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Help me find the arc length of a parametric equation...

  1. May 1, 2017 #1
    1. The problem statement, all variables and given/k
    nown data

    [itex] x = (sin(t))^2 [/itex] [itex] y = (cos(t))^2 [/itex] t goes from 0 to 3 pi

    2. Relevant equations
    ∫[itex] \sqrt{ {(dx/dt)}^2 + {(dy/dt)}^2 } dt[/itex]

    3. The attempt at a solution
    ∫[itex] \sqrt{ {(2sin(t)cost)}^2 + {(-2cos(t)sin(t))}^2 } dt[/itex]

    ∫[itex] \sqrt{ 4(sin(t)cos(t))^2 + {(-2cos(t)sin(t))}^2 } dt[/itex]

    ∫[itex] \sqrt{ 4(sin(t)cos(t))^2 + 4(cos(t)sin(t))^2 } dt[/itex]

    ∫[itex] \sqrt{ 8(sin(t)cos(t))^2 } dt[/itex]

    [itex]2\sqrt 2 [/itex]∫[itex] \sqrt{ (sin(t)cos(t))^2 } dt[/itex]

    [itex]2\sqrt 2 [/itex]∫[itex] { (sin(t)cos(t)) } dt[/itex]

    keep in mind my bounds are from 0 to 3 pi right now

    so I integrate this using u = sin (t) du = cos(t) dt
    and get
    (root 2)(sin(t))^2 from 0 to pi/3

    now when I evaluate I get 0-0

    where did my integration go wrong?
     
  2. jcsd
  3. May 1, 2017 #2

    Ray Vickson

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    Never, never just write ##\sqrt{X^2} = X## because half the time it is wrong. What IS always correct is ##\sqrt{X^2} = |X|.##
     
  4. May 1, 2017 #3
    so it should be absolute value sin(t)cos(t) ??
     
  5. May 1, 2017 #4

    Ray Vickson

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    I thought that was what I said.
     
  6. May 2, 2017 #5
    So where do I go from here? can I just call that the absolute value of 2sin since sin and cos have the same range?
     
  7. May 2, 2017 #6

    Ray Vickson

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    I have already offered all the help I feel I can according to PF rules.
     
  8. May 2, 2017 #7
    Thanks for the help.
     
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