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Math Help - Rate of change problem.

  1. #1
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    Rate of change problem.

    At a given instant, the radii of two concentric circles aree 8cm and 12cm. The radius of the outer circle is increasing at a rate of  1cms^{-1}

    and the radius of the inner circle is increasing at a rate of  2cms^{-1} . Find the rate of change of the area enclosed by the two circles.

      r = 12, \frac{dr}{dt} = 1

     r = 8 \frac{dr}{dt} = 2

     \frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt}

     A = \pi r^{2}

     \frac{dA}{dr} = 2\pi r

    is this all correct? So I would just add the two da/dt for each circle?

    I get  56\pi

    the correct answer is  8\pi

    Can someone tell me where I am going wrong? Thanks.
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  2. #2
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    Hey,

    First write down the equation for the enclosed area:

    A=pi*(r_1^2-r_2^2)

    Then take derivative with respect to t, you'll get:

    dA/dt = 2*pi*(r1*dr1/dt-r2*dr2/dt)

    Then sub in the values to get dA/dt:

    dA/dt = 2*pi*(12*1 - 8*2) = -8*pi

    So the rate of change is 8*pi (negative because the enclosed area is decreasing)
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  3. #3
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    Quote Originally Posted by arash View Post
    Hey,

    First write down the equation for the enclosed area:

    A=pi*(r_1^2-r_2^2)

    Then take derivative with respect to t, you'll get:

    dA/dt = 2*pi*(r1*dr1/dt-r2*dr2/dt)

    Then sub in the values to get dA/dt:

    dA/dt = 2*pi*(12*1 - 8*2) = -8*pi

    So the rate of change is 8*pi (negative because the enclosed area is decreasing)

    so thats  2\pi ( r^{1} \frac{dr^{1}}{dt} - r^{2} \frac{dr^{2}}{dt} )

    where did the 2pi come from?
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  4. #4
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    2*Pi comes from differentiating pi*(r1^2-r2^2).

    A=pi*(r1^2-r2^2) = pi*r1^2 - pi*r2^2

    dA/dt = 2*pi*r1*dr1/dt - 2*pi*r2*dr2/dt = 2*pi*(r1*dr1/dr - r2*dr2/dt)
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  5. #5
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    Quote Originally Posted by arash View Post
    2*Pi comes from differentiating pi*(r1^2-r2^2).

    A=pi*(r1^2-r2^2) = pi*r1^2 - pi*r2^2

    dA/dt = 2*pi*r1*dr1/dt - 2*pi*r2*dr2/dt = 2*pi*(r1*dr1/dr - r2*dr2/dt)

    Yes sorry I realised after I had posted,

    thanks for your help.
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