We have angle=arctan(dy/dx), but what happens when dx=0? Here is a formula: $\text{angle}=\arctan(dy/dx)$.
I can find an angle with my calculator for any value except $dx=0$.
My question is: is there no angle or, is there something that says when $dx=0$ the angle is found differently? 
 A: If $dx$ is $0$, if you look in the Cartesian plane, you are standing in the y-axis, so the angle would be $\frac{\pi}{2}$ or $\frac{3\pi}{2}$, depending if $dy > 0$ or $dy < 0$.
A: Since none of the answers have mentioned this, I'm putting this for completeness' sake: advanced calculators and computing environments provide for a "two-argument" arctangent function $\arctan(x,y)$ (denoted as atan2() in some environments, and with the order of the arguments sometimes reversed) that is especially intended for polar coordinate conversions. Briefly, $\arctan(x,y)$ gives the same results as $\arctan\frac{y}{x}$, adjusted when necessary so that the result is within $(-\pi,\pi]$, taking into account which quadrant the point $(x,y)$ is in. When $x=0$ and $y\neq 0$, $\arctan(0,y)=\frac{\pi}{2}\mathrm{sign}\,y$.
A: Actually, the $angle$ that a point in $R^2$ with cartesian coordinates $(x,y)\neq (0,0)$ makes with the positive x-axis is defined as the value $\theta (x,y)$ such that 
$\cos \theta (x,y) = \frac{x}{\sqrt{x^2+y^2}}$ and 
$\sin \theta (x,y) = \frac{y}{\sqrt{x^2+y^2}}$,
with $\theta\in [0, 2\pi)$. Note that, defined in this way, there is a unique $\theta$ for each point different from $(0,0)$. 
If $x>0$ and $y>0$, these formulas combined give $\theta (x,y) = \arctan(y/x)$.
If $x=0$ and $y>0$, it follows from the definition that $\theta=\pi/2$.
You can also consider $\theta \in[-\pi,\pi)$, and in this case the formula $\theta (x,y) = \arctan(y/x)$ is valid for all $y$ provided $x>0$.
A: You are correct in pointing out that
$$
\theta = \arctan{\frac{y}{x}}
$$
is not such a great formula, since it
will not work when $x = 0$.
In fact, $\arctan$ will always return a value in the interval $(-\pi/2, \pi/2)$, so it will give the wrong answer for the angle when $x < 0$!
(Unless you don't care if the angle is pointing in the opposite direction from the vector.)
What you really want
the unique angle $\theta$, say on the interval $[0, 2\pi)$, such that
\begin{align*}
\sin \theta = \frac{y}{\sqrt{x^2 + y^2}} \\
\cos \theta = \frac{x}{\sqrt{x^2 + y^2}}.
\end{align*}
What is the best way to calculate this?
It's maybe not as elegant as you would like, but you can use
$$
\theta =
\begin{cases}
\arccos\left( \frac{x}{\sqrt{x^2 + y^2}} \right) &\text{if } y \ge 0 \\
2\pi - \arccos\left( \frac{x}{\sqrt{x^2 + y^2}} \right) &\text{if } y \le 0.
\end{cases}
$$
A: A formula of atan that is working when dx or dy=0.
when dx=dy=0 the result is undefined.
Let"s  put dx=x and dy=y.
f(x,y)=pi()-pi()/2*(1+sign(x))* (1-sign(y^2))-pi()/4*(2+sign(x))*sign(y)
    -sign(x*y)*atan((abs(x)-abs(y))/(abs(x)+abs(y)))

The angle is from 0 to 2pi.
