Recover an amount from a starting value
Fuel in a generator
The dashed curve shows how fast a generator burns fuel, in liters per hour. The tank holds 42 liters at t = 0.
Move the End slider to the right. The red region is the fuel burned so far, and the lower graph shows the fuel left in the tank.
At t = 2 the red region has an area of 10 liters, so the tank holds the starting 42 liters minus 10, which is 32 liters. The red bracket below shows the 10 liters lost, and the teal bracket the 32 liters left.
The area gives only the change, and the starting amount is where the change begins.
Any starting amount
The top curve is the net rate r, inflow minus outflow, and the lower graph is the amount A. The dashed line marks the amount at the start, the starting amount.
Move the end x. The green or red bracket at the point shows the change since the start, and the teal bracket shows the amount at x. The amount at x is the starting amount plus this change.
The change is the area between r and the axis: green area above it raises the amount, and red area below lowers it.
Move the Starting amount slider. The lower graph moves up or down with it, because the rate, and so the change, is the same from any start.
Notation
The question gives the amount at the start of the interval, \(t = a\). The dashed line marks it.
\[\cq{A(a)}\]Nothing flows in, so the net rate \(r(t)\) is the burn rate with a minus sign, and the curve moves below the axis. The change from \(a\) to \(x\) is the area between \(r\) and the t-axis, and area below the axis counts as negative.
\[\cb{\int_a^x} \cy{r(t)}\,dt\]The amount at \(x\) equals the starting amount plus the change.
\[A(x) = A(a) + \int_a^x r(t)\,dt\]The same \(a\) must appear in both places.
When it fails
Run the end past empty. The formula gives 42 minus the fuel burned, which is −22 liters at \(t = 8\), so the red dashed curve falls below 0.
A tank cannot hold −22 liters. The tank is empty at t = 6, so from then on the generator burns no fuel and the red region ends. The solid teal curve, the true amount, stays at 0. Compare every answer with what is physically possible.
Example 1: a growing sapling
A sapling is 40 cm tall at \(t = 0\). It grows at \(g(t) = 12 - 2t\) centimeters per month, where \(t\) is in months. How tall is the sapling at \(t = 4\)?
The starting amount is the height at \(t = 0\).
\[A(0) = 40\]The change from 0 to 4 is the green area, the integral of the growth rate.
\[\begin{aligned} \cb{\int_0^4} (12 - 2t)\,dt &= \Big[\, 12t - t^2 \,\Big]_0^4 \\ &= 48 - 16 = 32 \end{aligned}\]Add the change to the starting height. The sapling is 72 cm tall.
\[\cg{40 + 32 = 72 \text{ cm}}\]
Example 2: a heated metal block
A metal block is 80 degrees Celsius at \(t = 2\) minutes. It heats at \(h(t) = 24 - 3t\) degrees per minute. Find its temperature at \(t = 6\).
The temperature is known at \(t = 2\), and the question asks for it at \(t = 6\).
\[T(2) = 80\]A tempting wrong move: Integrate the rate from 0, and add 80.
\[\begin{aligned} &80 + \int_0^6 (24 - 3t)\,dt \\ &= 80 + 90 = 170 \end{aligned}\]The 80 degrees is the temperature at \(t = 2\), so it already includes the 42 degrees of heating from 0 to 2, marked in red. The integral from 0 counts those 42 degrees a second time.
Start the change at \(t = 2\), the time of the known amount.
\[\begin{aligned} T(6) &= 80 + \cb{\int_2^6} (24 - 3t)\,dt \\ &= 80 + \Big[\, 24t - \tfrac32 t^2 \,\Big]_2^6 \\ &= 80 + (90 - 42) = 80 + 48 \end{aligned}\]The block is at 128 degrees Celsius at \(t = 6\).
\[\cg{128}\]
Practice
A bin holds 48 kilograms of flour at \(t = 0\) hours. Bakers use flour at \(u(t) = 8 + 2t\) kilograms per hour. How much flour is in the bin at \(t = 5\)?
Move the end to 5. Can a bin hold the amount the lower graph shows there? When is the bin empty?
Solution
The flour used from 0 to 5 is the red area, the integral of the rate. Subtract it from the starting 48.
\[\begin{aligned} &48 - \int_0^5 (8 + 2t)\,dt \\ &= 48 - \Big[\, 8t + t^2 \,\Big]_0^5 \\ &= 48 - 65 = -17 \end{aligned}\]A bin cannot hold −17 kilograms. The bin is empty when the flour used, \(8t + t^2\), reaches 48, which is at \(t = 4\).
\[8(4) + 4^2 = 48\]After \(t = 4\) there is no flour to use, so the amount stays at 0. The bin holds 0 kilograms at \(t = 5\).
\[0 \text{ kg}\]
Questions
The graph shows how fast the number of fish in a pond changes, in fish per month. The pond holds 120 fish at month 0.
How many fish does the pond hold at month 4?
- 200
- 80
- 120
- 160
Answer
200
The pond holds 200 fish at month 4. How many fish does it hold at month 8?
- 120
- 280
- 80
- 200
Answer
120
A cistern holds 100 liters at \(t = 0\). Water flows in at \(12t\) liters per hour. How much water is in the cistern at \(t = 5\)?
- 250 liters
- 150 liters
- 160 liters
- 100 liters
Answer
250 liters
Summary
You find the change as the integral of the rate and add it to the amount at the start of the interval.
This works because the integral gives only the change, and the starting amount is where that change begins.
You use it when a problem gives a starting amount and a rate, and the result must stay within what is physically possible.
This is one of five skills in topic 8.3 of Which & Why: AP Calculus.