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Chapter Four
Applications of Differentiation
Section 4.1
Related Rates
Goals
Learn to apply the Chain Rule to problems in which two or more rates…
are related.
Introduction
If…
two variables are connected by an equation,
and at a given moment one of them is changing at a known rate,
then using the Chain Rule we can find
the rate at which the other variable is changing.
Introduction (cont’d)
To solve a problem like this is to relate two rates of change; thus, such a problem is called a related rates problem.
Its solution always consists of two steps:
1. Find an equation that relates the quantities;
2. Use the Chain Rule to differentiate both sides with respect to time.
Example
Air is being pumped into a spherical balloon so that its volume increases at a rate of 100 cm3/s .
How fast is the radius of the balloon increasing when the diameter is 50 cm?
Solution We first identify
the given information and
the unknown, as given above.
Solution (cont’d)
Next we introduce notation: We let
V be the volume of the balloon, and
r be its radius.
Since rates of change are derivatives,
the given information is
the unknown is 3100 cm /s , and
dV
dt
when 25 cmdr
rdt
Solution (cont’d)
Now we relate V and r by the formula
Next we differentiate both sides of this equation with respect to t ; the right side calls for the Chain Rule:
34 for the volume of a sphere.
3V r
24dV dV dr dr
rdt dr dt dt
Solution (cont’d)
Now we solve for the unknown quantity:
Putting r = 25 and dV/dt = 100 gives
so the radius of the balloon is increasing at the rate of 1/(25π) cm/s .
2
1
4
dr dV
dt r dt
2
1 1100 ,
4 25 25
dr
dt
Example
A water tank has the shape of an inverted circular cone with base radius 2 m and height 4 m.
If water is being pumped into the tank at a rate of 2 m3/min, find the rate at which the water level is rising when the water is 3 m deep.
Solution The cone is shown on the next slide:
Solution (cont’d)
Solution (cont’d)
We are…
given that dV/dt = 2 m3/min and we are
asked to find dh/dt when h is 3 m.
The quantities V and h are related by the
equation
to express V as a function of h alone.
21 , but it is very useful
3V r h
Solution (cont’d)
To eliminate r, we use the similar triangles in Fig. 3 to write r/h = 2/4, or r = h/2 .
Thus the expression for V becomes
Now we can differentiate each side with respect to t :
231
3 2 12
hV h h
Solution (cont’d)
Substituting h = 3 m and dV/dt = 2 m3/min gives
The water level is rising at a rate of 8/(9π) m/min.
2
2
4, so
4
dV dh dh dVh
dt dt dt h dt
2
4 82
3 9
dh
dt
Strategy
We can apply some of the problem-solving principles from Chapter 1 to related rates problems specifically:
1. Read the problem carefully.
2. Draw a diagram if possible.
3. Introduce notation. Assign symbols to all quantities that are functions of time.
Strategy (cont’d)
4. Express the given information and the required rate in terms of derivatives.
5. Write an equation that relates the various quantities of the problem.
If necessary, use the geometry of the situation to eliminate one of the variables by substitution (as in the preceding example).
Strategy (cont’d)
6. Use the Chain Rule to differentiate both sides of the equation with respect to t .
7. Substitute the given information into the resulting equation and solve for the unknown rate.
Make sure that Step 7 follows Step 6! A common error is to substitute the given numerical information too early.
Example
A man walks along a straight path at a speed of 4 ft/s. A searchlight is located on the ground 20 ft from the path and is kept focused on the man.
At what rate is the searchlight rotating when the man is 15 ft from the point on the path closest to the searchlight?
Solution See the diagram on the next slide:
Solution (cont’d)
Solution (cont’d)
As in the diagram, we let
x be the distance from the man to the point on the path closest to the searchlight, and
θ be the angle between the beam of the searchlight and the perpendicular to the path.
We are…
given that dx/dt = 4 ft/s and are
asked to find dθ/dt when x = 15.
Solution (cont’d)
The equation that relates x and θ can be written from Fig. 5, namely
x = 20 tan θ
Differentiating each side with respect to t
2gives 20sec , so
dx d
dt dt
2 2 21 1 1cos cos 4 cos
20 20 5
d dx
dt dt
Solution (cont’d)
When x = 15, the length of the beam is 25, so cos θ = 4/5 and
The searchlight is rotating at a rate of 0.128 rad/s.
21 4 16= 0.128
5 5 125
d
dt
Review
Nature of a related rates problem
Role of the Chain Rule in solving a related rates problem
Strategy for solving a related rates problem
Section 4.2
Maximum and Minimum Values
Goals
Solve problems requiring the minimum or maximum value of a quantity
Study absolute vs. local maxima/minima of a function
Introduce the Extreme Value Theorem and Fermat’s Theorem, as well as critical points.
Optimization Problems
These are problems in which we are required to find the optimal (best) way of doing something.
Some examples:
What is the shape of a can that minimizes manufacturing costs?
At what angle should blood vessels branch so as to minimize the energy expended by the heart in pumping blood?
Absolute Maxima/Minima
A function f has an absolute (or global) maximum at c if f(c) ≥ f(x) for all x in the domain D of f . The number f(c) is called the maximum value
of f on D .
Similarly for absolute minimum.
The maximum and minimum values of f are called the extreme values of f .
Absolute Maxima/Minima (cont’d)
The figure on the next slide shows the graph of a function f with…
absolute maximum at d and
absolute minimum at a .
Note that…
(d, f(d)) is the highest point on the graph and
(a, f(a)) is the lowest point.
Absolute Maxima/Minima (cont’d)
Local Maxima/Minima
In Fig. 1…
if we consider only values of x near b ,
then f(b) is the largest of those values of f(x) ,
and is called a local maximum value of f :
A function f has a local (or relative) maximum at c if f(c) ≥ f(x) when x is near c .
We define local minimum similarly.
Example
The function f(x) = cos x takes on its (local and absolute) maximum value of 1 infinitely many times, since
cos 2nπ = 1 for any integer n and
–1 ≤ cos x ≤ 1 for all x .
Likewise, cos(2n + 1) = –1 is its minimum
value, where n is any integer.
Example
If f(x) = x2 , then f(x) ≥ f(0) because x2 ≥ 0 for all x .
Therefore, f(0) = 0 is the absolute (and local) minimum value of f .
However, there is no highest point on the parabola and so this function has no maximum value, as the next slide shows:
Example (cont’d)
Example
From the graph of f(x) = x3 , shown on the next slide, we see that this function has neither an…
absolute maximum value, nor an
absolute minimum value.
In fact, it has no local extreme values either:
Example (cont’d)
Example
The graph of the function
f(x) = 3x4 – 16x3 + 18x2 , –1 ≤ x ≤ 4
is shown on the next slide. We see that…
f(1) = 5 is a local maximum, whereas
the absolute maximum is f(–1) = 37.
f(0) = 0 is a local minimum and
f(3) = –27 is both a local and an absolute
minimum.
Example (cont’d)
Extreme Value Theorem
This theorem gives conditions under which a function is guaranteed to possess extreme values:
Note that an extreme value can be taken on more than once, as illustrated on the next slide:
Extreme Value Theorem (cont’d)
The figures on the next slide show that we cannot omit our assumptions that
f be continuous on the interval, and that
the interval be closed:
Extreme Value Theorem (cont’d)
Finding Extreme Values
The Extreme Value Theorem does not tell us how to find the extreme values whose existence it predicts!
We start by looking for local extreme values; the next slide illustrates the following theorem:
Extreme Values (cont’d)
Remarks on Fermat’s Theorem
As the next slide shows,
The converse of the theorem is false in general: If f(x) = x3 , then f (0) = 0 ,
but f has no maximum or minimum.
There may be an extreme value where f (c)
does not exist: If f(x) = |x| , then f(0) = 0 is a minimum value,
but f (0) does not exist.
Remarks (cont’d)
Critical Points
Fermat’s does suggest that we should at
least start looking for extreme values at the numbers c where… f (c) = 0 , or
f (c) does not exist.
Such numbers are given a special name:
Example
Find the critical numbers of f(x) = x3/5(4 – x).
Solution The Product Rule gives
(We get the same result by first writing f(x) = 4x3/5 – x3/5 .)
Solution (cont’d)
Therefore f (x) …
is 0 if 12 – 8x = 0 , that is, if x = 3/2 , and so
the graph has a horizontal tangent here;
does not exist when x = 0 ; note the graph has a vertical tangent here:
Finding Absolute Max/Minima
Note that in terms of critical numbers, Fermat’s Theorem can be rephrased:
If f has a local maximum or minimum at c , then c is a critical number of f .
To find an absolute maximum or minimum of a continuous function on a closed interval, we note either the maximum or minimum is
Closed Interval Method
local, in which case it occurs at a critical number, or
occurs at an endpoint of the interval.
This leads to the following procedure:
Example For the function f(x) = x – 2sin x, 0≤x≤2π
find the exact minimum and maximum values.
Solution
Solution (cont’d)
We can also solve this problem using calculus: The function f(x) = x – 2 sin x is continuous on
[0, 2π] .
Since f (x) = 1 – 2cos x , we have f (x) = 0
when cos x = ½ , that is, x = π/3 or 5π/3 .
The values of f at these critical points are
2sin 3 0.684853 and3 3 3 3
f
Solution (cont’d) Using calculus (cont’d):
The values of f at the endpoints are
f(0) = 0 and f(2π) = 2π ≈ 6.28
By the Closed Interval Method, the absolute
minimum value is
maximum value is
5 5 5 52sin 3 6.968039
3 3 3 3f
33 3
f
5 53
3 3f
Review
Definitions of absolute and local maximum/minimum
Extreme Value Theorem
Fermat’s Theorem
Critical numbers
Closed Interval Method
Section 4.3
Derivatives and the Shape of Curves Goals
Apply the Mean Value Theorem to finding where functions are increasing and decreasing
Discuss the first derivative test and
second derivative test
for local max/minima
Mean Value Theorem
This theorem is the key to connecting derivatives with the shape of curves:
Geometric Interpretation
The next slide shows the points A(a, f(a)) and B(b, f(b)) on the graphs of two differentiable functions.
The slope of the secant line AB is
the same expression as in the Theorem.
Also f (c) is the slope of the tangent at (c,
f(c)) .
,AB
f b f am
b a
Geometric (cont’d)
Geometric (cont’d)
Thus Equation 1 of the Theorem says that there is at least one point P(c, f(c)) on the graph where
the slope of the tangent line is the same as
the slope of the secant line AB .
In other words, there is a point P where the tangent line is parallel to the secant line AB .
Example
If an object moves in a straight line with position function s = f(t) , then the average velocity between t = a and t = b is
and the velocity at t = c is f (c) .
Thus the Mean Value Theorem says that…
f b f a
b a
Example (cont’d)
…at some time t = c between a and b the instantaneous velocity f (c) is equal to the
average velocity.
For instance, if a car traveled 180 km in 2 h, then the speedometer must have read 90 km/h at least once.
Increasing/Decreasing Functions Earlier we observed from graphs that a
function with a positive derivative is increasing.
This fact can also be deduced from the Mean Value Theorem:
Example
Find where the function
f(x) = 3x4 – 4x3 – 12x2 + 5
is increasing and decreasing.
Solution First we compute
f (x) = 12x3 – 12x2 – 24x = 12x(x – 2)(x + 1)
To find where f (x) > 0 and where f (x) < 0
we arrange our work in a chart.
Solution (cont’d)
The chart shows the sign of each factor of f (x) over
each interval between the critical numbers – 1,
0, and 2 :
The next slide shows the graph of f :
Solution (cont’d)
First Derivative Test
Recall that at a critical point a function can have a
a local maximum,
a local minimum, or
neither.
We can look at whether, and how, f changes sign at the critical point to decide which of the above possibilities is the case:
First Derivative Test (cont’d)
The figures on the next two slides illustrate the First Derivative Test:
First Derivative Test (cont’d)
First Derivative Test (cont’d)
Concavity
We recall the following definition of concavity from Chapter 2:
In the figure on the next slide, the
slopes of the tangent lines increase from left to right on the interval (a, b) , and so
f is increasing, and f is concave upward.
Concavity (cont’d)
Likewise, the
slopes of the tangent lines decrease from left to right on (b, c) , and so
f is decreasing, and f is concave downward:
Concavity (cont’d)
A point where a curve changes its direction of concavity is called an inflection point.
On the preceding graph, both P and Q are inflection points.
The fact that f = (f ) leads to a test for
concavity based on the second derivative:
Concavity (cont’d)
Thus there is a point of inflection at any point where the second derivative changes sign.
This leads to the following test for maximum and minimum values:
Concavity (cont’d)
As the next slide illustrates, part (a) is true because f (x) > 0 near c , and so f is
concave upward near c .
This means the graph of f lies above its horizontal tangent at c , and so f has a local minimum at c :
Concavity (cont’d)
Example
Discuss y = x4 – 4x3 with respect to
concavity,
points of inflection, and
local maxima and minima.
Solution If f(x) = x4 – 4x3 , then
f (x) = 4x3 – 12x2 = 4x2(x – 3)
and
f (x) = 12x2 – 24x = 12x(x – 2) .
Solution (cont’d)
Setting f (x) = 0 gives x = 0 and x = 3 .
To apply the Second Derivative Test we evaluate f at these critical numbers:
f (0) = 0 ; f (3) = 36 > 0
Thus f(3) = –27 is a local minimum, whereas
the Second Derivative Test gives no information about the critical number 0 .
Solution (cont’d)
However, since f (x) < 0 for
x < 0 and also for
0 < x < 3 ,
the First Derivative Test says that f does not have a maximum or minimum at 0 .
Next, since f (x) = 0 when x = 0 or 2 , we
complete the following chart:
Solution (cont’d)
The points (0, 0) and (2, –16) are
inflection points since the curve changes concavity at each of these points.
On the next slide is a sketch of f :
Solution (cont’d)
Example
Sketch the graph of f(x) = x2/3(6 – x)1/3 .
Solution Calculation gives
Since f (x) = 0 when x = 4 and f (x) does
not exist when x = 0 or 6, the critical numbers are 0, 4, and 6.
Solution (cont’d)
Here is a chart of the sign of f :
By the First Derivative Test, f has
a local minimum at x = 0 with f(0) = 0 ;
a local maximum at x = 4 with f(4) = 25/3 ;
neither at x = 6 .
Solution (cont’d)
Studying f (x) in a similar way shows
f (x) < 0 for x > 0 and for 0 < x < 6 ;
f (x) > 0 for x > 6 .
Therefore f is concave downward on (–∞, 0) and (0, 6) , and
upward on (6, ∞) , with one
inflection point at (6, 0) .
On the next slide is the graph of f :
Solution (cont’d)
Review
Mean Value Theorem
Increasing/Decreasing Test
First Derivative Test
Concavity Test
Second Derivative Test
Section 4.5
Indeterminate Forms and l’Hospital’s Rule Goals
Introduce the various types of indeterminate forms
Find limits of indeterminate forms using l’Hospital’s Rule.
Introduction
Suppose we want to analyze the behavior
of the function
Since…
the limit of the denominator as x 1 is 0 , we
cannot use Limit Laws to find
the limit of the numerator is also 0 as x 1 , it
is not even clear whether the limit exists at all.
ln near 1 .
1
xF x x
x
1
lim ;xF x
The Indeterminate Form 0/0
In general, a limit of the form
where both f(x) 0 and g(x) 0 as x
a may or may not exist.
Such a limit is called an indeterminate
form of type similarly, we also
consider the indeterminate form
limx a
f x
g x
0;
0
.
L’Hospital’s Rule
Here is a way to evaluate such limits:
L’Hospital’s Rule (cont’d)
Some remarks: L’Hospital’s Rule says that the limit of a
quotient of functions is equal to the limit of the quotient of their derivatives.
Be sure to verify the required conditions on f and g before using l’Hospital’s Rule.
L’Hospital’s Rule is also valid for one-sided
limits and for limits at ∞ or -∞ .
Example
Find the limit
Solution Since
we can apply l’Hospital’s Rule:
1
lnlim mentioned earlier.
1x
x
x
1 1
limln ln1 0 and lim 1 0x x
x x
Example
Calculate
Solution We have
so l’Hospital’s Rule gives
Since
the limit on the right is still indeterminate.
2lim .x
x
e
x
2lim and lim ,x
x xe x
2lim lim
2
x x
x x
e e
x x and 2 as ,xe x x
Solution (cont’d)
However, a second application of l’Hospital’s Rule gives
Remark Note that when using l’Hospital’s
Rule we differentiate the numerator and denominator separately.
We do not use the Quotient Rule!
2lim lim lim
2 2
x x x
x x x
e e e
x x
Example
Find
Solution Noting that both
we use l’Hospital’s Rule:
30
tanlim .x
x x
x
3tan 0 and 0 as 0 ,x x x x
2
3 20 0
tan sec 1lim lim
3x x
x x x
x x
Solution (cont’d)
Since the limit on the right is still indeterminate of type 0/0, we apply L’Hospital’s Rule again:
Because
calculation by writing
2 2
20 0
sec 1 2sec tanlim lim
3 6x x
x x x
x x
20lim sec 1 , we simplify thex x
Solution (cont’d)
We could evaluate this last limit using yet another application of l’Hospital’s Rule.
The final result is
Indeterminate Products
If f(x) 0 and g(x) ∞ (or –∞) as x a ,
then it isn’t clear what the value of limxa f(x)g(x) will be.
Such a limit is called an indeterminate form of type 0 ∙ ∞ .
We can deal with it by writing the product fg as a quotient:
Products (cont’d)
This converts the given limit into an indeterminate form of type 0/0 or ∞/∞ so that we can use l’Hospital’s Rule.
For example, we evaluate
and use information from derivatives to sketch the graph of y = x ln x :
or 1/ 1/
f gfg fg
g f
0lim ln ,
xx x
Solution
The given limit is indeterminate because, as x 0+ , the…
first factor (x) approaches 0 while the
second factor (ln x) approaches –∞ .
Writing x = 1/(1/x) , we have 1/x 0 as x 0+ , so l’Hospital’s Rule gives
Solution (cont’d)
If f(x) = xln x ,
So f (x) = 0 when ln x = –1 , or x = e-1 .
In fact, f (x) is positive when x > e-1 and
negative when x < e-1 , so f is
increasing on (1/e, ∞) and
decreasing on (0, 1/e) .
So by the First Derivative Test, f(1/e) = –
1/e is a local (and absolute) minimum.
1
ln 1 ln .f x x x xx
Solution (cont’d)
Also, f (x) = 1/x > 0 , so
f is concave downward on (0, ∞) .
We use this information to obtain the graph of f shown:
Indeterminate Differences
If
then the limit
indeterminate form of type ∞ – ∞ .
To evaluate the limit, we try to convert the difference into a quotient so that we have an indeterminate form of type 0/0 or ∞/∞ .
lim and lim ,x a x af x g x
lim is called anx a
f x g x
Example
Find
Solution First notice that sec x ∞ and tan x ∞ as x (π/2)– , so the limit is
indeterminate.
On the next slide we use a common denominator to show that the limit exists and is 0 :
/ 2lim sec tan .
xx x
Solution (cont’d)
Note that the use of l’Hospital’s Rule is
justified because 1 – sin x 0 and
cos x 0
as x (π/2)– .
Indeterminate Powers
The following indeterminate forms all arise
from the limit
lim :
g x
x af x
Powers (cont’d)
Each of these cases can be treated either…
by taking the natural logarithm:
let y = [f(x)]g(x) , then ln y = g(x) ln f(x)
or by writing the function as an exponential:
[f(x)]g(x) = eg(x) ln f(x)
Both of these methods were used earlier in differentiating such functions.
Example
Find
Solution First notice that as x 0+ ,
1 + sin 4x 1 and
cot x ∞ ,
so the given limit is indeterminate.
Let y = (1 + sin 4x)cot x , so
ln y = ln[(1 + sin 4x)cot x] = cot x ln(1 + sin 4x)
cot
0lim 1 sin4 .xx
x
Solution (cont’d)
So l’Hospital’s Rule gives
Now we want the limit of y itself, not ln y .
To find this we use the fact that y = eln y :
Review
L’Hospital’s Rule for the indeterminate
forms 0/0 and ∞/∞
Other types of indeterminate form
Using l’Hospital’s Rule with these other
indeterminate forms