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Math Help - Simple related rates prob

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    Simple related rates prob

    Helium is pumped into a spherical balloon at 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 \frac{dV}{dt} , and that \frac{dV}{dt} = \frac {dV}{dr} * \frac {dr}{dt} .

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

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

    I know that the rate the volume increases with respect to time is 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 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  120 = 4PIr^2*2 or \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  r = sqrt(\frac{120}{8PI})
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    Re: Simple related rates prob

    We are looking for \frac{dr}{dt}, and we know:

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

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

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

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

    We are given:

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

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

    Can you put this together to finish?
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    Re: Simple related rates prob

    Quote Originally Posted by MarkFL2 View Post

    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  2\frac{ft^3}{s} = 186.8\frac{dr}{dt} which gave 0.01071 \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.

    Thanks much.
    Last edited by AZach; October 7th 2012 at 04:41 PM.
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    Re: Simple related rates prob

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

    assuming V(0) = 0 ...

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

    solve for r as a function of t
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    Re: Simple related rates prob

    Quote Originally Posted by skeeter View Post
    V = \frac{4\pi}{3} r^3

    assuming V(0) = 0 ...

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

    solve for r as a function of t
    Oh! I kept looking at dV/dr.
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