Vtyjkyuk

Given a curved line, and any point on that line is a radial vector with only a known radius $r$ and known tangent angle $a$, how does one express the point $(r,a)$ as vector coordinates $(x, y)$?

I have a spiral defined by its changing tangent angle and radius. I want to know the angle from any given point $(r,a)$ to the $x$ axis (the $arg(x+iy)$ on the complex plane for instance) or the vector coordinates $(x,y)$ that corresponds to point $(r,a)$, as either will provide the other.

Click here to view the geometric representation

The usual method for finding the tangent angle $a$ of a radial vector follows

$$tan(a)=r(dr/d arg(z))^-1.$$ However, on the complex plane for this curve/spiral, the argument of $z$ cancels out entirely (in how $r$ is defined), such that the antiderivative of $darg(z)$,

$$int jt^-1 dt=arg(z),$$

is multivalued (in multiples of the inverse of the real part of $z$), as the integral $$-int_1^e^k jt^-1 dt=arg(z): k=e^arg(z)/j $$ is indeterminant: $$j ln(e^arg(z)/j) = j, e^arg(z)/j>=0.$$

Yet, there is a single argument of $z$, it's just that I do not know how to solve for it even given the radius and the tangent angle. This is not an Archimedean spiral in that the angle changes per a constant value and it is not quite a logarithmic spiral either (somewhere between the two). So if the spiral could be thought of as a particle curling in a field toward zero, it would be accelerating...not having a constant radial speed.

Any help would be greatly appreciated.

Consider a particle starting from the center
at t=0. The angular position of the particle is given by $theta=omega t$, where $omega$ is a constant. We also have a change in radial distance (radial speed) $dotr=u$ where $u$ is a constant. Now the position in the xy-plane at time t is given by

$$mathbfr(t)=(x(t),y(t))=(utcos(omega t), utsin(omega t))$$

Spiral t=0 to 9, $u=1$, $omega=2$

Edit: It seems you edited your question while I was answering, hence no mention of the complex plane, but I think you can still use the information to figure out what to do :)

The $(r,a)$ point z is

$$z=x+iy=r(cos a, isin a)$$