Question

(1 point) Suppose that water is pouring into a swimming pool in the shape of a right circular cylinder at a constant rate of 6 cubic feet per minute. If the pool has radius 3 feet and height 6 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 2 feet

Answer 1

The rate of change of the height of the water in the pool is approximately 0.2124 feet per minute.

Step-by-step explanation:

To find the rate of change of the height of the water in the pool, we can use the formula for the volume of a cylinder:
`V = \pi r^2h.`

The rate of change of the height, dh/dt, can be found by differentiating both sides of this equation with respect to time. Since the rate of change of the volume, dV/dt, is given to be 6 cubic feet per minute, we can substitute the values of r, h, and dV/dt into the equation and solve for dh/dt.

`V = \pi (3^2)(6)`

`= 54\pi cubic feet`

dV/dt = 6 cubic feet per minute

`54\pi = \pi (3^2)h`

h = 6 feet

Now, we can use the equation: dV/dt = π(r^2)(dh/dt) and solve for dh/dt.

`6 = \pi (3^2)(dh/dt)`

`dh/dt = 6/(9\pi )`

= 0.2124 feet per minute

Answer 2

dh/dt = 2/3*π ft/min

Step-by-step explanation:

We have a right circular cylinder, and water is pouring at a constant rate, we must expect the height of the water will rise a constant rate, therefore it does not matter the height

V(c) = π*r²*h ⇒ V(c) = π*(3)²*h ⇒ V(c) =9*π*h

DV(c)/dh = 9*π

DV(c)/dh = DV(c)/dt * dt/ dh ⇒ 9*π = DV(c)/dt * dt/ dh

dh/dt = DV(c)/dt / 9*π

DV(c)/dt = 6 ft³/m ( from problem statement )

Then

dh/dt = 6/9*π ft/min

dh/dt = 2/3*π ft/min