Calling Upon The Wizard

A lot of details of approaching the matrix exponential can be found in the solution to this problem .

Essentially:

$\mathrm{e}^{\left[ \begin{matrix} 0&-t^2\\t^2&0 \end{matrix}\right]} \left[ \begin{matrix} 1\\ 3\end{matrix}\right] = \left[ \begin{matrix} \cos(t^2)&-\sin(t^2)\\ \sin(t^2)&\cos(t^2) \end{matrix}\right]\left[ \begin{matrix} 1\\ 3\end{matrix}\right] = \left[ \begin{matrix} \cos(t^2)-3\sin(t^2)\\ \sin(t^2)+3\cos(t^2)\end{matrix}\right]$

Now:

$x = \int (\cos(t^2)-3\sin(t^2)) \ dt\implies \frac{dx}{dt} = \cos(t^2)-3\sin(t^2)$ $y =\int (\sin(t^2)+3\cos(t^2)) \ dt[\implies \frac{dy}{dt} =\sin(t^2)+3\cos(t^2)$

The arc length in the given interval is therefore:

$L=\int_{0}^{5}\sqrt{\left( \frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \ dt = \sqrt{10}\int_{0}^{5} \ dt= 5\sqrt{10}$