

∆y ≈ dy = f′(x)dx.

And this looks a lot like theLeibniz-Newton identity

dydx

= f′(x) . .x

.y

.

.x .x+∆x

.dx = ∆x

.∆y.dy

Linear approximation means ∆y ≈ dy = f′(x0)dx near x0.

Page 45: Lesson 12: Linear Approximation

. . . . . .

Using differentials to estimate error

If y = f(x), x0 and ∆x is known,and an estimate of ∆y isdesired:

I Approximate: ∆y ≈ dyI Differentiate: dy = f′(x)dxI Evaluate at x = x0 and

dx = ∆x.

. .x

.y

.

.x .x+∆x

.dx = ∆x

.∆y.dy

Page 46: Lesson 12: Linear Approximation

. . . . . .

Example

A 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

Write A(ℓ) =12ℓ2. We want to know ∆A when ℓ = 8 ft and ∆ℓ = 1 in.

(I) A(ℓ+∆ℓ) = A(9712

)=

9409288

So ∆A =9409288

− 32 ≈ 0.6701.

(II) dAdℓ

= ℓ, so dA = ℓdℓ, which should be a good estimate for ∆ℓ.

When ℓ = 8 and dℓ = 112 , we have dA = 8

12 = 23 ≈ 0.667. So we

get estimates close to the hundredth of a square foot.

Page 47: Lesson 12: Linear Approximation

. . . . . .

Example

Solution

(I) A(ℓ+∆ℓ) = A(9712

)=

9409288

So ∆A =9409288

− 32 ≈ 0.6701.

(II) dAdℓ

12 = 23 ≈ 0.667. So we

Page 48: Lesson 12: Linear Approximation

. . . . . .

Example

Solution

(I) A(ℓ+∆ℓ) = A(9712

)=

9409288

So ∆A =9409288

− 32 ≈ 0.6701.

(II) dAdℓ

12 = 23 ≈ 0.667. So we

Page 49: Lesson 12: Linear Approximation

. . . . . .

Example

Solution

(I) A(ℓ+∆ℓ) = A(9712

)=

9409288

So ∆A =9409288

− 32 ≈ 0.6701.

(II) dAdℓ

12 = 23 ≈ 0.667. So we

Page 50: Lesson 12: Linear Approximation

. . . . . .

Why?

Why use linear approximations dy when the actual difference ∆y isknown?

I Linear approximation is quick and reliable. Finding ∆y exactlydepends on the function.

I These examples are overly simple. See the “Advanced Examples”later.

I In real life, sometimes only f(a) and f′(a) are known, and not thegeneral f(x).

Page 51: Lesson 12: Linear Approximation

. . . . . .

Outline

Advanced Examples

Page 52: Lesson 12: Linear Approximation

. . . . . .

GravitationPencils down!

Example

I Drop a 1 kg ball off the roof of the Silver Center (50m high). Weusually say that a falling object feels a force F = −mg from gravity.

I In fact, the force felt is

F(r) = −GMmr2

,

where M is the mass of the earth and r is the distance from thecenter of the earth to the object. G is a constant.

I At r = re the force really is F(re) =GMmr2e

= −mg.

I What is the maximum error in replacing the actual force felt at thetop of the building F(re +∆r) by the force felt at ground levelF(re)? The relative error? The percentage error?

Page 53: Lesson 12: Linear Approximation

. . . . . .

GravitationPencils down!

Example

I Drop a 1 kg ball off the roof of the Silver Center (50m high). Weusually say that a falling object feels a force F = −mg from gravity.

I In fact, the force felt is

F(r) = −GMmr2

,

where M is the mass of the earth and r is the distance from thecenter of the earth to the object. G is a constant.

I At r = re the force really is F(re) =GMmr2e

= −mg.

I What is the maximum error in replacing the actual force felt at thetop of the building F(re +∆r) by the force felt at ground levelF(re)? The relative error? The percentage error?

Page 54: Lesson 12: Linear Approximation

. . . . . .

Gravitation Solution

SolutionWe wonder if ∆F = F(re +∆r)− F(re) is small.

I Using a linear approximation,

∆F ≈ dF =dFdr

∣∣∣∣redr = 2

GMmr3e

dr

=

(GMmr2e

)drre

= 2mg∆rre

I The relative error is∆FF

≈ −2∆rre

I re = 6378.1 km. If ∆r = 50m,

∆FF

≈ −2∆rre

= −250

6378100= −1.56× 10−5 = −0.00156%

Page 55: Lesson 12: Linear Approximation

. . . . . .

Systematic linear approximation

I√2 is irrational, but

√9/4 is rational and 9/4 is close to 2.

So

√2 =

√9/4 − 1/4 ≈

√9/4 +

12(3/2)

(−1/4) =1712

I This is a better approximation since (17/12)2 = 289/144

I Do it again!

√2 =

√289/144 − 1/144 ≈

√289/144 +

12(17/12)

(−1/144) = 577/408

Now(577408

)2=

332,929166,464

which is1

166,464away from 2.

Page 56: Lesson 12: Linear Approximation

. . . . . .

√9/4 is rational and 9/4 is close to 2. So

√2 =

√9/4 − 1/4 ≈

√9/4 +

12(3/2)

(−1/4) =1712

I Do it again!

√2 =

√289/144 − 1/144 ≈

√289/144 +

12(17/12)

(−1/144) = 577/408

Now(577408

)2=

332,929166,464

which is1

166,464away from 2.

Page 57: Lesson 12: Linear Approximation

. . . . . .

√2 =

√9/4 − 1/4 ≈

√9/4 +

12(3/2)

(−1/4) =1712

I Do it again!

√2 =

√289/144 − 1/144 ≈

√289/144 +

12(17/12)

(−1/144) = 577/408

Now(577408

)2=

332,929166,464

which is1

166,464away from 2.

Page 58: Lesson 12: Linear Approximation

. . . . . .

√2 =

√9/4 − 1/4 ≈

√9/4 +

12(3/2)

(−1/4) =1712

I Do it again!

√2 =

√289/144 − 1/144 ≈

√289/144 +

12(17/12)

(−1/144) = 577/408

Now(577408

)2=

332,929166,464

which is1

166,464away from 2.

Page 59: Lesson 12: Linear Approximation

. . . . . .

Illustration of the previous example

.

.2

.

(94 ,32)

..(2, 1712)

Page 60: Lesson 12: Linear Approximation

. . . . . .

.

.2

.

(94 ,32)

..(2, 1712)

Page 61: Lesson 12: Linear Approximation

. . . . . .

..2

.

(94 ,32)

..(2, 1712)

Page 62: Lesson 12: Linear Approximation

. . . . . .

..2

.

(94 ,32)

..(2, 1712)

Page 63: Lesson 12: Linear Approximation

. . . . . .

..2

.

(94 ,32)

..(2, 1712)

Page 64: Lesson 12: Linear Approximation

. . . . . .

..2

.

(94 ,32)

..(2, 1712)

Page 65: Lesson 12: Linear Approximation

. . . . . .

..2

.

(94 ,32)

..(2, 1712)

Page 66: Lesson 12: Linear Approximation

. . . . . .

..2

..(94 ,

32)..(2, 17/12)

..(289144 ,

1712

)..(2, 577408

)

Page 67: Lesson 12: Linear Approximation

. . . . . .

..2

..(94 ,

32)..(2, 17/12) .

.(289144 ,

1712

)

..(2, 577408

)

Page 68: Lesson 12: Linear Approximation

. . . . . .

..2

..(94 ,

32)..(2, 17/12) .

.(289144 ,

1712

)..(2, 577408

)

Page 69: Lesson 12: Linear Approximation

. . . . . .

Summary

I Linear approximation: If f is differentiable at a, the best linearapproximation to f near a is given by

Lf,a(x) = f(a) + f′(a)(x− a)

I Differentials: If f is differentiable at x, a good approximation to∆y = f(x+∆x)− f(x) is

∆y ≈ dy =dydx

· dx =dydx

·∆x

I Don’t buy plywood from me.

lesson 12: linear approximation

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