Answer
We verify that $\kappa \left( t \right) = \left| t \right|$
by using the Fundamental Theorem of Calculus and Eq. (11).
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( {x\left( t \right),y\left( t \right)} \right)$, where
$x\left( t \right) = \mathop \smallint \limits_0^t \sin \frac{{{u^2}}}{2}{\rm{d}}u$, ${\ \ }$ $y\left( t \right) = \mathop \smallint \limits_0^t \cos \frac{{{u^2}}}{2}{\rm{d}}u$
By the Fundamental Theorem of Calculus:
$x'\left( t \right) = \sin \frac{{{t^2}}}{2}$, ${\ \ }$ $y'\left( t \right) = \cos \frac{{{t^2}}}{2}$
The second derivatives are
$x{\rm{''}}\left( t \right) = t\cos \frac{{{t^2}}}{2}$, ${\ \ }$ $y{\rm{''}}\left( t \right) = - t\sin \frac{{{t^2}}}{2}$
Substituting these in Eq. (11) of Exercise 28:
$\kappa \left( t \right) = \frac{{\left| {x'\left( t \right)y{\rm{''}}\left( t \right) - x{\rm{''}}\left( t \right)y'\left( t \right)} \right|}}{{{{\left( {x'{{\left( t \right)}^2} + y'{{\left( t \right)}^2}} \right)}^{3/2}}}}$
gives
$\kappa \left( t \right) = \frac{{\left| { - t{{\sin }^2}\frac{{{t^2}}}{2} - t{{\cos }^2}\frac{{{t^2}}}{2}} \right|}}{{{{\left( {{{\sin }^2}\frac{{{t^2}}}{2} + {{\cos }^2}\frac{{{t^2}}}{2}} \right)}^{3/2}}}} = \frac{{\left| { - t\left( {{{\sin }^2}\frac{{{t^2}}}{2} + {{\cos }^2}\frac{{{t^2}}}{2}} \right)} \right|}}{{{{\left( {{{\sin }^2}\frac{{{t^2}}}{2} + {{\cos }^2}\frac{{{t^2}}}{2}} \right)}^{3/2}}}}$
Since ${\sin ^2}\frac{{{t^2}}}{2} + {\cos ^2}\frac{{{t^2}}}{2} = 1$, so $\kappa \left( t \right) = \left| { - t} \right| = \left| t \right|$.
