Rate of change of water depth in a triangular trough

Set up the cross-section: the triangular end has height $1$ m and top width $2$ m, so at depth $h$ the water's cross-sectional width $w$ satisfies $\frac{w}{h} = \frac{2}{1}$ by similar triangles, giving $w = 2h$. The cross-sectional area of water is a triangle with base $w=2h$ and height $h$, so $A = \frac{1}{2}(2h)(h) = h^2$. Since the trough has length $5$ m, the volume of water is $V = 5h^2$.

Differentiate both sides of $V = 5h^2$ with respect to $t$: $\frac{dV}{dt} = 10h\frac{dh}{dt}$. Substitute $\frac{dV}{dt} = 3$ m$^3$/min and $h = 0.5$ m: $3 = 10(0.5)\frac{dh}{dt} = 5\frac{dh}{dt}$, so $\frac{dh}{dt} = 0.6$ m/min.