I am trying to find a parametric equation for the intersection line of the surface of two orthogonal cylinders, $\vec{P}$ is a point that belongs to this intersection:
$$\vec{P(t)} = \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} r_1 cos \alpha(t) \\ - \\ r_1 sin \alpha(t) \end{bmatrix} = \begin{bmatrix} - \\ r_2 cos \beta(t) \\ z_0 - r_2 sin \beta(t)\end{bmatrix}$$
I tried to solve this by adding the following constraint:
$$\lVert\frac{d}{dt}\vec{P(t)}\rVert = 1$$
It does not matter where we start and which direction we start, so to pick one, one could say:
$$\left\{\begin{align*} &\alpha(0) = 0\\ &\beta(0) \ge 0\\ &\frac{d\alpha}{dt}\big|_{t=0} \gt 0 \end{align*}\right.$$
But I arrive at - to me - seemingly unsolvable equations, see below.
$$r_1^2\left(\frac{d\alpha}{dt}\right)^2+r_2^2sin^2\beta\left(\frac{d\beta}{dt}\right)^2 = 1$$
$$r_2\left(\frac{cos^2\beta}{\left(1-\left(z_0^2-2 z_0 r_2 sin\beta + r_2^2 cos^2\beta\right)\right)} + sin^2\beta\right) d\beta = dt$$
(I must have made a mistake as there is no $r_1$ in the last equation, but I think this is pretty close to what needs solving).
So first I'd have to find solve this integral, then I'd have to inverse the function - I do not think that either of those is going to happen.
Did I do anything wrong, or is there a simpler way to go about it?