Chapter 14 - Calculus of Vector-Valued Functions - 14.4 Curvature - Exercises - Page 735: 36

The normal vector ${\bf{N}}\left( \theta \right)$ of radius $R$ is ${\bf{N}}\left( \theta \right) = - \left( {\cos \theta ,\sin \theta } \right)$. Since the normal vector has negative sign, it points inside the circle. Please see the figure attached.

We have ${\bf{r}}\left( \theta \right) = R\left( {\cos \theta ,\sin \theta } \right)$. The tangent vector is ${\bf{r}}'\left( \theta \right) = R\left( { - \sin \theta ,\cos \theta } \right)$, where prime denotes differentiation with respect to $\theta$. We find the unit tangent vector ${\bf{T}}\left( \theta \right)$: ${\bf{T}}\left( \theta \right) = \frac{{{\bf{r}}'\left( \theta \right)}}{{||{\bf{r}}'\left( \theta \right)||}} = \frac{{R\left( { - \sin \theta ,\cos \theta } \right)}}{{\sqrt {{R^2}\left( { - \sin \theta ,\cos \theta } \right)\cdot\left( { - \sin \theta ,\cos \theta } \right)} }}$ ${\bf{T}}\left( \theta \right) = \frac{{\left( { - \sin \theta ,\cos \theta } \right)}}{{\sqrt {{{\sin }^2}\theta + {{\cos }^2}\theta } }} = \left( { - \sin \theta ,\cos \theta } \right)$ So, ${\bf{T}}'\left( \theta \right) = \left( { - \cos \theta , - \sin \theta } \right)$. By Eq. (6), the normal vector ${\bf{N}}\left( \theta \right)$ is given by ${\bf{N}}\left( \theta \right) = \frac{{{\bf{T}}'\left( \theta \right)}}{{||{\bf{T}}'\left( \theta \right)||}}$ ${\bf{N}}\left( \theta \right) = \frac{{\left( { - \cos \theta , - \sin \theta } \right)}}{{\sqrt {\left( { - \cos \theta , - \sin \theta } \right)\cdot\left( { - \cos \theta , - \sin \theta } \right)} }}$ ${\bf{N}}\left( \theta \right) = \frac{{\left( { - \cos \theta , - \sin \theta } \right)}}{{\sqrt {{{\cos }^2}\theta + {{\sin }^2}\theta } }} = - \left( {\cos \theta ,\sin \theta } \right)$ Since the normal vector has negative sign, it points inside the circle. At $\theta = \frac{\pi }{4}$, the normal vector is ${\bf{N}}\left( {\frac{\pi }{4}} \right) = - \left( {\frac{1}{2}\sqrt 2 ,\frac{1}{2}\sqrt 2 } \right)$. For $R=4$, we have ${\bf{r}}\left( {\frac{\pi }{4}} \right) = \left( {2\sqrt 2 ,2\sqrt 2 } \right)$.