math  calculus
posted by Justine .
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?

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 
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