Graduate Natural parametrization of a curve

[eva_92]

Hello,
I need the natural parametrization or a geodesic curve contained in the surface z=x^2+y^2, that goes through the origin, with x(s=0)=0, y(s=0)=0, dx/ds (s=0)=cos(a) and dy/ds(s=0)=sin(a), with "a" constant, expressed as a function of the arc length, i.e., I need r(s)=r(x(s),y(s)).
Thank you very much!
Eva

 

[anuttarasammyak]

Hi. The surface is made by rotating the graph of $z=x^2$ around z axis. The shortest curve between the Origin and Point $(X,Y,Z)$ on the surface is presented by parameter $\xi$ as $(X \xi,Y \xi,Z\xi^2)$ where $Z=X^2+Y^2,0<\xi<1$.

ds^2=X^2 d\xi^2 + Y^2 d\xi^2 + (X^2+Y^2)^2 4\xi^2 d\xi^2=(X^2 + Y^2)[ (X^2+Y^2)4\xi^2+1 ]d\xi^2
s(\xi)=\sqrt{X^2 + Y^2} \int_0^\xi \sqrt{ (X^2+Y^2)4\eta^2+1}\ d\eta
After calculating $s(\xi)$ you can get its inverse $\xi(s)$ and thus get $(X\xi(s),Y\xi(s),Z\xi(s)^2)$ as geodesic from Origin to end point (X,Y,Z) expressed by parameter s, the length from Origin.

 

[eva_92]

Hello, thank you for the answer. I have tried with the parametrization x=rcos(a), y=rsin(a), z=r^2, reaching to a similar expression: ds^2=1+4*r^2 (similar to yours with X=cos(a), Y=sin(a)). The solution is s = (1/4)*ln⁡(2*r+sqrt(1+4*r^2))+(1/2)*r*sqrt(1+4*r^2), which is an equation in which I can not calculate r as a function of s to perform the natural parametrization. This is the reason of my question, maybe another parametrization could work, buy it does not happen to me. Could you help me with this?

 

[anuttarasammyak]

You show points on geodesic
(r \cos a, r \sin a, r^2 )
where $0<r<R$ corresponding to the end point and
s = \frac{1}{4}ln⁡(2r+\sqrt{1+4r^2})+\frac{1}{2}r\sqrt{1+4r^2}
I would like to know what more you want.

 

[eva_92]

The idea is to write x,y,x as a function of s, i.e., x(s),y(s),z(s). If I can calculate r(s), then I can write x(r(s)),y(r(s)),z(r(s)), but I can't calculate r(s) from that equation

 

[anuttarasammyak]

I am afraid it is a tough thing to get the formula of r(s) from s(r) formula.