Chapter 4.6, Problem 11E

(a)

To determine

To find : How fast balloon’s radius increasing at the instant the radius of $5ft$ .

Answer to Problem 11E

The answer: $\frac{dr}{dt}=1\text{ft}/\text{min}$

Explanation of Solution

Given information:

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

Calculation:

Provided values volume variation with relation to time the current radius at that point.

  $\begin{array}{c}\frac{dV}{dt}=100\pi \\ r=5\end{array}$

Implicitly distinguish in relation to $t$ , utilzing the spherical volume equation $V$ the Rule of Power.

  $\begin{array}{c}V=\frac{4}{3}\pi r^3\\ \Rightarrow \frac{dV}{dt}=3\cdot \frac{4}{3}\pi r^2\frac{dr}{dt}\\ \Rightarrow \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\\ \Rightarrow \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\\ $ \Rightarrow 100\pi =4\pi {(5)}^2\frac{dr}{dt}$\Rightarrow 100\pi =100\pi \frac{dr}{dt}\\ \Rightarrow \frac{dr}{dt}=1\end{array}$

Therefore the required $\frac{dr}{dt}=1\text{ft}/\text{min}$ fast balloon’s radius increasing at the instant the radius of $5ft$ .

(b)

To determine

To find: How fast balloon’s surface area increasing at the instant.

Answer to Problem 11E

The answer: $\frac{dS}{dt}=40\pi \frac{f{t}^2}{\min }$

Explanation of Solution

Given information:

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

Calculation:

Applying the Power Rule, implicitly differentiate the spherical surface area equation $S$

To determine the solution, enter each of the values given.

  $\begin{array}{c}\frac{dV}{dt}=100\pi \frac{f{t}^3}{\min }\\ \Rightarrow \frac{dr}{dt}=1\frac{ft}{\min }\\ \Rightarrow r=5\end{array}$

So,

  $\begin{array}{c}S=4\pi r^2\\ \Rightarrow \frac{dS}{dt}=8\pi r\frac{dr}{dt}\\ =8\pi (5)\left(1\frac{ft}{\min }\right)\\ =40\pi \frac{f{t}^2}{\min }\end{array}$

Therefore the required $\frac{dS}{dt}=40\pi \frac{f{t}^2}{\min }$ fast balloon’s surface area increasing at the instant.