How to find the function $\phi(u)$ fullfills the condition

Question

I am facing problem in computing the function $\phi(u)$ such that the normal unit vector of the curve is parallel to the plane XY. The curve is $$\vec{r}=(a \cos(u), a \sin(u), \phi(u))$$ thanks in advance if any one will help properly.

Answer 1

First find the tangent vector $\mathbf{t} = \frac{d\vec{r}}{du}/ ||\frac{d\vec{r}}{du}||$ for your curve and then find the unit normal by $$ \vec{n} = \frac{\mathbf{\dot{t}}}{||\mathbf{\dot{t}}||}, \quad \text{with} \quad \mathbf{\dot{t}} = \frac{d\mathbf{t}}{du} $$ for $\vec{n}$ to parallel to $XY-$plane, then its $z-$component must be zero. You can find $\phi(u)$ from above condition.

$\textbf{EDIT :}$

There are many (families) of such curve. One quick (non trivial) guess that probably work is $\phi(u) = au+b$.