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I am having a lot of trouble reading some questions for Differential geometry.

The question states.

Compute the normal and geodesic curvature of these curves on the cylinder C.

$C= \{ (x,y,z) \in \mathbb {R^{3}} : x^2 + y^2 =1 \} $

$ \alpha_{1} (t)= (cos (t), sin (t),0) $

The thing thats confusing me is i have that $K_n = <\alpha^{''} (t),\dot n > $ (the normal curvature is the second derivative of alpha dot product the normal vector) But for some reason in my notes all the examples $ \dot n = \alpha (t) $

i.e

$K_n = <\alpha^{''} (t),\alpha (t) > $

$K_g = \alpha^{''} - \alpha (t) (<\alpha^{''} (t),\alpha (t) >) $

Why is the normal vector always just $\alpha (t) $ and why bother even writing $ \dot n $ if that's always the case?
Also why define the surface that the curve is on if it has nothing to do with the calculations?

Anyway im misunderstanding something any help much appreciated.

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    You want to see how much of the "curvature vector" $\kappa N$ of the curve is *normal* to the surface (that will tell you normal curvature, up to sign) and how much of it is *tangent* to the surface (that's the geodesic curvature). Geodesics on the surface are characterized by geodesic curvature = 0. – 2017-02-16
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    i Actually understand the words in what your saying... but it doesn't answer my question or if it does i don't know why. Why is the normal vector to the curve on the surface just $ \alpha_1 (t) $ ? – 2017-02-17
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    The unit normal to the cylinder (pointing outward) is the position vector of that circle, yes. Either draw the picture or just do the calculation with a parametrization of the cylinder. And the principal normal of the circle is the negative of that. – 2017-02-17

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