Total change from a rate

In 2 of 20 recent exams
Paper 2, calculator
3 to 6 marks

How it is set in the exam

Paper 2 (calculator) (GDC), 3–6 marks, inside a context question. Rare (2 of 20 exam versions), but kinematics (displacement and distance from a velocity) uses the same idea.

Spot it in the exam

Exam phrases to spot (3)

“water flows into a tank at a rate of $R(t)$ litres per minute; find the volume of water in the tank after 10 minutes”
“the rate of change of the population is …; find the population at the start of 2030”
“find the total amount …”

In disguise (4)

"will the tank overflow?", "is there enough …?": find the amount, then compare it with the capacity, and say so
"find the time at which the tank is full": the unknown is the upper limit, $\int_0^T R(t)\,dt = V$ , solved on the GDC
the total over any time interval of a fixed length, when the rate is a constant plus a sine or cosine: if the length is a whole number of periods, the trig part integrates to $0$ , so the total is the same wherever the interval starts, the constant $\times$ the length. Whatever the length, a bound can also decide: the total is at least the least rate $\times$ the length
the capacity comes from a cross-section made of a segment and a rectangle, times the length (Chords, segments and regions made of sectors and triangles)

Looks like it, but isn't (3)

the rate at one instant ("how many litres per minute at $t = 5$ ?"), or how fast the rate itself is changing: Rate of change in context
when the rate is greatest: Concavity and points of inflexion
distance travelled from a velocity: Displacement, position and distance travelled, with the method of Area between a curve and the x-axis

Decide first: is a rate asked for (differentiate), or is a rate given and an amount asked for (integrate)? The units decide: litres per minute is a rate, litres is an amount.

Worked example

A tank: the total, an overflow, the time it is full

Paper 2 (calculator)

Question

Water is pumped into a tank at a rate of $R(t) = 20 + 15\sin(0.5t)$ litres per minute, where $t$ is the time in minutes after the pump is switched on, for $0 \le t \le 30$ . When $t = 0$ the tank contains $150$ litres. The tank holds $800$ litres.
(a) Find the volume of water pumped into the tank in the first 10 minutes. [2]
(b) Will the tank overflow before the pump is switched off at $t = 30$ ? Justify your answer. [3]
(c) Find the time at which the tank becomes full. [3]

Try it yourself first, then compare.

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Steps

Name the rate, its units and the time variable.
The integral of a rate over a time interval is the change in the amount over that interval: litres per minute $\times$ minutes $=$ litres.
Turn dates or times into limits:
"the start of 2030", with $t$ counted from the start of 2020, is $t = 10$ (markschemes give this its own mark).
Write the integral with its limits, then its value
(GDC on : definite integral).
The amount at a time
$=$ the starting amount $+ \int$ of the rate from the start to that time. Or integrate with $+c$ and use the starting amount to find $c$ (Find f from f′ and a point): markschemes accept both.
Compare
with a capacity or a target when asked, and write the conclusion as a sentence. Unknown time: starting amount $+ \int_0^T R(t)\,dt =$ target, solved for $T$ on the GDC (definite integral).

Common mistakes

Differentiating a rate that is given
$R'(t)$ is how fast the flow rate itself changes, not an amount.
Leaving out the starting amount
when the amount at a time is asked for.
The wrong upper limit for a date
from the start of 2020, the start of 2030 is $t = 10$ , not $t = 2030$ and not $t = 11$ .

More mistakes (2)

Rounding before a comparison
$802.79\ldots$ and $800$ are close.
Units
an amount has the units of the rate times the units of time.

Easy to confuse with

Next typeDefinite integrals without the function

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