# Rate of Change: Depth of Water in a Tank

• Feb 16th 2010, 06:36 PM
-DQ-
Rate of Change: Depth of Water in a Tank
The base of a pyramid-shaped tank is a square with sides
of length 12 feet, and the vertex of the pyramid is 10 feet
above the base. The tank is filled to a depth of 4 feet, and
water is flowing into the tank at the rate of 2 cubic feet
per minute. Find the rate of change of the depth of water
in the tank.

I don't even know how to approach it.
• Feb 17th 2010, 03:15 AM
-DQ-
Help?
• Feb 17th 2010, 05:10 AM
HallsofIvy
Volume is changing at a constant rate: "water is flowing into the tank at the rate of 2 cubic feet per minute" so dV/dt= 2.

Write the volume as a function of height of the water: if V= f(h) the dV/dt= f'(h) dh/dt= 2 and you can solve for dh/dt.

To find the volume as a function of the height, volume of a pyramid of height h with base area A is $\frac{1}{3}Ah$. Here, of course, the base will be a square with side length depending on height. You can get the length of an edge of that square as a function of height by remembering that the pyramid of water will be similar to the entire pyramid. It has edge length 12 and height 10 so x/h= 12/10 or x= (6/5)h. The area of the base is [/tex]A= (36/25)h^2[/tex] so the volume is $V= \frac{12}{25}h^3$.
• Feb 17th 2010, 03:34 PM
ione
Quote:

Originally Posted by HallsofIvy
Volume is changing at a constant rate: "water is flowing into the tank at the rate of 2 cubic feet per minute" so dV/dt= 2.

Write the volume as a function of height of the water: if V= f(h) the dV/dt= f'(h) dh/dt= 2 and you can solve for dh/dt.

To find the volume as a function of the height, volume of a pyramid of height h with base area A is $\frac{1}{3}Ah$. Here, of course, the base will be a square with side length depending on height. You can get the length of an edge of that square as a function of height by remembering that the pyramid of water will be similar to the entire pyramid. It has edge length 12 and height 10 so x/h= 12/10 or x= (6/5)h. The area of the base is [/tex]A= (36/25)h^2[/tex] so the volume is $V= \frac{12}{25}h^3$.

Is the pyramid standing on its base or is it inverted?