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In this section, we will learn: How to compute the rate of change of one quantity with respect to another quantity.

1.
3

DIFFERENTIATION RULES

DIFFERENTIATION RULES

2.
DIFFERENTIATION RULES

If we are pumping air into a balloon, both

the volume and the radius of the balloon are

increasing and their rates of increase are

related to each other.

However, it is much easier to measure directly

the rate of increase of the volume than the rate of

increase of the radius.

If we are pumping air into a balloon, both

the volume and the radius of the balloon are

increasing and their rates of increase are

related to each other.

However, it is much easier to measure directly

the rate of increase of the volume than the rate of

increase of the radius.

3.
DIFFERENTIATION RULES

3.9

Related Rates

In this section, we will learn:

How to compute the rate of change of one quantity

in terms of that of another quantity.

3.9

Related Rates

In this section, we will learn:

How to compute the rate of change of one quantity

in terms of that of another quantity.

4.
RELATED RATES

In a related-rates problem, the idea is

to compute the rate of change of one quantity

in terms of the rate of change of another

quantity—which may be more easily

The procedure is to find an equation that relates

the two quantities and then use the Chain Rule to

differentiate both sides with respect to time.

In a related-rates problem, the idea is

to compute the rate of change of one quantity

in terms of the rate of change of another

quantity—which may be more easily

The procedure is to find an equation that relates

the two quantities and then use the Chain Rule to

differentiate both sides with respect to time.

5.
RELATED RATES Example 1

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?

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?

6.
RELATED RATES Example 1

We start by identifying two things:

Given information: The rate of increase

of the volume of air is 100 cm3/s.

Unknown: The rate of increase of the radius

when the diameter is 50 cm

We start by identifying two things:

Given information: The rate of increase

of the volume of air is 100 cm3/s.

Unknown: The rate of increase of the radius

when the diameter is 50 cm

7.
RELATED RATES Example 1

To express these quantities

mathematically, we introduce some

suggestive notation:

Let V be the volume of the balloon and

let r be its radius.

To express these quantities

mathematically, we introduce some

suggestive notation:

Let V be the volume of the balloon and

let r be its radius.

8.
RELATED RATES Example 1

The key thing to remember is that rates

of change are derivatives.

In this problem, the volume and the radius are

both functions of the time t.

The rate of increase of the volume with respect

to time is the derivative dV / dt.

The rate of increase of the radius is dr / dt.

The key thing to remember is that rates

of change are derivatives.

In this problem, the volume and the radius are

both functions of the time t.

The rate of increase of the volume with respect

to time is the derivative dV / dt.

The rate of increase of the radius is dr / dt.

9.
RELATED RATES Example 1

Thus, we can restate the given and

the unknown as follows:

dV

Given: 100cm3 / s

dt

dr

Unknown: when r 25 cm

dt

Thus, we can restate the given and

the unknown as follows:

dV

Given: 100cm3 / s

dt

dr

Unknown: when r 25 cm

dt

10.
RELATED RATES Example 1

To connect dV / dt and dr / dt, first

we relate V and r by the formula for

the volume of a sphere:

3

V r 4

3

To connect dV / dt and dr / dt, first

we relate V and r by the formula for

the volume of a sphere:

3

V r 4

3

11.
RELATED RATES Example 1

To use the given information, we

differentiate each side of the equation

with respect to t.

To differentiate the right side, we need to use

the Chain Rule:

dV dV dr 2 dr

4 r

dt dr dt dt

To use the given information, we

differentiate each side of the equation

with respect to t.

To differentiate the right side, we need to use

the Chain Rule:

dV dV dr 2 dr

4 r

dt dr dt dt

12.
RELATED RATES Example 1

Now, we solve for the unknown quantity:

dr 1 dV

2

dt 4 dt

If we put r = 25 and dV / dt = 100 in this equation,

we obtain: dr 1 1

2

100

dt 4 (25) 25

The radius of the balloon is increasing at the rate

of 1/(25π) ≈ 0.0127 cm/s.

Now, we solve for the unknown quantity:

dr 1 dV

2

dt 4 dt

If we put r = 25 and dV / dt = 100 in this equation,

we obtain: dr 1 1

2

100

dt 4 (25) 25

The radius of the balloon is increasing at the rate

of 1/(25π) ≈ 0.0127 cm/s.

13.
RELATED RATES Example 2

A ladder 10 ft long rests against a vertical wall.

If the bottom of the ladder slides away

from the wall at a rate of 1 ft/s, how fast is

the top of the ladder sliding down the wall

when the bottom of the ladder is 6 ft from

the wall?

A ladder 10 ft long rests against a vertical wall.

If the bottom of the ladder slides away

from the wall at a rate of 1 ft/s, how fast is

the top of the ladder sliding down the wall

when the bottom of the ladder is 6 ft from

the wall?

14.
RELATED RATES Example 2

We first draw a diagram and label it

as in the figure.

Let x feet be the distance from the bottom of the ladder

to the wall and y feet the distance from the top of the

ladder to the ground.

Note that x and y are

both functions of t

(time, measured in

seconds).

We first draw a diagram and label it

as in the figure.

Let x feet be the distance from the bottom of the ladder

to the wall and y feet the distance from the top of the

ladder to the ground.

Note that x and y are

both functions of t

(time, measured in

seconds).

15.
RELATED RATES Example 2

We are given that dx / dt = 1 ft/s

and we are asked to find dy / dt

when x = 6 ft.

We are given that dx / dt = 1 ft/s

and we are asked to find dy / dt

when x = 6 ft.

16.
RELATED RATES Example 2

In this problem, the relationship between

x and y is given by the Pythagorean

Theorem: x2 + y2 = 100

In this problem, the relationship between

x and y is given by the Pythagorean

Theorem: x2 + y2 = 100

17.
RELATED RATES Example 2

Differentiating each side with respect to t

using the Chain Rule, we have:

dx dy

2x 2 y 0

dt dt

Solving this equation for the desired rate,

we obtain: dy x dx

dt y dt

Differentiating each side with respect to t

using the Chain Rule, we have:

dx dy

2x 2 y 0

dt dt

Solving this equation for the desired rate,

we obtain: dy x dx

dt y dt

18.
RELATED RATES Example 2

When x = 6 , the Pythagorean Theorem gives

y = 8 and so, substituting these values and

dy 6 3

dx / dt = 1, we have: (1) ft / s

dt 8 4

The fact that dy / dt is negative means that

the distance from the top of the ladder to

the ground is decreasing at a rate of ¾ ft/s.

That is, the top of the ladder is sliding down

the wall at a rate of ¾ ft/s.

When x = 6 , the Pythagorean Theorem gives

y = 8 and so, substituting these values and

dy 6 3

dx / dt = 1, we have: (1) ft / s

dt 8 4

The fact that dy / dt is negative means that

the distance from the top of the ladder to

the ground is decreasing at a rate of ¾ ft/s.

That is, the top of the ladder is sliding down

the wall at a rate of ¾ ft/s.

19.
RELATED RATES Example 3

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.

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.

20.
RELATED RATES Example 3

We first sketch the cone and label it.

V is the volume of the water.

r is the radius of the surface.

h is the height of

the water at time t,

where t is measured

in minutes.

We first sketch the cone and label it.

V is the volume of the water.

r is the radius of the surface.

h is the height of

the water at time t,

where t is measured

in minutes.

21.
RELATED RATES Example 3

We are given that dV / dt = 2 m3/min

and we are asked to find dh / dt when

h is 3 m.

We are given that dV / dt = 2 m3/min

and we are asked to find dh / dt when

h is 3 m.

22.
RELATED RATES Example 3

The quantities V and h are related by

2

the equation V r h

1

3

However, it is very useful to express V as

a function of h alone.

The quantities V and h are related by

2

the equation V r h

1

3

However, it is very useful to express V as

a function of h alone.

23.
RELATED RATES Example 3

To eliminate r, we use the similar triangles

r 2 h

in the figure to write: r

h 4 2

The expression for V

becomes:

2

1 h 3

V h h

3 2 12

To eliminate r, we use the similar triangles

r 2 h

in the figure to write: r

h 4 2

The expression for V

becomes:

2

1 h 3

V h h

3 2 12

24.
RELATED RATES Example 3

Now, we can differentiate each side with

respect to t:

dV 2 dh

h

dt 4 dt

So, dh 4 dV

2

dt h dt

Now, we can differentiate each side with

respect to t:

dV 2 dh

h

dt 4 dt

So, dh 4 dV

2

dt h dt

25.
RELATED RATES Example 3

Substituting h = 3 m and dV / dt = 2 m3/min,

we have: dh 4 8

3

2

dt (3) 9

The water level is rising at a rate of

8/(9π) ≈ 0.28 m/min.

Substituting h = 3 m and dV / dt = 2 m3/min,

we have: dh 4 8

3

2

dt (3) 9

The water level is rising at a rate of

8/(9π) ≈ 0.28 m/min.

26.
It is useful to recall some of the problem-

solving principles from Chapter 1 and adapt

them to related rates in light of our experience

in Examples 1–3.

1. Read the problem carefully.

2. Draw a diagram if possible.

3. Introduce notation. Assign symbols to

all quantities that are functions of time.

solving principles from Chapter 1 and adapt

them to related rates in light of our experience

in Examples 1–3.

1. Read the problem carefully.

2. Draw a diagram if possible.

3. Introduce notation. Assign symbols to

all quantities that are functions of time.

27.
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 Example 3).

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.

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 Example 3).

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.

28.
The following examples

are further illustrations of

the strategy.

are further illustrations of

the strategy.

29.
RELATED RATES Example 4

Car A is traveling west at 50 mi/h and car B

is traveling north at 60 mi/h. Both are headed

for the intersection of the two roads.

At what rate are the cars approaching

each other when car A is 0.3 mi and car B

is 0.4 mi from the intersection?

Car A is traveling west at 50 mi/h and car B

is traveling north at 60 mi/h. Both are headed

for the intersection of the two roads.

At what rate are the cars approaching

each other when car A is 0.3 mi and car B

is 0.4 mi from the intersection?

30.
RELATED RATES Example 4

We draw this figure, where C is

the intersection of the roads.

At a given time t, let x be the distance from car A to C.

Let y be the distance from car B to C.

Let z be the distance

between the cars—

where x, y, and z are

measured in miles.

We draw this figure, where C is

the intersection of the roads.

At a given time t, let x be the distance from car A to C.

Let y be the distance from car B to C.

Let z be the distance

between the cars—

where x, y, and z are

measured in miles.

31.
RELATED RATES Example 4

We are given that dx / dt = –50 mi/h

and dy / dt = –60 mi/h.

The derivatives are negative because x and y

are decreasing.

We are asked to

find dz / dt.

We are given that dx / dt = –50 mi/h

and dy / dt = –60 mi/h.

The derivatives are negative because x and y

are decreasing.

We are asked to

find dz / dt.

32.
RELATED RATES Example 4

The equation that relates x, y, and z is given

by the Pythagorean Theorem: z2 = x2 + y2

Differentiating each side with respect to t,

we have:

dz dx dy

2z 2 x 2 y

dt dt dt

dz 1 dx dy

x y

dt z dt dt

The equation that relates x, y, and z is given

by the Pythagorean Theorem: z2 = x2 + y2

Differentiating each side with respect to t,

we have:

dz dx dy

2z 2 x 2 y

dt dt dt

dz 1 dx dy

x y

dt z dt dt

33.
RELATED RATES Example 4

When x = 0.3 mi and y = 0.4 mi,

the Pythagorean Theorem gives z = 0.5 mi.

So, dz 1 0.3( 50) 0.4( 60)

dt 0.5

78 mi / h

The cars are approaching each other at

a rate of 78 mi/h.

When x = 0.3 mi and y = 0.4 mi,

the Pythagorean Theorem gives z = 0.5 mi.

So, dz 1 0.3( 50) 0.4( 60)

dt 0.5

78 mi / h

The cars are approaching each other at

a rate of 78 mi/h.

34.
RELATED RATES Example 5

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

At what rate is the searchlight rotating

when the man is 15 ft from the point on

the path closest to the searchlight?

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

At what rate is the searchlight rotating

when the man is 15 ft from the point on

the path closest to the searchlight?

35.
RELATED RATES Example 5

We draw this figure and let x be the distance

from the man to the point on the path closest

to the searchlight.

We let θ be the angle

between the beam of

the searchlight and

the perpendicular to

the path.

We draw this figure and let x be the distance

from the man to the point on the path closest

to the searchlight.

We let θ be the angle

between the beam of

the searchlight and

the perpendicular to

the path.

36.
RELATED RATES Example 5

We are given that dx / dt = 4 ft/s and

are asked to find dθ / dt when x = 15.

The equation that

relates x and θ can be

written from the figure:

x

tan x 20 tan

20

We are given that dx / dt = 4 ft/s and

are asked to find dθ / dt when x = 15.

The equation that

relates x and θ can be

written from the figure:

x

tan x 20 tan

20

37.
RELATED RATES Example 5

Differentiating each side with respect to t,

we get:

dx 2 d

20sec

dt dt

So, d 1 cos 2 dx

dt 20 dt

1 2 1 2

cos (4) cos

20 5

Differentiating each side with respect to t,

we get:

dx 2 d

20sec

dt dt

So, d 1 cos 2 dx

dt 20 dt

1 2 1 2

cos (4) cos

20 5

38.
RELATED RATES Example 5

When x = 15, the length of the beam is 25.

2

So, cos θ = 4 and d 1 4 16 0.128

5

dt 5 5 125

The searchlight is rotating at a rate of 0.128 rad/s.

When x = 15, the length of the beam is 25.

2

So, cos θ = 4 and d 1 4 16 0.128

5

dt 5 5 125

The searchlight is rotating at a rate of 0.128 rad/s.