| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2017 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Model comparison/critique |
| Difficulty | Standard +0.3 This is a straightforward differential equations question requiring separation of variables and using initial conditions. Part (a)-(b) involve simple linear modeling, while (c)-(e) require standard integration of a separable DE with no conceptual challenges. The algebra is routine and the context provides clear guidance at each step, making this easier than average for A-level. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y)4.10a General/particular solutions: of differential equations4.10b Model with differential equations: kinematics and other contexts |
| Answer | Marks | Guidance |
|---|---|---|
| Uses \(x = kt\) or \(t = cx\) and \(x = 1.5\) when \(t = 2\), so \(k=\) or \(c=\) | M1 | May be result of differential equation \(\frac{dx}{dt} = k\) |
| \(t = \frac{4}{3}x\) | A1 | oe such as \(t = \frac{x}{0.75}\); just this with no working is M1A1 |
| Answer | Marks |
|---|---|
| \(t = 4\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dx}{dt} = \frac{\lambda}{(2x+1)}\) so separate variables to give \(\int(2x+1)\,dx = \int\lambda\,dt\) | M1 | Correct separation; condone missing integral signs |
| \(x^2 + x = \lambda t(+c)\) or \(\frac{(2x+1)^2}{4} = \lambda t(+c)\) | M1 | Correct form for both integrals |
| When \(t=0\), \(x=0\) so \(c = \frac{1}{4}\) or \(c'=0\), so \(t = \frac{x^2+x}{\lambda}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Uses \(x = 1.5\) when \(t = 2\) to give \(\lambda = \frac{15}{8}\) | B1 | or decimal 1.875 |
| Answer | Marks | Guidance |
|---|---|---|
| \(t = \frac{x^2+x}{\lambda} = \frac{12}{\lambda} = 6.4\) hours later | M1 | Substitutes \(x=3\) into expression for \(t\); implied by \(t = \frac{12}{"\lambda"}\) |
| 10.24pm or 22:24 | A1 | 10.24pm or 22:24 only |
# Question 12:
## Part (a)
| Uses $x = kt$ or $t = cx$ and $x = 1.5$ when $t = 2$, so $k=$ or $c=$ | M1 | May be result of differential equation $\frac{dx}{dt} = k$ |
| $t = \frac{4}{3}x$ | A1 | oe such as $t = \frac{x}{0.75}$; just this with no working is M1A1 |
## Part (b)
| $t = 4$ | B1 | |
## Part (c)
| $\frac{dx}{dt} = \frac{\lambda}{(2x+1)}$ so separate variables to give $\int(2x+1)\,dx = \int\lambda\,dt$ | M1 | Correct separation; condone missing integral signs |
| $x^2 + x = \lambda t(+c)$ or $\frac{(2x+1)^2}{4} = \lambda t(+c)$ | M1 | Correct form for both integrals |
| When $t=0$, $x=0$ so $c = \frac{1}{4}$ or $c'=0$, so $t = \frac{x^2+x}{\lambda}$ | A1 | |
## Part (d)
| Uses $x = 1.5$ when $t = 2$ to give $\lambda = \frac{15}{8}$ | B1 | or decimal 1.875 |
## Part (e)
| $t = \frac{x^2+x}{\lambda} = \frac{12}{\lambda} = 6.4$ hours later | M1 | Substitutes $x=3$ into expression for $t$; implied by $t = \frac{12}{"\lambda"}$ |
| 10.24pm or 22:24 | A1 | 10.24pm or 22:24 only |
---\begin{enumerate}
\item In freezing temperatures, ice forms on the surface of the water in a barrel. At time $t$ hours after the start of freezing, the thickness of the ice formed is $x \mathrm {~mm}$. You may assume that the thickness of the ice is uniform across the surface of the water.
\end{enumerate}
At 4 pm there is no ice on the surface, and freezing begins.\\
At 6pm, after two hours of freezing, the ice is 1.5 mm thick.\\
In a simple model, the rate of increase of $x$, in mm per hour, is assumed to be constant for a period of 20 hours.
Using this simple model,\\
(a) express $t$ in terms of $x$,\\
(b) find the value of $t$ when $x = 3$
In a second model, the rate of increase of $x$, in mm per hour, is given by the differential equation
$$\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { \lambda } { ( 2 x + 1 ) } \text { where } \lambda \text { is a constant and } 0 \leqslant t \leqslant 20$$
Using this second model,\\
(c) solve the differential equation and express $t$ in terms of $x$ and $\lambda$,\\
(d) find the exact value for $\lambda$,\\
(e) find at what time the ice is predicted to be 3 mm thick.
\hfill \mbox{\textit{Edexcel C34 2017 Q12 [9]}}