 Solution

Now let the basic buoyancy beThermal-wind balance requires the basic along-front velocity to have vertical shear $V_z=M^2/f$, soThe basic absolute vorticity and buoyancy gradient areand hence

For the prescribed disturbance the buoyancy perturbation vanishes, so the linearized buoyancy equation and incompressibility giveThe wavevector must therefore satisfythe disturbance velocity lies along a basic isopycnal. The along-front momentum equation becomesProjecting the remaining momentum equations onto the divergence-free direction eliminates the pressure and yieldsThus this isopycnal disturbance grows if and only if $fQ<0$. Geometrically, $Q$ is the component of the absolute vorticity along the buoyancy gradient, multiplied by $|\nabla B|$; the horizontal buoyancy gradient reduces that component through the term $-M^4/f$. When $M=0$, the result reduces to the most unstable, nearly horizontal-wavevector limit of part i.