The document appears to be lecture notes for a calculus class that covers several topics:
- Defining the derivative and discussing rates of change, tangent lines, velocity, population growth, and marginal costs.
- Worked examples are provided for finding the slope of a tangent line, instantaneous velocity, instantaneous population growth rate, and marginal cost of production.
- Numerical approximations are used to estimate some values when exact solutions are difficult to obtain analytically.
1. . . . . . .
Section 2.1
The Derivative and Rates of Change
V63.0121.034, Calculus I
September 23, 2009
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2. . . . . . .
Regarding WebAssign
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3. . . . . . .
Explanations
From the syllabus:
Graders will be expecting you to express your ideas
clearly, legibly, and completely, often requiring
complete English sentences rather than merely just a
long string of equations or unconnected mathematical
expressions. This means you could lose points for
unexplained answers.
4. . . . . . .
Rubric
Points Description of Work
3 Work is completely accurate and essentially perfect.
Work is thoroughly developed, neat, and easy to read.
Complete sentences are used.
2 Work is good, but incompletely developed, hard to
read, unexplained, or jumbled. Answers which are
not explained, even if correct, will generally receive 2
points. Work contains “right idea” but is flawed.
1 Work is sketchy. There is some correct work, but most
of work is incorrect.
0 Work minimal or non-existent. Solution is completely
incorrect.
5. . . . . . .
Outline
Rates of Change
Tangent Lines
Velocity
Population growth
Marginal costs
The derivative, defined
Derivatives of (some) power functions
What does f tell you about f′
?
How can a function fail to be differentiable?
Other notations
The second derivative
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.
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 = x2
at the point
(2, 4).
8. . . . . . .
Graphically and numerically
. .x
.y
.
.2
..4 .
x m
9. . . . . . .
Graphically and numerically
. .x
.y
.
.2
..4 .
.
.
.3
..9
x m
3 5
10. . . . . . .
Graphically and numerically
. .x
.y
.
.2
..4 .
.
.
.2.5
..6.25
x m
3 5
2.5 4.25
11. . . . . . .
Graphically and numerically
. .x
.y
.
.2
..4 .
.
.
.2.1
..4.41
x m
3 5
2.5 4.25
2.1 4.1
12. . . . . . .
Graphically and numerically
. .x
.y
.
.2
..4 ..
.
.2.01
..4.0401
x m
3 5
2.5 4.25
2.1 4.1
2.01 4.01
18. . . . . . .
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 = x2
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
mtangent = lim
x→a
f(x) − f(a)
x − a
19. . . . . . .
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 − 5t2
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?
29. . . . . . .
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 − 5t2
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
v = lim
t→1
(50 − 5t2) − 45
t − 1
= lim
t→1
5 − 5t2
t − 1
= lim
t→1
5(1 − t)(1 + t)
t − 1
= (−5) lim
t→1
(1 + t) = −5 · 2 = −10
30. . . . . . .
Upshot
If the height function is given
by h(t), the instantaneous
velocity at time t0 is given by
v = lim
t→t0
h(t) − h(t0)
t − t0
= lim
∆t→0
h(t0 + ∆t) − h(t0)
∆t
. .t
.y = h(t)
.
.
.
.t0
.
.t
.∆t
31. . . . . . .
Population growth
Problem
Given the population function of a group of organisms, find the
rate of growth of the population at a particular instant.
32. . . . . . .
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
P(t) =
3et
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)?
33. . . . . . .
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
lim
∆t→0
∆P
∆t
= lim
∆t→0
1
∆t
(
3et+∆t
1 + et+∆t
−
3et
1 + et
)
34. . . . . . .
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
lim
∆t→0
∆P
∆t
= lim
∆t→0
1
∆t
(
3et+∆t
1 + et+∆t
−
3et
1 + et
)
Too hard! Try a small ∆t to approximate.
41. . . . . . .
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
P(t) =
3et
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
The estimated rates of growth are 0.000136, 0.75, and 0.000136.
42. . . . . . .
Upshot
The instantaneous population growth is given by
lim
∆t→0
P(t + ∆t) − P(t)
∆t
43. . . . . . .
Marginal costs
Problem
Given the production cost of a good, find the marginal cost of
production after having produced a certain quantity.
44. . . . . . .
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) = q3
− 12q2
+ 60q
We are currently producing 5 tons a year. Should we change that?
57. . . . . . .
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) = q3
− 12q2
+ 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.
58. . . . . . .
Upshot
The incremental cost
∆C = C(q + 1) − C(q)
is useful, but depends on units.
59. . . . . . .
Upshot
The incremental cost
∆C = C(q + 1) − C(q)
is useful, but depends on units.
The marginal cost after producing q given by
MC = lim
∆q→0
C(q + ∆q) − C(q)
∆q
is more useful since it’s unit-independent.
60. . . . . . .
Outline
Rates of Change
Tangent Lines
Velocity
Population growth
Marginal costs
The derivative, defined
Derivatives of (some) power functions
What does f tell you about f′
?
How can a function fail to be differentiable?
Other notations
The second derivative
61. . . . . . .
The definition
All of these rates of change are found the same way!
62. . . . . . .
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) = lim
h→0
f(a + h) − f(a)
h
exists, the function is said to be differentiable at a and f′
(a) is the
derivative of f at a.
63. . . . . . .
Derivative of the squaring function
Example
Suppose f(x) = x2
. Use the definition of derivative to find f′
(a).
64. . . . . . .
Derivative of the squaring function
Example
Suppose f(x) = x2
. Use the definition of derivative to find f′
(a).
Solution
f′
(a) = lim
h→0
f(a + h) − f(a)
h
= lim
h→0
(a + h)2 − a2
h
= lim
h→0
(a2 + 2ah + h2
) − a2
h
= lim
h→0
2ah + h2
h
= lim
h→0
(2a + h) = 2a.
65. . . . . . .
Derivative of the reciprocal function
Example
Suppose f(x) =
1
x
. Use the
definition of the derivative to
find f′
(2).
66. . . . . . .
Derivative of the reciprocal function
Example
Suppose f(x) =
1
x
. Use the
definition of the derivative to
find f′
(2).
Solution
f′
(2) = lim
x→2
1/x − 1/2
x − 2
= lim
x→2
2 − x
2x(x − 2)
= lim
x→2
−1
2x
= −
1
4
. .x
.x
.
67. . . . . . .
The Sure-Fire Sally Rule (SFSR) for adding Fractions
In anticipation of the question, “How did you get that?”
a
b
±
c
d
=
ad ± bc
bd
So
1
x
−
1
2
x − 2
=
2 − x
2x
x − 2
=
2 − x
2x(x − 2)
68. . . . . . .
The Sure-Fire Sally Rule (SFSR) for adding Fractions
In anticipation of the question, “How did you get that?”
a
b
±
c
d
=
ad ± bc
bd
So
1
x
−
1
2
x − 2
=
2 − x
2x
x − 2
=
2 − x
2x(x − 2)
69. . . . . . .
What does f tell you about f′
?
If f is a function, we can compute the derivative f′
(x) at each
point x where f is differentiable, and come up with another
function, the derivative function.
What can we say about this function f′
?
70. . . . . . .
What does f tell you about f′
?
If f is a function, we can compute the derivative f′
(x) at each
point x where f is differentiable, and come up with another
function, the derivative function.
What can we say about this function f′
?
If f is decreasing on an interval, f′
is negative (well,
nonpositive) on that interval
71. . . . . . .
Derivative of the reciprocal function
Example
Suppose f(x) =
1
x
. Use the
definition of the derivative to
find f′
(2).
Solution
f′
(2) = lim
x→2
1/x − 1/2
x − 2
= lim
x→2
2 − x
2x(x − 2)
= lim
x→2
−1
2x
= −
1
4
. .x
.x
.
72. . . . . . .
What does f tell you about f′
?
If f is a function, we can compute the derivative f′
(x) at each
point x where f is differentiable, and come up with another
function, the derivative function.
What can we say about this function f′
?
If f is decreasing on an interval, f′
is negative (well,
nonpositive) on that interval
If f is increasing on an interval, f′
is positive (well,
nonnegative) on that interval
74. . . . . . .
What does f tell you about f′
?
Fact
If f is decreasing on (a, b), then f′
≤ 0 on (a, b).
Proof.
If f is decreasing on (a, b), and ∆x > 0, then
f(x + ∆x) < f(x) =⇒
f(x + ∆x) − f(x)
∆x
< 0
75. . . . . . .
What does f tell you about f′
?
Fact
If f is decreasing on (a, b), then f′
≤ 0 on (a, b).
Proof.
If f is decreasing on (a, b), and ∆x > 0, then
f(x + ∆x) < f(x) =⇒
f(x + ∆x) − f(x)
∆x
< 0
But if ∆x < 0, then x + ∆x < x, and
f(x + ∆x) > f(x) =⇒
f(x + ∆x) − f(x)
∆x
< 0
still!
76. . . . . . .
What does f tell you about f′
?
Fact
If f is decreasing on (a, b), then f′
≤ 0 on (a, b).
Proof.
If f is decreasing on (a, b), and ∆x > 0, then
f(x + ∆x) < f(x) =⇒
f(x + ∆x) − f(x)
∆x
< 0
But if ∆x < 0, then x + ∆x < x, and
f(x + ∆x) > f(x) =⇒
f(x + ∆x) − f(x)
∆x
< 0
still! Either way,
f(x + ∆x) − f(x)
∆x
< 0, so
f′
(x) = lim
∆x→0
f(x + ∆x) − f(x)
∆x
≤ 0
77. . . . . . .
Outline
Rates of Change
Tangent Lines
Velocity
Population growth
Marginal costs
The derivative, defined
Derivatives of (some) power functions
What does f tell you about f′
?
How can a function fail to be differentiable?
Other notations
The second derivative
78. . . . . . .
Differentiability is super-continuity
Theorem
If f is differentiable at a, then f is continuous at a.
79. . . . . . .
Differentiability is super-continuity
Theorem
If f is differentiable at a, then f is continuous at a.
Proof.
We have
lim
x→a
(f(x) − f(a)) = lim
x→a
f(x) − f(a)
x − a
· (x − a)
= lim
x→a
f(x) − f(a)
x − a
· lim
x→a
(x − a)
= f′
(a) · 0 = 0
80. . . . . . .
Differentiability is super-continuity
Theorem
If f is differentiable at a, then f is continuous at a.
Proof.
We have
lim
x→a
(f(x) − f(a)) = lim
x→a
f(x) − f(a)
x − a
· (x − a)
= lim
x→a
f(x) − f(a)
x − a
· lim
x→a
(x − a)
= f′
(a) · 0 = 0
Note the proper use of the limit law: if the factors each have a
limit at a, the limit of the product is the product of the limits.
81. . . . . . .
How can a function fail to be differentiable?
Kinks
. .x
.f(x)
82. . . . . . .
How can a function fail to be differentiable?
Kinks
. .x
.f(x)
. .x
.f′
(x)
.
83. . . . . . .
How can a function fail to be differentiable?
Kinks
. .x
.f(x)
. .x
.f′
(x)
.
.
84. . . . . . .
How can a function fail to be differentiable?
Cusps
. .x
.f(x)
85. . . . . . .
How can a function fail to be differentiable?
Cusps
. .x
.f(x)
. .x
.f′
(x)
86. . . . . . .
How can a function fail to be differentiable?
Cusps
. .x
.f(x)
. .x
.f′
(x)
87. . . . . . .
How can a function fail to be differentiable?
Vertical Tangents
. .x
.f(x)
88. . . . . . .
How can a function fail to be differentiable?
Vertical Tangents
. .x
.f(x)
. .x
.f′
(x)
89. . . . . . .
How can a function fail to be differentiable?
Vertical Tangents
. .x
.f(x)
. .x
.f′
(x)
90. . . . . . .
How can a function fail to be differentiable?
Weird, Wild, Stuff
. .x
.f(x)
This function is differentiable
at 0.
91. . . . . . .
How can a function fail to be differentiable?
Weird, Wild, Stuff
. .x
.f(x)
This function is differentiable
at 0.
. .x
.f′
(x)
But the derivative is not
continuous at 0!
92. . . . . . .
Outline
Rates of Change
Tangent Lines
Velocity
Population growth
Marginal costs
The derivative, defined
Derivatives of (some) power functions
What does f tell you about f′
?
How can a function fail to be differentiable?
Other notations
The second derivative
93. . . . . . .
Notation
Newtonian notation
f′
(x) y′
(x) y′
Leibnizian notation
dy
dx
d
dx
f(x)
df
dx
These all mean the same thing.
94. . . . . . .
Meet the Mathematician: Isaac Newton
English, 1643–1727
Professor at Cambridge
(England)
Philosophiae Naturalis
Principia Mathematica
published 1687
95. . . . . . .
Meet the Mathematician: Gottfried Leibniz
German, 1646–1716
Eminent philosopher as
well as mathematician
Contemporarily
disgraced by the
calculus priority dispute
96. . . . . . .
Outline
Rates of Change
Tangent Lines
Velocity
Population growth
Marginal costs
The derivative, defined
Derivatives of (some) power functions
What does f tell you about f′
?
How can a function fail to be differentiable?
Other notations
The second derivative
97. . . . . . .
The second derivative
If f is a function, so is f′
, and we can seek its derivative.
f′′
= (f′
)′
It measures the rate of change of the rate of change!
98. . . . . . .
The second derivative
If f is a function, so is f′
, and we can seek its derivative.
f′′
= (f′
)′
It measures the rate of change of the rate of change! Leibnizian
notation:
d2
y
dx2
d2
dx2
f(x)
d2
f
dx2