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Thread: Directional Derivative Help Needed!

  1. #1
    Member thaopanda's Avatar
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    Directional Derivative Help Needed!

    Let f: $\displaystyle R^2 \rightarrow R$ be given by

    f(x,y) :=

    $\displaystyle \frac{xy^2}{x^2+y^4}$ if (x,y) $\displaystyle \neq$ (0,0)
    0 otherwise

    (a) Show that $\displaystyle d_u f(0,0)$ exists for all the vectors $\displaystyle u \in R^2$ and that $\displaystyle d_u f(0,0)$ = $\displaystyle \frac{b^2}{a}$ if u = (a,b) with a $\displaystyle \neq$ 0.

    (b) Show that f is not differentiable at (0,0)

    We just started this topic and I am completely lost... please help
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  2. #2
    Senior Member Shanks's Avatar
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    (1)what is the diefinition of directional derivative? make sure you understand the definition !
    If u=(a,b), then (x,y)=(at,bt) tend to (0,0) as t tend to 0, then what is the limit ?
    (2) if (x,y) tend to 0 with the path x=ky^2, then what can you say about the limit?
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  3. #3
    Member thaopanda's Avatar
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    So... I got that the limit as h $\displaystyle \rightarrow$ 0 of $\displaystyle \frac{f((0,0)+hu) - f(0,0)}{h} = 0$ meaning $\displaystyle d_uf(0,0)$ exists.

    for u = (a,b),
    it would be $\displaystyle \frac{(at)(bt)^2}{(at)^2+(bt)^4}$ and take that as t $\displaystyle \rightarrow$ 0, but I'm not sure how to do that...

    I don't know what you mean by the path...
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  4. #4
    Senior Member Shanks's Avatar
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    $\displaystyle \frac{(at)(bt)^2}{(at)^2+(bt)^4}\to 0, \mbox{as} t\to 0, provide that a \neq 0$.
    If (x,y) tend to (0,0) along the path $\displaystyle x=ky^2$,then

    $\displaystyle \frac{xy^2}{x^2+y^4}=\frac{ky^4}{k^2y^4+y^4}=\frac {k}{k^2+1} \to \frac{k}{k^2+1}, \mbox{as} y\to 0$
    the limit depent on the value of k, it is not continous at (0,0).



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  5. #5
    Member thaopanda's Avatar
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    for the second part, if f was $\displaystyle \frac{xy^2}{x^2+y^2}$ (still 0, if (x,y) = (0,0)), what would I set x equal to?
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