= Solution
{c}
The momentum of a <material volume> $V(t)$ is $\int_{V(t)}\rho\mathbf u\,dV$. <Newton's second law> equates its time derivative to the <pressure> force and <gravity>:
$$
\frac d{dt}\int_{V(t)}\rho\mathbf u\,dV
=-\int_{\partial V(t)}p\mathbf n\,dS+\int_{V(t)}\rho\mathbf g\,dV.
$$
Apply the given <Reynolds transport theorem> componentwise to the left side, and the <divergence theorem> to the <pressure> term. Then
$$
\int_{V(t)}\left[\rho(\partial_t\mathbf u+(\mathbf u\cdot\nabla)\mathbf u)+\nabla p-\rho\mathbf g\right]dV=0.
$$
Since this holds for arbitrary material volumes and the fields are continuous, the integrand vanishes pointwise. Dividing by positive <density> gives the <Euler equations for an inviscid fluid>
$$
\boxed{\partial_t\mathbf u+(\mathbf u\cdot\nabla)\mathbf u=-\rho^{-1}\nabla p+\mathbf g.}
$$
At a fixed physical time $t_0$, a <streamline> is a curve tangent to $\mathbf u(\cdot,t_0)$, which may be parametrized by $d\mathbf x/ds=\mathbf u(\mathbf x(s),t_0)$. The <chain rule> gives $d f(\mathbf x(s),t_0)/ds=\mathbf u\cdot\nabla f=0$, proving constancy along it. The time is held fixed; a time-dependent streamline statement does not by itself assert material conservation along particle trajectories.
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