Calculate the curvature and vectors T,N,B

[skrat]

Homework Statement

Let $\Gamma$ be trajectory which we got from $z=xy$ and $x^2+y^2=4$.

Calculate the curvature $\kappa$ and vectors T, N and B (B is perpendicular to T and N).

Homework Equations

The Attempt at a Solution

Well, the hardest part here is of curse to find a parametrization of $\Gamma$.
Any ideas how can I do that? I tried using polar coordinates $x=rcos\varphi$, $y=rsin\varphi$ and $z=z$

buuut, this is obviously not good...

 

[HallsofIvy]

From x^2+ y^2= 4, an obvious parameterization is x= 2cos(t), y= 2sin(t). Then, from z= xy, z= 4 sin(t)cos(t).

 

[skrat]

well... that is exactly what I did but some really nasty expressions lead to the result. I guess it's ok.

One question, If the cylinder would be moved along x axis, let's say $(x-2)^2+y^2=4$ would than the parameterization be $x=2cos(t)+2$, $y=2sin(t)$ and $z=xy$ ?

 

[skrat]

Also one more:
Catenary $z=acosh(x/a)$, $y=0$ and $x\in \left [ -a,a \right ]$

I tried with $x=a(1-t)$ that would mean that $t\in \left [ 0,2\right ]$ but I just have no idea what to do with z coordinate? :/

 

[haruspex]

Also one more:
Catenary $z=acosh(x/a)$, $y=0$ and $x\in \left [ -a,a \right ]$

I tried with $x=a(1-t)$ that would mean that $t\in \left [ 0,2\right ]$ but I just have no idea what to do with z coordinate? :/

Why not just x = at? Can you calculate the curvature for that?

 

[skrat]

Haha, good point... it can easily be just x=at.

Actually, I have to calculate the moment of inertia for this catenary $z=acosh(x/a), y=0 and$x \in \left [ -a,a \right ]## but I didn't open another topic because the main problem here is still the parametrization.

But if you're asking about the original (first) post in this topic, than yes. I was able to calculate the curvature and vectors T,N, and B.