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Arc Length of a circle

  • Thread starter Colts
  • Start date
  • #1
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Homework Statement


Find an arc length parametrization of the circle in the plane z=5 with radius 6 and center (4,1,5)


Homework Equations


||r'(t)||=r'(u)
s=integral r'(u)du


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?
 

Answers and Replies

  • #2
gabbagabbahey
Homework Helper
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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?
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.
 
  • #3
77
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Got it. Thank you
 

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