Let $M$ be a surface of revolution given by the parameterization $r: I \times (0,2\pi) \to \mathbb R^3$, $r(t,\theta) = (x(t)\cos\theta,x(t)\sin\theta, z(t))$, and $x(t) \ne 0, z'(t) \ne 0$ for all $t \in I$ ($I$ is an open interval in $\mathbb R$). Show that for any $p \in M$, the normal to $M$ at $p$ intersects the $z$-axis.
I find it hard to think of a way to use the definitions I've seen in class to find the normal to begin with. This is what we covered so far:
- The normal vector to a k-dimensional smooth manifold at point $p \in M$ is a vector $v \in \mathbb R^n$ such that for any smooth $\gamma : (-\epsilon, \epsilon) \to M$ such that $\gamma(0)=p$ we have $\langle v, \gamma'(0) \rangle=0$ .
- If $M$ is a $n-1$ dimensional smooth manifold given implicitly with $F(x)=0$, then the normal at $p$ is $\nabla F(p)$.
- If $M$ is the graph of a smooth function $f:\mathbb R^{n-1}\to \mathbb R^n$, then the normal at $(x_1,x_2,\dots,x_{n-1},f(x_1,\dots,x_{n-1})$ is the vector $(-\partial_1f(x_1,\dots,x_{n-1}),\dots,-\partial_{n-1}f(x_1,\dots,x_{n-1}), 1)$.
- The cross product of $v_1,\dots,v_{n-1}$, $v_1 \times \dots \times v_{n-1}$, is the only vector such that $\langle v_1 \times \dots \times v_{n-1} , w\rangle$ equals the determinant of matrix $A$ with columns $v_1, \dots , v_{n-1}, w$ for all $w \in \mathbb R^n$ as an implication from Riesz representation theorem. In particular, $v_1 \times \dots \times v_{n-1} \in \operatorname{span}(v_1, \dots, v_{n-1})^{\perp}$.
I'm not sure how I can utilize these notions to solve the problem. Also I don't know if this particular case being a surface of revolution and not any surface has any significant meaning to the solution. From previous courses I took I know that in $\mathbb R^3$ the corss product of the partial derivatives of the parameterization does in fact yield a normal vector at a point, but I don't want to use this blindly.