Geodesic curvature of a line on a surface Problem:
Find the geodesic curvature of the line $t = s ^ 2$ on the surface $$r = p(s) + t  p'(s)$$ where $s$ is a natural parameter.
P.S.
I know how to find geodesic curvature for non natural parameter, but I do not understand how to use this formulas in this case. Any close example? Or any formulas and how to use them.
Here is formula $$k = (1/|r'|^3)*(r'',r',n)$$ where n - normal vector.(Maybe it will help)
 A: Hint 1: The geodesic curvature is the component of acceleration  in the tangent plane of the surface for  a particle travelling at unit speed along the curve.
Hint 2: Derivative of curve with Arc lenght parameterization gives unit tangent.
Arc length is given as $S(t) = \int_{t=0}^t  \sqrt{v(t) \cdot v(t)} dt$, meaning $ S'(t) = \sqrt{v(t) \cdot v(t)}$ . If t is arc length, then $S'$ means derivative of arc length with respect to arc length which is one. This makes $\sqrt{  v \cdot v} = 1$ meaning $|v|=1$

For understanding the calculation , let's take $p(s)=  \left( x(s) , y(s) , z(s) \right)$, then:
$$ r(s,t) = \left( x(s) + t x'(s) , y(s) + t y'(s) , z(s) + t z'(s)  \right)$$
To find the equation of curve on the surface, we replace plug in the relation of input variable in domain:
$$ F(s)= = r(s,s^2)  = \left( x(s) + s^2 x'(s) , y(s) + s^2 y'(s) , z(s) + s^2 z(s) \right)= p(s) + s^2 p'(s)$$
We can now differentiate:
$$ \frac{dF}{ds} =(1+2s) p'(s) + s^2 p''(s) = (1+2s) (x'(s) , y'(s) , z'(s) )  + s^2 ( x''(s), y''(s) , z''(s) )  \tag{1}$$
$$ \frac{d^2 F}{ds^2} = (1+4s)p''(s) + s^2 p'''(s) + 2 p'(s) \tag{2}$$
For the normal, go back to the equation of surface:
$$ r(s,t)= p(s) + t p'(s)$$
$$ \partial_s r \times \partial_t r = (p'(s) + t p''(s) ) \times ( p'(s) ) = t p''(s) \times p'(s)$$
The unit normal is given as:
$$ n= \frac{p''(s) \times p'(s) }{| p''(s) \times p'(s) |}$$
(1) is velocity, (2) is acceleration and above is normal.
