# Question on rates of change problem

1. Jul 26, 2013

### krackers

1. The problem statement, all variables and given/known data

A spherical balloon is being inflated. Suppose the radius of the balloon is increasing at a rate of 2 centimeters per second.

a. Express the radius r of the balloon as a function of time t.
b. Express the volume V of the balloon as a function of time t

2. Relevant equations

$\frac{dr}{dt}\; =\; 2$

3. The attempt at a solution

Part A.

$\frac{dr}{dt}\; =\; 2$

Therefore, $dr=2\cdot dt$

$\int_{}^{}{}dr=\int_{}^{}{}2\cdot dt$

$r\left( t \right)=2t$

Part B.

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

$\frac{dV}{dt}\; =\; 4\pi r^{2}\cdot \frac{dr}{dt}$

Substituting dr/dt = 2 gives:

$\frac{dV}{dt}\; =\; 8\pi r^{2}$

Since $r=2t$

$\frac{dV}{dt}=32\pi \cdot t^{2}$

$dV=32\pi \cdot t^{2}\cdot dt$

$V=\frac{32\pi \cdot t^{3}}{3}$

2. Jul 26, 2013

### verty

Hint: what is the radius at time t = 0?

3. Jul 26, 2013

### krackers

That would also be zero, right?

4. Jul 26, 2013

### Zondrina

Your solution looks fine to me.

5. Jul 26, 2013

### 1MileCrash

I don't see anything wrong either, and you wrote your solution very nicely. What's your question, or were you just not confident that it was correct?

6. Jul 26, 2013

### krackers

I wasn't sure if that was correct. This was actually a precalc problem, so I was not sure if integration/differentiation was needed and whether my answer was correct.

7. Jul 26, 2013

### haruspex

You're right, it isn't. It's sufficiently obvious that the radius is simply 2t cm after t seconds, so V = 4πr3/3 = 4π(2t)3/3.

8. Jul 27, 2013

### verty

This second line does not follow from the first without the additional assumption, nowhere written and not given in the question, that f(0) = 0. The question says, express the radius as "a" function of time t. The answer is correct but what is written here is not. I'm picking out the literal fallacy here.

9. Jul 27, 2013

### 1MileCrash

Good point.

An '$r_{0}$' should appear and be defined.

10. Jul 27, 2013

### krackers

That's a lot simpler.

You're right - I always forget to add the constant after anti-differentiation.