What does directional derivative zeros imply when directional vector is not zero? This question might sound stupid but I want to confirm an answer from it.
I saw somewhere online that it means that when the directional derivative of function $f$ along the none zero vector $v$ at certain point is equal to $0$, it means that the function $f$ is constant in that direction. But what does "constant in direction" mean? can anyone give me an example of it such as $f(x,y)$ to explain this?
Thanks!
 A: If $D_v f(x_0)=0$, then $f$ is constant to first order in the direction $v$ - that is, if you consider the values of $f$ along the line in the direction $v$, you find that there is no linear-order term:
$$ f(x_0+tv) = f(x_0) + t D_vf(x_0) + o(t) = f(x_0) + o(t). $$
(Here $o(t)$ is some function such that $o(t)/t \to 0$ as $t \to 0$, which is what we mean by zero to first order.)
Instead of considering a line, you can consider the level set $\{x : f(x) = f(x_0)\}$. So long as $x_0$ isn't a critical point of $f$, this level set will (at least in some neighbourhood of $x_0$) be a curve. Thus $f$ is constant along a curve in the direction $v$.
A: Let's start with the simplest case first. $f: \mathbb{R} \to \mathbb{R}$ given by $f(x) = x^2$. Then the directional derivative
$$\frac{\partial f}{\partial x} = \frac{df}{dx} = 2x$$
which is zero at the origin. If you look at a graph of a parabola, you see that the closer you zoom in on the origin, the more flat the graph looks (check this yourself on an online graph calculator).
Now take $g: \mathbb{R}^2 \to \mathbb{R}$ to be $g(x,y) = x^2 + y$, which is a parabolic cylinder, as can be seen here: http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e9800998ecf8427empntaacb4q.
If you take the directional derivative along $y$ (also known as the partial derivative with respect to $y$), you get $1$. Thus, this function is increasing as you fix $x$ and increase $y$. However, the directional derivative along $x$ is $2x$, which is zero at the origin. Thus, if you zoom in really close to the origin, what you will see is something that looks like a tilted plane. As you run along the $x$ axis, the plane stays at the same height (constant), but it changes as you run along the $y$ axis.
A: Let's break our understanding in several pieces and glue them together to understand the puzzle:


*

*$f(x,y)$ is a number.

*$\nabla f(x,y)$ is a vector.

*$\vec{v}$ is also a vector.

*The directional derivative is a number that measures increase or decrease if you consider points in the direction given by $\vec{v}$.

*Therefore if $\nabla f(x,y) \cdot \vec{v} = 0$ then nothing happens. The function does not increase (nor decrease) when you consider points in the direction of $\vec{v}$. 

A: Suppose that $f$ is differentiable and that the directional derivative of $f$ along the vector $v=(a_1,a_2,...,a_n)$ is zero i.e. $\nabla f\cdot v=0$ or $a_1f_1+a_2f_2+...+a_nf_n=0$. 
Claim: $f$ is constant along any line having direction $v$
$L$ be any such line which passes through $p=(b_1,b_2,...,b_n)$. Now the parametric equations of $L$ are: $x_1(s)=b_1+a_1s,x_2(s)=b_2+a_2s,...,x_n(s)=b_n+a_ns$. Keeping this in mind, we define $F(s)=f(x_1(s),x_2(s),...,x_n(s))$. The derivative $F'(s)=f_1\dot x_1(s)+f_2\dot x_2(s)+...+f_n\dot x_n(s)=f_1a_1+f_2a_2+...+f_na_n=0$. So $F$ must be a constant function. This shows that $f$ is constant along $L$.
