Q 45. - Calculus

Question

Consider a large helium balloon that is being inflated at the rate of $120$ cubic inches per second. 

How fast is the surface area of the balloon increasing at the instant that the radius of the balloon is $15$ inches? 

Step-by-Step Solution

Answer

The surface area of a balloon is increasing at a rate $16 i{n}^2/sec$.

Step 1. Given Information

It is given that

$$\begin{gathered} \frac{dV}{dt}=120 i{n}^3/sec \\ r=15 in \end{gathered}$$

Step 2. Finding the rate of change in radius

The volume of the balloon formula is 

$V=\frac{4}{3}\pi r^3$

Find derivative with respect to t

$$\begin{gathered} \frac{dV}{dt}=\frac{4}{3}\pi \left(3r^2\right)\frac{dr}{dt} \\ \frac{dV}{dt}=4\pi r^2\frac{dr}{dt} \end{gathered}$$

Plug values

$$\begin{gathered} 120=4\pi {\left(15\right)}^2\frac{dr}{dt}\frac{dr}{dt}=\frac{120}{4\pi {\left(15\right)}^2} \\ \frac{dr}{dt}=0.04244 in/sec \end{gathered}$$

Step 3. Find the rate of change in surface area

The surface area formula is

$S=4\pi r^2$

Find derivative with respect to t

$$\begin{gathered} \frac{dS}{dt}=4\pi \left(2r\right)\frac{dr}{dt} \\ \frac{dS}{dt}=8\pi r\frac{dr}{dt} \end{gathered}$$

Plug values

$$\begin{gathered} \frac{dS}{dt}=8\pi \left(15\right)(0.04244) \\ \frac{dS}{dt}=16 i{n}^2/sec \end{gathered}$$