| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2002 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Deriving the differential equation |
| Difficulty | Standard +0.3 This is a standard logistic growth modelling question requiring straightforward application of given conditions to derive a differential equation constant, routine partial fractions, and separation of variables. The 'show that' part involves simple substitution of initial conditions, and the solution method is a textbook exercise in separable equations—slightly easier than average due to clear scaffolding and standard techniques. |
| Spec | 1.02y Partial fractions: decompose rational functions1.07t Construct differential equations: in context1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Content | Mark | Guidance |
| (i) State or imply that \(\frac{da}{dt} = k(10-a)\) | B1 | |
| Justify \(k = 0.004\) | B1 | Max 2 marks |
| (ii) Resolve \(\frac{1}{a(10-a)}\) into partial fractions \(\frac{A}{a} + \frac{B}{10-a}\) and obtain values \(A = B = \frac{1}{10}\) | B1 | |
| Separate variables containing \(\int \frac{da}{a(10-a)} = \int k dt\) and attempt to integrate both sides | M1 | |
| Obtain \(\frac{1}{10} \ln a - \frac{1}{10} \ln(10-a)\) | A1✓ | |
| Obtain \(0.004t\), or equivalent | A1 | |
| Evaluate a constant, or use limits \(t = 0, a = 5\) | M1 | |
| Obtain answer \(t = 25 \ln\left(\frac{a}{10-a}\right)\), or equivalent | A1 | Max 6 marks |
| (iii) Substitute \(a = 9\) and calculate \(t\) | M1 | |
| Obtain answer \(t = 54.9\) or \(55\) | A1 | [Substitution of \(a = 0.9\) scores M0.] |
| Content | Mark | Guidance |
|---------|------|----------|
| **(i)** State or imply that $\frac{da}{dt} = k(10-a)$ | B1 | |
| Justify $k = 0.004$ | B1 | Max 2 marks |
| **(ii)** Resolve $\frac{1}{a(10-a)}$ into partial fractions $\frac{A}{a} + \frac{B}{10-a}$ and obtain values $A = B = \frac{1}{10}$ | B1 | |
| Separate variables containing $\int \frac{da}{a(10-a)} = \int k dt$ and attempt to integrate both sides | M1 | |
| Obtain $\frac{1}{10} \ln a - \frac{1}{10} \ln(10-a)$ | A1✓ | |
| Obtain $0.004t$, or equivalent | A1 | |
| Evaluate a constant, or use limits $t = 0, a = 5$ | M1 | |
| Obtain answer $t = 25 \ln\left(\frac{a}{10-a}\right)$, or equivalent | A1 | Max 6 marks |
| **(iii)** Substitute $a = 9$ and calculate $t$ | M1 | |
| Obtain answer $t = 54.9$ or $55$ | A1 | [Substitution of $a = 0.9$ scores M0.] | Max 2 marks |
---9 In an experiment to study the spread of a soil disease, an area of $10 \mathrm {~m} ^ { 2 }$ of soil was exposed to infection. In a simple model, it is assumed that the infected area grows at a rate which is proportional to the product of the infected area and the uninfected area. Initially, $5 \mathrm {~m} ^ { 2 }$ was infected and the rate of growth of the infected area was $0.1 \mathrm {~m} ^ { 2 }$ per day. At time $t$ days after the start of the experiment, an area $a \mathrm {~m} ^ { 2 }$ is infected and an area $( 10 - a ) \mathrm { m } ^ { 2 }$ is uninfected.\\
(i) Show that $\frac { \mathrm { d } a } { \mathrm {~d} t } = 0.004 a ( 10 - a )$.\\
(ii) By first expressing $\frac { 1 } { a ( 10 - a ) }$ in partial fractions, solve this differential equation, obtaining an expression for $t$ in terms of $a$.\\
(iii) Find the time taken for $90 \%$ of the soil area to become infected, according to this model.
\hfill \mbox{\textit{CAIE P3 2002 Q9 [10]}}