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Math Help - differentiation

  1. #1
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    differentiation

    differentiate
    (X^2 + 200^2)^(1/2) multiplied by (1/4), and substitute X=200tanY

    and

    ((500-X)^2 + 200^2)^(1/2) multiplied by (1/3), and substitute
    X= -200tanZ+500

    then combine equations and show that the minimum time occurs when (sinY)/m = (sinZ)/n where m and n are constants
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  2. #2
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    Quote Originally Posted by asian5
    differentiate
    (X^2 + 200^2)^(1/2) multiplied by (1/4), and substitute X=200tanY

    and

    ((500-X)^2 + 200^2)^(1/2) multiplied by (1/3), and substitute
    X= -200tanZ+500

    then combine equations and show that the minimum time occurs when (sinY)/m = (sinZ)/n where m and n are constants
    So after making the substitution we have equations 1 and 2.

    #1) \frac{\sqrt{(200\tan(y)^2+200^2}}{4}

    #2) \frac{\sqrt{200\tan(z)^2+200^2}}{3} (The 500's cancel out.)

    Now a little bit of trig comes in.

    #1) \frac{1}{4} \times \sqrt{400\tan^2(y)+400}
    Factor out the 400, then take it out of the square-root function.

    \frac{20}{4} \times \sqrt{\tan^2(y)+1}

    Now make the trig substitution \tan^2(x)+1=\sec^2(x)

    This should simplify things a lot and use the same process for equation #2.
    Last edited by Jameson; August 9th 2006 at 12:14 PM.
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  3. #3
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    the other part of the question is tricky.. could you continue
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