| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2016 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Exponential growth/decay - direct proportionality (dN/dt = kN) |
| Difficulty | Moderate -0.8 This is a standard exponential decay question requiring separation of variables, integration, and applying initial conditions—all routine C4 techniques. Part (b) involves straightforward logarithmic manipulation and unit conversion. The question is more procedural than conceptually challenging, making it easier than average for A-level. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int\frac{1}{x}\,dx = \int -\frac{5}{2}\,dt\) | B1 | Correct separation of variables; \(dx\) and \(dt\) not in wrong positions (can be implied by later working) |
| \(\ln x = -\frac{5}{2}t + c\) | M1, A1 | M1: integrates both sides to give \(\pm\frac{\alpha}{x}\to\pm\alpha\ln x\) or \(\pm k \to \pm kt\); A1: includes \(+c\) |
| \(\{t=0, x=60\} \Rightarrow \ln 60 = c\); \(x = 60e^{-\frac{5}{2}t}\) or \(x = \frac{60}{e^{\frac{5}{2}t}}\) | A1 cso | Correct algebra to achieve result with no incorrect working seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(20 = 60e^{-\frac{5}{2}t}\) or \(\ln 20 = -\frac{5}{2}t + \ln 60\) | M1 | Substitutes \(x=20\) into equation of form \(x = \pm\lambda e^{\pm\mu t \pm\beta}\) or \(x = \pm\lambda e^{\pm\mu t \pm \alpha\ln\delta x}\) or equivalent forms; \(\alpha,\lambda,\mu,\delta \neq 0\), \(\beta\) can be 0 |
| \(t = -\frac{2}{5}\ln\left(\frac{20}{60}\right) = 0.4394...\text{ (days)}\) | dM1 | Dependent on previous M; correct algebra to achieve \(t = A\ln\!\left(\frac{60}{20}\right)\) or \(A\ln\!\left(\frac{20}{60}\right)\) or \(A\ln 3\) or \(A\ln\!\left(\frac{1}{3}\right)\) o.e., \(t>0\) |
| \(t = 632.8006... \approx 633\) (to nearest minute) | A1 cso | awrt 633 or 10 hours and awrt 33 minutes |
# Question 4:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int\frac{1}{x}\,dx = \int -\frac{5}{2}\,dt$ | B1 | Correct separation of variables; $dx$ and $dt$ not in wrong positions (can be implied by later working) |
| $\ln x = -\frac{5}{2}t + c$ | M1, A1 | M1: integrates both sides to give $\pm\frac{\alpha}{x}\to\pm\alpha\ln x$ or $\pm k \to \pm kt$; A1: includes $+c$ |
| $\{t=0, x=60\} \Rightarrow \ln 60 = c$; $x = 60e^{-\frac{5}{2}t}$ or $x = \frac{60}{e^{\frac{5}{2}t}}$ | A1 cso | Correct algebra to achieve result with no incorrect working seen |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $20 = 60e^{-\frac{5}{2}t}$ or $\ln 20 = -\frac{5}{2}t + \ln 60$ | M1 | Substitutes $x=20$ into equation of form $x = \pm\lambda e^{\pm\mu t \pm\beta}$ or $x = \pm\lambda e^{\pm\mu t \pm \alpha\ln\delta x}$ or equivalent forms; $\alpha,\lambda,\mu,\delta \neq 0$, $\beta$ can be 0 |
| $t = -\frac{2}{5}\ln\left(\frac{20}{60}\right) = 0.4394...\text{ (days)}$ | dM1 | Dependent on previous M; correct algebra to achieve $t = A\ln\!\left(\frac{60}{20}\right)$ or $A\ln\!\left(\frac{20}{60}\right)$ or $A\ln 3$ or $A\ln\!\left(\frac{1}{3}\right)$ o.e., $t>0$ |
| $t = 632.8006... \approx 633$ (to nearest minute) | A1 cso | awrt 633 **or** 10 hours and awrt 33 minutes |
---4. The rate of decay of the mass of a particular substance is modelled by the differential equation
$$\frac { \mathrm { d } x } { \mathrm {~d} t } = - \frac { 5 } { 2 } x , \quad t \geqslant 0$$
where $x$ is the mass of the substance measured in grams and $t$ is the time measured in days.\\
Given that $x = 60$ when $t = 0$,
\begin{enumerate}[label=(\alph*)]
\item solve the differential equation, giving $x$ in terms of $t$. You should show all steps in your working and give your answer in its simplest form.
\item Find the time taken for the mass of the substance to decay from 60 grams to 20 grams. Give your answer to the nearest minute.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2016 Q4 [7]}}