| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2009 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Modelling with rate proportional to difference |
| Difficulty | Moderate -0.8 This is a standard Newton's Law of Cooling problem with straightforward separation of variables. Part (i) requires basic integration and applying initial conditions, part (ii) involves simple logarithm manipulation, and part (iii) is direct substitution. The question is entirely routine with no problem-solving insight required, making it easier than average but not trivial due to the algebraic manipulation with parameters A and k. |
| Spec | 1.06i Exponential growth/decay: in modelling context4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Separate variables correctly | B1 | |
| Integrate and obtain term \(\ln(\theta - A)\), or equivalent | B1 | |
| Integrate and obtain term \(-kt\), or equivalent | B1 | |
| Use \(\theta = 4A, t = 0\) to determine a constant, or as limits | M1 | |
| Obtain correct answer in any form, e.g. \(\ln(\theta-A) = -kt + \ln 3A\), with no errors seen | A1 | [5] |
| (ii) Substitute \(\theta = 3A, t = 1\) and justify the given statement | B1 | [1] |
| (iii) Substitute \(t = 2\) and solve for \(\theta\) in terms of \(A\) | M1 | |
| Remove logarithms | M1 | |
| Obtain answer \(\theta = \frac{7}{9}A\), or equivalent, with no errors seen | A1 | [3] |
**(i)** Separate variables correctly | B1 |
Integrate and obtain term $\ln(\theta - A)$, or equivalent | B1 |
Integrate and obtain term $-kt$, or equivalent | B1 |
Use $\theta = 4A, t = 0$ to determine a constant, or as limits | M1 |
Obtain correct answer in any form, e.g. $\ln(\theta-A) = -kt + \ln 3A$, with no errors seen | A1 | [5]
**(ii)** Substitute $\theta = 3A, t = 1$ and justify the given statement | B1 | [1]
**(iii)** Substitute $t = 2$ and solve for $\theta$ in terms of $A$ | M1 |
Remove logarithms | M1 |
Obtain answer $\theta = \frac{7}{9}A$, or equivalent, with no errors seen | A1 | [3]
[The M marks are only available if the solution to part (i) contains terms $a\ln(\theta - A)$ and $bt.$]9 The temperature of a quantity of liquid at time $t$ is $\theta$. The liquid is cooling in an atmosphere whose temperature is constant and equal to $A$. The rate of decrease of $\theta$ is proportional to the temperature difference $( \theta - A )$. Thus $\theta$ and $t$ satisfy the differential equation
$$\frac { \mathrm { d } \theta } { \mathrm {~d} t } = - k ( \theta - A )$$
where $k$ is a positive constant.\\
(i) Find, in any form, the solution of this differential equation, given that $\theta = 4 A$ when $t = 0$.\\
(ii) Given also that $\theta = 3 A$ when $t = 1$, show that $k = \ln \frac { 3 } { 2 }$.\\
(iii) Find $\theta$ in terms of $A$ when $t = 2$, expressing your answer in its simplest form.
\hfill \mbox{\textit{CAIE P3 2009 Q9 [9]}}