A conical tank (with vertex down) is 10 feet across the top and 12 feet deep. If water is flowing into the tank at a rate of 10 cubic feet per minute, find the rate of change of the depth of the water when the water is 8 feet deep.

radius,

r

height,

h

Need to know the volume,

V

The volume of a cone is given by:

V

 1 3 

r

2

h

Product Rule There are 3 variables,

V, r, h dV dt

 1 3   

r

2

dh

 2

r dt dr dt h

 

dV dt

 1 3   

r

2

dh

 2

r dt dr dt h

  We want to find the “rate of change of the depth of the water”.

We want

dh/dt

This is still the volume of a cone

dV dt

 1 3   

r

2

dh

 2

r dt dr dt h

  To find

dh/dt

, we need to know how many things?

4 From the problem:

dV dt

 10

ft

3 / min

h

 8

ft

We still need the radius,

r

, and its rate of change

We have to review Geometry and similar triangles:

5 12 r

Water level

h = 8

From the problem, we know the radius of the tank is 5 feet, the height of the tank is 12 feet and the water is 8 feet deep. From this we can find the radius of the water in the tank 5 

r

12 8

The radius of the water,

r

, is 10/3 feet.

Now, we need to find

dr/dt

. Again, we need to use similar triangles.

No matter what the level of the water, we can relate the radius,

r

, to the depth,

h h

12 

r

5

r

 5

h

12

r

 5

h

12

dr dt

 5 12

dh dt

Now, take the derivative of each side with respect to

t

This gives the last thing we need since we know

dh/dt dV dt

 1 3   

r

2

dh

 2

r dt dr dt h

 

dV dt

 1 3   

r

2

dh

 2

r dt dr dt h

  Make the substitution

dV dt

 1 3   

r

2

dh dt

 5 2

r

12

dh dt h

  Plug in the known numbers and solve for

dh/dt

dV dt

 1 3   

r

2

dh dt

 5 2

r

12

dh dt h

  10  1 3      10 3 2

dh

 2

dt

10 3 5 12

dh dt

( 8 )   

dh

 .

29

ft

/ min

dt

Of course, this is left for the student to do.