$$\gamma (t)= (R\cos (t/R), R\sin (t/R))$$ $$\dot {\gamma (t)}=(-\sin (t/R), \cos (t/R))$$ $$n_s= (-\cos (t/R), -\sin (t/R))$$
where $n_s$ is the signed normal.
the instructor has found the $n_s$. but i didnt know how to find $n_s$.
please can someone show me how to find it? thank you
the signed normaol curve is obtained by rotating $t$ by 90 in the ccw direction
also i know that $$ \ddot {\gamma}=\kappa_s n_s$$
First we note that the curve $\gamma(t)$ is parametrized by arc length, i.e., $\|\gamma'(t)\|=1$. This is important, because the usual definition for the tangent vector and the normal vector makes use of this fact.
Now, as you have said on the comments, $n_s$ is $\gamma'$ rotated by a angle of $90$ degrees. This implies that $$\langle \gamma'(t),n_s(t)\rangle=0,\ \forall\ t\tag{1}$$
For each $t$ the above equation has two solutions and we choose the solutions in such a way that $n_s$ is continuous. I suggest you to draw a graphic and see what is going on.
It is a exercise of Linear Algebra show that $$ \begin{pmatrix} 0 & -1 \\ 1 & 0 \\ \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ -1 & 0 \\ \end{pmatrix} $$
are the only two matrices of rotation by a angle og 90 degree. The first one conrresponds to a rotation of $\pi/2$ and the second one $-\pi/2$. The usual definition consists in choosing the first matrix to apply the rotation. This implies that $$n_s(t)=\begin{pmatrix} 0 & -1 \\ 1 & 0 \\ \end{pmatrix}\begin{pmatrix} -\sin\left({t/R}\right) \\ \cos{\left(t/R\right)} \end{pmatrix}$$