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.

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."

  • $\begingroup$ 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

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