Posted by **Justine** on Monday, March 12, 2012 at 10:16pm.

Suppose that water is pouring into a swimming pool in the shape of a right circular cylinder at a constant rate of 9 cubic feet per minute. If the pool has radius 3 feet and height 11 feet, what is the rate of change of the height of the water in the pool when the depth of the water in the pool is 9 feet?

- math - calculus -
**Reiny**, Monday, March 12, 2012 at 11:00pm
Let the height of water be h ft

Volume of water = π(3^2)h = 9πh

dV/dt = 18π dh/dt

9/(18π) = dh/dt

dh/dt = 1/(2π) , a constant no matter what the height is

- math - calculus -
**Tom**, Tuesday, March 13, 2012 at 7:21pm
The above answer is incorrect..

everything is right except v=9pi(h)

differentiate both sides. gives you...

dv/dt=9pi dh/dt

9/9pi= dh/dt.

dh/dt= 1/pi

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