# Directional derivative

• Jul 18th 2011, 01:14 PM
Ulysses
Directional derivative
Hi there. I have to find the directional derivative for: $\sqrt{|xy|}$ at the point $P_0(0,0)$ in the direction of $\vec{u}\left (5/13,12/13 \right )$.

The thing is that the derivatives for this function are not defined at the point P0. So, I can't use the proyection of the gradient over u. But using the definition I've found an answer, I think its right, but I'm not pretty sure, because the function is not well defined at the point it's asked.

This is what I did:
$\displaystyle\lim_{t \to{0}}{\displaystyle\frac{\sqrt{|t^2\displaystyle \frac{5}{13} \displaystyle\frac{12}{13} |} }{t}}= \displaystyle\frac{\sqrt[ ]{60}}{13}$
So, the thing is if this is the right answer, or if I should say that the derivative at that point doesn't exist.
• Jul 18th 2011, 11:22 PM
Drexel28
Re: Directional derivative
Quote:

Originally Posted by Ulysses
Hi there. I have to find the directional derivative for: $\sqrt{|xy|}$ at the point $P_0(0,0)$ in the direction of $\vec{u}\left (5/13,12/13 \right )$.

The thing is that the derivatives for this function are not defined at the point P0. So, I can't use the proyection of the gradient over u. But using the definition I've found an answer, I think its right, but I'm not pretty sure, because the function is not well defined at the point it's asked.

This is what I did:
$\displaystyle\lim_{t \to{0}}{\displaystyle\frac{\sqrt{|t^2\displaystyle \frac{5}{13} \displaystyle\frac{12}{13} |} }{t}}= \displaystyle\frac{\sqrt[ ]{60}}{13}$
So, the thing is if this is the right answer, or if I should say that the derivative at that point doesn't exist.

Well, assuming that you did this right then the directional derivative doesn't exist because that limit you wrote doesnt.
• Jul 18th 2011, 11:30 PM
FernandoRevilla
Re: Directional derivative
The limit does not exist, $\lim_{t\to 0^+}\frac{\sqrt{t^2}}{t}=\lim_{t\to 0^+}\frac{t}{t}=1\;,\quad\lim_{t\to 0^-}\frac{\sqrt{t^2}}{t}=\lim_{t\to 0^-}\frac{-t}{t}=-1$

Edited: Sorry, I didn't see Drexel28's post.
• Jul 18th 2011, 11:49 PM
CaptainBlack
Re: Directional derivative
Quote:

Originally Posted by Ulysses
Hi there. I have to find the directional derivative for: $\sqrt{|xy|}$ at the point $P_0(0,0)$ in the direction of $\vec{u}\left (5/13,12/13 \right )$.

The thing is that the derivatives for this function are not defined at the point P0. So, I can't use the proyection of the gradient over u. But using the definition I've found an answer, I think its right, but I'm not pretty sure, because the function is not well defined at the point it's asked.

This is what I did:
$\displaystyle\lim_{t \to{0}}{\displaystyle\frac{\sqrt{|t^2\displaystyle \frac{5}{13} \displaystyle\frac{12}{13} |} }{t}}= \displaystyle\frac{\sqrt[ ]{60}}{13}$
So, the thing is if this is the right answer, or if I should say that the derivative at that point doesn't exist.

Suppose you calculated the required directional derivative at the point $(\varepsilon, \delta)$ where $\varepsilon, \delta>0$ then took the limit as $\varepsilon$ and $\delta \to 0$.

CB
• Jul 19th 2011, 01:49 PM
Ulysses
Re: Directional derivative
Thank you all, I forgot the absolute value :P
• Jul 20th 2011, 05:27 AM
Ulysses
Re: Directional derivative
I've been thinking about this, shouldn't I take the limit only in the positive direction considering than that direction is the one pointed by the vector given? I mean taking this limit only:
$\displaystyle\lim_{t \to{0^+}}{\displaystyle\frac{\sqrt{|t^2 \displaystyle\frac{5}{13} \displaystyle\frac{12}{13} |} }{t}}$
• Jul 20th 2011, 05:37 AM
FernandoRevilla
Re: Directional derivative
No, it is $t\to 0$ . :)