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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.....

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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, $$ 2\left(\frac{Q}{100 + t}\right) = \frac{\textrm{gallons}}{\textrm{minutes}}\frac{\textrm{pounds}}{\textrm{gallons}} = \frac{\textrm{pounds}}{\textrm{minutes}}. $$

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  • $\begingroup$ This is clearXD $\endgroup$ – Palautot Ka Aug 21 '17 at 16:07

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