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Math Help - Highly Annoying Related Rates Problem: Spherical Ballon

  1. #1
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    Highly Annoying Related Rates Problem: Spherical Ballon

    This is just a tutorial on how to solve a problem like this.

    The problem is:
    A Spherical balloon is to be deflated so that its radius decreases at a constant rate of 10 cm/min. At what rate must air be removed when the radius is 8 cm?

    Air must be removed at ________ cm^3/min
    So right off the bat I know the following.

    It's only annoying because I can't find a video tutorial on it.

    Equation:

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

    I bring it into it's derivative form.

    (just so you know the threes cancel)
    <br />
\frac{dV}{dt}=4\pi*3r^2\frac{dr}{dt}

    Given:


    \frac{dr}{dt} = \frac{-10cm}{1min}

    (note: Any one reading this and wondering why it's negative ten is because it is deflating)

    My attempt:


    \frac{dV}{dt}=4\pi*(8cm)^{2}*\frac{-10cm}{1min}

    Just punch that into a caculator and you'll get the answer of:

    \frac{dV}{dt} \approx -8042.5 \frac{cm^{3}}{min}

    This basically will work for any problem that's like this it's just a matter of changing the numbers.
    Last edited by Zanderist; March 21st 2010 at 12:04 AM. Reason: This is now considered to be the correct answer.
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  2. #2
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    Quote Originally Posted by Zanderist View Post
    The problem is:
    So right off the bat I know the following.

    It's only annoying because I can't find a video tutorial on it.

    Equation:

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

    I bring it into it's derivative form.

    (just so you know the threes cancel)
    <br />
\frac{dV}{dt}=4\pi*3r^2\frac{dr}{dt}

    Given:

    \frac{dr}{dt} = \frac{-10cm}{1min}

    (note: Any one reading this and wondering why it's negative ten is because it is deflating)

    My Failed attempt:

    \frac{dV}{dt}=4\pi*(8cm)^{2}*\frac{-10cm}{1min}

    Just punch that into a caculator and you'll get the wrong answer of:

    \frac{dV}{dt} \approx -8042.5 \frac{cm^{3}}{min}

    Everything is right up until the attempt, that rate is just ridiculous.
    I got the same answer as you did.
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  3. #3
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    Well I would like the record to show that this problem is in fact right.

    My evidence:

    Calculus: related rates, spherical balloon, formula for the volume of a sphere

    So this one is for archives.

    Just one question on theory:

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

    Why doesn't \frac{4\pi}{3} turn out to be 0 or at least something else?
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