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Math Help - Please Check: Rate of Change problem

  1. #1
    Junior Member
    Joined
    Nov 2008
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    41

    Please Check: Rate of Change problem

    For Part B of the attached question;

    Terms: E = Change d = Partial derivative D = Derivative

    a = sqrt (h^2 - b^2) = (h^2 - b^2) ^1/2

    Ea = da/dh . Eh + da/db . Eb

    Da/Dt = da/dh . Da/Dt + da/db . Da/Dt

    Therefore;

    da/dh = 1/2 * (h^2 - b^2) ^ -1/2 . (2h)

    da/db = 1/2 * (h^2 - b^2) ^ -1/2 . (-2b)


    i.e:

    Da/Dt = h / sqrt (h^2 - b^2) * Dh/Dt - b / sqrt (h^2 - b^2) * Db/Dt

    Values are:

    h = 5cm Dh/Dt = (+3 cm/s)
    b = 3cm Db/Dt = (-2 cm/s)

    Da/Dt = 5 / sqrt (5^2 - 3^2) * 3 - 3 / sqrt (5^2 - 3^2) * (-2)


    Da/Dt = 3.75 -(-1.5) = 5.25 cm/s
    Attached Thumbnails Attached Thumbnails Please Check: Rate of Change problem-capture.jpg  
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  2. #2
    Senior Member
    Joined
    Jan 2010
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    354
    Your answer is correct, but it seems you took a very convoluted path to get to the answer. Here is a much simpler method:

    You are given:

    \frac{dh}{dt}=3

    \frac{db}{dt}=-2

    h=5

    b=3

    a=\sqrt{h^2-b^2} = 4

    So, differentiate with respect to t:

    h^2=a^2+b^2

    \implies 2h \frac{dh}{dt} = 2a \frac{da}{dt} + 2b  \frac{db}{dt}

    Then substitute in all known values and you are left with your solution.
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  3. #3
    Junior Member
    Joined
    Nov 2008
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    thats a great help. many thanks
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