2.6 Related Rates

After this lesson, you should be able to:

Find a related rate.To determine how variables change with respect to time. Use related rates to solve real-life problems.

Related Rates

In real-life, we meet those kind of questions often:1. How fast is the ladder slipping down the wall?2. How fast is the shadow moving?

3. How quickly is the angle decreasing?4. How fast is the height changing in the water tank?5. How fast is the area/volume changing?

When water is drained out of a conical tank, the volume V, the radius r, and the height h of the water level are all functions of time t.

We know that all the variables are related by the volume formula:

hrV 2

3

)()(3

)( 2 thtrtV

)]()([3

)]([ 2 thtrdt

dtV

dt

d

2'( ) [2 '( ) ( ) ( ) ( ) '( )]3

V t r t r t h t r t h t

The rate of change of V is related to the rates of change both r and h

Procedure:Procedure:1. Sketch a picture and label constants and all

values that vary

2. List all given rates and note if they are increasing (+) or decreasing (-).

3. Write an equation relating the quantities with the unknown rate of change with the given rates of change.

4. CR – Differentiate with respect to time & solve for the unknown rate

5. Substitute the given values in and simplify

6. Don’t forget to use units!!!Change in distance? m/sChange in area? m2/sChange in volume? m3/s

Also note:Also note: If the quantity increases: + answerIf the quantity decreases: - answer

Example 1 Suppose x and y are both differentiable functions of t and are related by the equation y = x2 + 3. Find dy/dt, given that dx/dt = 2 when x = 1

Solution

]3[][ 2 xdt

dy

dt

d

dt

dxx

dt

dy2

When x = 1 and dx/dt = 2, then

4)2)(1(2 dt

dy

Related Rates

Related Rates

Example 2 Suppose air is being pumped into a spherical balloon at the rate of 10 cubic centimeters per minute. How fast is the radius of the balloon increasing when the radius is 5 cm?Solution

Let V be the volume of the balloon and r be its radius. We know that all the variables are related by the volume formula:

3

3

4rV

)]([)(33

4)]([ 2 tr

dt

dtrtV

dt

d

dt

drr

dt

dV 2 4dt

dV

rdt

dr2 4

1

or,

Related Rates

Example 2 Suppose air is being pumped into a spherical balloon at the rate of 10 cubic centimeters per minute. How fast is the radius of the balloon increasing when the radius is 5 cm?Solution

10dt

dV

Since the volume is increasing at a rate of 10 cm3/min, the rate of change of the volume is

When r = 5, the rate of change of the radius is

10

1)10(

100

1

dt

dr

dt

dV

rdt

dr2 4

1

Related Rates

Note that the disturbed water area A and radius r are related by the formula:

Example 3 A pebble is dripped into a calm pond, causing in the form of concentric circles. The radius r of the outer ripple is increasing at a rate of 1 feet per second. When the radius is 4 feet, at what rate is the total area A of the disturbed water changing?Solution

2 rA ] [][ 2rdt

dA

dt

d

dt

drr

dt

dA 2

Related Rates

1dt

dr

Example 3 A pebble is dripped into a calm pond, causing in the form of concentric circles. The radius r of the outer ripple is increasing at a rate of 1 feet per second. When the radius is 4 feet, at what rate is the total area A of the disturbed water changing?SolutionSince the radius of outer ripple is increasing at a rate of 1 ft3/sec, the rate of change of the radius is

Then, 81)(4)( 2

dt

dA

dt

drr

dt

dA 2

Related Rates

x

50

Example 4 A patrol car is parked 50 feet from a long warehouse. The revolving light on top of the car turns at a rate of 30 revolutions per minute. How fast is the light beam moving along the wall when the beam makes angles of (a) = 30o

(b) = 60o and(c) = 70o

with the line perpendicular from the light to the wall?

Related Rates

x

50

Solution

rad/sec rad/min 60)2(30

dt

d

The information that “30 revolution per minute” means

The relationship among the variables is

50tan

x

Therefore,

dt

dx

dt

d

50

1sec2

or, dt

d

dt

dx 2sec50

Related Rates

x

50

Solution

(a)

dt

d

dt

dx 2sec50

ft/sec 3

200)(30sec50 02

dt

dx

= 30o

(b)

ft/sec 200)(60sec50 02 dt

dx

= 60o

(c)

= 70o

ft/sec 43.427)(70sec50 02 dt

dx

Related Rates

Read and understand the Example 6 on P 153

Related Rates

Example 5 A winch at the tip of a 12-meter building pulls pipe of the same length to a vertical position. The winch pulls in a rope at a rate of – 0.2 meter/sec. Find the rate of vertical change and the rate of horizontal change at the end of the pipe when y = 6

s

secm/ 2.0dt

ds

(x, y)

12

12

Solution

The relationships among the three variables are:222 12 yx 222 )12( syx and

When y = 6 =12/2,

36612 22 x 12sand

Related Rates

s

secm/ 5

12.0

dt

ds

(x, y)

12

12

Solution

Taking the derivative with respect of t to the above 2 equations, we have

222 12 yx222 )12( syx

and

36612 22 x 12sand

,022 dt

dyy

dt

dxx

,2)1)(12(22dt

dss

dt

dyy

dt

dxx ,)12(

dt

dss

dt

dyy

dt

dxx

0,dx dyx ydt dt

( 12)dx dy dsx y sdt dt dt

0dx dyx ydt dt

(1)

(2)

Related Rates

s

secm/ 5

12.0

dt

ds

(x, y)

12

12

Solution

36612 22 x 12sand

Equation (1) – (2), we have

12dy ds

sdt dt

or12

dy s ds

dt dt

12 1

12 5

dy ds

dt dt m/s

From Equation (1), we havedx y dy

dt x dt

6 1

56 3 3

15 m/s

HomeworkSection 2.6 page 154 #xxxx

Related Rates

s

secm/ 5

12.0

dt

ds

(x, y)

12

12

Solution

Taking the derivative with respect of t to the above 2 equations, we have

222 12 yx222 )12( syx

and

36612 22 x 12sand

,022 dt

dyy

dt

dxx

dt

dx

y

x

dt

dy

,2)1)(12(22dt

dss

dt

dyy

dt

dxx ,)12(

dt

dss

dt

dyy

dt

dxx

,)12(dt

dss

dt

dx

y

xy

dt

dxx

dt

ds

x

sy

dt

dx

12

12(6) 1 3 m/sec

12 5 1512(6 3)

dx sy ds

dt x dt

Related Rates

s

secm/ 5

12.0

dt

ds

(x, y)

12

12

Solution

36612 22 x and m/sec 15

3

dt

dx

m/sec 5

1

15

3

6

36

dt

dx

y

x

dt

dy