Question

T 0 2 4 6 8 10

P(t) 0 36 43 47 52 60
Kunyu's family has an above ground swimming pool in the shape of a cylinder, with a radius of 10 feet and a height of 5 feet. The pool contains 1000 cubic feet of water at time t=0. During the time interval 0≤t≤10 hours, water is pumped into the pool at the rate () cubic feet per hour. The table above gives values of () for selected values of . During the same time interval, water is leaking from the pool at the rate of () cubic feet per hour, where ()=18−0.04.

(Note: The volume V of a cylinder with radius r and height h is given by =2ℎ .)

Find the rate at which the volume of water in the pool is increasing at time t=6 hours. How fast is the water level in the pool rising at t=6 hours? Indicate units of measure in both answers.

Answer 1

a. 24.12 ft³/hr b. 0.0768 ft/hr

Explanation:

a. Find the rate at which the volume of water in the pool is increasing at time t=6 hours.

The net rate of change of volume of the cylinder dV/dt = volume flow rate in - volume flow rate out

Since volume flow rate in = P(t) and volume flow rate out = R(t),

dV/dt = P(t) - R(t)

`(dV)/(dt) = P(t) - 18e^(0.04t)`

We need to find the rate of change of volume when t = 6.

From the table when t = 6, P(6) = 47 ft³/hr

Also, substituting t = 6 into R(t), we have R(6)

`(dV)/(dt) = P(t) - 18e^(0.04t)`

`(dV)/(dt) = 47 - 18e^(0.04X6)\\(dV)/(dt) = 47 - 18e^(0.24)\\(dV)/(dt) = 47 - 18 X 1.27125\\(dV)/(dt) = 47 - 22.882\\(dV)/(dt) = 24.118 ft^(3)/hr`

dV/dt ≅ 24.12 ft³/hr

So, the rate at which the water level in the pool is increasing at t = 6 hours is 24.12 ft³/hr

b. How fast is the water level in the pool rising at t=6 hours?

Since the a rate at which the water level is rising is dV/dt and the volume of the cylinder is V = πr²h where r = radius of cylinder = 10 ft and h = height of cylinder = 5 feet

dV/dt = d(πr²h)/dt = πr²dh/dt since the radius is constant and dh/dt is the rate at which the water level is rising.

So, dV/dt = πr²dh/dt

dh/dt = dV/dt ÷ πr²

Since dV/dt = 24.12 ft³/hr and r = 10 ft,

Substituting the values of the variables into the equation, we have that

dh/dt = dV/dt ÷ πr²

dh/dt = 24.12 ft³/hr ÷ π(10 ft)²

dh/dt = 24.12 ft³/hr ÷ 100π ft²

dh/dt = 0.2412 ft³/hr ÷ π ft²

dh/dt = 0.2412 ft³/hr

dh/dt = 0.0768 ft/hr

So, the arate at which the water level is rising at t = 6 hours is 0.0768 ft/hr