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Math Help - Arclength

  1. #1
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    Arclength

    Last question of the day. Been working on math for about 8 hours now....

    Consider the curve r=(e^(2t)*cos(−2t),e^(2t)*sin(−2t),e^(2t)).
    Compute the arclength function s(t): (with initial point t=0).

    Now the derivatives are

    r'(t)= (2e^2t[cos(2t)-sin(2t)] , -2e^(2t)[sin(2t)+cos(2t)], 2e^2t)

    I know that s(t)= Integral from 0 to t of |r'(t)|

    Can someone show me the work of finding |r'(t)|. It should be simple but no matter how many times I work it, I keep getting the wrong answers.

    |r'(t)|= sqrt(4e^(4t)*(1-sin(4t))+4e^(4t)*(1+sin(4t))+4e^(4t)

    This is the simplified version, don't know if it's anywhere near correct, because I get a different answer every time I work the problem, and it's always the wrong answer. :\ Getting frustrated.
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  2. #2
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    Re: Arclength

    I think you're using the wrong formula.

    If you are wanting to find the arclength of a function \displaystyle \mathbf{r}(t) = \left(x(t), y(t), z(t)\right)

    the arclength is given by

    \displaystyle s = \int_a^b{\sqrt{\left[x'(t)\right]^2 + \left[y'(t)\right]^2 + \left[z'(t)\right]^2}\,dt}
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  3. #3
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    Re: Arclength

    Yea....That could be the problem. I think I've been staring at this math book way to long today. Thanks for the help. I guess this goes along with the problem I was having trouble with a few minutes ago....Apparently, 4^2 is 14....I need some sleep. :|
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