Modeling with First Order DE’s — Question 6

Question 6 – Solution

Strategy. The water surface area changes with depth. Convert a volume-loss law into a depth equation before separating variables.

Step 1: Use the conical geometry. Similarity gives $r=h/2$, so $V=\pi h^3/12$ and $$\frac{\pi h^2}{4}h'=-0.05\sqrt h,\qquad h'=-\frac{0.20}{\pi}h^{-3/2},\qquad h(0)=4.$$ The first equality has units m$^3$/s. Division by $h^2$ is valid only while water remains.

Step 2: Integrate to emptying. Integrating $h^{3/2}\,dh=-(0.20/\pi)\,dt$ gives $h^{5/2}=32-t/(2\pi)$, hence $$\boxed{h(t)=(32-t/(2\pi))^{2/5}},\qquad 0\le t<t_e,\qquad \boxed{t_e=64\pi\approx 201.06\text{ s}}.$$ Equivalently, $h=4(1-t/t_e)^{2/5}$. Differentiation verifies the depth equation; the physical depth is set to zero for $t\ge t_e$.

Step 3: Compare depth and volume targets. At $h=2$ m, $h/4=1/2$, giving $$\boxed{t_h/t_e=1-2^{-5/2}\approx 0.8232}.$$ At half volume, $(h/4)^3=1/2$, so $$\boxed{t_V/t_e=1-2^{-5/6}\approx 0.4388}.$$ Half depth leaves only one eighth of the original volume, so it takes substantially longer than losing half the volume.

Step 4: Interpret the endpoint. As $t\uparrow t_e$, $h'\to-\infty$, so the zero continuation of depth is continuous but not differentiable at emptying. However, $V'=-0.05\sqrt h\to 0$, agreeing with $V'=0$ afterward; volume has a continuously differentiable zero continuation. The divided depth equation is a wet-tank model and is undefined at $h=0$. It must not be continued algebraically into a fictitious refilling branch.