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.

a. 24.12 ft³/hr b. 0.0768 ft/hr

Step-by-step 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)

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 ≅ 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

x = 23.78

step-by-step explanation:

answer 1 : option "c"

answer 2 : option "d"

answer 3 : option "b"

answer 4 : option "d"

answer 5 : option "b"

answer 6 : option "a"

answer 7 : option "c"

answer 8 : option "d"

answer 9 : option "c"

answer 10 : option "d"

answer 11 : option "c"

answer 12 option "b"

answer 13 option "c"

answer 14 option "c"

answer 15 option "d"

answer 16 option "a"

answer 17 option "b"

answer 18 option "c"

answer 19 option "a"

answer 20 option "a"

answer 21 option "c"

d