# Find the critical points of $f(x,y) = x \sin y$

Find local extrema and saddle points of the function $$f(x,y)=x \sin y$$

I just want to make sure that I'm understanding this correctly.

The critical points of $f(x,y)$ is at

$$\nabla f= \begin{pmatrix}\sin y \\ x \cos y\end{pmatrix}=0$$

which gives $(x,y)= (0,0)$. That is the only critical point.

Using the second derivative test gives us the following

$$f_{xx}=0, \,\,\,\,\, f_{yy}=-x\sin y, \,\,\,\,\, f_{xy}=\cos y$$

Hence

\begin{align} D & =f_{xx}(0,0)f_{yy}(0,0) - (f_{xy}(0,0))^2\\ & = 0 - 1^2 \\ & = -1 \\ \end{align}

Since $D<0 \implies$ this is a saddlepoint with no extrema. Is this correct?

EDIT: Now knowing that $$\sin y = 0 \iff y=0 \, \pm \, n\pi$$ $$x\cos y =0 \iff y=\frac{\pi}{2} \, \pm \, n\pi \,\, \text{and } \, \, x=0$$

Doing the second derivative test again gives

\begin{align} D & =f_{xx}(0,\pm n\pi)f_{yy}(0,\pm n\pi) - (f_{xy}(0,\pm n\pi))^2\\ & = 0 - [(\pm1)^n]^2 \\ & = -[(1)^{2n}] \\ \end{align}

$D<0$ for $\forall n \in \mathbb{Z} \implies$ this is a saddlepoint with no extrema.

• What about $(x,y)=(0,\pi)$?
– 5xum
Oct 11, 2017 at 11:41

Recall that $$\sin(y+2n \pi)= \sin (y),\qquad \forall n\in \mathbb{Z} \\ \sin(y+n \pi)= -\sin (y),\qquad \forall n\in \mathbb{Z} \\ \cos(y+2n \pi)= \cos (y),\qquad \forall n\in \mathbb{Z} \\ \cos(y+n \pi)= -\cos (y),\qquad \forall n\in \mathbb{Z}$$ implies $$\sin(y)=0 \Leftrightarrow y=0\pm n\pi \\ x\cos(y)=0\Leftrightarrow y=\frac{\pi}{2}\pm n\pi \mbox{ or } x=0$$ Then the critical points of $f$ is $$\{(x,y)\in\mathbb{R}^2: x=0 \mbox{ and } y=\pm n\pi, \forall n\in \mathbb{Z}\}$$
$(0,0)$ is not the only critical point !
The set of critical points is $\{(0,k \pi): k \in \mathbb Z\}$
Your considerations concerning $(0,0)$ are correct.
• I totally forgot about $\pi$. Thanks very much! Oct 11, 2017 at 11:46