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Math Help - Rate of a decreasing spherical balloon

  1. #1
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    Rate of a decreasing spherical balloon



    I know:
    • Formula for a sphere's volume: V = \frac{4}{3}*\pi*r^3
    • Diameter of balloon is 20cm = radius of balloon is 10cm.
    • \frac{4}{3}*\pi*10^3 = 1333.3*\pi
    • Difference of 5cm in the radius is:  \frac{4}{3}*pi*5^3 = 166.67*\pi
    • 1333.3 - 166.67 = 1166.63
    • This is not a choice, so I am obviously wrong.
    • Why I think I am wrong: 1166.63*\pi cm^3 is not in the form of a rate. It needs to be in the form of  \frac{x cm^3}{y seconds}.
    • I can't think of how to do this.
    • Lastly, my practice test states that this question will see if I understand the historical development and application of calculus. I clearly did not use Calculus, and quite honestly, I'm don't know how.
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  2. #2
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    Quote Originally Posted by EMyk01 View Post


    I know:
    • Formula for a sphere's volume: V = \frac{4}{3}*\pi*r^3
    • Diameter of balloon is 20cm = radius of balloon is 10cm.
    • \frac{4}{3}*\pi*10^3 = 1333.3*\pi
    • Difference of 5cm in the radius is:  \frac{4}{3}*pi*5^3 = 166.67*\pi
    • 1333.3 - 166.67 = 1166.63
    • This is not a choice, so I am obviously wrong.
    • Why I think I am wrong: 1166.63*\pi cm^3 is not in the form of a rate. It needs to be in the form of  \frac{x cm^3}{y seconds}.
    • I can't think of how to do this.
    • Lastly, my practice test states that this question will see if I understand the historical development and application of calculus. I clearly did not use Calculus, and quite honestly, I'm don't know how.
    1. The rate of change of the volume wrt the time t is:

    \dfrac{dV}{dt}=\dfrac{d\left(\frac43 \pi r^3\right)}{dt}

    2. The rate of change of the radius is \dfrac{dr}{dt} = 5\ \frac{cm^3}{s}

    3. Calculate dV using the equation from 1. and the value of 2.

    4. The answer is D.
    Last edited by earboth; March 9th 2011 at 07:14 AM.
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