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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