Subject: Mathematics — Triangles
Question: $A$ and $B$ are two separate reservoirs of water. Capacity of reservoir $A$ is double the capacity of reservoir $B$. Both the reservoirs are filled completely with water, their inlets are closed and then the water is released simultaneously from both the reservoirs. The rate of flow of water out of each reservoir at any instant of time is proportional to the quantity of water in the reservoir at the time. One hour after the water is released, the quantity of water in reservoirs $A$ is $1\frac{1}{2}$ times the quantity of water in reservoir $B$. After how many hours do both the reservoirs have the same quantity of water?
Answer: $\frac{\log 2}{\log(4/3)}$
Solution: . Let, at any instant $t$, the volume of the reservoirs of
$A$ be $V_{A}$ and that of $B$ be $V_{B}$.
The initial capacity of $A$ be $V_{A}^{\circ}$ and $B$ be
$V_{B}^{\circ}$.
$$\therefore\frac{V_{A}^{\circ}}{V_{B}^{\circ}} = 2.$$
Also,
$$\frac{dV_{A}}{dt} = - K_{A}V_{A}\quad\text{and}\quad\frac{dV_{B}}{dt} = - K_{B}V_{B}$$
$$\Rightarrow \frac{dV_{A}}{V_{A}} = - K_{A}\, dt\quad\text{and}\quad\frac{dV_{B}}{V_{B}} = - K_{B}\, dt$$
$$\Rightarrow logV_{A} = - K_{A}t + C\quad\text{and}\quad\log V_{B} = - K_{B}t + C.$$
At $t = 0$, $V_{A} = V_{A}^{\circ}$, $V_{B} = V_{B}^{\circ}$.
$$\therefore log\frac{V_{A}}{V_{A}^{\circ}} = - K_{A}t\quad\text{and}\quad\log\frac{V_{B}}{V_{B}^{\circ}} = - K_{B}t.$$
Given that at $t = 1$,
$$\frac{V_{A}}{V_{B}} = \frac{3}{2},\quad\frac{V_{A}^{\circ}}{V_{B}^{\circ}} = 2$$
$$\Rightarrow \frac{V_{A}}{V_{B}} \cdot \frac{V_{B}^{\circ}}{V_{A}^{\circ}} = \frac{3}{4}.$$
Putting the value at $t = 1$, we get
$$\log\frac{V_{A}}{V_{A}^{\circ}} = - K_{A}$$
$$\log\frac{V_{B}}{V_{B}^{\circ}} = - K_{B}$$
$$\Rightarrow log\frac{V_{A}}{V_{A}^{\circ}} - log\frac{V_{B}}{V_{B}^{\circ}} = K_{B} - K_{A}$$
$$\Rightarrow log\left( \frac{V_{A}/V_{A}^{\circ}}{V_{B}/V_{B}^{\circ}} \right) = K_{B} - K_{A}$$
$$\Rightarrow log\left( \frac{3}{4} \right) = K_{B} - K_{A}.$$
Now, at $t = t_{1}$, say, $V_{A} = V_{B}$.
$$\Rightarrow log\frac{V_{A}}{V_{A}^{\circ}} = - K_{A}t_{1}$$
$$\log\frac{V_{B}}{V_{B}^{\circ}} = - K_{B}t_{1}$$
$$\Rightarrow log\left( \frac{V_{A}/V_{A}^{\circ}}{V_{B}/V_{B}^{\circ}} \right) = \left( K_{B} - K_{A} \right)t_{1} = \left\lbrack \log\left( \frac{3}{4} \right) \right\rbrack t_{1}$$
$$\Rightarrow log\left( \frac{1}{2} \right) = \left( \log\frac{3}{4} \right)t_{1}$$
$$\Rightarrow t_{1} = \frac{\log(1/2)}{\log(3/4)}$$
$$\therefore t_{1} = \frac{\log 2}{\log(4/3)}.$$
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