Highly Annoying Related Rates Problem: Spherical Ballon

*This is just a tutorial on how to solve a problem like this*.

The problem is:

Quote:

A Spherical balloon is to be deflated so that its radius decreases at a constant rate of 10 cm/min. At what rate must air be removed when the radius is 8 cm?

Air must be removed at ________ cm^3/min

So right off the bat I know the following.

It's only annoying because I can't find a video tutorial on it.

**Equation**:

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

I bring it into it's derivative form.

(just so you know the threes cancel)

$\displaystyle

\frac{dV}{dt}=4\pi*3r^2\frac{dr}{dt}$

Given:

$\displaystyle \frac{dr}{dt} = \frac{-10cm}{1min}$

(note: Any one reading this and wondering why it's negative ten is because it is deflating)

My attempt:

$\displaystyle \frac{dV}{dt}=4\pi*(8cm)^{2}*\frac{-10cm}{1min}$

Just punch that into a caculator and you'll get the answer of:

$\displaystyle \frac{dV}{dt} \approx -8042.5 \frac{cm^{3}}{min}$

*This basically will work for any problem that's like this it's just a matter of changing the numbers.*