Parametric curve on a surface

Question

Having trouble interpreting the given.

Let $M$ be a surface with $x(u,v) = (u,v,u^2+v^2)$. Calculate the normal curvature of $\alpha (t) = (\cos t, \sin t, 1)$ on the surface $M$.

Is $\alpha (t)$ to be considered a mapping $\alpha : \ldots \to M$? What does $t$ mean for $u,v$? Is $u = u(t), v= v(t)$?

I am only given the definition of the normal curvature of surface $M$ at point $p\in M$ in the direction of the unit tangent $\overline{s}\in T_pM$ via the Weingarten operator: $$k_p(\overline{s}) = \langle \mathfrak S _p(\overline{s}), \overline{s} \rangle $$ Is normal curvature a function, then? Evaluated at every point? There are potentially infinitely many tangent vectors at a point. What makes this definition correct?
And probably most important: what is the definition of normal curvature of a parametrized curve on a surface? Is it a function, a number? How does one calculate it?

Answer 1

Your $\alpha$ describes a curve embedded inside your surface $M$. The function $\alpha : t \mapsto (\cos t, \sin t, 1)$ is a function from $\mathbb R$ to $\mathbb R^3$, but fortunately, as you pointed out, the image of $\alpha$ is contained inside $M$, which is why it makes sense to think of $\alpha$ not just as a curve in $\mathbb R^3$ but also as a curve on $M$.

The definition you wrote down is the definition of the normal curvature of the surface $M$. However, your question asks you to find the normal curvature of the curve $\alpha$ on the surface $M$. This is a quantity defined at every point on your curve. To compute it at a given value of $t$, you should take the $p$ in your formula to be $\alpha(t)$, and you should take the $\bar s$ in your formula to be the unit tangent vector $ \widehat{\alpha ' (t)} = \alpha'(t)/|\alpha(t)|$ to the curve at time $t$. Thus the normal curvature to the curve $\alpha$ on $M$ is the function,$$t \mapsto \left\langle \mathfrak S_{\alpha(t)} \left( \widehat{\alpha ' (t)}\right), \widehat{\alpha ' (t)} \right\rangle.$$ This measures the rate at which the normal to $M$ "tilts over" in the direction of the tangent to the curve $\alpha$, as you move along $\alpha$. Here you can find an explicit formula in terms of the first and second fundamental forms.

Answer 2

Yes, your understanding is correct (notice that the points of $M$ satisfy the equation $x^2 + y^2 = z$, so $M$ is a round paraboloid around the $z$ axis). The components of $\alpha$ satisfy the same equation; $\alpha$ is seen to be a horizontal circle (it lives at constant height $z=1$) on this paraboloid, similar to how parallels are on the globe.