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Complicated Arclength Problem

  1. Dec 16, 2006 #1

    JJ6

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    1. The problem statement, all variables and given/known data

    Find the arclength of the curve given by y= integral from -pi/2 to x of sqrt(cost)dt. X is restricted between -pi/2 and pi/2.

    2. Relevant equations

    L = Integral from a to b of sqrt((dy/dx)^2 + 1)dx
    L = Integral from a to b of sqrt((dy/dt)^2 + (dx/dt)^2)dt

    3. The attempt at a solution

    I'm not even sure how to start this problem since integral of sqrt(cost) has no simple function representing it. Can somebody please give me some direction?
     
  2. jcsd
  3. Dec 16, 2006 #2

    quasar987

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    It is an application of the fondamental theorem of calculus.

    [tex]\frac{d}{dx}\int_{x_0}^xf(t)dt=f(x)[/tex]
     
  4. Dec 16, 2006 #3

    JJ6

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    So does that mean that d/dx of the integral from -pi/2 to x of sqrt(cost)dt = sqrt(cosx)?

    Does the -pi/2 just disappear because it is a constant?
     
    Last edited: Dec 16, 2006
  5. Dec 16, 2006 #4

    quasar987

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    Yep.

    Disapear from where?
     
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