51.5k views
1 vote
An upright cylinderical shaped above ground pool is being filled at a rate of 5ft/sec^3. the pool has a 6ft radius and is 6ft tall. how fast is the height of the water increasing

User Eddie Deng
by
8.2k points

1 Answer

1 vote

Answer: 5 / (36π) ft/sec.

Step-by-step explanation: To find how fast the height of the water in the pool is increasing, we need to use the formula for the volume of a cylinder.

The formula for the volume of a cylinder is V = πr^2h, where V is the volume, r is the radius, and h is the height.

Given that the radius of the pool is 6ft and the pool is being filled at a rate of 5ft^3/sec, we can find how fast the height of the water is increasing by differentiating the volume equation with respect to time.

Differentiating both sides of the equation V = πr^2h with respect to time (t), we get:

dV/dt = d(πr^2h)/dt

The derivative of the volume with respect to time (dV/dt) represents the rate at which the volume is changing over time, which is given as 5ft^3/sec.

Now, let's substitute the given values into the equation:

5 = π(6^2)(dh/dt)

Simplifying this equation, we get:

5 = 36π(dh/dt)

To find how fast the height of the water (dh/dt) is increasing, we can solve for it:

dh/dt = 5 / (36π)

Therefore, the height of the water is increasing at a rate of 5 / (36π) ft/sec.

User Georg Richter
by
7.9k points