Chapter 4.1, Problem 11E

To determine

To find: The rate at which diameter decreases when diameter is $10$ cm.

Answer to Problem 11E

The diameter decreases at the rate of $\frac{1}{20\pi }$ cm/min.

Explanation of Solution

Given information:

The given rate of decreasing area is $1$ cm ${}^2$ /min.

Calculation:

Let the surface area be S, then $\frac{dS}{dt}=-1$ cm ${}^2$ /min

Now, we have to find the rate at which diameter decreases.

Let the diameter be D. then, we have to find $\frac{dD}{dt}$

If the radius of the ball is R then, surface area of the ball is

  $\begin{array}{c}S=4\pi R^2\\ S=4\pi {\left(\frac{D}{2}\right)}^2\\ S=\pi D^2\end{array}$

Now, we have $S=\pi D^2$

Differentiating with respect to $t$ , implicitly

  $\frac{dS}{dt}=2\pi D\frac{dD}{dt}$

Or, $\frac{dD}{dt}=\frac{1}{2\pi D}.\frac{dS}{dt}$

Now, $\frac{dS}{dt}=-1$ cm ${}^2$ /min and D $=10$ cm

Then, $\frac{dD}{dt}=\frac{1}{2\pi \times 10cm}$ ( $-1$ cm ${}^2$ /min)

Or $\frac{dD}{dt}=\frac{-1}{20\pi }$ cm/min $\approx -0.0159$ cm/min .

Therefore, the diameter decreases at the rate of $\frac{1}{20\pi }$ cm/min.