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Math Help - checking rate of change answer

  1. #1
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    checking rate of change answer

    hello,


    the volume of a sphere is given by V =\frac{4}{3}\pi r^3. an elastic balloon is being blown up so that the radius is increasing at the rate of 1cm s^{-s} calculate the rate at which the volume of the baloon is increasing when the radius is 5 cm

    this is what i did

    \frac {dV}{dt} = 1cm s^{-s}
    find  \frac {dr}{dt}

    V = \frac{4}{3}\pi r^3 therefore \frac {dV}{dr}= 4\pi r^2

    aplying chain rule \frac {dV}{dt}= \frac {dV}{dr} .  <br />
\frac {dr}{dt}<br /> <br />

    1= 4\pi r^2 x \frac {dr}{dt}

    \frac {dr}{dt}= \frac {1}{4\pi r^2}

    when r = 5
    \frac {dr}{dt}= \frac {1}{4\pi 5^2}

    =0.0032 cm/sec
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  2. #2
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    Is \frac{\mathrm{d}V}{\mathrm{d}t} = 1 cm/s or is \frac{\mathrm{d}r}{\mathrm{d}t}? I think you have mixed up the radius and the volume.....
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  3. #3
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    seems like i have, can you show me the right way to work this problem.
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  4. #4
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    Quote Originally Posted by sigma1 View Post
    seems like i have, can you show me the right way to work this problem.
    Just substitute the correct things into the correct parts of your expression for dV/dt !
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