Combine inflow and outflow
A tank that fills and drains
The solid curve shows how fast water flows into a tank, in liters per minute, and the dashed curve how fast it flows out.
Move the End slider to the right. The region between the curves is green where the inflow is larger, so the tank gains water. It is red where the outflow is larger, so the tank loses water. Below, the water in the tank rises while the region is green and falls while it is red.
The orange segment is the net rate, the inflow minus the outflow. The green region from 0 to 5 is a gain of 50 liters. The red region from 5 to 8 is a loss of 18 liters, so the net change is \(50 - 18 = 32\) liters. The tank ends with \(60 + 32 = 92\) liters.
Any inflow and outflow
The solid curve is any rate in, E, and the dashed curve is any rate out, L. Both rates are positive, because each one measures how much moves, not which way.
Press the button. Both curves move down by the outflow at each time, so the dashed curve lies on the axis. The gap between them does not change, so the green and red regions keep their shape. The solid curve that remains is the net rate.
Above the axis the net rate is positive, so the amount below grows. Below the axis it is negative, so the amount shrinks.
Notation
The net rate is the rate in minus the rate out. A positive net rate adds to the amount, and a negative one removes from it.
\[\cy{r(t) = E(t) - L(t)}\]The net change from \(t = a\) to \(t = b\) is the area between the net rate and the t-axis. The part below the axis counts as negative.
\[\cb{\int_a^b} \cy{\big(E(t) - L(t)\big)}\,dt\]You can also integrate each rate and subtract, because the integral of a difference is the difference of the integrals.
\[\int_a^b E(t)\,dt - \int_a^b L(t)\,dt\]Over the 8 minutes, 288 liters flow in and 256 flow out, and \(288 - 256 = 32\).
When it fails
Add the two rates by mistake. The red dashed amount climbs off the top of the graph, and after 8 minutes it shows a gain of 544 liters. The tank gains only 32 liters.
Water that leaves lowers the amount, so the outflow enters with a minus sign.
Example 1: a grain bin
Trucks bring grain to a bin at \(E(t) = 18 + 3t\) tons per hour, and grain is shipped out at \(L(t) = 8 + 2t\) tons per hour. Find the net change in the grain in the first 6 hours.
Subtract the outflow from the inflow to find the net rate. It is positive throughout, so the region is all green.
\[\begin{aligned} r(t) &= (18 + 3t) - (8 + 2t) \\ &= 10 + t \end{aligned}\]Integrate the net rate from 0 to 6.
\[\begin{aligned} \cb{\int_0^6} (10 + t)\,dt &= \Big[\, 10t + \tfrac{t^2}{2} \,\Big]_0^6 \\ &= 60 + 18 \end{aligned}\]The bin gains 78 tons.
\[\cg{78 \text{ tons}}\]As a check, 162 tons arrive and 84 tons ship out, and \(162 - 84 = 78\).
Example 2: a parking garage
Cars enter a garage at \(E(t) = 50 - 2t\) cars per hour and leave at \(L(t) = 10 + 3t\) cars per hour. Find the net change in the number of cars from \(t = 0\) to \(t = 12\) hours.
Subtract to find the net rate. It is 0 at \(t = 8\). The region is green before 8, where cars are gained, and red after 8, where cars are lost.
\[\begin{aligned} r(t) &= (50 - 2t) - (10 + 3t) \\ &= 40 - 5t \end{aligned}\]The green part adds cars. The red part removes cars, so its area counts as negative.
\[\begin{aligned} &\cg{\int_0^8 (40 - 5t)\,dt = 160} \\ &\cr{\int_8^{12} (40 - 5t)\,dt = -40} \end{aligned}\]A tempting wrong move: Add the two areas as sizes.
\[160 + 40 = 200\]The red dashed curve is the amount if the red region counted as a gain. The teal amount falls after \(t = 8\), because the net rate is negative there and cars leave.
The red area counts as negative, so subtract it from the green area.
\[\cb{160 - 40 = 120 \text{ cars}}\]
Practice
A savings account receives deposits at \(D(t) = 50\) dollars per week and pays withdrawals at \(W(t) = 2t^2\) dollars per week. Find the net change in the account in the first 6 weeks.
Move the end to 6 weeks. Where does the region between the curves change color, and what does each color mean?
Solution
The net rate is the deposits minus the withdrawals. It is 0 at \(t = 5\), so it is positive before 5 and negative after 5.
\[r(t) = 50 - 2t^2\]Integrate the net rate from 0 to 6. This adds the green part and the red part with their signs.
\[\Big[\, 50t - \tfrac{2t^3}{3} \,\Big]_0^6 = 300 - 144\]The account gains 156 dollars.
\[\cb{156 \text{ dollars}}\]
Questions
The graph shows the inflow, solid, and the outflow, dashed, of a tank in liters per minute for the first 8 minutes. The curves cross at \(t = 4\).
What does the red region from \(t = 4\) to \(t = 8\) show?
- The tank loses water, because the outflow is larger than the inflow.
- The tank gains water, because the outflow is larger than the inflow.
- The tank holds the same amount, because the curves cross.
Answer
The tank loses water, because the outflow is larger than the inflow.
What is the net rate at \(t = 2\)?
- 6 liters per minute
- 14 liters per minute
- −6 liters per minute
- 10 liters per minute
Answer
6 liters per minute
Water enters a reservoir at \(5t\) cubic meters per hour and leaves at \(20 + t\) cubic meters per hour. What is the net change from \(t = 0\) to \(t = 8\)?
- −32 cubic meters
- 32 cubic meters
- 352 cubic meters
- 160 cubic meters
Answer
−32 cubic meters
Summary
You subtract the outflow from the inflow to find the net rate, and you integrate it over the interval.
This works because water that leaves lowers the amount, so its area counts as negative.
You use it when one rate adds to a quantity and another removes from it.
This is one of five skills in topic 8.3 of Which & Why: AP Calculus.