Section 2.6: Related Rates

Introduction to Related RatesWe have seen a lot of relations (whether implicit or

explicit) that involve two variables (frequently x and y). It is possible these two variables are themselves functions of another variable, such as t. For instance:

x 2 y 2 1

if

x cos t

y sin t

Introduction to Related RatesLet’s investigate what occurs when t changes:

x 2 y 2 1

if

x cos t

y sin t

t x y Equation

0

1

0

12 02

1

4

22

22

22 2

22 2

1

0

1

02 12

1

2

23

12

32

12 2

32 2

1

76

32 2

12 2

1

12

32

Notice how when t changes, both the x and y change in relation to the value of t.

As x and y change, their rates of change are related to each other. But how are

they related?

Introduction to Related RatesIn order to take the derivative of the relation using x and y , it must be done with the respect to t. For instance:

x 2 y 2 1

if

x cos t

y sin t

Chain Rule Twice

u1

f1 u1

u1'

f1' u1

x

u12

dxdt

2u1

ddt x

2 y 2 ddt 1

ddt x

2 ddt y

2 ddt 1

u2

f2 u2

y

u22

u2 '

f2 ' u2

dydt

2u2

2u1dxdt 2u2

dydt 0

Differentiate both sides

2xdxdt 2ydydt 0In our exercises, we will not need to know the exact relations.

x 2 y 2 1

if

x f t y g t

Now we know how the rate of change for x and y are related to each other.

Example 1Suppose x and y are both differentiable functions of t and are related

by . Find when x = 10, if when x = 10.

Chain Rule

dxdt 15

ddt 5x 2 y d

dt 100

ddt 5x 2 d

dt y ddt 100

5 ddt x

2 ddt y d

dt 100

52x dxdt dydt 0

52 10 15 dydt 0

Find the derivative by differentiating

both sides.

5x 2 y 100

1500 dydt 0

dydt 1500

dydt

Substitute the known information

Solve for the unknown

Example 2Suppose x and y are both differentiable functions of t and are related

by . Find when x = 9, if when x = 9 and y>0.

Chain Rule

dydt 5

ddt y

2 ddt 2 x 2

ddt y

2 ddt 2 x d

dt 22 1 22 2d d d

dt dt dty x

2y dydt 212 x

1 2 dxdt 0

2y 5 212 9 1 2 dx

dt

Find the derivative by differentiating

both sides.

y 2 2 x 2

1320 dx

dt 60dx

dt

dxdt

Substitute the known information

Solve for the unknown

y 2 2 9 2

y 2 42y

Find other important values:

x

1 2122 2 5 2 9 dx

dt

Example 3A spherical balloon is being filled with a gas in such a way that when the

radius is 2ft, the radius is increasing at the rate 1/6 ft/min. How fast is the volume ( ) changing at this time?

Chain Rule

ddtV d

dt43 r

3

ddtV 4

3 ddt r

3

dVdt 4

3 3r2 drdt

dVdt 4 2 2 1

6

Find the derivative by differentiating

both sides.

V 43 r

3

83

dVdt

Substitute the known information

Solve for the unknown ft3 per minute

dVdt 4r2 dr

dt

Related Rates Guidelines

1.1. Draw a figure, if appropriate, and assign variables to the Draw a figure, if appropriate, and assign variables to the quantities that vary. (Be careful not to label a quantity quantities that vary. (Be careful not to label a quantity with a number unless it never changes in the problem)with a number unless it never changes in the problem)

2.2. Find a formula or equation that relates the variables. Find a formula or equation that relates the variables. (Eliminate unnecessary variables)(Eliminate unnecessary variables)

3.3. Differentiate the equations. (typically implicitly)Differentiate the equations. (typically implicitly)

4.4. Substitute specific numerical values and solve Substitute specific numerical values and solve algebraically for any required rate. (The only unknown algebraically for any required rate. (The only unknown value should be the one that needs to be solved for.)value should be the one that needs to be solved for.)

Example 1A person 6 ft tall is walking away from a streetlight 20 ft

high at the rate of 7 ft/s. At what rate is the length of the person’s shadow increasing?

20 f

t

6 ft

xyChain Rule

xy20 x

6

ddtxy20 d

dtx6

120

ddt x y 1

6ddt x

120

ddt x d

dt y 16ddt x

120

dxdt

dydt 1

6dxdt

Find the rates by

differentiating both sides.

120dxdt

720 1

6dxdt

dxdt 3

Substitute the known information

Solve for the

unknown

ft/s

120

dxdt 7 1

6dxdt

720 7

60dxdt

Using similar triangles, the equation is:

Example 2A bag is tied to the top of a 5 m ladder resting against a vertical wall. Suppose the

ladder begins sliding down the wall in such a way that the foot of the ladder is moving away from the wall. How fast is the bag descending at the instant the foot of the ladder is 4 m from the wall and the foot is moving away at the rate of 2 m/s?

5 mLadder

x

y Chain Rule

x 2 y 2 52

ddt x

2 y 2 ddt 25

ddt x

2 ddt y

2 ddt 25

2xdxdt 2ydydt 0

2 4 2 2ydydt 0

Find the rates by

differentiating both sides.

16 6dydt 0

dydt 8

3

Substitute the known information

Solve for the

unknown

m/s

2 4 2 2 3 dydt 0

6dydt 16

Using The Pythagorean Theorem, the equation

is:

42 y 2 25

y 2 9

y 3

Find other important values:

x

Example 3A trough 10 ft long has a cross section that is an isosceles triangle 3 ft

deep and 8 ft across. If water flows in at the rate 2 ft3/min, how fast is the surface rising when the water is 2 ft deep?

Nothing is known

about b…

Vwater Abhprism

ddt V d

dt12 bh10

ddt V d

dt 583 hh

ddt V 40

3ddt h

2

2 803 2 dhdt

Find the rates by

differentiating both sides.

dhdt 3

80

Substitute the known information

Solve for the unknown

ft/min

2 1603dhdt

Using the volume of a prism, the equation is:

83 b

h

b 83 h

Using similar triangles:

10 ft

b8 ft

h3 ft

Vwater 12 bh10

Chain Rule

dVdt 40

3 2h dhdt

Example 4A rocket launches with a velocity of 550 miles per hour. 25 miles away

there is a photographer filming the launch. At what rate is the angle of elevation of the camera changing when the rocket achieves an altitude of 25 miles?

25 mi

Θ

x Chain Rule

25tan x

25tand d xdt dt

125tand d

dt dt x 2 1

25sec d dxdt dt

2 125sec 550d

dt

Find the rates by

differentiating both sides.

2

2 22ddt

11ddt

Substitute the known information

Solve for the

unknown

rad/h

24sec 22d

dt

2 22ddt

Using The Trigonometry, the equation is:

2525tan

1tan 1 4

Use “x” to find other important values:

This is “x” and there is no “x” in the derivative…