| Exam Board | Edexcel |
|---|---|
| Module | CP1 (Core Pure 1) |
| Year | 2021 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Tank/reservoir mixing problems |
| Difficulty | Standard +0.3 This is a standard first-order linear differential equation problem with a straightforward real-world context. Part (a) requires setting up a rate equation (routine for CP1), part (b) involves solving a separable DE using standard techniques, and part (c) is a simple model evaluation. The mathematics is entirely procedural with no novel insight required, making it slightly easier than average for A-level. |
| Spec | 4.10c Integrating factor: first order equations |
| Answer | Marks |
|---|---|
| 8(a) | Volume of paint = 30 litres therefore |
| Answer | Marks | Guidance |
|---|---|---|
| 3 0 | M1 | 3.3 |
| Answer | Marks | Guidance |
|---|---|---|
| dt 10 | A1 | 1.1b |
| Answer | Marks |
|---|---|
| (b) | d r r |
| Answer | Marks |
|---|---|
| IF = e 10 =... | Separates the variables |
| Answer | Marks | Guidance |
|---|---|---|
| . . . | M1 | 3.1a |
| Answer | Marks |
|---|---|
| r e 0 2 e 0 d t r e 0 e 0 ( c ) = = + | Integrates to the form |
| Answer | Marks | Guidance |
|---|---|---|
| 1 0 | M1 | 1.1b |
| Answer | Marks |
|---|---|
| r e 0 = 2 0 e 0 + c | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| 1 0 | A1ft | 1.1b |
| t = 0 , r = 1 0 c = ... | M1 | 3.4 |
| Answer | Marks |
|---|---|
| value for t | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Leading to a value for t | M1 | 3.4 |
| t =awrt 7 seconds | A1 | 2.2b |
| Answer | Marks | Guidance |
|---|---|---|
| (c) | The model predicts 7 seconds but it actually takes 9 seconds so (over) | |
| 2 seconds out (over 20%), therefore it is not a good model | B1ft | 3.5a |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Scheme | Marks |
Question 8:
--- 8(a) ---
8(a) | Volume of paint = 30 litres therefore
r
Rate of paint out = 3 litres per second
3 0 | M1 | 3.3
dr r
=2−
dt 10 | A1 | 1.1b
(2)
(b) | d r r
Rearranges + = 2 and attempts
d t 1 0
integrating factor
1
dt
IF = e 10 =... | Separates the variables
1 1
d r = d t
2 − 0 r 1 0
. . . | M1 | 3.1a
t1 t1 t1 t1
r e 0 2 e 0 d t r e 0 e 0 ( c ) = = + | Integrates to the form
1
l n ( 2 0 r ) t ( c ) − = +
1 0 | M1 | 1.1b
t1 t1
r e 0 = 2 0 e 0 + c | 1
− l n ( 2 0 − r ) = t + c
1 0 | A1ft | 1.1b
t = 0 , r = 1 0 c = ... | M1 | 3.4
t
2 0 e − 1 0 1 0
r = = 1 5 rearranges to
t1
0 e
t1
achieve e 0 = and solves to find a
value for t
or
t
−
r =20−10e 10 =15 rearranges to
t
−
achieve e 10 = and solves to find a
value for t | 1
− l n ( 2 0 − 1 5 ) = t − l n 1 0
1 0
Leading to a value for t | M1 | 3.4
t =awrt 7 seconds | A1 | 2.2b
(6)
(c) | The model predicts 7 seconds but it actually takes 9 seconds so (over)
2 seconds out (over 20%), therefore it is not a good model | B1ft | 3.5a
(1)
(9 marks)
Notes:
(a)
r
M1: Clearly identifies that Rate of paint out =3 . It is a “show that” question so
their volume
there must be clearly reasoning. Just answer with no reasoning scores M0.
A1: Puts all the components together to form the correct differential equation.
(b)
M1: Identifies as a first order differential equation and finds the integrating factor or separates the
d r r
variables and integrates. Allow if there are sign slips in rearranging (e.g. to − = 2 ) or in the
d t 1 0
integrating factor and allow with their value for a or with a as an unknown.
M1: Multiplies through by the IF and attempts to integrate or integrates to the form
1
l n ( 2 a r ) t c − = + oe
a
A1ft: Correct integration, including constant of integration. Follow through on their value of a, but
ta ta 1
not sign slips from rearrangement. So allow for r e = 2 a e + c or −ln(2a−r)= t+c oe with a
a
or their a.
M1: Uses the initial conditions to find the constant of integration. Must see substitution or can be
implied by the correct value for their equation. Allow for finding in terms of a if separation of
variables used.
t
t −
M1: Sets r = 1 5 , achieves e10 =0 or e 10 =0as appropriate and solves to find a value for t.
Separates the variable method sets r = 15 and rearranges to find a value for t. Note: For this mark a
value of a is needed, but need not be the correct one.
A1cso: t = a w r t 7 seconds from fully correct work.
(c)
B1ft: See scheme, follow through on their answer to part (b). Accept any reasonable comparative
comment but must have a reason, not just a statement of good or not good. So e.g. look for finding
the difference between their answer and 9, or the percentage difference. If their answer is close to 9,
then accept a conclusion of being a good model if a suitable reason is given. May substitute 9 into
their equation and obtain a value to compare with 15 and make a similar conclusion.
Question | Scheme | Marks | AOsTwo different colours of paint are being mixed together in a container.
The paint is stirred continuously so that each colour is instantly dispersed evenly throughout the container.
Initially the container holds a mixture of 10 litres of red paint and 20 litres of blue paint.
The colour of the paint mixture is now altered by
\begin{itemize}
\item adding red paint to the container at a rate of 2 litres per second
\item adding blue paint to the container at a rate of 1 litre per second
\item pumping fully mixed paint from the container at a rate of 3 litres per second.
\end{itemize}
Let $r$ litres be the amount of red paint in the container at time $t$ seconds after the colour of the paint mixture starts to be altered.
\begin{enumerate}[label=(\alph*)]
\item Show that the amount of red paint in the container can be modelled by the differential equation
$$\frac{dr}{dt} = 2 - \frac{r}{a}$$
where $a$ is a positive constant to be determined.
[2]
\item By solving the differential equation, determine how long it will take for the mixture of paint in the container to consist of equal amounts of red paint and blue paint, according to the model. Give your answer to the nearest second.
[6]
It actually takes 9 seconds for the mixture of paint in the container to consist of equal amounts of red paint and blue paint.
\item Use this information to evaluate the model, giving a reason for your answer.
[1]
\end{enumerate}
\hfill \mbox{\textit{Edexcel CP1 2021 Q8 [9]}}