at a time of t minutes, let the height of the water be h ft
V = π(6^2)(h) = 36π h
dV/dt = 36π dh/dt
subbing in our values
3 = 36π dh/dt
dh/dt = 1/(12π) ft/min
Did you notice that the rate was constant (it did not depend on what h was) ?
Suppose that water is pouring into a swimming pool in the shape of a right circular cylinder at a constant rate of 3 cubic feet per minute. If the pool has radius 6 feet and height 10 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?
1 answer
1 answer