The tangent line to the graph of a function at a point can be thought of as a function itself. As such, it is the best linear function which agrees with the given function at the point. The function and its linear approximation will probably diverge away from the point at which they agree, but this "error" can be measured using the differential notation.

1. Section 2.8
Linear Approximation and
Differentials
V63.0121.027, Calculus I
October 13, 2009
Announcements
Midterm Thursday on Sections 1.1–2.4
. . . . . .

2. Outline
The linear approximation of a function near a point
Examples
Differentials
The not-so-big idea
Using differentials to estimate error
Midterm Review
Advanced Examples
. . . . . .

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

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

5. 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))?
. . . . . .

6. 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)
. . . . . .

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

8. Example
Example
Estimate sin(61◦ ) 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
. . . . . .

9. Example
Example
Estimate sin(61◦ ) by using a linear approximation
(i) about a = 0 (ii) about a = 60◦ = π/3.
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
So the linear approximation
near 0 is
L(x) = 0 + 1 · x = x.
Thus
( )
61π 61π
sin ≈ ≈ 1.06465
180 180
. . . . . .

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

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

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

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

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

15. Example
Example
Estimate sin(61◦ ) by using a linear approximation
(i) about a = 0 (ii) about a = 60◦ = π/3.
Solution (i) Solution (ii)
( ) √
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
Thus ( )
( ) 61π
61π 61π sin ≈ 0.87475
sin ≈ ≈ 1.06465 180
180 180
. . . . . .

16. Example
Example
Estimate sin(61◦ ) by using a linear approximation
(i) about a = 0 (ii) about a = 60◦ = π/3.
Solution (i) Solution (ii)
( ) √
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
Thus ( )
( ) 61π
61π 61π sin ≈ 0.87475
sin ≈ ≈ 1.06465 180
180 180
Calculator check: sin(61◦ ) ≈
. . . . . .

17. Example
Example
Estimate sin(61◦ ) by using a linear approximation
(i) about a = 0 (ii) about a = 60◦ = π/3.
Solution (i) Solution (ii)
( ) √
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
Thus ( )
( ) 61π
61π 61π sin ≈ 0.87475
sin ≈ ≈ 1.06465 180
180 180
Calculator check: sin(61◦ ) ≈ 0.87462.
. . . . . .

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

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

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

21. 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
. . . . . .

22. 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
. . . . . .

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

24. Another Example
Example
√
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 = 9 to estimate f(10) = 10.
. . . . . .

25. Another Example
Example
√
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 = 9 to estimate f(10) = 10.
√ √d√
10 ≈ 9+x (1)
dx x=9
1 19
=3+ (1 ) = ≈ 3.167
2·3 6
. . . . . .

26. Another Example
Example
√
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 = 9 to estimate f(10) = 10.
√ √ d√
10 ≈ 9+ x (1)
dx x=9
1 19
=3+ (1 ) = ≈ 3.167
2·3 6
( )2
19
Check: =
6
. . . . . .

27. Another Example
Example
√
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 = 9 to estimate f(10) = 10.
√ √ d√
10 ≈ 9+ x (1)
dx x=9
1 19
=3+ (1 ) = ≈ 3.167
2·3 6
( )2
19 361
Check: = .
6 36
. . . . . .

28. 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
. . . . . .

29. 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. . . . . . .

30. Outline
The linear approximation of a function near a point
Examples
Differentials
The not-so-big idea
Using differentials to estimate error
Midterm Review
Advanced Examples
. . . . . .

31. 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?
. . . . . .

32. Answers
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?
. . . . . .

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

34. 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?
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?
. . . . . .

35. Answers
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?
Answer
$100
$150
$600 (?)
. . . . . .

36. 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?
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?
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?
. . . . . .

37. Answers
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?
. . . . . .

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

39. 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
. . . . . .

40. 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 .
. . . . . .

41. 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
. . . . . .

42. 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?
. . . . . .

43. 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
1
Write A(ℓ) = ℓ2 . We want to know ∆A when ℓ = 8 ft and
2
∆ℓ = 1 in.
. . . . . .

44. 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
1
Write A(ℓ) = ℓ2 . We want to know ∆A when ℓ = 8 ft and
2
∆ℓ = 1 in. ( )
97 9409
(I) A(ℓ + ∆ℓ) = A = So
12 288
9409
∆A = − 32 ≈ 0.6701.
288
. . . . . .

45. 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
1
Write A(ℓ) = ℓ2 . We want to know ∆A when ℓ = 8 ft and
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.
. . . . . .

46. 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).
. . . . . .

47. Outline
The linear approximation of a function near a point
Examples
Differentials
The not-so-big idea
Using differentials to estimate error
Midterm Review
Advanced Examples
. . . . . .

48. Midterm Facts
Covers sections 1.1–2.4
(Limits, Derivatives,
Differentiation up to
Quotient Rule)
Calculator free
Has about 7 problems
each could have
multiple parts
Some fixed-response,
some free-response
To study:
outline
do problems
metacognition
ask questions!
. . . . . .

49. Outline
The linear approximation of a function near a point
Examples
Differentials
The not-so-big idea
Using differentials to estimate error
Midterm Review
Advanced Examples
. . . . . .

50. 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.
. . . . . .

51. 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?
. . . . . .

52. 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
. . . . . .

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

54. 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
. . . . . .

55. 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
. . . . . .

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
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
. . . . . .

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

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

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

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

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

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

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

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

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

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