Answer
$${\bf{T}}\left( t \right) = \left\langle { - \cos t,\sin t} \right\rangle {\text{ and }}{\bf{N}}\left( t \right) = \left\langle {\sin t,\cos t} \right\rangle $$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \left\langle {{{\cos }^3}t,{{\sin }^3}t} \right\rangle \cr
& {\text{Calculate }}{\bf{v}}\left( t \right){\text{ and }}\left| {{\bf{v}}\left( t \right)} \right| \cr
& {\bf{v}}\left( t \right) = {\bf{r}}'\left( t \right) \cr
& {\bf{v}}\left( t \right) = \left\langle { - 3{{\cos }^2}t\sin t,3{{\sin }^2}t\cos t} \right\rangle \cr
& \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {{{\left( { - 3{{\cos }^2}t\sin t} \right)}^2} + {{\left( {3{{\sin }^2}t\cos t} \right)}^2}} \cr
& \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {9{{\cos }^4}t{{\sin }^2}t + 9{{\sin }^4}t{{\cos }^2}t} \cr
& \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {9{{\sin }^2}t{{\cos }^2}t\left( {{{\cos }^2}t + {{\sin }^2}t} \right)} \cr
& \left| {{\bf{v}}\left( t \right)} \right| = 3\sin t\cos t \cr
& \cr
& {\text{Find the unit tangent vector }}{\bf{T}}\left( t \right) \cr
& {\bf{T}}\left( t \right) = \frac{{{\bf{v}}\left( t \right)}}{{\left| {{\bf{v}}\left( t \right)} \right|}} = \frac{{\left\langle { - 3{{\cos }^2}t\sin t,3{{\sin }^2}t\cos t} \right\rangle }}{{3\sin t\cos t}} \cr
& {\bf{T}}\left( t \right) = \left\langle { - \cos t,\sin t} \right\rangle \cr
& \cr
& {\text{Find the principal unit normal vector }}{\bf{N}}\left( t \right) \cr
& {\bf{N}}\left( t \right) = \frac{{{\bf{T}}'\left( t \right)}}{{\left| {{\bf{T}}'\left( t \right)} \right|}} \cr
& {\bf{T}}'\left( t \right) = \frac{d}{{dt}}\left\langle { - \cos t,\sin t} \right\rangle \cr
& {\bf{T}}'\left( t \right) = \left\langle {\sin t,\cos t} \right\rangle \cr
& \left| {{\bf{T}}'\left( t \right)} \right| = \sqrt {{{\sin }^2}t + {{\cos }^2}t} \cr
& \left| {{\bf{T}}'\left( t \right)} \right| = 1 \cr
& {\bf{N}}\left( t \right) = \frac{{{\bf{T}}'\left( t \right)}}{{\left| {{\bf{T}}'\left( t \right)} \right|}} = \left\langle {\sin t,\cos t} \right\rangle \cr
& \cr
& {\bf{T}}\left( t \right) = \left\langle { - \cos t,\sin t} \right\rangle {\text{ and }}{\bf{N}}\left( t \right) = \left\langle {\sin t,\cos t} \right\rangle \cr
& \cr} $$
