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Math Help - Rate of Change

  1. #1
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    Rate of Change

    Recall that the volume of a sphere of radius r is V=(4/3)*pi*r^3. We have a balloon which is in the shape of a sphere and is leaking air at the rate of 24 cubic centimeters per second. Find the derivative of the radius with respect to time when the radius is 7 centimeters.
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    We first restate all the information given in mathematical notation. We'll call the volume V and the radius r. We know, first of all, that

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

    To say that the balloon is leaking at a rate of 24 cubic centimeters per second is to say that

    \frac{dV}{dt}=-24.

    Substituting the formula above for V, we find that

    \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)=-24.

    Using the Chain Rule when needed, what does \frac{dr}{dt} turn out to be when r=7?
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  3. #3
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    4*pi*r^2?
    Then it would be 196*pi
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    We must remember that r is a function of t. Therefore, using the Chain Rule, we find that

    \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)=4\pi r^2\frac{dr}{dt}.

    Here, we see it applied as D_t\,V(r(t))=V'(r(t))r'(t).
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  5. #5
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    Quote Originally Posted by Velvet Love View Post
    Recall that the volume of a sphere of radius r is V=(4/3)*pi*r^3. We have a balloon which is in the shape of a sphere and is leaking air at the rate of 24 cubic centimeters per second. Find the derivative of the radius with respect to time when the radius is 7 centimeters.
    Another way using the chain rule:

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

    \frac{dV}{dt} = -24

    Use the chain rule

    \frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt}

    \frac{dr}{dt} = \frac{dV}{dt} \div \frac{dV}{dr}

    \frac{dr}{dt} = -24 \div 4\pi r^2 = -\frac{6}{\pi r^2}

    Sub in r = 7 to get \frac{dr}{dt} = -\frac{6}{49\pi}
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