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.

1. Section 2.8
Linear Approximation and
Differentials
V63.0121, Calculus I
February 26/March 2, 2009
Announcements
Midterm I Wednesday March 4 in class.
OH this week: MT 1–2pm, W 2–3pm
Get half of additional ALEKS points through March 22, 11:59pm
.
Image credit: cobalt123
. . . . . . .

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

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
(ii) about a = 60◦ = π/3.
(i) about a = 0
. . . . . .

8. 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
. . . . . .

9. 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
. . . . . .

10. 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
. . . . . .

11. 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
. . . . . .

12. 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
. . . . . .

13. 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
. . . . . .

14. 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
. . . . . .

15. 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
. . . . . .

16. 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◦ ) ≈
. . . . . .

17. 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.
. . . . . .

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

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

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

21. 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
.
. . . . . .

22. 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
.
. . . . . .

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

24. 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.
. . . . . .

25. 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
. . . . . .

26. 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
. . . . . .

27. 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
. . . . . .

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

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

31. 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
. . . . . .

32. 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 .
. . . . . .

33. 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?
. . . . . .

34. 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
. . . . . .

35. 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
. . . . . .

36. 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.
. . . . . .

37. 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.
. . . . . .

38. 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?
. . . . . .

39. 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
. . . . . .

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

41. 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
. . . . . .

42. 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
. . . . . .

43. 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
. . . . . .

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

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

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

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

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

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

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

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

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

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