level curves and gradient vectors consider a function of two variables, $f(x,y)$. It is stated that at level curves (i.e. f(x,y) = k), it follows that $\nabla f(x,y)$ is perpendicular to $f(x,y) = k$ at every point $(x,y)$. Firstly, $\nabla f(x,y)$ is a vector, so does this mean that $\nabla f(x,y)$ is perpendicular to the tangential vector of the level curve at point $(x,y)$?
Secondly, if I want to find a direction from a point $(x_{0},y_{0})$ that results in $f(x,y)$ neither increasing or decreasing, then how would I find this direction? I know that this direction would be along the the level set and perpendicular to $\nabla f(x_{0},y_{0})$. Could I get a hint regarding this? Could consider an example $f(x,y) = 50 - \frac{x^{2} + y^{2}}{10}$ at point $(2,4)$ if that makes explanation easier. 
Thanks.
 A: *

*Yes. The total differential is
$$
0 = df = \mbox{grad } f \cdot dr
$$
and the tangent unit vector $T$ is
$$
T = \dot{r}(t)/\lVert \dot{r}(t) \rVert
$$
thus
$$
0 = \mbox{grad } f \cdot T
$$

*The gradient would vanish at that point $(x,y)$, so a small step in any direction gives no change.
Assuming you want some iteration from an initial point $(x_0, y_0)$ towards a root of the gradient: 
Using the example: $G(x,y) = \mbox{grad } f = (-x/5, -y/5)$ and Newton-Raphson iteration to find a root:
\begin{align}
r_{n+1} 
&= 
r_n - (J(r_n))^{-1} G(r_n) \\
&=
r_n -
\left(
\begin{matrix}
-5 & 0 \\
0 & -5
\end{matrix}
\right)
\left(
\begin{matrix}
-x_n/5 \\
-y_n/5
\end{matrix}
\right)
\\
&= 0
\end{align}
So it immediately chooses $0$.
Note: I suppose you want an answer here using the level sets. It would result in moving orthogonal to the level sets to move to the extremal level, and that will involve the gradient in some form (as this is orthogonal to the tangent vectors of the level curves). So we get gradient descent or similar.
