# Simple related rates prob

• Oct 7th 2012, 10:39 AM
AZach
Simple related rates prob
Helium is pumped into a spherical balloon at $\displaystyle 2 \frac{ft^3}{s}$. How fast is the radius increasing after 2 minutes?

I've searched some of the other related rates probs posted on this forum, so I apologize if a question like this has been brought up already.

Basically, I understand I'm looking for $\displaystyle \frac{dV}{dt}$ , and that $\displaystyle \frac{dV}{dt} = \frac {dV}{dr} * \frac {dr}{dt}$.

Now, the rate is 2 $\displaystyle \frac{ft^3}{s}$ and the volume of a sphere is $\displaystyle V = \frac{4}{3} * {PI}r^3$.

Implicitly differentiating V with respect to T gets $\displaystyle \frac {dV}{dt} = 4 * {PI}r^2 \frac{dr}{dt}$

I know that the rate the volume increases with respect to time is $\displaystyle 2 \frac{ft^3}{s}$, and I'm not sure what to do with my 2 minutes (or 120 seconds). Is the volume with respect to time 120 seconds and $\displaystyle 2 \frac{ft^3}{s}$ is the rate with respect to time? That would make more sense I think because the dimensions of cubic feet with a factor of 2 is like a rate and seconds is most definitely a part of time.

However, when I try to solve $\displaystyle 120 = 4PIr^2*2$ or $\displaystyle \frac{dV}{dt} = \frac {dV}{dr} * \frac {dr}{dt}$ and that's incorrect. Since I don't know the volume with respect to time, how can I properly go about solving for the rate with respect to time?

I get $\displaystyle r = sqrt(\frac{120}{8PI})$
• Oct 7th 2012, 10:59 AM
MarkFL
Re: Simple related rates prob
We are looking for $\displaystyle \frac{dr}{dt}$, and we know:

$\displaystyle dV=4\pi r^2\,dr$ hence:

$\displaystyle \frac{dV}{dt}=4\pi r^2\,\frac{dr}{dt}$

If we assume $\displaystyle V(0)=0$ then $\displaystyle V(t)=2t \text{ ft}^3$ and so:

$\displaystyle r(t)=\left(\frac{3t}{2\pi} \right)^{\frac{1}{3}}$

We are given:

$\displaystyle \frac{dV}{dt}=2\,\frac{\text{ft}^3}{\text{s}}$

$\displaystyle t=2\text{ min}=120\text{ s}$

Can you put this together to finish?
• Oct 7th 2012, 04:19 PM
AZach
Re: Simple related rates prob
Quote:

Originally Posted by MarkFL2

$\displaystyle r(t)=\left(\frac{3t}{2\pi} \right)^{\frac{1}{3}}$

Can you put this together to finish?

Do you mind if I ask where you derived that?

Well, I took the steps you laid out. since the problem asks for the rate of change with respect to time when t = 120 seconds, I plugged 120 into r(t) which == 3.86. Since we know the rate of change with respect to the volume we can plug that into dv/dr which gets 4pi(3.86)^2. And dV/dt is 2 cubic feet per second.

So the final equation came out to be $\displaystyle 2\frac{ft^3}{s} = 186.8\frac{dr}{dt}$ which gave 0.01071$\displaystyle \frac{ft^3}{s}$, and that means the growth of the balloon's volume has decreased significantly by the 2 minute mark. I'm just still unclear how you got the r(t) equation. (Shake)

Thanks much.
• Oct 7th 2012, 06:00 PM
skeeter
Re: Simple related rates prob
$\displaystyle V = \frac{4\pi}{3} r^3$

assuming $\displaystyle V(0) = 0$ ...

$\displaystyle 2t = \frac{4\pi}{3} r^3$

solve for $\displaystyle r$ as a function of $\displaystyle t$
• Oct 7th 2012, 08:21 PM
AZach
Re: Simple related rates prob
Quote:

Originally Posted by skeeter
$\displaystyle V = \frac{4\pi}{3} r^3$

assuming $\displaystyle V(0) = 0$ ...

$\displaystyle 2t = \frac{4\pi}{3} r^3$

solve for $\displaystyle r$ as a function of $\displaystyle t$

Oh! I kept looking at dV/dr.