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Math Help - Calculus and balloon inflation rates

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    Calculus and balloon inflation rates

    Hey guys, I've been stuck on a problem for a while now and the more I think about it, the harder it becomes to wrap my head around it.

    It goes something like this,
    Air is pumped into a spherical balloon at a rate of 2 cm3/s (two cubic centimetres per
    second). Denote by V the volume of the balloon, r the radius of the balloon, and S the
    surface area of the balloon.

    (a) Find \frac{dr}{dt} when V = 30 cm3.

    (b) Find \frac{dS}{dt} when V = 30 cm3.

    You may need these formulas: V = \frac{4}{3}\pi r^3, S = 4\pi r^2.

    While I found a similar question, it dealt more with a change in radius and surface area, rather than volume. Hoping you guys can put me on the right track; I'm not sure what equation I need to find, or how to get there.

    Thanks!
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    Re: Calculus and balloon inflation rates

    Quote Originally Posted by bradbury17 View Post
    Hey guys, I've been stuck on a problem for a while now and the more I think about it, the harder it becomes to wrap my head around it.

    It goes something like this,[INDENT]Air is pumped into a spherical balloon at a rate of 2 cm3/s (two cubic centimetres per
    second). Denote by V the volume of the balloon, r the radius of the balloon, and S the
    surface area of the balloon.

    (a) Find \frac{dr}{dt} when V = 30 cm3.
    Well
    \frac{dV}{dt} = 2

    and
    \frac{dV}{dt} = \frac{dV}{dr} \frac{dr}{dt}

    So what does dr/dt equal?

    Quote Originally Posted by bradbury17 View Post
    (b) Find \frac{dS}{dt} when V = 30 cm3.
    Do this the same as the other. Let us know if you run into any problems.

    -Dan
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