Question

Gas is escaping from a spherical balloon at the rate of 12 ft3/hr. At what rate (in feet per hour) is the radius of the balloon changing when the volume is 300 ft3?

Answer 1

Answer:

This is the rate at which the radius of the balloon is changing when the volume is 300 $ft^3$ $\frac{dr}{dt}=-\frac{3}{225^{\frac{2}{3}}\pi ^{\frac{1}{3}}} \:\frac{ft}{h} \approx -0.05537 \:\frac{ft}{h}$

Step-by-step explanation:

Let $r$ be the radius and $V$ the volume.

We know that the gas is escaping from a spherical balloon at the rate of $\frac{dV}{dt}=-12\:\frac{ft^3}{h}$ because the volume is decreasing, and we want to find $\frac{dr}{dt}$

The two variables are related by the equation

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

taking the derivative of the equation, we get

$\frac{d}{dt}V=\frac{d}{dt}(\frac{4}{3}\pi r^3)\\\\\frac{dV}{dt}=\frac{4}{3}\pi (3r^2)\frac{dr}{dt} \\\\\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}$

With the help of the formula for the volume of a sphere and the information given, we find $r$  

$V=\frac{4}{3}\pi r^3\\\\300=\frac{4}{3}\pi r^3\\\\r^3=\frac{225}{\pi }\\\\r=\sqrt[3]{\frac{225}{\pi }}$

Substitute the values we know and solve for $\frac{dr}{dt}$

$\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}\\\\\frac{dr}{dt}=\frac{\frac{dV}{dt}}{4\pi r^2} \\\\\frac{dr}{dt}=-\frac{12}{4\pi (\sqrt[3]{\frac{225}{\pi }})^2} \\\\\frac{dr}{dt}=-\frac{3}{\pi \left(\sqrt[3]{\frac{225}{\pi }}\right)^2}\\\\\frac{dr}{dt}=-\frac{3}{\pi \frac{225^{\frac{2}{3}}}{\pi ^{\frac{2}{3}}}}\\\\\frac{dr}{dt}=-\frac{3}{225^{\frac{2}{3}}\pi ^{\frac{1}{3}}} \approx -0.05537 \:\frac{ft}{h}$