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?
To find the rate of change of the height of the water in the pool, we need to use the volume equation for a right circular cylinder and differentiate it with respect to time.
The volume V of a right circular cylinder is given by the formula: V = πr^2h, where r is the radius and h is the height.
Given that the water is pouring into the pool at a constant rate of 3 cubic feet per minute, we can set up the following equation for the rate of change of the volume of the water in the pool with respect to time (dh/dt):
dV/dt = 3
To find the rate of change of the height of the water in the pool, we need to determine how the height is related to the volume. We know that the radius of the pool is 6 feet, so we can rewrite the volume equation in terms of h only:
V = π(6^2)h = 36πh
Now, differentiate both sides of the equation with respect to time:
dV/dt = d(36πh)/dt
Using the product rule, the derivative on the right-hand side becomes:
dV/dt = 36π(dh/dt)
Substituting the value of dV/dt from the previous equation, we have:
3 = 36π(dh/dt)
Now, we can solve for dh/dt:
dh/dt = 3 / (36π)
Simplifying further, we get:
dh/dt = 1 / (12π)
Therefore, the rate of change of the height of the water in the pool when the depth of the water is 6 feet is 1 / (12π) feet per minute.