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