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?
I did this one yesterday
http://www.jiskha.com/display.cgi?id=1331604970
To find the rate of change of the height of the water in the pool, we need to differentiate the volume of the cylinder with respect to time.
Given:
- The radius of the cylinder, r = 3 feet
- The height of the cylinder, h = 11 feet
- The rate of water pouring into the pool, dV/dt = 9 cubic feet per minute
We know that the volume of a cylinder is given by the formula V = πr^2h, where π is a constant.
To determine the rate of change of the height of the water, we need to differentiate both sides of the equation with respect to time (t):
dV/dt = d/dt (πr^2h)
Now, let's differentiate each term.
Since r is a constant, d/dt (πr^2h) = d/dt (πh * r^2)
Derivative of a constant (πh) with respect to time is 0, so we can drop it.
The derivative of r^2 with respect to time is also 0 since radius is constant.
Therefore, dV/dt = πh * 2r * dr/dt
Given that r = 3 feet and dV/dt = 9 cubic feet per minute, we can now determine dr/dt when h = 9 feet.
dV/dt = πh * 2r * dr/dt
9 = π * 9 * 2(3) * dr/dt
9 = 54π * dr/dt
Dividing both sides by 54π:
1/6π = dr/dt
So, the rate of change of the height of the water in the pool when the depth of the water is 9 feet is 1/6π feet per minute.