We have the following scenario: There is a 100 gallon tank with 20 lb of salt dissolved in it. We have a pipe flowing into the tank at 4 gal/min delivering solution that has 0.005 lb/gal of salt, and we have an output that is flowing out of the tank at 5 gal/min, with salt content same as that of the tank the instant it drained out.
We call a variable $Q$ as the pounds of salt in the tank, and we want to find the function that describes $Q$. My calculus teacher says that we should do it using differential equations.
$$Q'(lb/min)= 4(gal/min)\cdot0.005 lb/gal (IN) – 5(gal/min)\cdot Q (lb)/V(gal)$$
where $V=100-t$ because of the way the tank drains. However, the solution to that doesn't really fit the logic of the problem that well. Why can't we just do it without differential equations, like this?
Because the solution flowing in is the same no matter what time, we can say that the pounds of salt coming in are
$$4(gal/min)\cdot0.005 lb/gal\cdot t (min)=0.02t(lb)$$
Similarly, the pounds of salt going out is just going to be
$$5(gal/min)\cdot \frac{Q}{100-t} (lb/gal)\cdot t (min)=\frac{5tQ}{100-t} (lb)$$
And have $Q$ equal to the difference of the two plus the original 20 pounds or salt?
$$Q=0.02t-\frac{5tQ}{100-t}+20$$
We can then solve for $Q$ to be
$$Q=\frac{(0.02t+20)(100-t)}{100+4t}$$
So, who's right? (1) Does my solution work, or do we have to use differential equations, and (2) how can we tell when to use what for future problems?
Best Answer
In your language, it should be "the pounds of salt that have come in by time $t$" to show that it is the total, not the rate. In your next sentence you are also trying to compute the pounds of salt that have left by time $t$. The problem is that the rate of salt leaving is not constant, so to compute the amount of salt that has left by time $t$ you need to integrate $\frac {5Q(t)}{100-t}$ with respect to $t$. To get rid of that integral you will differentiate the equation and get back to your teacher's differential equation.