| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2005 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Modelling with rate proportional to amount |
| Difficulty | Standard +0.3 This is a straightforward separable differential equation (not requiring integrating factor despite the topic label) with standard initial value problem setup. Students separate variables, integrate both sides (giving ln x and t²/2 terms), apply initial condition, then use given data to find k and solve for the required time. All steps are routine for P3 level with no conceptual surprises, making it slightly easier than average. |
| Spec | 1.07t Construct differential equations: in context1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks |
|---|---|
| Separate variables correctly and attempt to integrate both sides | M1 |
| Obtain term \(\ln x\), or equivalent | A1 |
| Obtain term \(-\frac{1}{2}kt^2\), or equivalent | A1 |
| Use \(t = 0, x = 100\) to evaluate a constant, or as limits | M1 |
| Obtain solution in any correct form, e.g. \(\ln x = -\frac{1}{2}kt^2 + \ln 100\) | A1 |
| Answer | Marks |
|---|---|
| Use \(t = 20, x = 90\) to obtain an equation in \(k\) | M1* |
| Substitute \(x = 50\) and attempt to obtain an unsimplified numerical expression for \(t^2\), such as \(t^2 = 400(\ln 100 - \ln 50)(\ln 100 - \ln 90)\) | M1(dep*) |
| Obtain answer \(t = 51.3\) | A1 |
**(i)**
Separate variables correctly and attempt to integrate both sides | M1 |
Obtain term $\ln x$, or equivalent | A1 |
Obtain term $-\frac{1}{2}kt^2$, or equivalent | A1 |
Use $t = 0, x = 100$ to evaluate a constant, or as limits | M1 |
Obtain solution in any correct form, e.g. $\ln x = -\frac{1}{2}kt^2 + \ln 100$ | A1 |
**Total for (i): [5]**
**(ii)**
Use $t = 20, x = 90$ to obtain an equation in $k$ | M1* |
Substitute $x = 50$ and attempt to obtain an unsimplified numerical expression for $t^2$, such as $t^2 = 400(\ln 100 - \ln 50)(\ln 100 - \ln 90)$ | M1(dep*) |
Obtain answer $t = 51.3$ | A1 |
**Total for (ii): [3]**
---8 In a certain chemical reaction the amount, $x$ grams, of a substance present is decreasing. The rate of decrease of $x$ is proportional to the product of $x$ and the time, $t$ seconds, since the start of the reaction. Thus $x$ and $t$ satisfy the differential equation
$$\frac { \mathrm { d } x } { \mathrm {~d} t } = - k x t$$
where $k$ is a positive constant. At the start of the reaction, when $t = 0 , x = 100$.\\
(i) Solve this differential equation, obtaining a relation between $x , k$ and $t$.\\
(ii) 20 seconds after the start of the reaction the amount of substance present is 90 grams. Find the time after the start of the reaction at which the amount of substance present is 50 grams.
\hfill \mbox{\textit{CAIE P3 2005 Q8 [8]}}