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Calculate the arc length of the vector function

  1. Oct 11, 2013 #1
    1. The problem statement, all variables and given/known data
    Calculate the arc length of [itex]<2t,t^2,lnt>[/itex] from [itex]1=<t=<e[/itex]

    2. Relevant equations
    Arc length=[itex]∫√{(x')^2 + (y')^2 + (z')^2}[/itex]

    3. The attempt at a solution
    So I have gotten to this point:
    [itex]∫√{4 + 4t^2 + \frac{1}{t^2}}[/itex]

    Am I on the right track, and if so, how do I integrate that?
     
  2. jcsd
  3. Oct 11, 2013 #2

    Mark44

    Staff: Mentor

    Missing dt in the integrand.

    Also, if you use \sqrt instead of √, it will look better.
    $$\int \sqrt{x'^2 + y'^2 + z'^2}dt $$
    Write the quantity in the radical as 4t2 + 4 + 1/t2, and then factor it.
     
  4. Oct 13, 2013 #3
    Thanks. Good advice so far. I think I might just be an idiot by how do you factor that? Haven't had to do that for a while.
     
  5. Oct 13, 2013 #4
    Nevermind, got it. multiply t^2
     
  6. Oct 13, 2013 #5
    Thanks for the help!
     
  7. Oct 13, 2013 #6

    Mark44

    Staff: Mentor

    You can't do that, but you can multiply by t2 over itself. OTOH, the expression can be factored directly, to (2t + 1/t)2.
     
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