Chapter 4.6, Problem 27E

To determine

To find: how fast is the outer surface area of ice decreasing when the outer diameter $($ ball plus ice $)$ is $20 \text{cm}$ .

Answer to Problem 27E

The answer: Decreasing at $1.6 {\text{cm}}^2/\text{min}$

Explanation of Solution

Given information:

An even layer of ice is spread on a sphere-shaped iron ball. If the rate of ice melting is $8 \text{mL}/\text{min}$ .

Calculation:

Get an equation relating the volume $V$ with the surface area $A$ by substituting for $r$ .

  $\begin{array}{c}\frac{dV}{dt}=-8 {\text{cm}}^3/\text{min}\\ A=4\pi r^2\\ r=\sqrt{\frac{A}{4\pi }}\\ =\frac{1}{2}\sqrt{\frac{A}{\pi }}\\ V=\frac{4}{3}\pi r^3\\ =\frac{1}{6\sqrt{\pi }}{A}^{3/2}\\ \frac{dV}{dt}=\frac{1}{4\sqrt{\pi }}\sqrt{A}\frac{dA}{dt}\end{array}$

When diameter $=20$

And

  $\begin{array}{c}r=10\\ A=400\pi \end{array}$

Plug in the values and solve for $\frac{dA}{dt}$ :

  $\begin{array}{c}\frac{dV}{dt}=\frac{1}{4\sqrt{\pi }}\sqrt{A}\frac{dA}{dt}\\ \Rightarrow -8=\frac{1}{4\sqrt{\pi }}\sqrt{400\pi }\frac{dA}{dt}\\ \Rightarrow -8=5\frac{dA}{dt}\\ \Rightarrow \frac{dA}{dt}=\frac{-8}{5}\\ =-1.6\end{array}$

Therefore, the required decreasing at $1.6 {\text{cm}}^2/\text{min}$ is the outer surface area of ice decreasing when the outer diameter $($ ball plus ice $)$ is $20 \text{cm}$ .