# Show that for all vector, directional derivative is zero

Consider the following function: $$f(x,y) = \begin{cases} 1 & \text{if y=x^2, x \neq 0} \\ 0 & \text{otherwise} \end{cases}$$

Show that for any vector $$v \neq 0$$, $$D_vf(0,0)$$, the directional derivative, is $$0$$.

So I mark $$v = [v_1, v_2]$$. By definition:

$$D_vf(0,0) = \lim\limits_{t \to 0} \frac{f((0,0) + tv) - f(0,0)}{t} = \lim\limits_{t \to 0} \frac{f(tv_1, tv_2)}{t}$$

since t goes to 0, and the vector is not the zero vector, I can assume $$x = tv_1 \neq 0$$, which means the limit $$\frac{1}{t}$$ as $$t \to 0$$ does not exist , and I get stuck.

How to solve this?

• How did you get $1$ in the numerator? This is only correct for (at most) one particular value of $t>0$ May 13 '19 at 18:14
• becuase $x = tv_1 \neq 0$.. or am I wrong? Should I split into cases? May 13 '19 at 18:15
• OK, $x=tv_1 \neq 0$, and...? Please continue to relate it to $f$, where is the 1? May 13 '19 at 18:16

Actually, for $$v\neq 0$$, $$f(tv_1, tv_2)$$ is not usually going to be 1 as you have written, since $$f(x,y)$$ is only 1 when $$x\neq 0$$ and $$y=x^2$$.
But since $$t\mapsto tv$$ is a line through the origin, it will only cross the parabola $$y=x^2$$ at a single point other than the origin at most. That means that there exists $$t'>0$$ such that $$f(tv_1,tv_2)=0$$ for $$0. Thus we actually have $\lim_{t\to0}\frac{f(tv_1,tv_2)}{t}=\lim_{t\to0}\frac{0}{t}=0.$
Consider the equality $$tv_2=(tv_1)^2$$. It is equivalent to $$tv_2=t^2{v_1}^2$$, which holds if and only if $$t=0$$ or $$t=\frac{v_2}{v_1}$$. But then, if $$t$$ is small enough and different from $$0$$, $$f(tv_1,tv_2)=0$$ and so$$\lim_{t\to0}\frac{f(tv_1,tv_2)}t=0.$$