Related rates problems involve finding a rate at which a quantity changes by relating that quantity to other quantities whose rates of change are known.

Because these problems involve rates, they must be differentiated with respect to

time.

Example

:

y

3 

y

2  5

y



x

2   4 3

y

2

dy dx

 2

y dy dx

 5

dy dx

 2

x

 0

dy

( 3

y

2

dt

 2

y

 5 )  2

x dy dx

 ( 3

y

2  2

x

2

y

 5 )

We will determine how to solve problems involving: I. Circles II.Spheres

III.Triangles

IV.Cones

V. Cylinders

The most common way to approach related rates problems is the following: 1. Identify the known variables, including rates of change and the rate of change that is to be found. **Drawing a picture or representation of the

problem can help to keep everything in order**

2. Construct an equation relating the quantities whose rates of change are known to the quantity whose rate of change is to be found. 3. Differentiate both sides of the equation with respect to time. (Often, the chain rule is employed at this step.) 4. Substitute the known rates of change and the known quantities into the equation. Solve for the wanted rate of change. **Errors in this procedure are often caused by plugging in the known values for the variables before (rather than after) finding the derivative with respect to time. Doing so will yield an incorrect result.**

Olympic swimmer Mr. Spitz leaps into a pool, causing ripples in the form of concentric circles. The radius is increasing at a constant of 4 feet per second. When the radius is 8 feet, at what rate is the total area of disturbed water changing (assuming the splash is negligible)?

The variables r and A are related by A=πr 2 . The rate of change of the radius r is dr/dt=4 Equation: A=πr 2 dr/dt=4 Given rate: Find: dA/dt when r=8

d dt

[

A

] 

d dt

[ 

r

2 ]

dA

 2 

r dr dt dt

dA

 2  ( 8 )( 4 )

dt

dA

 64 

dt ft

2 sec

Mr. Spitz, the circus clown, is traveling with the Barnum and Bailey circus, is told to inflate a balloon. If the volume increases at a constant rate of 50 cm 3 /sec, at what rate is the radius increasing when the volume is 972π cm 3 ?

Equation: V= 4/3πr 3 Given: dV/dt=50 V = 972π Find dr/dt

V

 972  4 3 

r

3  4 3 

r

3 729 

r

3

r

 9

cm dV dt

 4 

r

2

dr dt

50  4  ( 9 ) 2

dr dt dr dt

 25 162   .

0491

cm

sec

Major Spitz is flying a rescue mission to extract General Earl during a battle in WWI. He flies his plane at 200 mph at an altitude of 6 miles. What is the rate of change of the shortest distance between them when the horizontal distance is 8 miles? dx/dt = -200 dz/dt = ??

6 mi 8 mi

z

2 

x

2 

y

2

z



z

 ( 8 ) 2  ( 6 ) 2 100  10

mi

z

2 

x

2 

y

2 2

z dz dt

 2

x dx dt

2 ( 10 )

dz dt

 2 ( 8 )(  200 )

dz dt

 2 ( 8 )( 200 ) 2 ( 10 )   160

mph

General Earl is watching for his rescue plane with binoculars. Major Spitz slows down to 100 mph and descends to 3 miles to spot General Earl. What is the rate of change of the angle the General is watching from when the horizontal distance is 4 miles?

tan  

y

 3

x x

(sec 2  )

d



dt

  3

x

 2

dx dt d



dt

  3 (cos 2  )

x

2

dx dt d



dt

 (  3 ( 4 5 ) 2 4 2 )(  100 )

d



dt

 300 25  12

rad hr

Mr. Spitz was best friends with Augustus Caesar. While studying by candlelight, Spitz was testing his speed with implicit differentiation and timed himself with an hourglass, the most advanced technology at the time. He knows that the sand initially falls forming a cone, whose radius is twice its height. The hourglass fills at a constant rate of 3 cm 3 /min. When the volume is 36π cm 3 , what is the rate of change of the height?

V

 1 3 

r

2

h V

 1 3  ( 2

h

) 2

h V

 1 3  ( 4

h

3 )

dV dt

 1 3  ( 12 )(

h

2 )

dh dt

3  4  ( 3 ) 2

dh dt dh dt

 3 36   1 12 

cm

sec Note: When V = 36 π, h = 3

Mr. Spitz is filling his cylindrical water jug with some “good H 2 O” before he runs the iron man triathlon. The radius is 4 inches; as Mr. Spitz fills the jug, the height is changing at a rate of .5 in/sec. What rate is the volume changing at?

V

 

r

2

h V

 ( 4 ) 2 

h

dV dt

 16 

dh dt dV dt

 16  (.

5 )  8 

in

3 sec