Question

Suppose that water is pouring into a swimming pool in the shape of a right circular cylinder at a constant rate of 4 cubic feet per minute. If the pool has radius 4 feet and height 9 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 6 feet?

Answer 1

i get 0.0353677

Answer 2

\[\frac{ 1 }{ 4(\pi) }\]

Answer 3

lol i was multiplying pi

Answer 4

Ahh related rates. We know that the rate of change for the volume of the water is\[dV/dt=4 ft^{3}\]What we want to find is the insantaneous rate of change of height when the height is 6 ft. So we want to find \[dh/dt=?\]

We also know that volume of a right circular cylinder is\[V=\pi r^{2}h\]
So we do implicit derivation by taking the derivative of both sides.

I don't have time right now to finish explaining or finish the problem myself to check if you're right. I'll pop back on tmrw to see if you have it solved.