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Question: Show that the parametrized curve $\alpha(t)=(cos(at+b), sin(at+b), ct+d)$ is a geodesic for the cylinder $x_1^2+x_2^2=1$, $a,b,c,d\in{\mathbb{R}}$

I know that to be a geodesic, the acceleration $\ddot{\alpha(t)}$$\in{S_{\alpha(t)}^{\bot}}$

And $\alpha(t)=(cos(at+b), sin(at+b), ct+d)$

$\dot{\alpha(t)}=(-asin(at+b), acos(at+b), c)$

$\ddot{\alpha(t)}=(-a^2cos(at+b), -asin^2(at+b), 0)$

But $\ddot{\alpha(t)}\cdot{\alpha(t)}=-a^2\neq{0}$ So I fail to see how they're orthogonal.

The solution given says

$\ddot{\alpha(t)}=(-a^2cos(at+b), -asin^2(at+b), 0)=\pm{a^2\mathbb{N}(\alpha(t))}$

But $\mathbb{N}(\alpha(t))= \frac{\dot{\alpha(t)}}{||\dot{\alpha(t)}||}\neq{\ddot{\alpha(t)}}$

So I am still confused as to where my misunderstanding lies. Thanks for your time.

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    The condition $\ddot{\alpha}(t) \cdot \alpha(t) = 0$ does not express the fact that $\ddot{\alpha}(t)$ is orthogonal to the surface $S$ unless the position vector $\alpha(t)$ happens to be orthogonal to $S$ at $\alpha(t)$. – 2017-01-30
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    What condition then expresses that $\ddot{\alpha(t)}$ is orthogonal to the surface S? – 2017-01-31
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    If $p$ is a point of $S$ and $N$ denotes a unit normal field on $S$, then any vector orthogonal to $N(p)$ is orthogonal to $S$ at $p$. – 2017-01-31

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