Chapter 5.6, Problem 11E

a.

To determine

To find: the balloon’s radius increasing at the instant.

Answer to Problem 11E

The value of $\frac{dr}{dt}$ is $\text{1 ft}\text{.}/\min .$

Explanation of Solution

Given information:

A spherical balloon is inflated with helium at the rate of $100\text{ ft}{\text{.}}^3/\min .$

  $\frac{dv}{dt}=1\text{00}\pi$

The radius is 5ft.

All variables are differentiable functions of t .

Calculation :

We have to calculate the balloon’s radius increasing at the instant.

Therefore,

Volume of a spherical balloon-

  $\text{V}=\frac{4}{3}\pi r^3$

Differentiate with respect to t using product rule.

  $\frac{dv}{dt}=\frac{4}{3}\times 3\pi r^2\frac{dr}{dt}$

Putting the value of $\frac{dv}{dt}=1\text{00}\pi$ and $r=5$ in the above equation,

  $\begin{array}{l}100\pi =\frac{4}{3}\times 3\pi {\left(5\right)}^2\frac{dr}{dt}\\ 100\pi =4\times 25\pi \frac{dr}{dt}\\ 100\pi =100\pi \frac{dr}{dt}\\ \frac{dr}{dt}=\frac{100\pi }{100\pi }\\ \frac{dr}{dt}=1\text{ ft}\text{./min}\text{.}\end{array}$

Hence, the value of $\frac{dr}{dt}$ is $\text{1 ft}\text{.}/\min .$

b.

To determine

To find: the surface area increasing at that instant.

Answer to Problem 11E

The value of $\frac{ds}{dt}$ is $\text{40}\pi \text{ ft}{\text{.}}^2/\min .$

Explanation of Solution

Given information:

A spherical balloon is inflated with helium at the rate of $100\text{ ft}{\text{.}}^3/\min .$

  $\frac{dv}{dt}=1\text{00 ft}{\text{.}}^3\text{/min}\text{.}$

  $r=5\text{ ft}\text{.}$

All variables are differentiable functions of t .

Calculation :

We have to calculate the surface area increasing at that instant.

Therefore,

Surface area of a spherical balloon-

  $\text{S}=4\pi r^2$

Differentiate with respect to t using product rule.

  $\frac{ds}{dt}=4\pi \left[2r\frac{dr}{dt}\right]$

Putting the value of $\frac{dr}{dt}=1$ and $r=5$ in the above equation,

  $\begin{array}{l}\frac{ds}{dt}=4\pi \left[2\left(5\right)\left(1\right)\right]\\ \frac{ds}{dt}=4\pi \left[10\right]\\ \frac{ds}{dt}=40\pi \text{ ft}{\text{.}}^2\text{/min}\text{.}\end{array}$

Hence, the value of $\frac{ds}{dt}$ is $\text{40}\pi \text{ ft}{\text{.}}^2/\min .$