1
$\begingroup$

I have read a chapter about differentiability in two variables. I now have two questions:

  1. Why do we need the constraint that $|\vec{u}|=1$ when we calculate the directional derivative?
  2. Definition of gradient: $\nabla f = \frac{\partial f}{\partial x_1 }\mathbf{e}_1 + \cdots + \frac{\partial f}{\partial x_n }\mathbf{e}_n$ but why do we need f to be differetiable? Is it simply becuase we "take the partial derivatives"? I have learned the pure defn of differetiable, but maybe I don't have the "gut feeling" of what it really is, in simple language.
$\endgroup$
1
$\begingroup$

1) You are free to choose $u$ not unitary. But you have to be careful. For example, if $|u|=2$ you get double the amount you get with $|u|=1$, as it means that you are "moving twice as fast".

For example, consider a function $f(x,y)$, and consider the "curve" $(x=t, y=0)$, which is a straight line along the $x$ axis.

How do we "see" $f$ from "inside" the curve? We get: $$ f(x(t), y(t)) = f(t,0). $$

The derivative of $f(t,0)$ with respect to $t$ is exactly the directional derivative in the direction of $x$.

Now pick the "curve" $(x=2t, y=0)$, which is a straight line like before, but now the "speed" is double. What happens now if you calculate: $$ f(x(t), y(t)) = f(2t,0), $$

and derive it with respect to $t$?

2) "Differentiable" intuitively means that "its partial derivatives exist and behave well". Which translates to: "if $f$ is not differentiable, it makes no sense to talk about its gradient."

$\endgroup$
  • $\begingroup$ can you elaborate on 1) ? $\endgroup$ – jacob Apr 19 '14 at 15:13
  • $\begingroup$ @jacob Edited the question, adding an example...Let me know if it is clear! $\endgroup$ – geodude Apr 19 '14 at 15:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.