Rate In, Rate Out: Accumulation in Context

The amount is the start plus the net accumulation

Water in a tank, people in a queue, sand on a beach: whenever the problem gives rates, the amount at time $t$ is the start plus the integral of the net rate. On its own, $\displaystyle\int_a^b R_{\text{in}}(t)\,dt$ is "the total amount that entered between $t=a$ and $t=b$" — a standard interpretation question, units included.

Four questions, one derivative

Because $A'(t)=R_{\text{in}}(t)-R_{\text{out}}(t)$:

• Increasing at $t_0$? Yes if $R_{\text{in}}(t_0)\gt R_{\text{out}}(t_0)$.
• Greatest amount? Solve $R_{\text{in}}=R_{\text{out}}$, then argue $A'$ changes from positive to negative, or compare candidates including endpoints.
• Amount at $T$? The formula above; on the calculator section store intermediate values, round only the final answer to three decimals.
• Empty? Solve $A(t)=0$.

Worked example: a tank holds $50$ L at $t=0$; water enters at $R_{\text{in}}(t)=6$ L/min and leaves at $R_{\text{out}}(t)=\dfrac{t^2}{3}$ L/min for $0\le t\le6$. Amount at $t=6$: $50+\displaystyle\int_0^6\left(6-\frac{t^2}{3}\right)dt=50+\left[6t-\frac{t^3}{9}\right]_0^6=50+36-24=62$ L. Maximum: $6=\dfrac{t^2}{3}\Rightarrow t=\sqrt{18}\approx4.243$; $A'$ changes from positive to negative there, so the maximum is $A(\sqrt{18})=50+4\sqrt{18}\approx66.971$ L.