explanation of differential geometry concept The  derivative of paremeterized curve $r(t)$  gives the attached tangent vector at that point , if $\dot{r}(t)=\vec t$ is further differentiated , $r''(t)=\dot{\vec{t}}=\kappa n$. where n is the normal attached at that point 
but in the given illustration the $\dot{\vec{t}}$ is not in the direction of n .
please help i am confused and quite naive in this topic.
is the principal normal vector $\vec{p}$ different from the normal vector n?
reference page 5 https://www.cmu.edu/biolphys/deserno/pdf/diff_geom.pdf

 A: $\newcommand{\Vec}[1]{\mathbf{#1}}$In a word, "yes": There are two distinct "normal vectors" in your situation.
(Notation below altered slightly in edit to match the diagram below rather than the diagram in the question.)


*

*The unit normal field to the surface $S$, labeled "$\Vec{n}$" in the diagram below.

*The principal normal $\Vec{N}$ of the path $\Vec{r}$, which satisfies $\Vec{r}''(t) = \kappa \Vec{N}$.
If the image of a path $\Vec{r}$ lies on a surface $S$, the second derivative of $\Vec{r}$ may be decomposed into components parallel to and orthogonal to the surface normal. The magnitudes of these components are the normal curvature $\kappa_{n}$ and geodesic curvature $\kappa_{g}$, respectively.
If $\Vec{T}(t) = \dfrac{\Vec{r}'(t)}{\|\Vec{r}'(t)\|}$ is the unit tangent field of $\Vec{r}$, then
$$
\Vec{r}''(t) = \kappa \Vec{N} = \kappa_{n} \Vec{n} + \kappa_{g} (\Vec{n} \times \Vec{T}).
$$
The diagram below shows a latitude on a sphere, parametrized at unit speed. The principal normal $\Vec{N}$ lies in the plane of the latitude circle, and points directly at the center. The acceleration $\Vec{r}''(t)$, which is parallel to the principal normal (but neither parallel nor orthogonal to the unit surface normal $\Vec{n}$ in this instance), decomposes uniquely into a component $\kappa_{n} \Vec{n}$ normal to $S$ and a component $\kappa_{g} (\Vec{n} \times \Vec{T})$ tangent to $S$. The magnitudes of these components are the normal curvature $\kappa_{n}$ and geodesic curvature $\kappa_{g}$ of the latitude. Neither quantity is equal to the curvature $\kappa$ of the latitude. (In fact, the Pythagorean theorem gives $\kappa = \sqrt{\kappa_{n}^{2} + \kappa_{g}^{2}}$.)

A: "The derivative of parameterized curve, r(t), gives a tangent vector"- there are many different tangent vectors, all pointing in the same direction with different lengths.  If the parameter, t, is "arclength" the derivative with respect to t gives the unit tangent vector an in that case, the second derivative is the normal vector.
