How to find a vector normal to a circle?

Question

Assume $S_1$ be the disk in the $ y = 1$ plane bounded by the circle $ x^2$ +$ z^2$ = $9$ . Prove that the rightward pointing unit normal to $ S_1$ is the vector $ (0, 1, 0)$.

I know that the gradient is orthogonal to level curves.

Answer 1

Call $f(x,y,z)=y-1$, then we have $\nabla f=(0,1,0)$ and $\|\nabla f\|=1$. This unit vector $\frac{\nabla f}{\|\nabla f\|}=\nabla f=(0,1,0)$ is normal to $S_1=f^{-1}(\{0\})$ due to the fact that

the gradient is orthogonal to level curves

and it is indeed rightward pointing.

The circle only puts a bound on $x$ and $z$, which is not used here.

Answer 2

Since the curve lies in the xz plane, it seems sort of trival to say that the vector normal to the curve is normal to the plane.

In a more general case you might do something like.

$x = 3\cos t\\ y = 1\\ z = 3\sin t$

$\frac {dr}{dt} = 3(-\sin t,0, \cos t)\\ T = \frac {dr}{dS} = (-\sin t,0, \cos t)\\ N = \frac {dT}{dS} = (-\cos t,0, -\sin t)\\ B = T\times N = (0,1,0)$