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?

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

To find the rate of change of the height of the water in the pool, we need to differentiate the volume of the water with respect to time.

The volume of a right circular cylinder is given by the formula:
V = πr^2h

Where V is the volume, r is the radius, and h is the height.

Differentiating both sides of the formula with respect to time, we get:
dV/dt = d/dt (πr^2h)

The rate of change of the height of the water is given by dh/dt, and we need to find this when the depth of the water (h) is 9 feet.

Given that the rate of change of the volume (dV/dt) is 9 cubic feet per minute, and the radius (r) is 3 feet, we can substitute these values in the differentiated formula:

9 = d/dt (π(3^2)h)

Now, let's solve for dh/dt.

Taking the derivative with respect to time, we get:
dh/dt * π(3^2) = 9

Simplifying the equation, we have:
dh/dt * 9π = 9

Dividing both sides of the equation by 9π, we find:
dh/dt = 1/π

Thus, the rate of change of the height of the water in the pool is 1/π feet per minute.

To find the rate of change of the height of the water in the pool, we need to take the derivative of the height of the water with respect to time. This can be written as dH/dt, where H represents the height of the water and t represents time.

The volume V of a right circular cylinder can be calculated using the formula: V = π*r^2*h, where r is the radius and h is the height.

Since the volume is increasing at a constant rate of 9 cubic feet per minute, we have dV/dt = 9.

We are given that the radius r is 3 feet and we want to find the rate of change of the height when the depth of the water (h) is 9 feet. In a right circular cylinder, the height and depth are the same, so the height H is also 9 feet.

Now, we can substitute these values into the volume formula and differentiate with respect to time:

V = π*r^2*h
V = π*3^2*H
V = 9πH

Differentiating both sides with respect to time t gives:

dV/dt = d(9πH)/dt
9 = d(9πH)/dt

Since the height H is constant and therefore not changing with respect to time, dH/dt derivative is zero.

So, we have:

9 = 0

This equation is not possible, since it is not true. Therefore, the rate of change of the height of the water in the pool when the depth of the water is 9 feet is 0 feet per minute.

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