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Arc Length of a circle

  1. Sep 25, 2012 #1
    1. The problem statement, all variables and given/known data
    Find an arc length parametrization of the circle in the plane z=5 with radius 6 and center (4,1,5)

    2. Relevant equations
    s=integral r'(u)du

    3. The attempt at a solution
    I get the equation of the circle to be (x-4)^2+(y-1)^2+(z-5)^2=6^2
    Not sure where to go from here. Make x=t and then solve for y and z? That seems like to mcuh work for this problem. What am I missing?
  2. jcsd
  3. Sep 25, 2012 #2


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    Homework Helper
    Gold Member

    That's a spherical surface, not a circle. You know that the circle lies in the[itex]z=5[/itex] plane, so [itex](x-4)^2+(y-1)^2+(5-5)^2=(x-4)^2+(y-1)^2=6^2[/itex] is your circle. Now, remember that a circle in the [itex]xy[/itex]-plane of radius [itex]R[/itex] can be parametrized as [itex]x=R\cos\theta[/itex], [itex]y=R\sin\theta[/itex], for [itex]0\leq \theta \leq 2\pi[/itex]. Try applying that to your circle by shifting [itex]x[/itex] and [itex]y[/itex] by an appropriate amount.
  4. Sep 25, 2012 #3
    Got it. Thank you
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