| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2003 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Exponential growth/decay - approach to limit (dN/dt = k(N - Nā)) |
| Difficulty | Standard +0.3 This is a straightforward exponential growth/decay problem with clear setup. Part (i) requires simple substitution of given values to find the constant of proportionality. Part (ii) is a standard separable differential equation (variables separable, integrate both sides). Part (iii) asks for a limiting behavior observation. All steps are routine for P3 level with no novel insight required, making it slightly easier than average. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y)1.08l Interpret differential equation solutions: in context |
| Answer | Marks | Guidance |
|---|---|---|
| (i) State or imply that \(\frac{dx}{dt} = k(100 - x)\) | B1 | Justify \(k = 0.02\) |
| (ii) Separate variables and attempt to integrate \(\frac{1}{100-x}\) | M1 | Obtain term \(-\ln(100 - x)\), or equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| (iii) State that x tends to 100 as t becomes very large | B1 | [1] |
**(i)** State or imply that $\frac{dx}{dt} = k(100 - x)$ | B1 | Justify $k = 0.02$ | B1 | **[2]**
**(ii)** Separate variables and attempt to integrate $\frac{1}{100-x}$ | M1 | Obtain term $-\ln(100 - x)$, or equivalent | A1 | Obtain term $0.02t$, or equivalent | A1 | Use $x = 5, t = 0$ to evaluate a constant, or as limits | M1 | Obtain correct answer in any form, e.g. $-\ln(100 - x) = 0.02t - \ln 95$ | A1 | Rearrange to give x in terms of t in any correct form, e.g. $x = 100 - 95\exp(-0.02t)$ | A1 | **[6]**
[SR: In $\ln(100-x)$ for -ln$(100-x)$. If no other error and $x = 100 - 95\exp(0.02t)$ or equivalent obtained, give M1A0A1M1A0A1ā.]
**(iii)** State that x tends to 100 as t becomes very large | B1 | **[1]**7 In a chemical reaction a compound $X$ is formed from a compound $Y$. The masses in grams of $X$ and $Y$ present at time $t$ seconds after the start of the reaction are $x$ and $y$ respectively. The sum of the two masses is equal to 100 grams throughout the reaction. At any time, the rate of formation of $X$ is proportional to the mass of $Y$ at that time. When $t = 0 , x = 5$ and $\frac { \mathrm { d } x } { \mathrm {~d} t } = 1.9$.\\
(i) Show that $x$ satisfies the differential equation
$$\frac { \mathrm { d } x } { \mathrm {~d} t } = 0.02 ( 100 - x ) .$$
(ii) Solve this differential equation, obtaining an expression for $x$ in terms of $t$.\\
(iii) State what happens to the value of $x$ as $t$ becomes very large.
\hfill \mbox{\textit{CAIE P3 2003 Q7 [9]}}