Directional Derivative Help Needed!

**thaopanda** · December 10th 2009, 01:34 PM

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

f(x,y) :=

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

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

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

We just started this topic and I am completely lost... please help

**Shanks** · December 11th 2009, 12:48 AM

(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?

**thaopanda** · December 13th 2009, 05:53 AM

So... I got that the limit as h $\rightarrow$ 0 of $\frac{f((0,0)+hu) - f(0,0)}{h} = 0$ meaning $d_uf(0,0)$ exists.

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

I don't know what you mean by the path...

**Shanks** · December 13th 2009, 07:06 AM

$\frac{(at)(bt)^2}{(at)^2+(bt)^4}\to 0, \mbox{as} t\to 0, provide that a \neq 0$ .
If (x,y) tend to (0,0) along the path $x=ky^2$ ,then

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

**thaopanda** · December 13th 2009, 07:49 AM

for the second part, if f was $\frac{xy^2}{x^2+y^2}$ (still 0, if (x,y) = (0,0)), what would I set x equal to?

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