. . . . . .

Section 2.8Linear Approximation and

Differentials

V63.0121, Calculus I

February 26/March 2, 2009

Announcements

I Midterm I Wednesday March 4 in class.I OH this week: MT 1–2pm, W 2–3pmI Get half of additional ALEKS points through March 22, 11:59pm

.

.Image credit: cobalt123

. . . . . .

Outline

The linear approximation of a function near a pointExamples

DifferentialsThe not-so-big idea

. . . . . .

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

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

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

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)

. . . . . .

Example

ExampleEstimate sin(61◦) 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(

61π

180

)≈ 61π

180≈ 1.06465

Solution (ii)

I We have f(

π3

)=

√3

2

andf′

(π3

)=

12

.

I So L(x) =

√3

2+

12

(x − π

3

)

I Thus

sin(

61π

180

)≈

0.87475

Calculator check: sin(61◦) ≈

0.87462.

. . . . . .

Example

ExampleEstimate sin(61◦) 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(

61π

180

)≈ 61π

180≈ 1.06465

Solution (ii)

I We have f(

π3

)=

√3

2

andf′

(π3

)=

12

.

I So L(x) =

√3

2+

12

(x − π

3

)

I Thus

sin(

61π

180

)≈

0.87475

Calculator check: sin(61◦) ≈

0.87462.

. . . . . .

Example

ExampleEstimate sin(61◦) 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(

61π

180

)≈ 61π

180≈ 1.06465

Solution (ii)

I We have f(

π3

)=

√3

2

andf′

(π3

)=

12

.

I So L(x) =

√3

2+

12

(x − π

3

)

I Thus

sin(

61π

180

)≈

0.87475

Calculator check: sin(61◦) ≈

0.87462.

. . . . . .

Example

ExampleEstimate sin(61◦) 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(

61π

180

)≈ 61π

180≈ 1.06465

Solution (ii)

I We have f(

π3

)=

√3

2 andf′

(π3

)=

12

.

I So L(x) =

√3

2+

12

(x − π

3

)

I Thus

sin(

61π

180

)≈

0.87475

Calculator check: sin(61◦) ≈

0.87462.

. . . . . .

Example

ExampleEstimate sin(61◦) 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(

61π

180

)≈ 61π

180≈ 1.06465

Solution (ii)

I We have f(

π3

)=

√3

2 andf′

(π3

)= 1

2 .

I So L(x) =

√3

2+

12

(x − π

3

)

I Thus

sin(

61π

180

)≈

0.87475

Calculator check: sin(61◦) ≈

0.87462.

. . . . . .

Example

ExampleEstimate sin(61◦) 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(

61π

180

)≈ 61π

180≈ 1.06465

Solution (ii)

I We have f(

π3

)=

√3

2 andf′

(π3

)= 1

2 .

I So L(x) =

√3

2+

12

(x − π

3

)I Thus

sin(

61π

180

)≈

0.87475

Calculator check: sin(61◦) ≈

0.87462.

. . . . . .

Example

ExampleEstimate sin(61◦) 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(

61π

180

)≈ 61π

180≈ 1.06465

Solution (ii)

I We have f(

π3

)=

√3

2 andf′

(π3

)= 1

2 .

I So L(x) =

√3

2+

12

(x − π

3

)

I Thus

sin(

61π

180

)≈

0.87475

Calculator check: sin(61◦) ≈

0.87462.

. . . . . .

Example

ExampleEstimate sin(61◦) 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(

61π

180

)≈ 61π

180≈ 1.06465

Solution (ii)

I We have f(

π3

)=

√3

2 andf′

(π3

)= 1

2 .

I So L(x) =

√3

2+

12

(x − π

3

)I Thus

sin(

61π

180

)≈

0.87475

Calculator check: sin(61◦) ≈

0.87462.

. . . . . .

Example

ExampleEstimate sin(61◦) 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(

61π

180

)≈ 61π

180≈ 1.06465

Solution (ii)

I We have f(

π3

)=

√3

2 andf′

(π3

)= 1

2 .

I So L(x) =

√3

2+

12

(x − π

3

)I Thus

sin(

61π

180

)≈ 0.87475

Calculator check: sin(61◦) ≈

0.87462.

. . . . . .

Example

ExampleEstimate sin(61◦) 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(

61π

180

)≈ 61π

180≈ 1.06465

Solution (ii)

I We have f(

π3

)=

√3

2 andf′

(π3

)= 1

2 .

I So L(x) =

√3

2+

12

(x − π

3

)I Thus

sin(

61π

180

)≈ 0.87475

Calculator check: sin(61◦) ≈

0.87462.

. . . . . .

Example

ExampleEstimate sin(61◦) 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(

61π

180

)≈ 61π

180≈ 1.06465

Solution (ii)

I We have f(

π3

)=

√3

2 andf′

(π3

)= 1

2 .

I So L(x) =

√3

2+

12

(x − π

3

)I Thus

sin(

61π

180

)≈ 0.87475

Calculator check: sin(61◦) ≈ 0.87462.

. . . . . .

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!

. . . . . .

Illustration

. .x

.y

.y = sin x

.61◦

.y = L1(x) = x

..0

.

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

√3

2 + 12

(x − π

3

)

.

.π/3

.

.very little difference!

. . . . . .

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!

. . . . . .

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!

. . . . . .

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!

. . . . . .

Another Example

ExampleEstimate

√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

.

. . . . . .

Another Example

ExampleEstimate

√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

.

. . . . . .

Another Example

ExampleEstimate

√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

.

. . . . . .

Another Example

ExampleEstimate

√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

.

. . . . . .

Another Example

ExampleEstimate

√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

.

. . . . . .

Dividing without dividing?ExampleSuppose 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.

SolutionLet 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.

. . . . . .

Dividing without dividing?ExampleSuppose 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.

SolutionLet 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.

. . . . . .

Outline

The linear approximation of a function near a pointExamples

DifferentialsThe not-so-big idea

. . . . . .

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.

. . . . . .

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.

. . . . . .

ExampleA sheet of plywood measures 8 ft × 4 ft. Suppose ourplywood-cutting machine will cut a rectangle whose width is exactlyhalf its length, but the length is prone to errors. If the length is off by1 in, how bad can the area of the sheet be off by?

SolutionWrite 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 = ℓ dℓ, 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.

. . . . . .

ExampleA sheet of plywood measures 8 ft × 4 ft. Suppose ourplywood-cutting machine will cut a rectangle whose width is exactlyhalf its length, but the length is prone to errors. If the length is off by1 in, how bad can the area of the sheet be off by?

SolutionWrite 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 = ℓ dℓ, 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.

. . . . . .

ExampleA sheet of plywood measures 8 ft × 4 ft. Suppose ourplywood-cutting machine will cut a rectangle whose width is exactlyhalf its length, but the length is prone to errors. If the length is off by1 in, how bad can the area of the sheet be off by?

SolutionWrite 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 = ℓ dℓ, 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.

. . . . . .

ExampleA sheet of plywood measures 8 ft × 4 ft. Suppose ourplywood-cutting machine will cut a rectangle whose width is exactlyhalf its length, but the length is prone to errors. If the length is off by1 in, how bad can the area of the sheet be off by?

SolutionWrite 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 = ℓ dℓ, 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.

. . . . . .

GravitationPencils down!

Example

I 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 fromgravity.

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) =GMm

r2e= −mg.

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

. . . . . .

GravitationPencils down!

Example

I 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 fromgravity.

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) =GMm

r2e= −mg.

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

. . . . . .

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

I Using a linear approximation,

∆F ≈ dF =dFdr

∣∣∣∣re

dr = 2GMm

r3edr

=

(GMm

r2e

)drre

= 2mg∆rre

I The relative error is∆FF

≈ −2∆rre

I re = 6378.1 km. If ∆r = 50 m,

∆FF

≈ −2∆rre

= −250

6378100= −1.56 × 10−5 = −0.00156%

. . . . . .

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.

. . . . . .

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.

. . . . . .

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.

. . . . . .

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.

. . . . . .

Illustration of the previous example

.

.2

.

( 94 , 3

2)

..(2, 17

12)

. . . . . .

Illustration of the previous example

.

.2

.

( 94 , 3

2)

..(2, 17

12)

. . . . . .

Illustration of the previous example

..2

.

( 94 , 3

2)

..(2, 17

12)

. . . . . .

Illustration of the previous example

..2

.

( 94 , 3

2)

..(2, 17

12)

. . . . . .

Illustration of the previous example

..2

.

( 94 , 3

2)

..(2, 17

12)

. . . . . .

Illustration of the previous example

..2

.

( 94 , 3

2)

..(2, 17

12)

. . . . . .

Illustration of the previous example

..2

.

( 94 , 3

2)

..(2, 17

12)

. . . . . .

Illustration of the previous example

..2

..( 9

4 , 32).

.(2, 17/12)

..( 289

144 , 1712

)..(2, 577

408

)

. . . . . .

Illustration of the previous example

..2

..( 9

4 , 32).

.(2, 17/12)..( 289

144 , 1712

)

..(2, 577

408

)

. . . . . .

Illustration of the previous example

..2

..( 9

4 , 32).

.(2, 17/12)..( 289

144 , 1712

)..(2, 577

408

)