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) \(V = \pi r^{2}h=\pi (1)^{2}h=\pi h\)
\(\frac{dV}{dt}|_{h=4}=\pi \frac{dh}{dt}|_{h=4}=\pi \left ( -\frac{1}{10}\sqrt{4} \right )=-\frac{\pi }{5}\) cubic feet per second
(b) \(\frac{d^{2}h}{dt^{2}}=-\frac{1}{20\sqrt{h}}\cdot \frac{dh}{dt}=-\frac{1}{20\sqrt{h}}\cdot \left ( -\frac{1}{10} \sqrt{h}\right )=\frac{1}{200}\)
Because \(\frac{d^{2}h}{dt^{2}}=\frac{1}{200}>0\) for h > 0, the rate of change of the height is increasing when the height of the water is 3 feet.
(c) \(\frac{dh}{\sqrt{h}}=-\frac{1}{10}dt\)
\(\int \frac{dh}{\sqrt{h}}=\int -\frac{1}{10}dt\)
\(2\sqrt{h}=-\frac{1}{10}t+C\)
\(2\sqrt{5}=-\frac{1}{10}\cdot 0+C\Rightarrow C = 2\sqrt{5}\)
\(2\sqrt{h}=-\frac{1}{10}t+2\sqrt{5}\)
\(h(t)=\left ( -\frac{1}{20}t+\sqrt{5} \right )^{2}\)