Answer
$${\bf{T}}\left( t \right) = \frac{{\left\langle {1,4t} \right\rangle }}{{\sqrt {1 + 16{t^2}} }}{\text{ and }}\kappa \left( t \right) = \frac{4}{{{{\left( {1 + 16{t^2}} \right)}^{3/2}}}}$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \left\langle {t,2{t^2}} \right\rangle \cr
& {\text{Calculate }}{\bf{v}}\left( t \right) \cr
& {\bf{v}}\left( t \right) = {\bf{r}}'\left( t \right) \cr
& {\bf{v}}\left( t \right) = \frac{d}{{dt}}\left\langle {t,2{t^2}} \right\rangle \cr
& {\bf{v}}\left( t \right) = \left\langle {1,4t} \right\rangle \cr
& {\text{Find the unit tangent vector }}{\bf{T}} \cr
& {\bf{T}}\left( t \right) = \frac{{{\bf{v}}\left( t \right)}}{{\left| {{\bf{v}}\left( t \right)} \right|}} \cr
& {\bf{T}}\left( t \right) = \frac{{\left\langle {1,4t} \right\rangle }}{{\left| {\left\langle {1,4t} \right\rangle } \right|}} = \frac{{\left\langle {1,4t} \right\rangle }}{{\sqrt {1 + 16{t^2}} }} \cr
& {\bf{T}}\left( t \right) = \frac{{\left\langle {1,4t} \right\rangle }}{{\sqrt {1 + 16{t^2}} }} \cr
& \cr
& {\text{Therefore}}{\text{, the curvature is}} \cr
& \kappa \left( t \right) = \frac{1}{{\left| {\bf{v}} \right|}}\left| {\frac{{d{\bf{T}}}}{{dt}}} \right| \cr
& \frac{{d{\bf{T}}}}{{dt}} = \frac{1}{{\sqrt {1 + 16{t^2}} }}\left| {\frac{d}{{dt}}\left[ {\frac{{\left\langle {1,4t} \right\rangle }}{{\sqrt {1 + 16{t^2}} }}} \right]} \right| \cr
& \frac{{d{\bf{T}}}}{{dt}} = \frac{1}{{\sqrt {1 + 16{t^2}} }}\left| {\left\langle { - \frac{{16t}}{{{{\left( {1 + 16{t^2}} \right)}^{3/2}}}},4{{\left( {1 + 16{t^2}} \right)}^{ - 5/2}}\left( {1 - 28{t^2}} \right)} \right\rangle } \right| \cr
& \kappa \left( t \right) = \frac{1}{{\sqrt {1 + 16{t^2}} }}\left( {\sqrt {\frac{{16}}{{{{\left( {1 + 16{t^2}} \right)}^2}}}} } \right) \cr
& \kappa \left( t \right) = \frac{1}{{\sqrt {1 + 16{t^2}} }}\left( {\frac{4}{{1 + 16{t^2}}}} \right) \cr
& \kappa \left( t \right) = \frac{4}{{{{\left( {1 + 16{t^2}} \right)}^{3/2}}}} \cr
& \cr
& {\bf{T}}\left( t \right) = \frac{{\left\langle {1,4t} \right\rangle }}{{\sqrt {1 + 16{t^2}} }}{\text{ and }}\kappa \left( t \right) = \frac{4}{{{{\left( {1 + 16{t^2}} \right)}^{3/2}}}} \cr} $$
