Archimede's spiral A point M is moving uniformly on a straight line ON, which is rotating with constant angular velocity around the point O. Find the equations of the trajectory of M.
 A: If we consider polar coordinates we have that:
$$\dot{r}=\text{constant}=v$$
This is the radial velocity. Then if you integrate you get $r=v\cdot t +r_0$.
The angular velocity is constant too.
$$\dot{\theta}=\text{constant}=\omega$$ Therefore you have $\theta=\omega\cdot t+\theta_0$.
You can convert it to cartesian coordinates by having that:
$$\vec{x}(t)=\begin{pmatrix} r\cos(\theta)\\ r\sin(\theta)\end {pmatrix}=\begin{pmatrix} (v\cdot t +r_0)\cos(\omega\cdot t+\theta_0)\\ (v\cdot t +r_0)\sin(\omega\cdot t+\theta_0)\end {pmatrix}$$
If  you want the equation describing it on the plane you also know that:
$$x^2+y^2=(vt+r_0)^2$$
$$\frac{x}{y}=\tan\left(\omega t+\theta_0\right)$$
$$\Rightarrow \quad t=\frac{\arctan\left(\frac{x}{y}\right)-\theta_0}{\omega}$$
As a consequence you finally get:
$$x^2+y^2=\left(\frac{v}{\omega}\left(\arctan\left(\frac{x}{y}\right)-\theta_0\right)+r_0\right)^2$$
A: A the motion is uniform, the position along the straight line is a linear function of time, hence a linear function of the angle.
In polar coordinates,
$$\rho=\lambda\theta+\rho_0.$$
The coefficient $\lambda$ measures the displacement along the line corresponding to a rotation of one radian.
In Cartesian coordinates,
$$\tan\left(\frac{\sqrt{x^2+y^2}-\rho_0}\lambda\right)=\frac yx,$$ which cannot really be simplified, or in a parametric form,
$$\begin{cases}x=(\lambda\theta+\rho_0)\cos\theta,\\y=(\lambda\theta+\rho_0)\sin\theta.\end{cases}$$
