Razzle Dazzle

Taking our force equation, it can be expressed as $F = m\frac{dv}{dt} = [e^{-t} sin(t + \frac{\pi}{2})] \cdot (-v_{y} i + v_{x} j)$ where $v_{x}, v_{y}$ are functions of time $t$ . Knowing that $\frac{dv}{dt} = \frac{dv_{x}}{dt} i + \frac{dv_{y}}{dt} j$ , we can now match the components in the force vector equation by the following:

$m\frac{dv_{x}}{dt} = (e^{-t} sin(t + \frac{\pi}{2})) \cdot -v_{y}; \,\,\,...(i)$

$m\frac{dv_{y}}{dt} = (e^{-t} sin(t + \frac{\pi}{2})) \cdot v_{x}\,\,\,...(ii)$

We can now equate (i) and (ii) according to:

$\frac{\frac{dv_{x}}{dt}}{-v_{y}} = \frac{\frac{dv_{y}}{dt}}{v_{x}};$

$v_{x} \cdot \frac{dv_{x}}{dt} = -v_{y} \cdot \frac{dv_{y}}{dt};$

$\frac{d}{dt} \left(\frac{v^{2}_{x}}{2} \right) = \frac{d}{dt} \left(-\frac{v^{2}_{y}}{2} \right);$

$\frac{v^{2}_{x}}{2} = -\frac{v^{2}_{y}}{2} + C.$

If we substitute our initial velocities $v_{x}(0) = 3, v_{y}(0) = 4$ , we now obtain:

$\frac{3^2}{2} = -\frac{4^2}{2} + C \Rightarrow C = \frac{25}{2}$

which now leaves us with $v^{2}_{x}(t) + v^{2}_{y}(t) = 25$ (iii). Finally, the required path length is calculated per the definite integral:

$L = \int_{0}^{1} \sqrt{v^{2}_{x}(t) + v^{2}_{y}(t)} dt = \int_{0}^{1} 5 dt = \boxed{5}.$