An ordinary differential equation problem

Question

Our teacher had asked that question in the exam but I couldn't solve it so I write it so you may help me. I really appreciate your help.

Water in a cylindrical tank is drained from a tap at the bottom of the tank. The volume $V$ of the water in the tank is expressed by the following differential equation

$$ \frac{dV}{dt} = -k\sqrt{V},$$

where $k$ is constant. There is initially $20$ liters of water in the tank. And the water is discharged initially at a rate of $1$L / min. Accordingly, how long does it take the tank to empty completely?

Here's an approach: https://math.stackexchange.com/questions/1549982/how-to-solve-the-differential-equation-y-sqrty — Matti P., Jul 10 '20 at 08:20
For this question, you need to separate the variables and then integrate both sides. Remember that $k$ is a constant so you can integrate it with respect to $t$. — Toby Mak, Jul 10 '20 at 08:30

score 2 · Answer 1 · answered Jul 10 '20 at 08:42

2

Use separation of variables and then integrate: $$\frac{dv}{dt} = -k\sqrt{v}\quad \Rightarrow \quad v^{-1/2}dv = -kdt \quad \Rightarrow \quad \int_{v_0}^{v_f}v^{-1/2}dv = -\int_0^{t_f}k\,dt.$$ Here $v_f$ is the final volume of water at time $t_f$, $v_0$ is the volume at time $t=0$. Using the initial rate of discharge you can find the right value of $k$ giving you $t_f$.

answered Jul 10 '20 at 08:42

Gibbs

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Which factor 2? – Gibbs Jul 10 '20 at 08:47
Sorry, misread. – Jul 10 '20 at 08:49

score 1 · Answer 2 · 2020-07-10T08:56:14.123

1

This is a so-called separable equation. Assuming $v\ne0$, you rewrite

$$\frac{dv}{2\sqrt v}=-\frac k2dt$$ and integrate

$$\int_{V_0}^V\frac{dv}{2\sqrt v}=-\frac{k}2\int_0^Tdt$$

or

$$\sqrt V-\sqrt{V_0}=-\frac{kT}2.$$

For emptying, just consider $V=0$ and

$$-\sqrt{V_0}=-\frac{kT}2.$$ The initial discharging rate, let $D_0$ tells you

$$D_0=k\sqrt{V_0}$$ and finally,

$$T=\frac{2V_0}{D_0}.$$

edited Jul 10 '20 at 08:56

answered Jul 10 '20 at 08:44

How many minutes does it take to fully discharge? – Little Prince Jul 10 '20 at 09:01
@LittlePrince: are you serious ? – Jul 10 '20 at 09:03

score 0 · Answer 3 · answered Jul 10 '20 at 08:47

Separate the variables as follows:

$$\frac{1}{\sqrt V} \ dV = -k \ dt \Rightarrow \int V^{-1/2}\ dV = \int -k \ dt \Rightarrow 2V^{1/2} = -kt+C \tag{1}$$

At $t = 0$, $V = 20$. This means that $2 \cdot (20)^{1/2} = -k(0) +C$ so $C = 2 \sqrt{20}$.

The other condition is that when $t = 0$, $\frac{dV}{dt} = -1$, where the negative sign indicates the amount of water in the tank is decreasing. When $t = 0$, $2V^{1/2} = 2 \sqrt{20}$ from equation $(1)$, so $V = 20$. This means that we can substitute back into the original condition $\frac{dV}{dt} = -k \sqrt{V}$ to get that $-1 = -k \sqrt{20}$. What must $k$ be then?

Finally, when the tank is empty, $V = 0$. Going back to equation $(1)$, you need that $0 = -kt+C$. You have the values of $k$ and $C$, so you just need a simple linear equation to solve for $t$.

An ordinary differential equation problem

3 Answers3