Could you plz help me with this question?
2 answers:
You might be interested in
Answer:
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