0
$\begingroup$

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?

$\endgroup$
3
  • 1
    $\begingroup$ How did you get $1$ in the numerator? This is only correct for (at most) one particular value of $t>0$ $\endgroup$ May 13, 2019 at 18:14
  • $\begingroup$ becuase $ x = tv_1 \neq 0$.. or am I wrong? Should I split into cases? $\endgroup$
    – Tegernako
    May 13, 2019 at 18:15
  • 1
    $\begingroup$ OK, $x=tv_1 \neq 0$, and...? Please continue to relate it to $f$, where is the 1? $\endgroup$ May 13, 2019 at 18:16

2 Answers 2

1
$\begingroup$

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 $0<t<t'$. Thus we actually have \[\lim_{t\to0}\frac{f(tv_1,tv_2)}{t}=\lim_{t\to0}\frac{0}{t}=0.\]

$\endgroup$
1
$\begingroup$

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$$\lim_{t\to0}\frac{f(tv_1,tv_2)}t=0.$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .