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Math Help - Directional derivative

  1. #1
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    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.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: Directional derivative

    Quote Originally Posted by Ulysses View Post
    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.
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    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.
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  4. #4
    Grand Panjandrum
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    Re: Directional derivative

    Quote Originally Posted by Ulysses View Post
    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
    Last edited by CaptainBlack; July 20th 2011 at 07:50 AM.
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  5. #5
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    Re: Directional derivative

    Thank you all, I forgot the absolute value :P
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  6. #6
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    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}}
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  7. #7
    MHF Contributor FernandoRevilla's Avatar
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    Re: Directional derivative

    No, it is t\to 0 .
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