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Math Help - rate of growth of a leaking oil tank

  1. #1
    Newbie allywallyrus's Avatar
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    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
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  2. #2
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    Quote Originally Posted by allywallyrus View Post
    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.
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  3. #3
    Newbie allywallyrus's Avatar
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    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?
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