Question

If the volume of a trough with height, h(x)=x^2 and length L, has a volume of (4L/3)h^(3/2), with a hole at the bottom, whose volumetric flow rate is Q; Q=kAsqrt(h), where k is a constant, A is the area of the hole, and h is the height of the fluid.

Show that the fluid depth changes in time such that:

(4L/3)d(h^(3/2))dt=-kAsqrt(h)

Can anyone help with this?