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Integrating a square root function

  1. Oct 12, 2015 #1
    Problem statement:
    Find the arc length of the curve defined by x = √t and y = 3t -1 on the interval 0 < t < 1


    attempted solution:
    dx/dt = 1/2t-1/2 , dy/dt = 3 and dx = dt/ 2√t
    dy/dx = 6√t
    length = ∫01 √(1 + (6√t)2) .dt/ 2√t
    = ∫01 √1 + 36t) dt/2√t
    now I'm stuck with a product that is very difficult to solve using integration by parts or by trig substitution. Please shed some light.
     
    Last edited by a moderator: Oct 12, 2015
  2. jcsd
  3. Oct 12, 2015 #2

    Ray Vickson

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    Science Advisor
    Homework Helper

    If I was doing the problem I would first perform a simple change-of-variables to make the problem easier. However, PF rules forbid me from saying more, except to suggest you look in your textbook and/or course notes (or on-line) to find similar integrals.

    And: please stop using bold fonts in your posts. It is annoying and a bit intimidating.
     
  4. Oct 12, 2015 #3

    Svein

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    Looking at the equation. I can see that y = 3x2-1. Furthermore t=0⇒x=0 and t=1⇒x=1.
     
  5. Oct 12, 2015 #4

    Mark44

    Staff: Mentor

    Fixed in original post...
    The excess bolding comes from deleting the template boilerplate, but missing the bold tags.
     
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