AP Calculus AB and BC glossary

Rate In Minus Rate Out

Also called: Net rate, Tank problem

Rate in minus rate out is the standard accumulation setup: the net rate is the rate entering minus the rate leaving. The amount at time t is the starting amount plus the integral of that difference, and it can only turn around where the rates cross, so the maximum is such a crossing or an endpoint.

A(t) = A(0) + \int_0^t \left[R_{\text{in}}(s) - R_{\text{out}}(s)\right] ds

Water flows into a tank at $R_{\text{in}}(t)$ and drains at $R_{\text{out}}(t)$ . Neither integral alone answers how much is in the tank. The net rate is the difference, and the amount present is the initial amount plus everything that difference has accumulated since then.

\begin{aligned} A'(t) &= R_{\text{in}}(t) - R_{\text{out}}(t) \\ A'(t) = 0 \ &\Longleftrightarrow\ R_{\text{in}}(t) = R_{\text{out}}(t) \end{aligned}

The amount peaks where its derivative turns from positive to negative, which is the moment the inflow rate falls below the outflow rate. Locating that crossing is the calculation, and stating that the net rate changes sign from positive to negative there is the justification. If the net rate never changes from positive to negative on the interval, the maximum sits at an endpoint, so the endpoints belong in the candidates list alongside every crossing.

The mistake

Not every crossing is a maximum. Where the net rate turns from negative to positive the amount is at its smallest, and on an interval where inflow never drops below outflow there is no interior candidate at all. Check the sign of $R_{\text{in}} - R_{\text{out}}$ on each side of the crossing, then compare the endpoint values before naming a maximum.

Appears in: Unit 6: Integration and Accumulation, Unit 8: Applications of Integration