# Are we unable to find the extrema of the function $f(x,y) = \frac{y}{x^2+y^2}$ using the second partial test?

\begin{align*} &f(x,y) = \frac{y}{x^2+y^2}\\ &f_{xx} = \frac{∂}{∂x}\left(-\frac{2xy}{(x^2+y^2)^2}\right)=-\frac{2y(y^2-3x^2)}{(x^2+y^2)^3}\\ &f_{yy} = \frac{∂}{∂x}\left(\frac{x^2-y^2}{(x^2+y^2)^2}\right)=\frac{2y(y^2-3x^2)}{(x^2+y^2)^3} \end{align*}

$a$ and $b$ must be $(0,0)$ or at least one of them has to be $0$ to be critical points of the function

if the point is $(0,0)$, $f_{xx}$ or $f_{yy}$ will be undefined and finding the point $(?,0)$ for which $f_x$ or $f_y$ would $= 0$ would be mighty hard.

Are we unable to find the extrema of this function using the second partial test?

-

First, you’ve some errors in your third line of calculations. The correct partials are:

\begin{align*} &f_x(x,y)=-\frac{2xy}{(x^2+y^2)^2}\\ &f_y(x,y)=\frac{x^2-y^2}{(x^2+y^2)^2}\\ &f_{xx}(x,y)=\frac{2y(3x^2-y^2)}{(x^2+y^2)^3}\\ &f_{xy}(x,y)=\frac{2x(3y^2-x^2)}{(x^2+y^2)^3}\\ &f_{yy}(x,y)=\frac{3y(y^2-2x^2)}{(x^2+y^2)^3} \end{align*}

Clearly $f_x(x,y)=0$ only if $xy=0$, i.e., at least one of $x$ and $y$ is $0$. Similarly, $f_y(x,y)=0$ only if $0=x^2-y^2=(x-y)(x+y)$, i.e., only if $y=x$ or $y=-x$. The only way to satisfy both of these conditions is to have $x=y=0$, and the function and its partial derivatives aren’t defined at $(0,0)$. Thus, you’re quite right: there is no point at which the second derivative test applies.

Rewriting the function in polar coordinates as

$$f(r,\theta)=\frac{r\sin\theta}{r^2}=\frac{\sin\theta}r$$

may help to explain what’s going on. As we travel around the circle $C_r$ of radius $r$ centred at the origin, the function value is $0$ where $C_r$ crosses the $x$-axis, reaches a maximum of $\frac1r$ where $C_r$ crosses the positive $y$-axis, and reaches a minimum of $-\frac1r$ where $C_r$ crosses the negative $x$-axis. But that high point with value $\frac1r$ on $C_r$ can’t be a local maximum of the function, because the value of the function gets larger as you move down the $y$-axis towards the origin: $\frac1r$ increases as $r$ decreases. Similarly, the low point on $C_r$ can’t be a local minimum of $f$, because $-\frac1r$ gets smaller (more negative) as you move up the negative $y$-axis and $r$ decreases.

-
So, taking from the theroem, would we say "f does not have continuous second partial derivatives anywhere, thus the second partial test does not apply" ? –  Matt Dec 15 '11 at 21:10
@Matt: That isn’t the problem: the second partials are continuous except at $(0,0)$, where they’re not defined. The problem is that there is no point where the first partials are both $0$, so there’s no point that is even a candidate to be a local maximum or minimum. –  Brian M. Scott Dec 15 '11 at 21:19