Question

A & B are two separate reservoirs of water. The capacity of reservoir A is double the capacity of reservoir B. Both the reservoirs are filled 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 that time. One hour after the water is released, the quantity of water in reservoir A is 1.5 times the quantity of water in reservoir B. After how many hours do both the reservoirs have the same quantity of water?
(Enter n if the answer is $T={{\log }_{\dfrac{4}{3}}}n$)

Answer 1

Hint: First find the rate of change of the volume of tank A, and integrate it. Do the same for tank B. Now, at start time, water in tank A is twice the water in tank B, substitute the values in the given condition, and find the relation between the constant term ${{C}_{1}}$ and ${{C}_{2}}$. After 1 hour, water in tank A is 1.5 times the water in tank B, substitute the values in a given condition, and find the difference of ${{k}_{1}}$ and ${{k}_{2}}$. For the time at which the water in tank A and tank B are the same, substitute the values and simplify to get the desired result.

Complete step-by-step answer:
Given: - Volume of A is twice the volume of B at the start.
After 1 hour, the volume of A is 1.5 times the volume of B.
Let the water present at any time in A be $x$ and in B be $y$.
For tank A,
$-\dfrac{dx}{dt}={{k}_{1}}x$
Multiply both sides by -1,
$\dfrac{dx}{dt}=-{{k}_{1}}x$
Move $dt$ to the right side and $x$ to the denominator of the left side,
$\dfrac{dx}{x}=-{{k}_{1}}dt$
Integrate both sides of the equation,
$\int{\dfrac{dx}{x}=-\int{{{k}_{1}}dt}}$
After integration, the equation will be,
$\log x=-{{k}_{1}}t+{{c}_{1}}$
Take exponential on both sides,
${{e}^{\log x}}={{e}^{-{{k}_{1}}t+{{c}_{1}}}}$
Cancel out exponential with log on the left side and factor the terms on the right side,
$x={{e}^{-{{k}_{1}}t}}\times {{e}^{{{c}_{1}}}}$
Let ${{C}_{1}}={{e}^{{{c}_{1}}}}$. Then,
$x={{C}_{1}}{{e}^{-{{k}_{1}}t}}$ ….. (1)
For tank B,
$-\dfrac{dy}{dt}={{k}_{2}}y$
Multiply both sides by -1,
$\dfrac{dy}{dt}=-{{k}_{2}}y$
Move $dt$ to the right side and $y$ to the denominator of the left side,
$\dfrac{dy}{y}=-{{k}_{2}}dt$
Integrate both sides of the equation,
$\int{\dfrac{dy}{y}=-\int{{{k}_{2}}dt}}$
After integration, the equation will be,
$\log y=-{{k}_{2}}t+{{c}_{2}}$
Take exponential on both sides,
${{e}^{\log y}}={{e}^{-{{k}_{2}}t+{{c}_{2}}}}$
Cancel out exponential with log on the left side and factor the terms on the right side,
$y={{e}^{-{{k}_{2}}t}}\times {{e}^{{{c}_{2}}}}$
Let ${{C}_{2}}={{e}^{{{c}_{2}}}}$. Then,
$y={{C}_{2}}{{e}^{-{{k}_{2}}t}}$ ….. (2)
At the time $t=0$,
$x=2y$
Substitute the values from equation (1) and (2),
${{C}_{1}}{{e}^{-{{k}_{1}}\times 0}}={{C}_{2}}{{e}^{-{{k}_{2}}\times 0}}$
As we know that, any number raised to the power of 0 equals 1. Then,
${{C}_{1}}=2{{C}_{2}}$ ….. (3)
At the time $t=1$,
$x=1.5y$
Substitute the values from equation (1) and (2),
${{C}_{1}}{{e}^{-{{k}_{1}}\times 1}}=\dfrac{3}{2}{{C}_{2}}{{e}^{-{{k}_{2}}\times 1}}$
Substitute the value of ${{C}_{1}}$ from equation (3),
$2{{C}_{2}}{{e}^{-{{k}_{1}}}}=\dfrac{3}{2}{{C}_{2}}{{e}^{-{{k}_{2}}}}$
Cancel out the common term,
$2{{e}^{-{{k}_{1}}}}=\dfrac{3}{2}{{e}^{-{{k}_{2}}}}$
Move variables on one side and constant on another side,
$\dfrac{{{e}^{-{{k}_{2}}}}}{{{e}^{-{{k}_{1}}}}}=\dfrac{4}{3}$
By the law of indices,
${{e}^{{{k}_{1}}-{{k}_{2}}}}=\dfrac{4}{3}$
Take log on both sides,
$\log {{e}^{{{k}_{1}}-{{k}_{2}}}}=\log \left( \dfrac{4}{3} \right)$
Cancel out ln with exponential,
${{k}_{1}}-{{k}_{2}}=\log \left( \dfrac{4}{3} \right)$ ….. (4)
Now, for $x=y$,
${{C}_{1}}{{e}^{-{{k}_{1}}t}}={{C}_{2}}{{e}^{-{{k}_{2}}t}}$
Substitute the value of ${{C}_{1}}$ from equation (3),
$2{{C}_{2}}{{e}^{-{{k}_{1}}t}}={{C}_{2}}{{e}^{-{{k}_{2}}t}}$
Cancel out the common term,
$2{{e}^{-{{k}_{1}}t}}={{e}^{-{{k}_{2}}t}}$
Move variables on one side and constant on another side,
$\dfrac{{{e}^{-{{k}_{2}}t}}}{{{e}^{-{{k}_{1}}t}}}=2$
By the law of indices,
${{e}^{\left( {{k}_{1}}-{{k}_{2}} \right)t}}=2$
Take log on both sides,
$\log {{e}^{\left( {{k}_{1}}-{{k}_{2}} \right)t}}=\log 2$
Cancel out ln with exponential,
$\left( {{k}_{1}}-{{k}_{2}} \right)t=\log 2$
Move $\left( {{k}_{1}}-{{k}_{2}} \right)$ to the denominator of the right side and substitute the value from equation (4),
$t=\dfrac{\log 2}{\log \left( \dfrac{4}{3} \right)}$
Apply log base change law,
$t={{\log }_{\dfrac{4}{3}}}2$
Hence, the time at which both the reservoirs have the same quantity of water is 2 hours.

The value of n is equal to n.
Note: A differential equation can be defined as an equation that consists of a function along with one or more derivatives. The functions of a differential equation usually represent the physical quantities whereas the rate of change of the physical quantities is expressed by its derivatives. A differential equation is a relationship between the function and its derivatives.