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Homework Help: Arc length of a curve (trigonometric identity)

  1. Sep 28, 2008 #1
    1. The problem statement, all variables and given/known data

    find arc length of the segment of the 2space curbe that is defined by the parametric equations

    x(t) = t-sin(t)
    y(t) = 1+cos(t)
    0 ≤ t ≤ 4π

    3. The attempt at a solution

    I've found dx/dt and dy/dt respectively and put them into the arc length equation, i.e. sqrt[(dx/dt)²+(dy/dt)²]

    dx/dt = 1-cost
    dy/dt = -sint

    therefore arc length L = sqrt[(1-cost)²+(-sint)²]
    this leads to L = sqrt[1-2cost+cos²t+sin²t]

    I am then told to use the double angle formula 2sin²t = 1-cos2t to simplify the integrand. I cannot see how this applies though!
    If someone could point me in the right direction o:)
  2. jcsd
  3. Sep 28, 2008 #2


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    Gold Member

    If cos 2t = 1-2sin²t, then cos t = 1-2sin²(t/2).

    Sub that into the integrand and it should make it easier, and use the pythagorean identity to reduce sin²t+cos²t to 1.
  4. Sep 29, 2008 #3
    so I used that and trig identity and got sqrt[4sin²(t/2)]

    I am given a second hint that says sqrt(sin²t) = |sint|,
    my question now is when I integrate |sin²(t/2)|, does anything change or will I perform integration as normal?
  5. Sep 29, 2008 #4


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    Gold Member

    I would think that you'd need to integrate over different parts of the interval separately.

    For 0 < t < 2π, sin(t/2) is positive, so that |sin(t/2)|=sin(t/2)

    However for 2π < t < 4π, sin(t/2) is negative so that |sin(t/2)|=-sin(t/2)
  6. Sep 29, 2008 #5
    ah yes, thank you! I went on to find an arc length of 16.
    Last edited: Sep 29, 2008
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