Calculating Arc Length of a Circle with Radius 6 and Center (4,1,5)

[Colts]

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?

 

[gabbagabbahey]

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$z=5$ plane, so $(x-4)^2+(y-1)^2+(5-5)^2=(x-4)^2+(y-1)^2=6^2$ is your circle. Now, remember that a circle in the $xy$-plane of radius $R$ can be parametrized as $x=R\cos\theta$, $y=R\sin\theta$, for $0\leq \theta \leq 2\pi$. Try applying that to your circle by shifting $x$ and $y$ by an appropriate amount.

 

[Colts]

Got it. Thank you