Question:
A cylindrical barrel with a diameter of 2 feet contains collected rainwater, as shown in the figure above. The water drains out through a valve (not shown) at the bottom of the barrel. The rate of change of the height h of the water in the barrel with respect to time t is modeled by \(\frac{dh}{dt}=-\frac{1}{10}\sqrt{h},\) where h is measured in feet and t is measured in seconds. (The volume V of a cylinder with radius r and height h is V = πr2h.)
(a) Find the rate of change of the volume of water in the barrel with respect to time when the height of the water is 4 feet. Indicate units of measure.
(b) When the height of the water is 3 feet, is the rate of change of the height of the water with respect to time increasing or decreasing? Explain your reasoning.
(c) At time t = 0 seconds, the height of the water is 5 feet. Use separation of variables to find an expression for h in terms of t.
▶️Answer/Explanation
Ans:
(a)
h = 4
\(\frac{dh}{dt}=\frac{-1}{10}\sqrt{4}=\frac{-1}{5}\) \(\frac{dv}{dt}=\pi \left [ r^{2}\frac{dh}{dt}+h(2r)\left ( \frac{dr}{dt} \right ) \right ]\)
\(\frac{dv}{dt}=?\) \(\frac{dv}{dt}=\pi \left [ (1)\left ( \frac{-1}{5} \right )+(4)(2)(0) \right ]\)
r = 1
\(\frac{dv}{dt}=0\) \(\frac{dv}{dt}=\frac{-\pi }{5}ft^{3}/sec\)
(b)
h = 3
\(\frac{d^{2}h}{dt^{2}}=?\) \(\frac{d^{2}h}{dt^{2}}=\frac{-1}{20}h^{-1/2}\frac{dh}{dt}=\frac{-1}{20\sqrt{3}}\left ( \frac{-1}{10}\sqrt{3} \right )=\frac{1}{200}\)
\(\frac{-1}{10}h^{1/2}\) The rate of change of height is increasing since \(\frac{d^{2}h}{dt^{2}}\) at h = 3 is positive.
1/20
(c)
\(\int h^{-1/2}dh=\int \frac{-1}{10}dt\)
\(\int h^{1/2}=\frac{-1}{10}t+c_{1}\)
\(h^{1/2}=\frac{-1}{20}t+c_{2}\)
\(h=\left ( \frac{-1}{20}t+c \right )^{2}\)
\(5=\left ( \frac{-1}{20}(0)+c \right )^{2}\)
\(c = \sqrt{5}\)
\(h = \left ( \frac{-1}{20}t+\sqrt{5} \right )^{2}\)