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Math Help - Trouble Calculating Length of a Curve Segment (Surface Integral?)

  1. #1
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    Trouble Calculating Length of a Curve Segment (Surface Integral?)

    So, I am trying to find the length of a segment of a curve that is given by an x equation and a y equation and a limit placed on the 3rd variable t. Sorry in advance for the lack of actual images. Apparently I can't use that extension here? The initial problem is:

    x = (e^t) + (e^-t) y = -2t + 5 0 <= t <= 3

    Since the equation for finding the length of a curve segment described like this is: { [ (dx/dt)^2 ] + [ (dy/dt)^2 ] }^(1/2)
    Using a pic from wolfram in the hopes of being a little clearer the formula is: http://www4a.wolframalpha.com/Calcul...=61&w=106&h=28

    Then I calculate:

    dx/dt = (e^t) - (e^-t)
    dy/dt = -2t + 5

    Which finally leads me to: { ( [ (e^t) - (e^-t) ]^2 ) + 4 }^(1/2) or http://www4a.wolframalpha.com/Calcul...=61&w=300&h=47

    Now, I know that the answer is clearly in that last image but really I'm not so concerned with that as I am with the actual steps to solve this problem. By "problem" I am referring to that final integral calculation. I am led to believe that that is just true similar to how \int (1/x) = ln x. Even that equality can be stretched out to include some steps though if I remember correctly. I am looking for the steps required for this problem. Anyone?
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  2. #2
    Behold, the power of SARDINES!
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    It is mostly algebra lets focus on the radicand to start

    (e^t-e^{-t})^2+(-2)^2=e^{2t}-2+e^{-2t}+4=e^{2t}+\frac{1}{e^{2t}}+2

    \frac{(e^{2t})^2+1+2e^{2t}}{e^{2t}}=\frac{(e^{2t})  ^2+2e^{2t}+1}{e^{2t}}=\frac{(e^{2t}+1)^2}{e^{2t}}

    Now if you take the square root we get

    \frac{e^{2t}+1}{e^{t}}=e^{t}+e^{-t}

    Now just integrate

    \int_{0}^{3} e^{t}+e^{-t} dt

    You should be able to finish from here.
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  3. #3
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    Thank you! I really appreciate this. Now, it's time for sleep.. I should be able to now that I can stop thinking about this
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  4. #4
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    You might find it simpler to write x= 2cosh(t), y= -2t+ 5 so that \frac{dx}{dt}= 2sinh(t), \frac{dy}{dt}= -2dt and then \sqrt{\left(\frac{dx}{dt}\right)^2+ \left(\frac{dy}{dt}\right)^2}= 2\sqrt{sinh^2(t)+ 1}= 2cosh(t)
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