Net Change and In-Out Rate Problems

A tank with water flowing in and out has two separate rate functions working against each other, so one rate alone never tells you what's actually in the tank. The inflow rate tells you how fast water arrives; the outflow rate tells you how fast it leaves. Subtract the second from the first and you get the net rate, the true speed at which the water level is rising or falling at each instant. That net rate is a derivative, so to undo it and get back to an actual amount, you integrate it over time and add whatever was there at the start.

This gives a simple template: amount(t) = starting amount + integral from 0 to t of (inflow rate minus outflow rate) dt. Plug in a specific time to get the amount at that moment. To find when the tank is fullest, you do not maximize the inflow rate by itself. You look for where the net rate changes from positive to negative, since that's where the amount stops rising and starts falling. Net rate equals zero exactly when inflow equals outflow, so solve inflow(t) = outflow(t) and check the sign change around that time.

This topic is the accumulation idea from earlier integration work applied to a physical situation with two competing rates instead of one. It leans on the same logic as total distance from velocity, but now a maximum or minimum of the accumulated quantity depends on a sign change in a difference of two functions rather than in a single rate.