1. ## Rate of Change

Recall that the volume of a sphere of radius r is V=(4/3)*pi*r^3. We have a balloon which is in the shape of a sphere and is leaking air at the rate of 24 cubic centimeters per second. Find the derivative of the radius with respect to time when the radius is 7 centimeters.

2. We first restate all the information given in mathematical notation. We'll call the volume $V$ and the radius $r$. We know, first of all, that

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

To say that the balloon is leaking at a rate of 24 cubic centimeters per second is to say that

$\frac{dV}{dt}=-24.$

Substituting the formula above for $V$, we find that

$\frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)=-24.$

Using the Chain Rule when needed, what does $\frac{dr}{dt}$ turn out to be when $r=7$?

3. 4*pi*r^2?
Then it would be 196*pi

4. We must remember that $r$ is a function of $t$. Therefore, using the Chain Rule, we find that

$\frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)=4\pi r^2\frac{dr}{dt}.$

Here, we see it applied as $D_t\,V(r(t))=V'(r(t))r'(t)$.

5. Originally Posted by Velvet Love
Recall that the volume of a sphere of radius r is V=(4/3)*pi*r^3. We have a balloon which is in the shape of a sphere and is leaking air at the rate of 24 cubic centimeters per second. Find the derivative of the radius with respect to time when the radius is 7 centimeters.
Another way using the chain rule:

$V = \frac{4}{3}\pi r^3 \: \rightarrow \: \frac{dV}{dr} = 4\pi r^2$

$\frac{dV}{dt} = -24$

Use the chain rule

$\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt}$

$\frac{dr}{dt} = \frac{dV}{dt} \div \frac{dV}{dr}$

$\frac{dr}{dt} = -24 \div 4\pi r^2 = -\frac{6}{\pi r^2}$

Sub in r = 7 to get $\frac{dr}{dt} = -\frac{6}{49\pi}$