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

    this is what i did

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

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

    aplying chain rule $\displaystyle \frac {dV}{dt}$= $\displaystyle \frac {dV}{dr}$ . $\displaystyle
    \frac {dr}{dt}

    $

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

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

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

    =0.0032 cm/sec
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  2. #2
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    Is $\displaystyle \frac{\mathrm{d}V}{\mathrm{d}t} = 1$ cm/s or is $\displaystyle \frac{\mathrm{d}r}{\mathrm{d}t}$? 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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