Is there a second derivative test in polar coordinates ? How to calculate concavity in the radial direction? I searched google for articles talking about second derivative test in polar coordinates and found nothing at all.
Could you please refer me to an article or a book about it?
I'm asking because I want to calculate the concavity of a surface at origin in the $radial$ direction as a function of $\theta$.
Thanks for help.
 A: The second derivative test can be used in polar coordinates as well to determine the convexity. Just note that you need to calculate $\frac{d^2 y}{dx^2}$ in terms of $r$ and $\theta$. Knowing that $x=r\cos\theta$ and $y=r\sin\theta$, one can write:
$$y'=\frac{d y}{d x}=\frac{d y}{d \theta}\frac{d \theta}{d x}=\frac{d y/d\theta}{d x/d\theta}=\frac{r\cos\theta+r'\sin\theta}{-r\sin\theta+r'\cos\theta}$$
where $r'$ means $\frac{dr}{d\theta}$. Hence:
$$y''=\frac{d y'}{d x}=\frac{d y'/d\theta}{d x/d\theta}=\frac{r^2+2r'r'-r r''}{(r'\cos\theta-r\sin\theta)^3}$$
Now you can use the second derivative test wherever you want. As a side-note there are plenty of online resources about this matter. For example, this and this were among the first results that google gave me.
And another side-note is, you mentioned you want to calculate the concavity of a surface. This means you would deal with 3-D polar coordinates, which is much more complex but pretty the same.
A: If I understand you correctly, you would like to have a discriminant test for polar coordinates, so that you can easily see whether the critical point of a function $f(r,\theta)$ is an extremum or a saddle point.
The transformation of the discriminant to polar coordinates is a straightforward if somewhat tedious exercise.  A direct calculation spawns a great number of terms, most of which conspire to annihilate one another.  The final result can then be written as the determinant of a symmetric matrix,
$$
\Delta(f) = \det\left[
 \begin{array}{cc}
 \frac{\partial^2f}{\partial r^2} & \frac{1}{r}\frac{\partial^2f}{\partial r\partial\theta}
 -\frac{1}{r^2}\frac{\partial f}{\partial\theta}\\
 \frac{1}{r}\frac{\partial^2f}{\partial \theta\partial r}
 -\frac{1}{r^2}\frac{\partial f}{\partial\theta} & \frac{1}{r^2}\frac{\partial^2 f}{\partial\theta^2}
 +\frac{1}{r}\frac{\partial f}{\partial r}
 \end{array}
\right]
$$
Note that the individual entries in the above matrix are ${\it not}$ equal to the corresponding entries in the cartesian version, but the result for the determinant is the same.  In particular, for any twice differentiable function $F(x,y)$, if you feed
$$
f(r,\theta) = F(r\cos\theta,r\sin\theta)
$$
into the above expression for $\Delta(f)$ and simplify, the result will be
$$
\Delta(f)
=
F_{xx}(r\cos\theta,r\sin\theta)F_{yy}(r\cos\theta,r\sin\theta)-\left[F_{xy}(r\cos\theta,r\sin\theta)\right]^2,
$$
just as you would wish.
