Accumulation in Context — AP® Calculus AB Notes

Many exam questions are about a container: a tank, a reservoir, a museum, a pile of gravel, a driveway under snow. Something comes in at one rate and goes out at another. The questions are always the same few. How much is there at a given time? Is it going up or down right now? When is there the most? This page gives one equation that answers all of them.

Rate in, rate out

Call the rate in $R(t)$ (water entering, people arriving, snow falling) and the rate out $D(t)$ (water draining, people leaving, snow melting). The amount $A(t)$ changes at the difference:

$A'(t) = R(t) - D(t).$

So at any moment, the amount is increasing when $R(t) > D(t)$ and decreasing when $R(t) < D(t)$ . Notice what does not matter: whether $R$ itself is rising, or how big $R$ is on its own. Only the comparison counts.

The amount at a time

Integrate the net rate and add what was there at the start:

$A(t) = A(a) + \int_a^t \big(R(s) - D(s)\big)\,ds.$

The integral is the net change; the first term is the starting amount. Leaving out the starting amount is the most common lost point in these questions.

Two rates on 0 ≤ t ≤ 12 minutes: water flows in at R(t) = 10 + 3 sin(πt/6) liters per minute (blue) and out at D(t) = 4 + t liters per minute (crimson). From t = 0 to 6 the rate in is larger and the gap is shaded green: the amount rises. The curves cross at t = 6, where both rates are 10. After that the rate out is larger and the gap is shaded vermillion: the amount falls. The amount is greatest at t = 6. — **Figure 1.** Rate in $R(t) = 10 + 3\sin\left(\frac{\pi t}{6}\right)$ and rate out $D(t) = 4 + t$ , in liters per minute. Green: the amount rises. Vermillion: it falls. LevlPrep original diagram.

Worked example 1 (calculator). A tank holds 50 liters at $t = 0$ . Water enters at $R(t) = 10 + 3\sin\left(\frac{\pi t}{6}\right)$ and leaves at $D(t) = 4 + t$ liters per minute, $0 \le t \le 12$ . (a) How much water is in the tank at $t = 9$ ? (b) Is the amount increasing or decreasing at $t = 9$ ? (c) What is the greatest amount?

(a) $A(9)$ $= 50$ $+ \int_0^9 \big(R(t) - D(t)\big)\,dt$ $\approx 50$ $+ 19.230$ $= 69.230$ liters.

(b) $A'(9)$ $= R(9) - D(9)$ $= 7 - 13$ $= -6 < 0$ , so the amount is decreasing, at 6 liters per minute.

(c) $A'(t) = R(t) - D(t) = 0$ on $(0, 12)$ only at $t = 6$ , where $A'$ changes from positive to negative. Candidates: $A(0) = 50$ , $A(6) \approx 79.459$ , $A(12) \approx 50$ . The greatest amount is about 79.459 liters, at $t = 6$ .

In (c), the sign change alone shows a relative maximum at $t = 6$ . For the absolute maximum on the closed interval, compare with the endpoints: the candidates test.

From a table

When the rate is only known at a few times, estimate the integral with a Riemann or trapezoidal sum, using the table's own widths. They are often unequal.

Worked example 2 (no calculator). Visitors enter a museum at the rates in the table. 25 visitors are inside at $t = 0$ , and visitors leave at a constant 70 people per hour. Estimate the number inside at $t = 8$ .

$t$ (hours)	0	2	3	5	8
$E(t)$ (people per hour)	60	110	150	90	30

Trapezoidal sum: $2 \cdot \frac{60 + 110}{2}$ $+ 1 \cdot \frac{110 + 150}{2}$ $+ 2 \cdot \frac{150 + 90}{2}$ $+ 3 \cdot \frac{90 + 30}{2}$ $= 170$ $+ 130$ $+ 240$ $+ 180$ $= 720$ people entered.

Leaving: $\int_0^8 70\,dt$ $= 560$ people.

Amount: $25 + 720 - 560 = 185$ people inside at $t = 8$ (an estimate).

Units and interpretations

An integral of a rate carries the accumulated unit. Gallons per hour integrated over hours gives gallons; people per hour over hours gives people. The derivative of an amount carries the rate unit; the derivative of a rate carries "per hour per hour".

When asked what an integral means, name three things: the quantity, its units and the interval. For example, $\int_0^6 D(t)\,dt$ $= 42$ means "42 liters of water leave the tank during the first 6 minutes." Not "the tank holds 42 liters", and not "the water drops by 42 liters": water was also coming in.

Expression	Meaning	Units (rates in liters per minute)
$R(t) - D(t)$	how fast the amount changes at time $t$	liters per minute
$\int_a^b R(t)\,dt$	amount that comes in from $a$ to $b$	liters
$\int_a^b (R(t) - D(t))\,dt$	net change in the amount from $a$ to $b$	liters
$A(a)$ $+ \int_a^b (R(t) - D(t))\,dt$	amount at time $b$	liters

Common slips

Dropping the starting amount.
Adding the rate out instead of subtracting it.
Deciding "increasing" from $R'(t) > 0$ instead of from $R(t) - D(t) > 0$ .
Finding the time of the maximum and stopping: the question may ask for the amount.
Rate units on an amount: an integral of a rate is never "per minute".