Directional derivative

**Ulysses** · July 18th 2011, 12:14 PM

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.

**Drexel28** · July 18th 2011, 10:22 PM

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.

**FernandoRevilla** · July 18th 2011, 10:30 PM

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.

**CaptainBlack** · July 18th 2011, 10:49 PM

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

**Ulysses** · July 19th 2011, 12:49 PM

Thank you all, I forgot the absolute value :P

**Ulysses** · July 20th 2011, 04:27 AM

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:
$\displaystyle\lim_{t \to{0^+}}{\displaystyle\frac{\sqrt{|t^2 \displaystyle\frac{5}{13} \displaystyle\frac{12}{13} |} }{t}}$

**FernandoRevilla** · July 20th 2011, 04:37 AM

No, it is $t\to 0$ .

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