CAIE P3 2020 Specimen — Question 10 11 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2020
SessionSpecimen
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFirst order differential equations (integrating factor)
TypeDeriving the differential equation
DifficultyStandard +0.5 This is a standard applied differential equations question requiring translation of a word problem into mathematical form, then solving using integrating factor method. Part (a) involves setting up the DE from the given conditions (routine modelling), part (b) is a standard integrating factor solution, and part (c) requires interpretation of the limiting behaviour. While it requires multiple steps and careful algebraic manipulation, it follows a well-established template for chemical reaction problems and doesn't require novel insight beyond applying standard techniques.
Spec1.08k Separable differential equations: dy/dx = f(x)g(y)1.08l Interpret differential equation solutions: in context

10 In a chemical reaction, a compound \(X\) is formed from two compounds \(Y\) and \(Z\).
The masses in grams of \(X , Y\) and \(Z\) present at time \(t\) seconds after the start of the reaction are \(x , 10 - x\) and \(20 - x\) respectively. At any time the rate of formation of \(X\) is proportional to the product of the masses of \(Y\) and \(Z\) present at the time. When \(t = 0 , x = 0\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = 2\).
  1. Show that \(x\) and \(t\) satisfy the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = 0.01 ( 10 - x ) ( 20 - x ) .$$
  2. Solve this differential equation and obtain an expression for \(x\) in terms of \(t\).
  3. State what happens to the value of \(x\) when \(t\) becomes large.

Question 10(a):
AnswerMarks Guidance
AnswerMark Guidance
State or imply \(\frac{dx}{dt} = k(10-x)(20-x)\) and show \(k = 0.01\)B1
Question 10(b):
AnswerMarks Guidance
AnswerMark Guidance
Separate variables and attempt integration of at least one sideM1
Carry out an attempt to find \(A\) and \(B\) such that \(\frac{1}{(10-x)(20-x)} \equiv \frac{A}{10-x} + \frac{B}{20-x}\)M1
Obtain \(A = \frac{1}{10}\) and \(B = -\frac{1}{10}\), or equivalentA1
Integrate and obtain \(-\frac{1}{10}\ln(10-x) + \frac{1}{10}\ln(20-x)\), or equivalentA1FT
Integrate and obtain term \(0.01t\), or equivalentA1
Evaluate a constant, or use limits \(t=0\), \(x=0\) in a solution containing terms of the form \(a\ln(10-x)\) and \(ct\)M1
Obtain answer in any form, e.g. \(-\frac{1}{10}\ln(10-x) + \frac{1}{10}\ln(20-x) = 0.01t + \frac{1}{10}\ln 2\)A1FT
Use laws of logarithms correctly to remove logarithmsM1
Rearrange and obtain \(x = \frac{20(e^{0.1t}-1)}{2e^{0.1t}-1}\), or equivalentA1
Total9
Question 10(c):
AnswerMarks Guidance
AnswerMark Guidance
State that \(x\) approaches 10B1
## Question 10(a):

| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply $\frac{dx}{dt} = k(10-x)(20-x)$ and show $k = 0.01$ | B1 | |

## Question 10(b):

| Answer | Mark | Guidance |
|--------|------|----------|
| Separate variables and attempt integration of at least one side | M1 | |
| Carry out an attempt to find $A$ and $B$ such that $\frac{1}{(10-x)(20-x)} \equiv \frac{A}{10-x} + \frac{B}{20-x}$ | M1 | |
| Obtain $A = \frac{1}{10}$ and $B = -\frac{1}{10}$, or equivalent | A1 | |
| Integrate and obtain $-\frac{1}{10}\ln(10-x) + \frac{1}{10}\ln(20-x)$, or equivalent | A1FT | |
| Integrate and obtain term $0.01t$, or equivalent | A1 | |
| Evaluate a constant, or use limits $t=0$, $x=0$ in a solution containing terms of the form $a\ln(10-x)$ and $ct$ | M1 | |
| Obtain answer in any form, e.g. $-\frac{1}{10}\ln(10-x) + \frac{1}{10}\ln(20-x) = 0.01t + \frac{1}{10}\ln 2$ | A1FT | |
| Use laws of logarithms correctly to remove logarithms | M1 | |
| Rearrange and obtain $x = \frac{20(e^{0.1t}-1)}{2e^{0.1t}-1}$, or equivalent | A1 | |
| **Total** | **9** | |

## Question 10(c):

| Answer | Mark | Guidance |
|--------|------|----------|
| State that $x$ approaches 10 | B1 | |
10 In a chemical reaction, a compound $X$ is formed from two compounds $Y$ and $Z$.\\
The masses in grams of $X , Y$ and $Z$ present at time $t$ seconds after the start of the reaction are $x , 10 - x$ and $20 - x$ respectively. At any time the rate of formation of $X$ is proportional to the product of the masses of $Y$ and $Z$ present at the time. When $t = 0 , x = 0$ and $\frac { \mathrm { d } x } { \mathrm {~d} t } = 2$.\\
(a) Show that $x$ and $t$ satisfy the differential equation

$$\frac { \mathrm { d } x } { \mathrm {~d} t } = 0.01 ( 10 - x ) ( 20 - x ) .$$

(b) Solve this differential equation and obtain an expression for $x$ in terms of $t$.\\
(c) State what happens to the value of $x$ when $t$ becomes large.\\

\hfill \mbox{\textit{CAIE P3 2020 Q10 [11]}}