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It's an old past paper with no mark scheme, this always makes me a slightly afraid of exploring unaided.

The function is: $f(x,y)=\frac{xy^2}{x^2+y^6}$ if $(x,y)\ne(0,0)$ else $0$.

I am to show all the directional derivatives exist at $(0,0)$

Intuitively, $(\nabla f)(0,y)=0=(\nabla f)(x,0)$ because clearly in either case $f$ is constant, at 0.

So using some linear algebra, and the linear map that it the derivative ($(Df(a))(v_x,v_y)$ is the derivative function at a in direction $v=(v_x,v_y)$, well as it is a linear map $(Df(a))(v)=(Df(a))(v_x(1,0)+v_y(0,1))=v_x(Df(a))(1,0)+v_y(Df(a))(0,1)$

By using definitions (namely: $[(Df(a))]_{i,j}=(\nabla f_i)_j$) we see
$(Df(a))(v)=(\nabla f)(v_x,0)+(\nabla f)(v_y,0)=0$

The question asks for:

1) State the exact definition of differentiability at $(0,0)$ for $f:\mathbb{R}^2\rightarrow\mathbb{R}$

2) Show all the directional derivatives of f exist at (0,0) and comput their values.

3) Show that f is not continuous at (0,0) I can (probably) do this - I'd show it isn't sequentially continuous $\implies$ not continuous

4) Show that $f$ is not differentiable at $(0,0)$ using the definition in part 1.

(Hint: you can take a small deviation away from the origin param by $h_1=t^3,h_2=t,t\in\mathbb{R},t\ne0$)

For part 4 I think I can use:

$f$ differentiable $\iff$ $$lim_{h->0}(\frac{f(a+h)-f(a)-(Df(a))(h)} {||h||})=0$$

Then show the limit from $t<0$ and $t>0$ disagree (or something)

I am confused because the curve $f(x,y)=xy$ also has derivatives of 0 along each axis, yet for the direction (1,1) the curve is certainly not doing nothing (staying constant), but the properties of a linear map....

I am also confused because $(t^3,t)$ is a direction!

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Continuity and differentiability here involve two dimensional limits that fail to exist at (0,0), which can be verified (I think) using the cubic path. The directional derivatives are one dimensional limits along straight lines (or rays) starting at (0,0), so they only involve paths (t,mt) and (0,t).

Added in edit: As you say, the partial derivatives at $(0,0)$ are both $0$ since $f$ is identically $0$ along both $x$ and $y$ axes, and so $grad(f)(0,0)=0$. Hence, if $f$ were differentiable at $(0,0)$, all directional derivatives at $(0,0)$ would equal $0$ by the usual formula (the dot product for the directional derivative at a point of differentiability).

To compute the directional derivative at $(0,0)$ in an arbitrary direction (i.e. unit vector) $(c,s)$ with $c <> 0$, find the following limit as $t -> 0$ :

$lim (f(0+tc,0+ts)-f(0,0))/t = lim ( (ct)(st)^2 )/( (t^2)(c^2+(t^4)(s^6))t ) = s^2/c$

which is obviously not $0$ at arbitrary unit vectors $(c,s)$. But the limit, and hence the directional derivative, DOES exist for all directions (the directions with $c=0$ are along the y axis and those directional derivatives are just (+/-) the partial w.r.t $y$ at the origin-- in this case, both are 0). [By the way $c$ and $s$ stand for cosine and sine].

Finally, if we approach the origin along the path $(t^3,t)$ then the one dimensional limit of the function itself is

$lim (t^5)/(t^6 + t^6) = lim 1/(2t)$ which is NOT $0$ and so $f$ is not continuous at the origin (since the full two dimensional limit at $(0,0)$ can't be $0$ if the one dimensional path limits don't all come out to $0$)

Sorry, I don't know how to format better. Also, I have forgotten the definition of differentiability -- but a consequence of it is the chain rule, from which is derived the formula for directional derivative in terms of the partial derivatives (and I think is what you were referring to with "linear..."). Since the formula fails here, the function must not be differentiable at the origin (and it fails to be continuous at the origin as well, which also implies non-differentiability).

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  • $\begingroup$ Could you flesh this out a bit please? Knowing if I get to the right answers is very important, so edit in just the answers, that'd be great $\endgroup$ – Alec Teal Jun 9 '14 at 1:14

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