1. rate of change

The base of a cone-shaped tank is a circle of radius 5 feet, and the vertex of the cone is 12 below the base. The tank is filled to a depth of 7 feet, and water is flowing out of the tank at a rate of 3 cubic feet per minute.

Find the rate of change of the depth of the water in the tank.

2. Mr_green... do u go to scn high?

This one has got me stumped too =[

3. Originally Posted by Mr_Green
The base of a cone-shaped tank is a circle of radius 5 feet, and the vertex of the cone is 12 below the base. The tank is filled to a depth of 7 feet, and water is flowing out of the tank at a rate of 3 cubic feet per minute.

Find the rate of change of the depth of the water in the tank.
...at the instant the height is 7 feet?...

4. h= 7

dV/dt= -3

r= (5h)/12

V= (1/3)*pi*(r)^2 * h = (pi*25h^3)/(432)

5. Originally Posted by DINOCALC09
h= 7

dV/dt= -3

r= (5h)/12

V= (1/3)*pi*(r)^2 * h = (pi*25h^3)/(432)
correct so far, now differentiate implicitly with respect to t

6. dont understand..

7. there are a few final steps that it fails to explain.

8. is the answer -0.11225 ft/sec

9. Refer back. It's been updated.