1. Section 2.1
The Derivative and Rates of Change
V63.0121.002.2010Su, Calculus I
New York University
May 20, 2010
Announcements
Written Assignment 1 is on the website
2. Announcements
Written Assignment 1 is on
the website
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 2 / 28
3. Format of written work
Please:
Use scratch paper and copy
your final work onto fresh
paper.
Use loose-leaf paper (not
torn from a notebook).
Write your name,
assignment number, and
date at the top.
Staple your homework
together.
See the website for more information.
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 3 / 28
4. Objectives
Understand and state the
definition of the derivative of
a function at a point.
Given a function and a point
in its domain, decide if the
function is differentiable at
the point and find the value
of the derivative at that
point.
Understand and give several
examples of derivatives
modeling rates of change in
science.
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 4 / 28
5. Outline
Rates of Change
Tangent Lines
Velocity
Population growth
Marginal costs
The derivative, defined
Derivatives of (some) power functions
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 5 / 28
6. The tangent problem
Problem
Given a curve and a point on the curve, find the slope of the line tangent
to the curve at that point.
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 6 / 28
7. The tangent problem
Problem
Given a curve and a point on the curve, find the slope of the line tangent
to the curve at that point.
Example
Find the slope of the line tangent to the curve y = x 2 at the point (2, 4).
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 6 / 28
8. Graphically and numerically
y
x 2 − 22
x m=
x −2
4
x
2
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
9. Graphically and numerically
y
x 2 − 22
x m=
x −2
3
9
4
x
2 3
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
10. Graphically and numerically
y
x 2 − 22
x m=
x −2
3 5
9
4
x
2 3
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
11. Graphically and numerically
y
x 2 − 22
x m=
x −2
3 5
2.5
6.25
4
x
2 2.5
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
12. Graphically and numerically
y
x 2 − 22
x m=
x −2
3 5
2.5 4.5
6.25
4
x
2 2.5
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
13. Graphically and numerically
y
x 2 − 22
x m=
x −2
3 5
2.5 4.5
2.1
4.41
4
x
2.1
2
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
14. Graphically and numerically
y
x 2 − 22
x m=
x −2
3 5
2.5 4.5
2.1 4.1
4.41
4
x
2.1
2
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
15. Graphically and numerically
y
x 2 − 22
x m=
x −2
3 5
2.5 4.5
2.1 4.1
2.01
4.0401
4
x
2.01
2
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
16. Graphically and numerically
y
x 2 − 22
x m=
x −2
3 5
2.5 4.5
2.1 4.1
2.01 4.01
4.0401
4
x
2.01
2
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
17. Graphically and numerically
y
x 2 − 22
x m=
x −2
3 5
2.5 4.5
2.1 4.1
2.01 4.01
4
1
1
x
1 2
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
18. Graphically and numerically
y
x 2 − 22
x m=
x −2
3 5
2.5 4.5
2.1 4.1
2.01 4.01
4
1 3
1
x
1 2
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
19. Graphically and numerically
y
x 2 − 22
x m=
x −2
3 5
2.5 4.5
2.1 4.1
2.01 4.01
4
1.5
2.25
1 3
x
1.5 2
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
20. Graphically and numerically
y
x 2 − 22
x m=
x −2
3 5
2.5 4.5
2.1 4.1
2.01 4.01
4
1.5 3.5
2.25
1 3
x
1.5 2
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
21. Graphically and numerically
y
x 2 − 22
x m=
x −2
3 5
2.5 4.5
2.1 4.1
2.01 4.01
4 1.9
3.61
1.5 3.5
1 3
x
1.9
2
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
22. Graphically and numerically
y
x 2 − 22
x m=
x −2
3 5
2.5 4.5
2.1 4.1
2.01 4.01
4
3.9601 1.9 3.9
3.61
1.5 3.5
1 3
x
1.99
1.9
2
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
23. Graphically and numerically
y
x 2 − 22
x m=
x −2
3 5
2.5 4.5
2.1 4.1
2.01 4.01
1.99
4
3.9601 1.9 3.9
1.5 3.5
1 3
x
1.99
2
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
24. Graphically and numerically
y
x 2 − 22
x m=
x −2
3 5
2.5 4.5
2.1 4.1
2.01 4.01
1.99 3.99
4 1.9 3.9
1.5 3.5
1 3
x
2
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
25. Graphically and numerically
y
x 2 − 22
x m=
x −2
3 5
9 2.5 4.5
2.1 4.1
2.01 4.01
6.25
limit 4
4.41 1.99 3.99
4.0401
4
3.9601 1.9 3.9
3.61
1.5 3.5
2.25
1 3
1
x
1 1.5 2.1 3
1.99
1.9 2.5
2.01
2
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
26. The tangent problem
Problem
Given a curve and a point on the curve, find the slope of the line tangent
to the curve at that point.
Example
Find the slope of the line tangent to the curve y = x 2 at the point (2, 4).
Upshot
If the curve is given by y = f (x), and the point on the curve is (a, f (a)),
then the slope of the tangent line is given by
f (x) − f (a)
mtangent = lim
x→a x −a
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 8 / 28
27. Velocity
Problem
Given the position function of a moving object, find the velocity of the
object at a certain instant in time.
Example
Drop a ball off the roof of the Silver Center so that its height can be
described by
h(t) = 50 − 5t 2
where t is seconds after dropping it and h is meters above the ground.
How fast is it falling one second after we drop it?
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 9 / 28
28. Numerical evidence
h(t) − h(1)
t vave =
t −1
2 − 15
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 10 / 28
29. Numerical evidence
h(t) − h(1)
t vave =
t −1
2 − 15
1.5
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 10 / 28
30. Numerical evidence
h(t) − h(1)
t vave =
t −1
2 − 15
1.5 − 12.5
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 10 / 28
31. Numerical evidence
h(t) − h(1)
t vave =
t −1
2 − 15
1.5 − 12.5
1.1
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 10 / 28
32. Numerical evidence
h(t) − h(1)
t vave =
t −1
2 − 15
1.5 − 12.5
1.1 − 10.5
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 10 / 28
33. Numerical evidence
h(t) − h(1)
t vave =
t −1
2 − 15
1.5 − 12.5
1.1 − 10.5
1.01
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 10 / 28
34. Numerical evidence
h(t) − h(1)
t vave =
t −1
2 − 15
1.5 − 12.5
1.1 − 10.5
1.01 − 10.05
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 10 / 28
35. Numerical evidence
h(t) − h(1)
t vave =
t −1
2 − 15
1.5 − 12.5
1.1 − 10.5
1.01 − 10.05
1.001
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 10 / 28
36. Numerical evidence
h(t) − h(1)
t vave =
t −1
2 − 15
1.5 − 12.5
1.1 − 10.5
1.01 − 10.05
1.001 − 10.005
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 10 / 28
37. Velocity
Problem
Given the position function of a moving object, find the velocity of the
object at a certain instant in time.
Example
Drop a ball off the roof of the Silver Center so that its height can be
described by
h(t) = 50 − 5t 2
where t is seconds after dropping it and h is meters above the ground.
How fast is it falling one second after we drop it?
Solution
The answer is
(50 − 5t 2 ) − 45 5 − 5t 2 5(1 − t)(1 + t)
v = lim = lim = lim
t→1 t −1 t→1 t − 1 t→1 t −1
V63.0121.002.2010Su, Calculus I (NYU) t) = Section 2.1 = −10
= (−5) lim (1 + −5 · 2 The Derivative May 20, 2010 11 / 28
38. y = h(t)
Upshot
If the height function is given by
h(t), the instantaneous velocity
at time t0 is given by
h(t) − h(t0 )
v = lim
t→t0 t − t0
h(t0 + ∆t) − h(t0 )
= lim
∆t→0 ∆t
∆t
t
t0 t
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 12 / 28
39. Population growth
Problem
Given the population function of a group of organisms, find the rate of
growth of the population at a particular instant.
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 13 / 28
40. Population growth
Problem
Given the population function of a group of organisms, find the rate of
growth of the population at a particular instant.
Example
Suppose the population of fish in the East River is given by the function
3e t
P(t) =
1 + et
where t is in years since 2000 and P is in millions of fish. Is the fish
population growing fastest in 1990, 2000, or 2010? (Estimate numerically)
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 13 / 28
41. Derivation
Let ∆t be an increment in time and ∆P the corresponding change in
population:
∆P = P(t + ∆t) − P(t)
This depends on ∆t, so we want
∆P 1 3e t+∆t 3e t
lim = lim −
∆t→0 ∆t ∆t→0 ∆t 1 + e t+∆t 1 + et
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 14 / 28
42. Derivation
Let ∆t be an increment in time and ∆P the corresponding change in
population:
∆P = P(t + ∆t) − P(t)
This depends on ∆t, so we want
∆P 1 3e t+∆t 3e t
lim = lim −
∆t→0 ∆t ∆t→0 ∆t 1 + e t+∆t 1 + et
Too hard! Try a small ∆t to approximate.
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 14 / 28
43. Numerical evidence
∆P ∆P
Use ∆t = 0.1, and use to approximate lim .
∆t ∆t→0 ∆t
P(−10 + 0.1) − P(−10)
r1990 ≈ =
0.1
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 15 / 28
44. Numerical evidence
∆P ∆P
Use ∆t = 0.1, and use to approximate lim .
∆t ∆t→0 ∆t
P(−10 + 0.1) − P(−10)
r1990 ≈ = 0.000143229
0.1
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 15 / 28
45. Numerical evidence
∆P ∆P
Use ∆t = 0.1, and use to approximate lim .
∆t ∆t→0 ∆t
P(−10 + 0.1) − P(−10)
r1990 ≈ = 0.000143229
0.1
P(0.1) − P(0)
r2000 ≈ =
0.1
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 15 / 28
46. Numerical evidence
∆P ∆P
Use ∆t = 0.1, and use to approximate lim .
∆t ∆t→0 ∆t
P(−10 + 0.1) − P(−10)
r1990 ≈ = 0.000143229
0.1
P(0.1) − P(0)
r2000 ≈ = 0.749376
0.1
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 15 / 28
47. Numerical evidence
∆P ∆P
Use ∆t = 0.1, and use to approximate lim .
∆t ∆t→0 ∆t
P(−10 + 0.1) − P(−10)
r1990 ≈ = 0.000143229
0.1
P(0.1) − P(0)
r2000 ≈ = 0.749376
0.1
P(10 + 0.1) − P(10)
r2010 ≈ =
0.1
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 15 / 28
48. Numerical evidence
∆P ∆P
Use ∆t = 0.1, and use to approximate lim .
∆t ∆t→0 ∆t
P(−10 + 0.1) − P(−10)
r1990 ≈ = 0.000143229
0.1
P(0.1) − P(0)
r2000 ≈ = 0.749376
0.1
P(10 + 0.1) − P(10)
r2010 ≈ = 0.0001296
0.1
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 15 / 28
49. Population growth
Problem
Given the population function of a group of organisms, find the rate of
growth of the population at a particular instant.
Example
Suppose the population of fish in the East River is given by the function
3e t
P(t) =
1 + et
where t is in years since 2000 and P is in millions of fish. Is the fish
population growing fastest in 1990, 2000, or 2010? (Estimate numerically)
Solution
We estimates the rates of growth to be 0.000143229, 0.749376, and
0.0001296. So the population is growing fastest in 2000.
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 16 / 28
50. Upshot
The instantaneous population growth is given by
P(t + ∆t) − P(t)
lim
∆t→0 ∆t
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 17 / 28
51. Marginal costs
Problem
Given the production cost of a good, find the marginal cost of production
after having produced a certain quantity.
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 18 / 28
52. Marginal costs
Problem
Given the production cost of a good, find the marginal cost of production
after having produced a certain quantity.
Example
Suppose the cost of producing q tons of rice on our paddy in a year is
C (q) = q 3 − 12q 2 + 60q
We are currently producing 5 tons a year. Should we change that?
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 18 / 28
53. Comparisons
q C (q)
4
5
6
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 19 / 28
54. Comparisons
q C (q)
4 112
5
6
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 19 / 28
55. Comparisons
q C (q)
4 112
5 125
6
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 19 / 28
56. Comparisons
q C (q)
4 112
5 125
6 144
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 19 / 28
57. Comparisons
q C (q) AC (q) = C (q)/q
4 112
5 125
6 144
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 19 / 28
58. Comparisons
q C (q) AC (q) = C (q)/q
4 112 28
5 125
6 144
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 19 / 28
59. Comparisons
q C (q) AC (q) = C (q)/q
4 112 28
5 125 25
6 144
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 19 / 28
60. Comparisons
q C (q) AC (q) = C (q)/q
4 112 28
5 125 25
6 144 24
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 19 / 28
61. Comparisons
q C (q) AC (q) = C (q)/q ∆C = C (q + 1) − C (q)
4 112 28
5 125 25
6 144 24
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 19 / 28
62. Comparisons
q C (q) AC (q) = C (q)/q ∆C = C (q + 1) − C (q)
4 112 28 13
5 125 25
6 144 24
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 19 / 28
63. Comparisons
q C (q) AC (q) = C (q)/q ∆C = C (q + 1) − C (q)
4 112 28 13
5 125 25 19
6 144 24
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 19 / 28
64. Comparisons
q C (q) AC (q) = C (q)/q ∆C = C (q + 1) − C (q)
4 112 28 13
5 125 25 19
6 144 24 31
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 19 / 28
65. Marginal costs
Problem
Given the production cost of a good, find the marginal cost of production
after having produced a certain quantity.
Example
Suppose the cost of producing q tons of rice on our paddy in a year is
C (q) = q 3 − 12q 2 + 60q
We are currently producing 5 tons a year. Should we change that?
Example
If q = 5, then C = 125, ∆C = 19, while AC = 25. So we should produce
more to lower average costs.
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 20 / 28
66. Upshot
The incremental cost
∆C = C (q + 1) − C (q)
is useful, but depends on units.
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 21 / 28
67. Upshot
The incremental cost
∆C = C (q + 1) − C (q)
is useful, but depends on units.
The marginal cost after producing q given by
C (q + ∆q) − C (q)
MC = lim
∆q→0 ∆q
is more useful since it’s unit-independent.
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 21 / 28
68. Outline
Rates of Change
Tangent Lines
Velocity
Population growth
Marginal costs
The derivative, defined
Derivatives of (some) power functions
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 22 / 28
69. The definition
All of these rates of change are found the same way!
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 23 / 28
70. The definition
All of these rates of change are found the same way!
Definition
Let f be a function and a a point in the domain of f . If the limit
f (a + h) − f (a) f (x) − f (a)
f (a) = lim = lim
h→0 h x→a x −a
exists, the function is said to be differentiable at a and f (a) is the
derivative of f at a.
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 23 / 28
71. Derivative of the squaring function
Example
Suppose f (x) = x 2 . Use the definition of derivative to find f (a).
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 24 / 28
72. Derivative of the squaring function
Example
Suppose f (x) = x 2 . Use the definition of derivative to find f (a).
Solution
f (a + h) − f (a) (a + h)2 − a2
f (a) = lim = lim
h→0 h h→0 h
(a2 + 2ah + h2 ) − a2 2ah + h2
= lim = lim
h→0 h h→0 h
= lim (2a + h) = 2a.
h→0
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 24 / 28
73. Derivative of the reciprocal function
Example
1
Suppose f (x) = . Use the
x
definition of the derivative to find
f (2).
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 25 / 28
74. Derivative of the reciprocal function
Example
1
Suppose f (x) = . Use the
x
definition of the derivative to find x
f (2).
Solution
1/x − 1/2 2−x
f (2) = lim = lim x
x→2 x −2 x→2 2x(x − 2)
−1 1
= lim =−
x→2 2x 4
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 25 / 28
75. The Sure-Fire Sally Rule (SFSR) for adding
Fractions
In anticipation of the question, “How did you get that?”
a c ad ± bc
± =
b d bd
So
1 1 2−x
−
x 2 = 2x
x −2 x −2
2−x
=
2x(x − 2)
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 26 / 28
76. The Sure-Fire Sally Rule (SFSR) for adding
Fractions
In anticipation of the question, “How did you get that?”
a c ad ± bc
± =
b d bd
So
1 1 2−x
−
x 2 = 2x
x −2 x −2
2−x
=
2x(x − 2)
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 26 / 28
78. Summary
The derivative measures instantaneous rate of change
The derivative has many interpretations: slope of the tangent line,
velocity, marginal quantities, etc.
The derivative is calculated with a limit.
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 28 / 28