Chapter 1.4, Problem 31E

Solution

To find: The equation that models the volume $V$ of the balloon at time $t$ and find the volume when $t=300$ seconds.

The equation that models the volume $V$ of the balloon at time $t$ is equal to $V=\frac{4}{3}\pi {\left(0.03t+48\right)}^3$ and the volume when $t=300$ seconds is equal to $751114$ cube inches..

Given information:

A high-altitude spherical weather balloon expands as it rises due to the drop in atmospheric pressure. Suppose that the radius $r$ increases at the rate of $0.03$ inches per second and that $r=48$ inches at time $t=0$ .

Calculation:

Since the rate of increase of radius with respect to of $0.03$ inches per second, so it can be expressed below:

  $\frac{dr}{dt}=0.03$

Rewrite the above equation:

  $dr=0.03dt$

On integration both sides:

  $\begin{array}{c}{\displaystyle \int dr}={\displaystyle \int 0.03}dt\\ r=0.03t+C\end{array}$

Substitute $t=0,r=48$ and calculate the value of $C$ :

  $\begin{array}{c}48=0.03\left(0\right)+C\\ C=48\end{array}$

Therefore, the radius function is:

  $r=0.03t+48$

Formula used:

The volume of the sphere is given by

  $V=\frac{4}{3}\pi r^3$

Apply the above formula and calculate the volume of spherical balloon below:

  $V=\frac{4}{3}\pi {\left(0.03t+48\right)}^3$

Now, Substitute $t=280$ and calculate the volume $V$ :

  $\begin{array}{c}V=\frac{4}{3}\pi {\left(0.03\left(280\right)+48\right)}^3\\ =\frac{4}{3}\pi {\left(8.4+48\right)}^3\\ =\frac{4}{3}\pi {\left(56.4\right)}^3\\ \approx 751114\end{array}$

Hence, the volume is $751114$ cube inches.