Chapter 5.6, Problem 8E

To determine

To find: the rate when the plate’s area is increasing.

Answer to Problem 8E

The value of $\frac{d\text{A}}{dt}$ is $\pi {\text{ cm}}^2\text{/sec}$ .

Explanation of Solution

Given information:

All variables are differentiable function of $t$ .

Area of circular plate is $\text{A}=\pi r^2$ .

  $\frac{dr}{dt}=0.01\text{ cm/sec}$ , $r=50\text{ cm}\text{.}$

Calculation :

We have to find the rate when the plate’s area is increasing.

Therefore,

  $\text{A}=\pi r^2$

Differentiate with respect to t .

  $\begin{array}{l}\frac{d\text{A}}{dt}=\pi \left[2r\frac{dr}{dt}\right]\\ \frac{d\text{A}}{dt}=2\pi \left[\left(50\right)\left(0.01\right)\right]\left[\begin{array}{l}\text{Since }r=50\text{ cm}\text{.}\\ \frac{dr}{dt}=0.01\text{ cm}\text{./sec}\end{array}\right]\\ \frac{d\text{A}}{dt}=\pi {\text{cm}}^2\text{/sec}\end{array}$

Hence, the value of $\frac{d\text{A}}{dt}$ is $\pi {\text{ cm}}^2\text{/sec}$ .