| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2002 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Chemical reaction kinetics |
| Difficulty | Standard +0.3 This is a separable differential equation (not requiring integrating factor despite the topic label) with straightforward algebra. Part (i) is simple substitution, part (ii) involves standard separation of variables with partial fractions, parts (iii-iv) are routine evaluation and limit-taking. The question is slightly easier than average due to its guided structure and standard techniques, though the algebraic manipulation requires care. |
| Spec | 1.02z Models in context: use functions in modelling1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| (i) State that \(\frac{dm}{dt} = k(50-m)^2\) | B1 | |
| Justify \(k = 0.002\) | B1 | Total: 2 marks |
| (ii) Separate variables and attempt to integrate \(\frac{1}{(50-m)^2}\) | M1 | |
| Obtain \(+\frac{1}{(50-m)}\) and \(0.002t\), or equivalent | A1 | |
| Evaluate a constant or use limits \(t = 0, m = 0\) | M1 | |
| Obtain any correct form of solution e.g. \(\frac{1}{(50-m)} = 0.002t + \frac{1}{50}\) | A1 | |
| Obtain given answer correctly | A1 | Total: 5 marks |
| (iii) Obtain answer \(m = 25\) when \(t = 10\) | B1 | |
| Obtain answer \(t = 90\) when \(m = 45\) | B1 | Total: 2 marks |
| (iv) State that \(m\) approaches \(50\) | B1 | Total: 1 mark |
**(i)** State that $\frac{dm}{dt} = k(50-m)^2$ | B1 |
Justify $k = 0.002$ | B1 | **Total: 2 marks**
**(ii)** Separate variables and attempt to integrate $\frac{1}{(50-m)^2}$ | M1 |
Obtain $+\frac{1}{(50-m)}$ and $0.002t$, or equivalent | A1 |
Evaluate a constant or use limits $t = 0, m = 0$ | M1 |
Obtain any correct form of solution e.g. $\frac{1}{(50-m)} = 0.002t + \frac{1}{50}$ | A1 |
Obtain given answer correctly | A1 | **Total: 5 marks**
**(iii)** Obtain answer $m = 25$ when $t = 10$ | B1 |
Obtain answer $t = 90$ when $m = 45$ | B1 | **Total: 2 marks**
**(iv)** State that $m$ approaches $50$ | B1 | **Total: 1 mark**
---7 In a certain chemical process a substance is being formed, and $t$ minutes after the start of the process there are $m$ grams of the substance present. In the process the rate of increase of $m$ is proportional to $( 50 - m ) ^ { 2 }$. When $t = 0 , m = 0$ and $\frac { \mathrm { d } m } { \mathrm {~d} t } = 5$.\\
(i) Show that $m$ satisfies the differential equation
$$\frac { \mathrm { d } m } { \mathrm {~d} t } = 0.002 ( 50 - m ) ^ { 2 }$$
(ii) Solve the differential equation, and show that the solution can be expressed in the form
$$m = 50 - \frac { 500 } { t + 10 }$$
(iii) Calculate the mass of the substance when $t = 10$, and find the time taken for the mass to increase from 0 to 45 grams.\\
(iv) State what happens to the mass of the substance as $t$ becomes very large.
\hfill \mbox{\textit{CAIE P3 2002 Q7 [10]}}