# Highly Annoying Related Rates Problem: Spherical Ballon

• Mar 20th 2010, 05:20 PM
Zanderist
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.
• Mar 20th 2010, 05:44 PM
ione
Quote:

Originally Posted by Zanderist
The problem is:
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 Failed attempt:

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

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

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

Everything is right up until the attempt, that rate is just ridiculous.

I got the same answer as you did.
• Mar 20th 2010, 11:01 PM
Zanderist
Well I would like the record to show that this problem is in fact right.

My evidence:

Calculus: related rates, spherical balloon, formula for the volume of a sphere

So this one is for archives.

Just one question on theory:

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

Why doesn't $\displaystyle \frac{4\pi}{3}$ turn out to be 0 or at least something else?