# 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!

• I think it simply means that the value of the function does not change as you move along the direction given by that vector. I sounds redundant, I guess. Hope this helps – Vishesh Sep 13 '13 at 12:38
• As an afterthought, this happens when you look at tangent vectors to level curves of the given function. Also if you know $D_{v}f = \nabla f . v$, where $v$ is your tangent vector. The gradient is always normal to a level curve of a function. – Vishesh Sep 13 '13 at 12:43
• But if the value long curve of that direction doesn't change, doesn't it imply that the curve is a constant? But hardly can I imagine a concrete function like this. – Cancan Sep 13 '13 at 12:54
• Other better answers have already been given. Anyway for the sake of completion, just take a look at $f(x,y)= 2e^{x}+3e^{y}$ at $(0,0)$ Then use the vector $(-3,2)$. Cheers – Vishesh Sep 13 '13 at 13:05
• What do you mean by saying a curve is a constant??? – Vishesh Sep 13 '13 at 13:06

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 $v$.

Let's break our understanding in several pieces and glue them together to understand the puzzle:

1. $f(x,y)$ is a number.
2. $\nabla f(x,y)$ is a vector.
3. $\vec{v}$ is also a vector.
4. The directional derivative is a number that measures increase or decrease if you consider points in the direction given by $\vec{v}$.
5. 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 $\vec{v}$.

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.

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 $$L$$.

• Why does $F'(s) = 0$ implies that $F$ is constant? – John Mars Apr 7 at 20:49
• @JohnMars By Lagrange's mean value theorem, $F(x)-F(y)=F'(z)(x-y)=0$ for all $x,y$. So $F$ is constant. – Hrit Roy Apr 8 at 23:13