The document contains a calculus class announcement detailing information about the midterm exam, including topics covered, exam format and preparation strategies. It emphasizes the importance of understanding problems and offers examples related to linear approximations and differentials. The content also includes guidelines on using tangent lines for approximations and various examples of estimating functions like sine and square roots.

1. Sec on 2.8
Linear Approxima ons and
Differen als
V63.0121.011: Calculus I
Professor Ma hew Leingang
New York University
March 2, 2011
.

2. Announcements
Quiz in recita on this
week on 1.5, 1.6, 2.1, 2.2
Midterm March 7 in class
on 1.1–2.5
No quiz in recita on next
week

3. Midterm FAQ
Ques on
What sec ons are covered on the midterm?

4. Midterm FAQ
Ques on
What sec ons are covered on the midterm?
Answer
The midterm will cover Sec ons 1.1–2.5 (The Chain Rule).

5. Midterm FAQ
Ques on
What sec ons are covered on the midterm?
Answer
The midterm will cover Sec ons 1.1–2.5 (The Chain Rule).
Ques on
Is Sec on 2.6 going to be on the midterm?

6. Midterm FAQ
Ques on
What sec ons are covered on the midterm?
Answer
The midterm will cover Sec ons 1.1–2.5 (The Chain Rule).
Ques on
Is Sec on 2.6 going to be on the midterm?
Answer
The midterm will cover Sec ons 1.1–2.5 (The Chain Rule).

7. Midterm FAQ, continued
Ques on
What format will the exam take?

8. Midterm FAQ, continued
Ques on
What format will the exam take?
Answer
There will be both fixed-response (e.g., mul ple choice) and
free-response ques ons.

9. Midterm FAQ, continued
Ques on
Will explana ons be necessary?

10. Midterm FAQ, continued
Ques on
Will explana ons be necessary?
Answer
Yes, on free-response problems we will expect you to explain
yourself. This is why it was required on wri en homework.

11. Midterm FAQ, continued
Ques on
Is (topic X) going to be tested?

12. Midterm FAQ, continued
Ques on
Is (topic X) going to be tested?
Answer
Everything covered in class or on homework is fair game for the
exam.
No topic that was not covered in class nor on homework will be
on the exam.
(This is not the same as saying all exam problems are similar to
class examples or homework problems.)

13. Midterm FAQ, continued
Ques on
Will there be a review session?

14. Midterm FAQ, continued
Ques on
Will there be a review session?
Answer
The recita ons this week will review for the exam.

15. Midterm FAQ, continued
Ques on
Will calculators be allowed?

16. Midterm FAQ, continued
Ques on
Will calculators be allowed?
Answer
No. The exam is designed for pencil and brain.

17. Midterm FAQ, continued
Ques on
How should I study?

18. Midterm FAQ, continued
Ques on
How should I study?
Answer
The exam has problems; study by doing problems. If you get
one right, think about how you got it right. If you got it wrong
or didn’t get it at all, reread the textbook and do easier
problems to build up your understanding.
Break up the material into chunks. (related) Don’t put it all off
un l the night before.
Ask ques ons.

19. Midterm FAQ, continued
Ques on
How many ques ons are there?

20. Midterm FAQ, continued
Ques on
How many ques ons are there?
Answer
Does this ques on contribute to your understanding of the
material?

21. Midterm FAQ, continued
Ques on
Will there be a curve on the exam?

22. Midterm FAQ, continued
Ques on
Will there be a curve on the exam?
Answer
Does this ques on contribute to your understanding of the
material?

23. Midterm FAQ, continued
Ques on
When will you grade my get-to-know-you and photo extra credit?

24. Midterm FAQ, continued
Ques on
When will you grade my get-to-know-you and photo extra credit?
Answer
Does this ques on contribute to your understanding of the
material?

25. Objectives
Use tangent lines to make linear
approxima ons to a func on.
Given a func on and a point in
the domain, compute the
lineariza on of the func on at
that point.
Use lineariza on to approximate
values of func ons
Given a func on, compute the
differen al of that func on
Use the differen al nota on to
es mate error in linear
approxima ons.

26. Outline
The linear approxima on of a func on near a point
Examples
Ques ons
Differen als
Using differen als to es mate error
Advanced Examples

27. The Big Idea
Ques on
What linear func on best approximates f near a?

28. The Big Idea
Ques on
What linear func on best approximates f near a?
Answer
The tangent line, of course!

29. The Big Idea
Ques on
What linear func on best approximates f near a?
Answer
The tangent line, of course!
Ques on
What is the equa on for the line tangent to y = f(x) at (a, f(a))?

30. The Big Idea
Ques on
What linear func on best approximates f near a?
Answer
The tangent line, of course!
Ques on
What is the equa on for the line tangent to y = f(x) at (a, f(a))?
Answer
L(x) = f(a) + f′ (a)(x − a)

31. tangent line = linear approximation
y
The func on
L(x)
L(x) = f(a) + f′ (a)(x − a)
f(x)
is a decent approxima on to f
near a. f(a)
x−a
. x
a x

32. tangent line = linear approximation
y
The func on
L(x)
L(x) = f(a) + f′ (a)(x − a)
f(x)
is a decent approxima on to f
near a. f(a)
x−a
How decent? The closer x is to
a, the be er the
approxima on L(x) is to f(x) . x
a x

33. Example
Example
Es mate sin(61◦ ) = sin(61π/180) by using a linear approxima on
(i) about a = 0 (ii) about a = 60◦ = π/3.

34. Example
Example
Es mate sin(61◦ ) = sin(61π/180) by using a linear approxima on
(i) about a = 0 (ii) about a = 60◦ = π/3.
Solu on (i)
If f(x) = sin x, then f(0) = 0
and f′ (0) = 1.
So the linear approxima on
near 0 is L(x) = 0 + 1 · x = x.
( )
61π 61π
sin ≈ ≈ 1.06465
180 180

35. Example
Example
Es mate sin(61◦ ) = sin(61π/180) by using a linear approxima on
(i) about a = 0 (ii) about a = 60◦ = π/3.
Solu on (i)
Solu on (ii)
( )
We have f π =
( ) 3
and
If f(x) = sin x, then f(0) = 0
f′ π = .
and f′ (0) = 1. 3
So the linear approxima on
near 0 is L(x) = 0 + 1 · x = x.
( )
61π 61π
sin ≈ ≈ 1.06465
180 180

36. Example
Example
Es mate sin(61◦ ) = sin(61π/180) by using a linear approxima on
(i) about a = 0 (ii) about a = 60◦ = π/3.
Solu on (i)
Solu on (ii)
( ) √
3
We have f π =
( ) 3 2
and
If f(x) = sin x, then f(0) = 0
f′ π = .
and f′ (0) = 1. 3
So the linear approxima on
near 0 is L(x) = 0 + 1 · x = x.
( )
61π 61π
sin ≈ ≈ 1.06465
180 180

37. Example
Example
Es mate sin(61◦ ) = sin(61π/180) by using a linear approxima on
(i) about a = 0 (ii) about a = 60◦ = π/3.
Solu on (i)
Solu on (ii)
( ) √
3
We have f π =
( ) 3 2
and
If f(x) = sin x, then f(0) = 0
f′ π = 1 .
and f′ (0) = 1. 3 2
So the linear approxima on
near 0 is L(x) = 0 + 1 · x = x.
( )
61π 61π
sin ≈ ≈ 1.06465
180 180

38. Example
Example
Es mate sin(61◦ ) = sin(61π/180) by using a linear approxima on
(i) about a = 0 (ii) about a = 60◦ = π/3.
Solu on (i)
Solu on (ii)
( ) √
3
We have f π =
( ) 3 2
and
If f(x) = sin x, then f(0) = 0
f′ π = 1 .
and f′ (0) = 1. 3 2
So the linear approxima on So the linear approxima on is
near 0 is L(x) = 0 + 1 · x = x. L(x) =
( )
61π 61π
sin ≈ ≈ 1.06465
180 180

39. Example
Example
Es mate sin(61◦ ) = sin(61π/180) by using a linear approxima on
(i) about a = 0 (ii) about a = 60◦ = π/3.
Solu on (i)
Solu on (ii)
( ) √
3
We have f π =
( ) 3 2
and
If f(x) = sin x, then f(0) = 0
f′ π = 1 .
and f′ (0) = 1. 3 2
So the linear approxima on So the linear approxima on is
√
3 1( π)
near 0 is L(x) = 0 + 1 · x = x. L(x) = + x−
( ) 2 2 3
61π 61π
sin ≈ ≈ 1.06465
180 180

40. Example
Example
Es mate sin(61◦ ) = sin(61π/180) by using a linear approxima on
(i) about a = 0 (ii) about a = 60◦ = π/3.
Solu on (i)
Solu on (ii)
( ) √
3
We have f π =
( ) 3 2
and
If f(x) = sin x, then f(0) = 0
f′ π = 1 .
and f′ (0) = 1. 3 2
So the linear approxima on So the linear approxima on is
√
3 1( π)
near 0 is L(x) = 0 + 1 · x = x. L(x) = + x−
( ) 2 2 3
61π 61π ( )
sin ≈ ≈ 1.06465 61π
180 180 sin ≈
180

41. Example
Example
Es mate sin(61◦ ) = sin(61π/180) by using a linear approxima on
(i) about a = 0 (ii) about a = 60◦ = π/3.
Solu on (i)
Solu on (ii)
( ) √
3
We have f π =
( ) 3 2
and
If f(x) = sin x, then f(0) = 0
f′ π = 1 .
and f′ (0) = 1. 3 2
So the linear approxima on So the linear approxima on is
√
3 1( π)
near 0 is L(x) = 0 + 1 · x = x. L(x) = + x−
( ) 2 2 3
61π 61π ( )
sin ≈ ≈ 1.06465 61π
180 180 sin ≈ 0.87475
180

42. Example
Example
Es mate sin(61◦ ) = sin(61π/180) by using a linear approxima on
(i) about a = 0 (ii) about a = 60◦ = π/3.
Solu on (i)
Solu on (ii)
( ) √
3
We have f π =
( ) 3 2
and
If f(x) = sin x, then f(0) = 0
f′ π = 1 .
and f′ (0) = 1. 3 2
So the linear approxima on So the linear approxima on is
√
3 1( π)
near 0 is L(x) = 0 + 1 · x = x. L(x) = + x−
( ) 2 2 3
61π 61π ( )
sin ≈ ≈ 1.06465 61π
180 180 sin ≈ 0.87475
180

43. Example
Example
Es mate sin(61◦ ) = sin(61π/180) by using a linear approxima on
(i) about a = 0 (ii) about a = 60◦ = π/3.
Solu on (i)
Solu on (ii)
( ) √
3
We have f π =
( ) 3 2
and
If f(x) = sin x, then f(0) = 0
f′ π = 1 .
and f′ (0) = 1. 3 2
So the linear approxima on So the linear approxima on is
√
3 1( π)
near 0 is L(x) = 0 + 1 · x = x. L(x) = + x−
( ) 2 2 3
61π 61π ( )
sin ≈ ≈ 1.06465 61π
180 180 sin ≈ 0.87475
180

44. Illustration
y
y = sin x
. x
61◦

45. Illustration
y
y = L1 (x) = x
y = sin x
. x
0 61◦

46. Illustration
y
y = L1 (x) = x
big difference! y = sin x
. x
0 61◦

47. Illustration
y
y = L1 (x) = x
√ ( )
y = L2 (x) = 2
3
+ 1
2 x− π
3
y = sin x
. x
0 π/3 61◦

48. Illustration
y
y = L1 (x) = x
√ ( )
y = L2 (x) = 2
3
+ 1
2 x− π
3
y = sin x
very li le difference!
. x
0 π/3 61◦

49. Another Example
Example
√
Es mate 10 using the fact that 10 = 9 + 1.

50. Another Example
Example
√
Es mate 10 using the fact that 10 = 9 + 1.
Solu on
√
The key step is to use a linear approxima on to f(x) = x near a = 9 to es mate
√
f(10) = 10.

51. Another Example
Example
√
Es mate 10 using the fact that 10 = 9 + 1.
Solu on
√
The key step is to use a linear approxima on to f(x) = x near a = 9 to es mate
√
f(10) = 10.
f(10) ≈ L(10) = f(9) + f′ (9)(10 − 9)
1 19
=3+ (1) = ≈ 3.167
2·3 6

52. Another Example
Example
√
Es mate 10 using the fact that 10 = 9 + 1.
Solu on
√
The key step is to use a linear approxima on to f(x) = x near a = 9 to es mate
√
f(10) = 10.
f(10) ≈ L(10) = f(9) + f′ (9)(10 − 9)
1 19
=3+ (1) = ≈ 3.167
2·3 6
( )2
19
Check: =
6

53. Another Example
Example
√
Es mate 10 using the fact that 10 = 9 + 1.
Solu on
√
The key step is to use a linear approxima on to f(x) = x near a = 9 to es mate
√
f(10) = 10.
f(10) ≈ L(10) = f(9) + f′ (9)(10 − 9)
1 19
=3+ (1) = ≈ 3.167
2·3 6
( )2
19 361
Check: = .
6 36

54. Dividing without dividing?
Example
A student has an irra onal fear of long division and needs to
es mate 577 ÷ 408. He writes
577 1 1 1
= 1 + 169 = 1 + 169 × × .
408 408 4 102
1
Help the student es mate .
102

55. Dividing without dividing?
Solu on
1
Let f(x) = . We know f(100) and we want to es mate 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.
408

56. 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?

57. 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?
Answer
100 mi
150 mi
600 mi (?) (Is it reasonable to assume 12 hours at the same
speed?)

58. Questions
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?

59. Questions
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 (?)

60. Questions
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 ver cal movement?

61. Questions
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 ver cal 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.

62. Outline
The linear approxima on of a func on near a point
Examples
Ques ons
Differen als
Using differen als to es mate error
Advanced Examples

63. Differentials are derivatives
The fact that the the tangent line is an
approxima on means that
y
f(x + ∆x) − f(x) ≈ f′ (x) ∆x
∆y dy
Rename ∆x = dx, so we can write this as
dy
∆y ≈ dy = f′ (x)dx. ∆y
dx = ∆x
Note this looks a lot like the Leibniz-Newton
iden ty
dy . x
= f′ (x) x x + ∆x
dx

64. Using differentials to estimate error
Es ma ng error with
differen als y
If y = f(x), x0 and ∆x is
known, and an es mate of ∆y
is desired:
Approximate: ∆y ≈ dy ∆y
dy
Differen ate:
dy = f′ (x) dx
dx = ∆x
Evaluate at x = x0 and . x
dx = ∆x. x x + ∆x

65. Using differentials to estimate error
Example
A regular sheet of plywood
measures 8 ft × 4 ft. Suppose
a defec ve plywood-cu ng
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?

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

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

68. Solution
Solu on
1
Write A(ℓ) = ℓ2 . We want to know ∆A when ℓ = 8 ft and
2
∆ℓ = 1 in. ( )
97 9409 9409
(I) A(ℓ + ∆ℓ) = A = So ∆A = − 32 ≈ 0.6701.
12 288 288
dA
(II) = ℓ, so dA = ℓ dℓ, which should be a good es mate for ∆ℓ.
dℓ

69. Solution
Solu on
1
Write A(ℓ) = ℓ2 . We want to know ∆A when ℓ = 8 ft and
2
∆ℓ = 1 in. ( )
97 9409 9409
(I) A(ℓ + ∆ℓ) = A = So ∆A = − 32 ≈ 0.6701.
12 288 288
dA
(II) = ℓ, so dA = ℓ dℓ, which should be a good es mate for ∆ℓ.
dℓ
When ℓ = 8 and dℓ = 12 , we have dA = 12 = 2 ≈ 0.667.
1 8
3

70. Solution
Solu on
1
Write A(ℓ) = ℓ2 . We want to know ∆A when ℓ = 8 ft and
2
∆ℓ = 1 in. ( )
97 9409 9409
(I) A(ℓ + ∆ℓ) = A = So ∆A = − 32 ≈ 0.6701.
12 288 288
dA
(II) = ℓ, so dA = ℓ dℓ, which should be a good es mate for ∆ℓ.
dℓ
When ℓ = 8 and dℓ = 12 , we have dA = 12 = 2 ≈ 0.667.
1 8
3
So we get es mates close to the hundredth of a square foot.

71. Why should we care?
Why use linear approxima ons dy when the actual difference ∆y is
known?
Linear approxima on is quick and reliable. Finding ∆y exactly
depends on the func on.
With more complicated func ons, linear approxima on much
simpler. See the “Advanced Examples” later.
In real life, some mes only f(a) and f′ (a) are known, and not
the general f(x).

72. Outline
The linear approxima on of a func on near a point
Examples
Ques ons
Differen als
Using differen als to es mate error
Advanced Examples

73. Gravitation
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.

74. Gravitation
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.
GMm
In fact, the force felt is F(r) = − 2 , where M is the mass of the earth and
r
r is the distance from the center of the earth to the object. G is a constant.

75. Gravitation
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.
GMm
In fact, the force felt is F(r) = − 2 , where M is the mass of the earth and
r
r is the distance from the center of the earth to the object. G is a constant.
GMm GM
At r = re the force really is F(re ) = 2
, and g is defined to be 2
re re

76. Gravitation
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.
GMm
In fact, the force felt is F(r) = − 2 , where M is the mass of the earth and
r
r is the distance from the center of the earth to the object. G is a constant.
GMm GM
At r = re the force really is F(re ) = 2
, and g is defined to be 2
re 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 rela ve
error? The percentage error?

77. Gravitation Solution
Solu on
We wonder if ∆F = F(re + ∆r) − F(re ) is small.
Using a linear approxima on,
dF GMm
∆F ≈ dF = dr = 2 3 dr
dr re
( re )
GMm dr ∆r
= = 2mg
r2
e re re
∆F ∆r
The rela ve error is ≈ −2
F re

78. Solution continued
re = 6378.1 km. If ∆r = 50 m,
∆F ∆r 50
≈ −2 = −2 = −1.56 × 10−5 = −0.00156%
F re 6378100

79. Systematic linear approximation
√ √
2 is irra onal, but 9/4 is ra onal and 9/4 is close to 2.

80. Systematic linear approximation
√ √
2 is irra onal, but 9/4 is ra onal and 9/4 is close to 2. So
√ √ √ 1 17
2= 9/4 − 1/4 ≈ 9/4 + (−1/4) =
2(3/2) 12

81. Systematic linear approximation
√ √
2 is irra onal, but 9/4 is ra onal 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 be er approxima on since (17/12)2 = 289/144

82. Systematic linear approximation
√ √
2 is irra onal, but 9/4 is ra onal 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 be er approxima on since (17/12)2 = 289/144
Do it again!
√ √ √ 1
2 = 289/144 − 1/144 ≈ 289/144 + 17 (−1/144) = 577/408
2( /12)
( )2
577 332, 929 1
Now = which is away from 2.
408 166, 464 166, 464

83. Illustration of the previous
example
.

84. Illustration of the previous
example
.

85. Illustration of the previous
example
.
2

86. Illustration of the previous
example
.
2

87. Illustration of the previous
example
.
2

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

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

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

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

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

93. Summary
Linear approxima on: If f is differen able at a, the best linear
approxima on to f near a is given by
Lf,a (x) = f(a) + f′ (a)(x − a)
Differen als: If f is differen able at x, a good approxima on to
∆y = f(x + ∆x) − f(x) is
dy dy
∆y ≈ dy = · dx = · ∆x
dx dx
Don’t buy plywood from me.