Answer
The normal vector at $t = \frac{\pi }{4}$ is
${\bf{N}}\left( {\frac{\pi }{4}} \right) = - \sqrt {\frac{2}{3}} \left( {\frac{1}{2}\sqrt 2 ,1} \right)$
The normal vector at $t = \frac{{3\pi }}{4}$ is
${\bf{N}}\left( {\frac{{3\pi }}{4}} \right) = \sqrt {\frac{2}{3}} \left( {\frac{1}{2}\sqrt 2 ,1} \right)$
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( {t,\cos t} \right)$.
The tangent vector is ${\bf{r}}'\left( t \right) = \left( {1, - \sin t} \right)$.
We find the unit tangent vector ${\bf{T}}\left( t \right)$:
${\bf{T}}\left( t \right) = \frac{{{\bf{r}}'\left( t \right)}}{{||{\bf{r}}'\left( t \right)||}} = \frac{{\left( {1, - \sin t} \right)}}{{\sqrt {\left( {1, - \sin t} \right)\cdot\left( {1, - \sin t} \right)} }}$
${\bf{T}}\left( t \right) = \frac{{\left( {1, - \sin t} \right)}}{{\sqrt {1 + {{\sin }^2}t} }} = \left( {\frac{1}{{\sqrt {1 + {{\sin }^2}t} }},\frac{{ - \sin t}}{{\sqrt {1 + {{\sin }^2}t} }}} \right)$
To evaluate the derivative of ${\bf{T}}\left( t \right)$:
1. Evaluate the derivative of $\frac{1}{{\sqrt {1 + {{\sin }^2}t} }}$
$\frac{d}{{dt}}\left( {{{\left( {1 + {{\sin }^2}t} \right)}^{ - 1/2}}} \right) = - \frac{{2\sin t\cos t}}{2}{\left( {1 + {{\sin }^2}t} \right)^{ - 3/2}}$
$\frac{d}{{dt}}\left( {{{\left( {1 + {{\sin }^2}t} \right)}^{ - 1/2}}} \right) = - \frac{{\cos t\sin t}}{{{{\left( {1 + {{\sin }^2}t} \right)}^{3/2}}}}$
2. Evaluate the derivative of $\frac{{ - \sin t}}{{\sqrt {1 + {{\sin }^2}t} }}$
$\frac{d}{{dt}}\left( {\frac{{ - \sin t}}{{\sqrt {1 + {{\sin }^2}t} }}} \right) = \frac{{\sqrt {1 + {{\sin }^2}t} \left( { - \cos t} \right) - \left( { - \sin t} \right)\left( {\frac{{2\sin t\cos t}}{{2\sqrt {1 + {{\sin }^2}t} }}} \right)}}{{1 + {{\sin }^2}t}}$
$ = \frac{{\left( {1 + {{\sin }^2}} \right)\left( { - \cos t} \right) + {{\sin }^2}t\cos t}}{{{{\left( {1 + {{\sin }^2}t} \right)}^{3/2}}}}$
$\frac{d}{{dt}}\left( {\frac{{ - \sin t}}{{\sqrt {1 + {{\sin }^2}t} }}} \right) = - \frac{{\cos t}}{{{{\left( {1 + {{\sin }^2}t} \right)}^{3/2}}}}$
Thus, the derivative of the tangent vector is
${\bf{T}}'\left( t \right) = \left( { - \frac{{\cos t\sin t}}{{{{\left( {1 + {{\sin }^2}t} \right)}^{3/2}}}}, - \frac{{\cos t}}{{{{\left( {1 + {{\sin }^2}t} \right)}^{3/2}}}}} \right)$
${\bf{T}}'\left( t \right) = - \frac{{\cos t}}{{{{\left( {1 + {{\sin }^2}t} \right)}^{3/2}}}}\left( {\sin t,1} \right)$
3. Evaluate $||{\bf{T}}'\left( t \right)||$
$||{\bf{T}}'\left( t \right)|{|^2} = \frac{{{{\cos }^2}t}}{{{{\left( {1 + {{\sin }^2}t} \right)}^3}}}\left( {\sin t,1} \right)\cdot\left( {\sin t,1} \right)$
$||{\bf{T}}'\left( t \right)|{|^2} = \frac{{{{\cos }^2}t}}{{{{\left( {1 + {{\sin }^2}t} \right)}^3}}}\left( {{{\sin }^2}t + 1} \right)$
$||{\bf{T}}'\left( t \right)|{|^2} = \frac{{{{\cos }^2}t}}{{{{\left( {1 + {{\sin }^2}t} \right)}^2}}}$
So, $||{\bf{T}}'\left( t \right)|| = \pm \frac{{\cos t}}{{1 + {{\sin }^2}t}}$
By Eq. (6), the normal vector ${\bf{N}}\left( t \right)$ is given by
${\bf{N}}\left( t \right) = \frac{{{\bf{T}}'\left( t \right)}}{{||{\bf{T}}'\left( t \right)||}}$
Substituting ${\bf{T}}'\left( t \right)$ and $||{\bf{T}}'\left( t \right)||$ in ${\bf{N}}\left( t \right)$ gives
${\bf{N}}\left( t \right) = \mp \frac{{\cos t}}{{{{\left( {1 + {{\sin }^2}t} \right)}^{3/2}}}}\left( {\sin t,1} \right)\left( {\frac{{1 + {{\sin }^2}t}}{{\cos t}}} \right)$
${\bf{N}}\left( t \right) = \mp \frac{1}{{\sqrt {1 + {{\sin }^2}t} }}\left( {\sin t,1} \right)$
At $t = \frac{\pi }{4}$, we get
${\bf{N}}\left( {\frac{\pi }{4}} \right) = \mp \frac{1}{{\sqrt {1 + {{\left( {\frac{1}{2}\sqrt 2 } \right)}^2}} }}\left( {\frac{1}{2}\sqrt 2 ,1} \right) = \mp \sqrt {\frac{2}{3}} \left( {\frac{1}{2}\sqrt 2 ,1} \right)$
At $t = \frac{\pi }{4}$, for the normal vector to point inside the curve, the correct choice is
${\bf{N}}\left( {\frac{\pi }{4}} \right) = - \sqrt {\frac{2}{3}} \left( {\frac{1}{2}\sqrt 2 ,1} \right)$
At $t = \frac{{3\pi }}{4}$, we get
${\bf{N}}\left( {\frac{{3\pi }}{4}} \right) = \mp \frac{1}{{\sqrt {1 + {{\left( {\frac{1}{2}\sqrt 2 } \right)}^2}} }}\left( {\frac{1}{2}\sqrt 2 ,1} \right) = \mp \sqrt {\frac{2}{3}} \left( {\frac{1}{2}\sqrt 2 ,1} \right)$
At $t = \frac{{3\pi }}{4}$, for the normal vector to point inside the curve, the correct choice is
${\bf{N}}\left( {\frac{{3\pi }}{4}} \right) = \sqrt {\frac{2}{3}} \left( {\frac{1}{2}\sqrt 2 ,1} \right)$
