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What I am doing wrong? arc length of parametric functions

  1. Mar 22, 2008 #1
    So this seems to be a pretty straightforward question but I keep getting the arc length to be 0 and I redid this question many times..

    Find the length of the parametrized curve given by

    x(t) =[tex]t^{2}-8t + 24[/tex]
    y(t) =[tex]t^{2}-8t -7 [/tex]

    How many units of distance are covered by the point P(t) =(x(t), y(t)) between t=0, and t =8?

    So my first step of course is to find dx/dt and dy/dt

    Then set up the arc length equation
    arc length = [tex]\int^{8}_{0}\sqrt{{\frac{dx}{dt}}^2+{\frac{dy}{dt}}^2}dt[/tex]
    = [tex]\int^{8}_{0}\sqrt{{(2t-8)}^2+{(2t-8)}^2}dt[/tex]
    = [tex]\int^{8}_{0}\sqrt{2{(2t-8)}^2}dt[/tex]
    = [tex]\int^{8}_{0}\sqrt{2}(2t-8)dt[/tex]

    Which give me an answer of zero when the answer is suppose to be 45.2548.

    What I am doing wrong? Thanks.
  2. jcsd
  3. Mar 22, 2008 #2


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    Science Advisor

    Have you forgotten that [itex]\sqrt{x^2}= |x|[/itex]?
  4. Mar 22, 2008 #3
    But it still give me zero though. I don't see how it changes things. Could you elaborate a bit more? Thanks
  5. Mar 22, 2008 #4
    you need [tex]\sqrt{2}[/tex][tex]\int[/tex][tex]^{8}_{0}[/tex]|2t-8|dt. to integrate this, notice that from 0 to 4, 2t - 8 is the negative of |2t - 8|. so we must multiply it by -1 to get 8 - 2t. from 4 to 8 2t - 8 = |2t - 8|, so w,e can leave it as is. so you must seperate this into two integrals, the 8 - 2t from 0 to 4, and 2t - 8 from 4 to 8, then add the results.
    Last edited: Mar 22, 2008
  6. Mar 22, 2008 #5
    Oh I see. Thanks so much
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