Question

homework6: problem 4
(3 points)
a potter forms a piece of clay into a right circular cylinder. as she rolls it, the height h of the cylinder increases and the radius r decreases. assume that no clay is lost in the process. suppose the height of the cylinder is increasing by 0.6 centimeters per second. what is the rate at which the radius is changing when the radius is 5 centimeters and the height is 9 centimeters?
(include help (units) with your answer.)
hint: look back at your answers to a previous question.
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Explanation:

Step1: Recall the volume formula of a cylinder

The volume \(V\) of a right - circular cylinder is \(V=\pi r^{2}h\). Since no clay is lost, the volume \(V\) is constant. Differentiate both sides with respect to time \(t\) using the product rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u = r^{2}\) and \(v = h\). So, \(\frac{dV}{dt}=\pi(2rh\frac{dr}{dt}+r^{2}\frac{dh}{dt})\). Since \(V\) is constant, \(\frac{dV}{dt} = 0\). Then \(2rh\frac{dr}{dt}+r^{2}\frac{dh}{dt}=0\).

Step2: Solve for \(\frac{dr}{dt}\)

We want to find \(\frac{dr}{dt}\). From \(2rh\frac{dr}{dt}+r^{2}\frac{dh}{dt}=0\), we can isolate \(\frac{dr}{dt}\). First, subtract \(r^{2}\frac{dh}{dt}\) from both sides: \(2rh\frac{dr}{dt}=-r^{2}\frac{dh}{dt}\). Then divide both sides by \(2rh\) (assuming \(r
eq0\) and \(h
eq0\)): \(\frac{dr}{dt}=-\frac{r}{2h}\frac{dh}{dt}\).

Step3: Substitute the given values

We are given that \(\frac{dh}{dt}=0.6\) cm/s, \(r = 5\) cm, and \(h = 9\) cm. Substitute these values into the formula for \(\frac{dr}{dt}\): \(\frac{dr}{dt}=-\frac{5}{2\times9}\times0.6\).
Calculate \(-\frac{5}{18}\times0.6=-\frac{5\times0.6}{18}=-\frac{3}{18}=-\frac{1}{6}\) cm/s.

Answer:

\(-\frac{1}{6}\text{ cm/s}\)