Q 43. - Calculus

Question

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

How fast is the radius of the balloon increasing at the instant that the balloon has a radius of $12$ inches? 

Step-by-Step Solution

Answer

The radius is increasing at a rate of $0.06631 in/sec$.

Step 1. Given Information

It is given that

$$\begin{gathered} \frac{dV}{dt}=120 i{n}^3/sec \\ r=12 in \end{gathered}$$

Step 2. Find the rate of increase in radius

The volume of a sphere is 

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

Find derivative with respect to t

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

Plug values $\frac{dV}{dt}=120 i{n}^3/min, r=12 in$

$$\begin{gathered} 120=4\pi {\left(12\right)}^2\frac{dr}{dt}\frac{dr}{dt}=\frac{120}{4\pi {\left(12\right)}^2} \\ \frac{dr}{dt}=\hspace{0.17em}0.06631 i{n}^3/min \end{gathered}$$