Show γ(t) = (a cos(t), a sin(t), t) is a geodesic of the cylinder . The first part of this question required me to compute the general geodesic of a surface of revolution. And I obtained a general equation using Christoffel's symbols.
The second part asked me to find use the first part to find the geodesic of a cylinder and I obtained that:
$$u''(t)=0$$$$
v''(t)=0$$
and then the third part is the question above. I believe that the aim here is to use what I obtained $(u''(t)=v''(t)=0)$.
But to me this makes 
$$u(t)=a+bt \text{ and } v(t)=c+dt $$
so I aim slightly confused as to how straight lines would take that form. 
I apologise if this is a duplicate, I did make sure to search beforehand but I could not find one. Furthemore, please be kind. 

The equation of the geodesic is given by $\phi(u,v)=(a\cos u,a\sin u, v)$ after you substitute $u=v=t$:
$$
\gamma(t)=\phi(u(t),v(t))=(a\cos t,a\sin t, t).
$$
