Q 44. - 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 contains $300$ cubic inches of air? 

Step-by-Step Solution

Answer

The radius is increasing at a rate of $0.5537 in/sec$.

Step 1. Given Information

It is given that

$$\begin{gathered} \frac{dV}{dt}=120 i{n}^3/sec \\ V=300 i{n}^3 \end{gathered}$$

Step 2. Find the radius of the balloon

The volume of a sphere is 

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

Plug $V=300 i{n}^3$ and solve for $r$

$$\begin{gathered} 300=\frac{4}{3}\pi r^3 \\ \frac{4}{3}\pi r^3=300 \\ r^3=\frac{3\times 300}{4\pi } \\ r=\sqrt[3]{\frac{3\times 300}{4\pi }} \\ r=4.15283 in \end{gathered}$$

Step 3. 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

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

Plug values

$$\begin{gathered} 120=4\pi {(4.15283)}^2 \times \frac{dr}{dt} \\ \frac{dr}{dt}=\frac{120}{4\pi {(4.15283)}^2} \\ \frac{dr}{dt}=0.5537 in/sec \end{gathered}$$