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Multivariable calculus: Length of curve

  1. Oct 26, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the length of the curve traced by the given vector function on the indicated interval.
    r(t)=<t, tcost, tsint> ; 0<t<pi


    2. Relevant equations

    s= ∫||r'(t)||dt

    3. The attempt at a solution

    r'(t) = <1, -tsint + cost, tcost + sint>
    s= ∫||r'(t)||dt

    ||r'(t)|| = sqrt(1^2+(-tsint+cost)^2+(tcost+sint)^2)
    ||r'(t)|| = sqrt(1+t^2sin^2t-2tsintcost+cos^2t+t^2cos^t+2tsintcost+sin^2t)
    ||r'(t)|| = sqrt(1 + t^2sin^2t+t^2cos^2t+sin^2t+cos^2t)

    I'm not sure where to go from here. I was thinking of using the trig identity sin^2x+cos^2x= 1, but I don't think I can do that.
     
  2. jcsd
  3. Oct 26, 2013 #2

    haruspex

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    Why not?
     
  4. Oct 26, 2013 #3
    I'd end up with sqrt(2+t^2), but then I don't know how to integrate that from 0 to pi. At first, I tried u-substitution, so u =2+t^2, du=2tdt, du/2=tdt, but there is no tdt outside the sqrt so that I can substitute.
     
  5. Oct 26, 2013 #4

    CAF123

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    Try integration by parts.
     
  6. Oct 26, 2013 #5
    :S isn't what I'm doing integration by parts? Or do you mean separate it into two integrals? I don't think you can do that with a sqrt sign
     
  7. Oct 26, 2013 #6

    CAF123

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    No, what you did before was substitution. Your integral is $$I= \int_0^\pi \sqrt{t^2 + 2}\,dt$$ Let ##u = \sqrt{t^2+ 2}## and ##dv = dt##. With one application of integration of parts, you will get a form that appears to not be any simpler: $$t \sqrt{t^2 + 2}|_0^\pi - \int_0^\pi t \frac{t}{\sqrt{t^2 + 2}}dt$$What I thought you could then do was use integ by parts again, but this only leads to the completely trivial statement ##I= I##
    I'll think over it some more - maybe a hyperbolic substitution would be useful.
     
  8. Oct 26, 2013 #7

    CAF123

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    Try the substitution t = √2 tan x.
     
  9. Oct 26, 2013 #8

    haruspex

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    I think √2 sinh x works out a little more easily.
     
  10. Oct 26, 2013 #9

    CAF123

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    Yes, I agree.
     
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