| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Newton's law of cooling |
| Difficulty | Standard +0.3 This is a standard C4 differential equations question with routine steps: forming a simple proportionality DE, verifying a solution by differentiation, then solving a separable DE with given boundary conditions. Part (c) requires more algebraic manipulation than typical but follows standard separation of variables technique with no novel insight required. Slightly easier than average due to clear structure and guidance. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dM}{dt} = -kM\), where \(k > 0\) | M1 A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dM}{dt} = \ln(0.98) \times 10(0.98)^t = -0.02M\) | B1, B1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int \frac{10 dM}{10M - 1} = -\int k \, dt\) | B1 | |
| \(\ln(10M - 1) = -kt + c\) | M1 A1 | |
| At \(t = 0, M = 10 \quad \therefore c = \ln 99\) | M1 A1 | |
| At \(t = 10, M = 8.5 \quad \therefore k = \frac{1}{10}\ln\frac{99}{84}\) (= 0.0164) | M1 A1 | |
| Uses \(10M - 1 = 99e^{-kt}\) with values for \(k\) and \(t = 15\) | M1 | |
| To give 7.8 grams (accept awrt 7.8) | A1 | (9) |
## (a)
$\frac{dM}{dt} = -kM$, where $k > 0$ | M1 A1 | (2)
## (b)
$\frac{dM}{dt} = \ln(0.98) \times 10(0.98)^t = -0.02M$ | B1, B1 | (2)
## (c)
$\int \frac{10 dM}{10M - 1} = -\int k \, dt$ | B1 |
$\ln(10M - 1) = -kt + c$ | M1 A1 |
At $t = 0, M = 10 \quad \therefore c = \ln 99$ | M1 A1 |
At $t = 10, M = 8.5 \quad \therefore k = \frac{1}{10}\ln\frac{99}{84}$ (= 0.0164) | M1 A1 |
Uses $10M - 1 = 99e^{-kt}$ with values for $k$ and $t = 15$ | M1 |
To give 7.8 grams (accept awrt 7.8) | A1 | (9)
**Total: (13 marks)**In an experiment a scientist considered the loss of mass of a collection of picked leaves. The mass $M$ grams of a single leaf was measured at times $t$ days after the leaf was picked.
The scientist attempted to find a relationship between $M$ and $t$. In a preliminary model she assumed that the rate of loss of mass was proportional to the mass $M$ grams of the leaf.
\begin{enumerate}[label=(\alph*)]
\item Write down a differential equation for the rate of change of mass of the leaf, using this model.
[2]
\item Show, by differentiation, that $M = 10(0.98)^t$ satisfies this differential equation.
[2]
\end{enumerate}
Further studies implied that the mass $M$ grams of a certain leaf satisfied a modified differential equation
$$10 \frac{dM}{dt} = -k(10M - 1), \quad (1)$$
where $k$ is a positive constant and $t \geq 0$.
Given that the mass of this leaf at time $t = 0$ is 10 grams, and that its mass at time $t = 10$ is 8.5 grams,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item solve the modified differential equation (1) to find the mass of this leaf at time $t = 15$.
[9]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q7 [13]}}