2.6 Related Rates2.6 Related RatesSolving Real-Life Problems

You pump air at a steady rate into a deflated balloon until the balloon bursts. Does the radius of the balloon change faster when you first start pumping the air, or just before the balloon bursts? Why?

Ex 1 p. 149 Two rates that are relatedSuppose x and y are two differentiable functions of t and are related by equation y = x3 – 2 Find dy/dt , given that dx/dt = 2 when x = 1

3 2y x Implicitly derive x and y with respect to t

23dy dx

xdt dt

Now substitute in what you know

23(1) (2)dy

dt 6

In this problem you were given the equation that relates y and x. Most times you have to create the equation that relates variables. Luckily we know geometry and trig!

Ex 2 p. 150 Ripples in a pondA pebble is dropped in a calm pond, causing ripples in the form of concentric circles. The radius r of the outer ripple is increasing at a constant rate of 1 foot per second. When the radius is 5 feet, at what rate is the total area A of the disturbed water changing?

A = r2

r

2A rEquation:

Given rate:

Find dA/dt when r = 5 ft

1 ft/secdr

dt

(2 )dA dr

rdt dt

2dA dr

rdt dt

Substitute what you know!

2 (5)(1) 10dA

dt

When radius is 5 ft, the Area is changing at rate of 10π ft2/sec

Ex 3 p. 151 An Inflating BalloonAir is being pumped into a spherical balloon at a rate of 5 cubic feet per minute. Find the rate of change of the radius when the radius is 1.5 feet.Volume is changing with time, and so is the radius

Given dV/dt = 5 ft3/min, Find dr/dt when r = 1.5 ft

34

3V r

243

3

dV drr

dt dt

Now plug in the given info after a cleanup

24dV dr

rdt dt

25 4 (1.5)dr

dt

5/ min 0.177 / min

9

drft ft

dt

Equation:

Ex 5 p.152 Changing Angle of ElevationFind the rate of change in the angle of elevation of the camera 10 seconds after lift-off

Solution: Let θ be angle of elevation. Equations:

Given t = 10 so s(10) = 5000 ft

tan2000

s 250s t

100ds

t velocitydt

2 1sec

2000

d ds

dt dt

2 (100 )cos

2000

d t

dt

2 2

2000cos

2000s

2 2 22000hyp s

2

2 2

2000 (100 )

20002000

d t

dt s

2 2

2000 100 10 2 rad/sec

5000 2000 29

Ex 6 Conical tanks and water going into or out of a tank

When water goes into a conical tank, the volume, the radius, and the height of the water are all a function of time. We need to find an equation that relates volume, radius and height, namely 2

3

r hV

(You might recall that a cylinder with same height and radius is 3 times as much volume as a cone.)

2

3

r hV

Implicitly derive with respect to time:

2 (2 )3

dV dh drr h r

dt dt dt

2 23

dV dh drr rh

dt dt dt

A conical tank is 8 inches high and 8inches across the top. If water is flowing into the tank at rate of 5 in3/min, find the rate of change of the depth when the water is 6 inches deep.

Given:35 / min

dVin

dt

4

8

r

hWhat you can observe:

or

What you are solving for: when h = 6 in.

dh

dt

1

2r h

1

2

dr dh

dt dt

Replace all r’s with expressions in h, and all dr/dt’s with expressions in dh/dt

Yuck!!!!

35 / mindV

indt

1

2r h

Solution:

or the height is changing

inches /minute

5

9

dh

dt

5

9

Let’s start again. We can figure out:

2

3

r hV

Substitute in for r

21( )23

h hV

3

12

h

2(3 )12

dV dhh

dt dt

2

4

h dh

dt

265

4

dh

dt

9

dh

dt

•Related rate problems often involve a situation in which you are asked to calculate the rate at which one quantity changes with respect to time from the rate at which a second quantity changes with respect to time.

•Related rate problems can be recognized because the rate of change of one or more quantities with respect to time is given and the rate of change with respect to time of another quantity is required.

http://www.mathdemos.org/mathdemos/relatedrates/relatedrates.html

DO NOT MAKE ANY SUBSTITUTIONS before you differentiate! That creates a problem where things can’t change with time.

If a rate is decreasing or getting smaller over time, it is a negative rate!

Geometry formulas are in the back of your book.

2.6 p. 154 #1-8, 13-27 odd, 35, 36, 45, 46