Find the parameterization of a curve

[skrat]

Homework Statement

Find a parameterization of a curve which we get from $x^2+y^2+z^2=4$ and $x^2+y^2=2x$.

Homework Equations

The Attempt at a Solution

I hope this doesn the job, I am just not sure, so if anybody could check my result I would be really happy.

I started with $x=1+\cos \vartheta$ for $\vartheta \in \left [ 0,2\pi \right ]$ and $y=\sin \vartheta$

Than from $x^2+y^2+z^2=4$, z as function of $\vartheta$ is $z=\sqrt{2(1-\cos \vartheta )}$

Or not?

 

[BruceW]

yeah, looks good to me. nice work there. you've chosen the positive root for z. so that gives one of two possible curves. There is another curve, but since they say just to find a parameterization of a curve, I guess you don't need to write down both curves.

 

[skrat]

Thank you!

 

[Ray Vickson]

Homework Statement

Find a parameterization of a curve which we get from $x^2+y^2+z^2=4$ and $x^2+y^2=2x$.

Homework Equations

The Attempt at a Solution

I hope this doesn the job, I am just not sure, so if anybody could check my result I would be really happy.

I started with $x=1+\cos \vartheta$ for $\vartheta \in \left [ 0,2\pi \right ]$ and $y=\sin \vartheta$

Than from $x^2+y^2+z^2=4$ z as function of $\vartheta$ is $z=\sqrt{2(1-\cos \vartheta )}$

Or not?

Please use correct punctuation: I first read your statement $x^2+y^2+z^2=4$ z as $``x^2 + y^2 + z^2 = 4z ''$ (which is exactly how it is typeset) but it should be $x^2 + y^2 + z^2 = 4$, z ... .

 

[skrat]

Please use correct punctuation: I first read your statement $x^2+y^2+z^2=4$ z as $``x^2 + y^2 + z^2 = 4z ''$ (which is exactly how it is typeset) but it should be $x^2 + y^2 + z^2 = 4$, z ... .

Thanks, I've edited the first post!