I was reading a calculus math text when I stumbled upon this question.
Consider a tank that initially contains 100 gallons of solution in which 50 pounds of salt are dissolved. Supposed that 3 gallons of brine, each containing 2 pounds of salt, run into the tank each minute, and the mixture, kept uniform by stirring, runs out at the rate of 2 gallons per minute. Find the amount of salt in the tank at time $t$
A part of the solution that would answer the question above looks like this:
Let Q denote the number of pounds of salt in the tank at time $t$ in minutes. Since the salt concentration at the time $t$ is $\frac{Q}{100+t}$, we have $$\frac{dQ}{dt} = 3(2) – 2\left( \frac{Q}{100+t} \right)$$
To better understand the equation above, I put units to the terms on the differential equation above. Hence:
$$\frac{dQ}{dt} \frac{\space pounds \space of \space salt}{minute} = 3\space gallons\space\space of \space brine\left(2\space \frac{\frac{pounds\space of \space salt}{gallon\space of\space brine}}{minutes}\right) – 2 \frac{\space gallons\space of\space brine\space}{minute} \left( \frac{Q \space pounds \space of \space salt}{100 \space gallons\space of\space brine+t \space minutes} \right)$$
After cancelling out those who needed cancelling, It end up looking like this:
$$\frac{dQ}{dt} \frac{\space pounds \space of \space salt}{minute} = 6 \frac{pounds\space of\space salt}{minute} – 2 \frac{\space gallons\space of\space brine\space}{minute} \left( \frac{Q \space pounds \space of \space salt}{100 \space gallons\space of\space brine+t \space minutes} \right)$$
The first two expressions were dimensionally-correct, but the last one has dimensions of $$ \frac{\space gallons\space of\space brine\space}{minute} \left( \frac{ \space pounds \space of \space salt}{ \space gallons\space of\space brine+ \space minutes} \right)$$ which seems you can add volume and time.
How did the author derived the differential equation above? Showing the proper units of the differential equations is of great help…..
Best Answer
The dimensions for each term in the expression $2\left(\dfrac{Q}{100 + t}\right)$ are: \begin{align*} 2 & = \frac{\textrm{gallons}}{\textrm{minutes}} \\ Q & = \textrm{pounds} \\ 100 & = \textrm{gallons} \\ t & = (3-2)t = \frac{\textrm{gallons}}{\textrm{minutes}}\cdot\textrm{minutes} = \textrm{gallons} \end{align*} The coefficient of $t$ in the denominator is the net flow rate, which has dimension of gallons/minutes. Consequently, $$ 2\left(\frac{Q}{100 + t}\right) = \frac{\textrm{gallons}}{\textrm{minutes}}\frac{\textrm{pounds}}{\textrm{gallons}} = \frac{\textrm{pounds}}{\textrm{minutes}}. $$