Lesson 12: Linear Approximation

Section 2.8
Linear Approximation and
Differentials

V63.0121.006/016, Calculus I

February 26, 2010

Announcements
Quiz 2 is February 26, covering §§1.5–2.3
Midterm is March 4, covering §§1.1–2.5
. . . . . .

Announcements

Quiz 2 is February 26, covering §§1.5–2.3
Midterm is March 4, covering §§1.1–2.5

. . . . . .

Outline

The linear approximation of a function near a point
Examples
Questions

Midterm Review

Differentials
Using differentials to estimate error

Advanced Examples

. . . . . .

The Big Idea

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

. . . . . .

The Big Idea

Question

Answer
The tangent line, of course!

. . . . . .

The Big Idea

Question

Answer

Question
What is the equation for the line tangent to y = f(x) at (a, f(a))?

. . . . . .

The Big Idea

Question

Answer

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)

. . . . . .

The tangent line is a linear approximation

y
.

L(x) = f(a) + f′ (a)(x − a)

is a decent approximation to L
. (x) .
f near a. f
.(x) .

f
.(a) .
.
x−a

. x
.
a
. x
.

. . . . . .

The tangent line is a linear approximation

y
.

L(x) = f(a) + f′ (a)(x − a)

is a decent approximation to L
. (x) .
f near a. f
.(x) .
How decent? The closer x is
to a, the better the f
.(a) .
.
x−a
approxmation L(x) is to f(x)

. x
.
a
. x
.

. . . . . .

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

. . . . . .

Example
Example
approximation

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

. . . . . .

Example
Example
approximation

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
near 0 is
L(x) = 0 + 1 · x = x.
Thus
( )
61π 61π
sin ≈ ≈ 1.06465
180 180

. . . . . .

Example
Example
approximation

( ) √
If f(x) = sin x, then f(0) = 0 We have f π = 3
and
and f′ (0) = 1. ( ) 3 2
f′ π = .
3
near 0 is
L(x) = 0 + 1 · x = x.
Thus
( )
61π 61π
sin ≈ ≈ 1.06465
180 180

. . . . . .

Example
Example
approximation

( ) √
and
and f′ (0) = 1. ( ) 3 2
f′ π = 1 .
3 2
near 0 is
L(x) = 0 + 1 · x = x.
Thus
( )
61π 61π
sin ≈ ≈ 1.06465
180 180

. . . . . .

Example
Example
approximation

( ) √
and
and f′ (0) = 1. ( ) 3 2
f′ π = 1 .
3 2
near 0 is So L(x) =
L(x) = 0 + 1 · x = x.
Thus
( )
61π 61π
sin ≈ ≈ 1.06465
180 180

. . . . . .

Example
Example
approximation

( ) √
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

. . . . . .

Example
Example
approximation

( ) √
and f′ (0) = 1. ( ) 3
f′ π = 1 .
3 2
√
2 2 3
L(x) = 0 + 1 · x = x. Thus
Thus ( )
( ) 61π
61π 61π sin ≈
sin ≈ ≈ 1.06465 180
180 180

. . . . . .

Example
Example
approximation

( ) √
and f′ (0) = 1. ( ) 3
f′ π = 1 .
3 2
√
2 2 3
L(x) = 0 + 1 · x = x. Thus
Thus ( )
( ) 61π
61π 61π sin ≈ 0.87475
sin ≈ ≈ 1.06465 180
180 180

. . . . . .

Example
Example
approximation

( ) √
and f′ (0) = 1. ( ) 3
f′ π = 1 .
3 2
√
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◦ ) ≈ . . . . . .

Example
Example
approximation

( ) √
and f′ (0) = 1. ( ) 3
f′ π = 1 .
3 2
√
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. . . . . . .

Illustration

y
.

y
. = sin x

. x
.
. 1◦
6

. . . . . .

Illustration

y
.
y
. = L1 (x) = x

y
. = sin x

. x
.
0
. . 1◦
6

. . . . . .

Illustration

y
.
y
. = L1 (x) = x

b
. ig difference! y
. = sin x

. x
.
0
. . 1◦
6

. . . . . .

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

. . . . . .

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

. . . . . .

Another Example

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

. . . . . .

Another Example

Example
√

Solution √
The key step is to use a linear approximation to f(x) =
√ x near
a = 9 to estimate f(10) = 10.

. . . . . .

Another Example

Example
√

Solution √
√ x near
√ √ d√
10 ≈ 9 + x (1)
dx x=9
1 19
=3+ (1 ) = ≈ 3.167
2·3 6

. . . . . .

Another Example

Example
√

Solution √
√ x near
√ √ d√
10 ≈ 9 + x (1)
dx x=9
1 19
=3+ (1 ) = ≈ 3.167
2·3 6
( )2
19
Check: =
6

. . . . . .

Another Example

Example
√

Solution √
√ x near
√ √ d√
10 ≈ 9 + x (1)
dx x=9
1 19
=3+ (1 ) = ≈ 3.167
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
Calculator check: ≈ 1.41422. . . . . . .

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?

. . . . . .

Answers

Example
By midnight?

. . . . . .

Answers

Example
By midnight?

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

. . . . . .

Questions

Example
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?

. . . . . .

Answers

Example

Answer
$100
$150
$600 (?)

. . . . . .

Questions

Example
By midnight?

Example

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?

. . . . . .

Answers

Example

. . . . . .

Answers

Example

Answer
The slope of the line is
rise
m=
run
We are given a “run” of dx, so the corresponding “rise” is m dx.

. . . . . .

Midterm Facts

Covers sections 1.1–2.5
(Limits, Derivatives,
Differentiation up to
Chain Rule)
Calculator free
20 multiple-choice
questions and 4
free-response questions
To study:
outline
do problems
metacognition
ask questions!
(maybe in recitation?)

. . . . . .

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

. . . . . .

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 .

. . . . . .

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

. . . . . .

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

Solution
1
Write A(ℓ) = ℓ2 . We want to know ∆A when ℓ = 8 ft and
2
∆ℓ = 1 in.

. . . . . .

Example

Solution
1
2
∆ℓ = 1 in. ( )
97 9409
(I) A(ℓ + ∆ℓ) = A = So
12 288
9409
∆A = − 32 ≈ 0.6701.
288

. . . . . .

Example

Solution
1
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.
. . . . . .

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).

. . . . . .

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.

. . . . . .

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?
. . . . . .

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

. . . . . .

Systematic linear approximation

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

. . . . . .


√ √
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

. . . . . .


√ √
√ √ √ 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

. . . . . .


√ √
√ √ √ 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

. . . . . .

Illustration of the previous example

.

. . . . . .


.
2
.

. . . . . .


.

.
2
.

. . . . . .


. 2, 17 )
( 12
. .

.
2
.

. . . . . .


.
. 2, 17/12)
(
. (9 2
. 4, 3)

. . . . . .


.
. 2, 17/12)
(
.. . 9, 3)
(
( 289 17 ) 4 2
. 144 , 12

. . . . . .


.
. 2, 17/12)
(
.. . 9, 3)
(
( 577 ) ( 289 17 ) 4 2
. 2, 408 . 144 , 12

. . . . . .

Lesson 12: Linear Approximation

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