= Solution
{c}
The <velocity> of a <fluid element> is the <time derivative> of its position with its <Lagrangian coordinates> $X,Y$ held fixed. Direct <differentiation> gives
$$
\begin{aligned}
\dot x&=-e^{-t}(X\cos\Omega t+Y\sin\Omega t)
+\Omega e^{-t}(Y\cos\Omega t-X\sin\Omega t)
=-x+\Omega e^{-2t}y,\\
\dot y&=e^t(Y\cos\Omega t-X\sin\Omega t)
-\Omega e^t(X\cos\Omega t+Y\sin\Omega t)
=y-\Omega e^{2t}x.
\end{aligned}
$$
Replacing the <Lagrangian coordinates> by the current <Eulerian coordinates> therefore gives the <velocity field>
$$
\boxed{\mathbf u(x,y,t)=(-x+\Omega e^{-2t}y,\ y-\Omega e^{2t}x)}.
$$
The distinction between following a <fluid element> and differentiating at fixed <Eulerian coordinates> is what connects the given <deformation map> to this <velocity field>.
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