I know that to find the angle between two curves in $\mathbb{R}^3$ at their intersection I differentiate both curves and evaluate at the intersection to find the slope of each tangent line and then do all the dot-product, cross-product, arccosine shenanigans.
But let's say I have some curve $C_1$ on the surface of a cylinder $x^2+y^2=a^2$. I want the parametric equation of $C_1$ such that the angle between the tangent of $C_1$ and the $Y axis$ at any point $P$ of $C_1$ is equal to the angle between the tangent of $C_1$ and a vector $\vec{u}$ starting at the base $(0,0,0)$ of the cylinder and passing through $P$.
I thought of writing both the $Y axis$ and $\vec{u}$ as parametric curves and making it so that both said angles are equal, but I don't know if this makes sense, and even if it did, I don't know how to proceed exactly and find the parametric equation of $C_1$ such that this condition is true for all points of $C_1$.
The parametric equations that constrain the curve to the cylinder would just be $t \mapsto \mathbb{R}^2: \langle a\cdot cos(\phi(t)), a\cdot sin(\phi(t))\rangle$. Then $z(t)$ can be any function:
$$ \vec{C}_1(t) = \left\langle a \cdot cos(\phi(t)), a \cdot sin(\phi(t)), z(t) \right\rangle \\ \frac{d\vec{C}_1}{dt} = \left \langle-a\dot{\phi}sin(\phi), a\dot{\phi} cos(\phi), \dot{z}\right\rangle $$
Now, we should set the $cos(\theta)$ of the two angles equal. I would do that by finding the dot product, dividing by the two magnitudes, then setting those two equal. The dot products are fairly easy to find, but we also need the magnitudes of the vectors (we'll just use $\langle 0, 1, 0 \rangle$ for the y-axis since that already has magnitude $1$:
$$ ||\vec{u}|| = ||\vec{C}_1|| = \sqrt{a^2 + z^2} \\ \left|\left|\frac{d\vec{C}_1}{dt}\right|\right| = \sqrt{a^2\dot{\phi}^2 + \dot{z}^2} \\ \frac{d\vec{C}_1}{dt}\circ \hat{y} = a\dot{\phi}cos(\phi) \\ \frac{d\vec{C}_1}{dt}\circ \vec{u} = a^2\dot{\phi}\left(cos(\phi)sin(\phi) - sin(\phi)cos(\phi)\right) + \dot{z}z = z\dot{z} $$
Now divide by the magnitudes and set them equal (and pray):
$$ \frac{a\dot{\phi}cos(\phi)}{\sqrt{a^2\dot{\phi}^2 + \dot{z}^2}} = \frac{z\dot{z}}{\sqrt{a^2\dot{\phi}^2 + \dot{z}^2}\sqrt{a^2 + z^2}} $$
This gives the following differential equation:
$$ \sqrt{a^2 + z^2}a\dot{\phi}cos(\phi) = z\dot{z} $$
So this looks bad, but one thing we can reason right away is that we really want a function $\phi(z)$, then we can simply say $z(t) = t$ and this should give the same curve (we can parameterize it any an infinite number of different ways--this is one particular way). If we make this substitution then we may be able to integrate to find $\phi(z)$ (note I'm not going to change much here because I'm really saying that I have $\phi(t)$ and $z(t) = t$ so that $\dot{z} = 1$):
$$ \sqrt{a^2 + t^2}acos(\phi)\frac{d\phi}{dt} = t \\ \int \left(a cos(\phi)d\phi = \frac{t}{\sqrt{a^2 + t^2}}dt \right)\\ asin(\phi) = \sqrt{a^2 + t^2} + C $$
Do you like inverse sines? I don't, so instead let's just find $t(\phi)$ (which is really $z(\phi)$:
$$ t^2 = \left(a\cdot sin(\phi) - C\right)^2 - a^2 \\ t(\phi) = \sqrt{\left(a\cdot sin(\phi) - C\right)^2 - a^2} $$
This finally gives a parameterization:
$$ \vec{C}_1(t) = \left\langle a\cdot cos(t), a\cdot sin(t), \sqrt{\left(a\cdot sin(t) - C\right)^2 - a^2}\right\rangle $$