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Math Help - related rates: help

  1. #1
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    related rates: help

    Couple word problems I'm having a little trouble getting.

    If a snowball melts so that its surface area decreases at a rate of 1 cm^2/min , find the rate at which the diameter decreases when the diameter is 10cm.

    and

    Brain weight B as a function of body weight W in fish has been modeled by the power function B = 0.002w^(2/3), where B and W are meased in grams. A model for the body, weight, as a function of body lenght L (measured in centimeters) is w = o.12L^2.53. If over 10 million years, the average length of a certain species of fish evolved from 15cm to 20cm at a constant rate, how fast was this species' brain growing when the average length was 18 cm?

    Whew, that problem kicks the heck outta me lol. Thanks for any help
    Last edited by ThePerfectHacker; March 15th 2007 at 12:45 PM.
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  2. #2
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    Quote Originally Posted by UMStudent View Post
    If a snowball melts so that its surface area decreases at a rate of 1 cm^2/min , find the rate at which the diameter decreases when the diameter is 10cm.
    I'll do a Physics thing:
    Assume a spherical snowball. Then the surface area of the snowball is
    SA = 4(pi)r^2 = (pi)x^2 (where x is the diameter)
    (Note: I'm avoiding d as the diameter because otherwise I'd be taking a "dd/dt" which is confusing!)

    So
    d(SA)/dt = 2(pi)x * dx/dt

    1 cm^2/min = 2(pi)(10 cm) * dx/dt

    Now solve for dx/dt.

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by UMStudent View Post
    Brain weight B as a function of body weight W in fish has been modeled by the power function B = 0.002w^(2/3), where B and W are meased in grams. A model for the body, weight, as a function of body lenght L (measured in centimeters) is w = o.12L^2.53. If over 10 million years, the average length of a certain species of fish evolved from 15cm to 20cm at a constant rate, how fast was this species' brain growing when the average length was 18 cm?
    B = 0.002w^(2/3)
    w = 0.12L^(2.53)

    Thus
    B = 0.002[0.12L^(2.53)]^(2/3)

    B = [0.002*0.12^(2/3)]*L^(2.53*2/3)

    B = 0.000486576*L^1.68667

    dB/dt = (0.000486576*1.68667)*L^(1.68667 - 1)*dL/dt

    dB/dt = 0.000820692*L^0.68667*dL/dt

    Now, we know that the rate of change of length is constant, so
    dL/dt = (20 cm - 15 cm)/10000000 yr = 5 x 10^(-7) cm/yr is constant
    and L = 18 cm

    Thus:
    dB/dt = 0.000820692*(18)^0.68667*(5 x 10^(-7)) = 2.9861 x 10^(-9) g/yr

    -Dan
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