Question

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 0.1 centimeters per second. what is the rate at which the radius is changing when the radius is 6 centimeters and the height is 9 centimeters?

Explanation:

Step1: Write the volume formula for a cylinder

The volume formula for a cylinder is \(V=\pi r^{2}h\). Since the volume of clay remains constant (no clay is lost), \(\frac{dV}{dt} = 0\).

Step2: Differentiate the volume formula 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 \(\frac{dV}{dt} = 0\), we have \(2rh\frac{dr}{dt}+r^{2}\frac{dh}{dt}=0\).

Step3: Solve for \(\frac{dr}{dt}\)

We are given that \(r = 6\), \(h = 9\), and \(\frac{dh}{dt}=0.1\).
Substitute these values into \(2rh\frac{dr}{dt}+r^{2}\frac{dh}{dt}=0\):

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Answer:

\(\frac{dr}{dt}=- \frac{1}{30}\text{ cm/s}\)