# Thread: rate of growth of a leaking oil tank

1. ## rate of growth of a leaking oil tank

A gas tank on a dock has a small puncture and is leaking gas at the rate of $1cm^3/min$ into a lake. It forms a circular slick that is 1mm thick on the surface of the water.

so as a function of time the amount of gas leaked is $V=1cm^3 * m$ but i also have to state the radius as a function of volume and as a function of time. im supposing i should approach the radius as volume as a very short cylinder, which i think i would be $r=\sqrt{\frac{\pi*h}{V}}$ and as a function of a time as $r=\sqrt{\frac{\pi*h}{1cm^3 * min}}$
this is my first time attempting a question like this and im not even sure its in the correct forum so any advice would be appreciated

2. Originally Posted by allywallyrus
A gas tank on a dock has a small puncture and is leaking gas at the rate of $1cm^3/min$ into a lake. It forms a circular slick that is 1mm thick on the surface of the water.

so as a function of time the amount of gas leaked is $V=1cm^3 * m$ but i also have to state the radius as a function of volume and as a function of time. im supposing i should approach the radius as volume as a very short cylinder, which i think i would be $r=\sqrt{\frac{\pi*h}{V}}$ and as a function of a time as $r=\sqrt{\frac{\pi*h}{1cm^3 * min}}$
For a cylinder of height h and radius r, $V= \pi r^2h$, then $r^2= \frac{V}{\pi h}$ so $r= \sqrt{\frac{V}{\pi h}}$. It looks to me like you have the fraction upside-down.

this is my first time attempting a question like this and im not even sure its in the correct forum so any advice would be appreciated
To find the rate at which r is increasing, differentiate $\frac{dr}{dt}= \sqrt{\frac{1}{\pi h}}V^{1/2}$ with respect to t. h is constant, h= 0.1 cm, and dV/dt= 1.

3. ah, so i wasnt that far off at least. thanks
also the question wasnt asking the rate that r was increasing, but just how to determine the radius after so much time had elapsed, so would i also need that formula?