| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2013 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Newton's law of cooling |
| Difficulty | Standard +0.3 This is a straightforward first-order linear differential equation with separation of variables, followed by a simple logarithm calculation. The question guides students through the solution by providing the target form, making it slightly easier than average. The techniques are standard C4 material with no novel problem-solving required. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int \frac{1}{120-\theta}\, d\theta = \int \lambda\, dt\) or \(\int \frac{1}{\lambda(120-\theta)}\, d\theta = \int dt\) | B1 | Separates variables correctly |
| \(-\ln(120-\theta) = \lambda t + c\) or \(-\frac{1}{\lambda}\ln(120-\theta) = t + c\) | M1 A1; M1 A1 | M1: integrates to \(\pm A\ln(120-\theta)\); A1: correct signs and coefficients |
| \(\{t=0, \theta=20\} \Rightarrow -\ln(120-20) = \lambda(0) + c\) | M1 | Substitutes \(t=0\) AND \(\theta=20\) into integrated equation containing \(c\) |
| \(c = -\ln 100\) | — | — |
| \(-\lambda t = \ln\!\left(\frac{120-\theta}{100}\right)\) or \(\lambda t = \ln\!\left(\frac{100}{120-\theta}\right)\) | — | — |
| \(e^{-\lambda t} = \frac{120-\theta}{100}\) or \(e^{\lambda t} = \frac{100}{120-\theta}\) leading to \(\theta = 120 - 100e^{-\lambda t}\) | dddM1 | Dependent on all 3 previous M marks; must use \(c\) value and fully eliminate logarithms |
| \(\theta = 120 - 100e^{-\lambda t}\) | A1* | Given answer; all previous marks must be scored, no errors in working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\{\lambda=0.01,\, \theta=100\} \Rightarrow 100 = 120 - 100e^{-0.01t}\) | M1 | Substitutes both values into printed equation or earlier equation connecting \(\theta\) and \(t\) |
| \(100e^{0.01t} = 120-100 \Rightarrow -0.01t = \ln\!\left(\frac{120-100}{100}\right)\) | dM1 | Correct order of operations moving from \(100=120-100e^{-0.01t}\) to \(t=\ldots\); requires \(t = A\ln B\) where \(B>0\) |
| \(t = \frac{1}{-0.01}\ln\!\left(\frac{1}{5}\right) = 100\ln 5\) | — | — |
| \(t = 160.94379\ldots \approx 161\) s (nearest second) | A1 | awrt 161 |
## Question 6:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int \frac{1}{120-\theta}\, d\theta = \int \lambda\, dt$ or $\int \frac{1}{\lambda(120-\theta)}\, d\theta = \int dt$ | B1 | Separates variables correctly |
| $-\ln(120-\theta) = \lambda t + c$ or $-\frac{1}{\lambda}\ln(120-\theta) = t + c$ | M1 A1; M1 A1 | M1: integrates to $\pm A\ln(120-\theta)$; A1: correct signs and coefficients |
| $\{t=0, \theta=20\} \Rightarrow -\ln(120-20) = \lambda(0) + c$ | M1 | Substitutes $t=0$ AND $\theta=20$ into integrated equation containing $c$ |
| $c = -\ln 100$ | — | — |
| $-\lambda t = \ln\!\left(\frac{120-\theta}{100}\right)$ or $\lambda t = \ln\!\left(\frac{100}{120-\theta}\right)$ | — | — |
| $e^{-\lambda t} = \frac{120-\theta}{100}$ or $e^{\lambda t} = \frac{100}{120-\theta}$ leading to $\theta = 120 - 100e^{-\lambda t}$ | dddM1 | Dependent on all 3 previous M marks; must use $c$ value and fully eliminate logarithms |
| $\theta = 120 - 100e^{-\lambda t}$ | A1* | Given answer; all previous marks must be scored, no errors in working |
**[8 marks]**
---
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\{\lambda=0.01,\, \theta=100\} \Rightarrow 100 = 120 - 100e^{-0.01t}$ | M1 | Substitutes both values into printed equation or earlier equation connecting $\theta$ and $t$ |
| $100e^{0.01t} = 120-100 \Rightarrow -0.01t = \ln\!\left(\frac{120-100}{100}\right)$ | dM1 | Correct order of operations moving from $100=120-100e^{-0.01t}$ to $t=\ldots$; requires $t = A\ln B$ where $B>0$ |
| $t = \frac{1}{-0.01}\ln\!\left(\frac{1}{5}\right) = 100\ln 5$ | — | — |
| $t = 160.94379\ldots \approx 161$ s (nearest second) | A1 | awrt 161 |
**[3 marks]**
---6. Water is being heated in a kettle. At time $t$ seconds, the temperature of the water is $\theta ^ { \circ } \mathrm { C }$.
The rate of increase of the temperature of the water at any time $t$ is modelled by the differential equation
$$\frac { \mathrm { d } \theta } { \mathrm {~d} t } = \lambda ( 120 - \theta ) , \quad \theta \leqslant 100$$
where $\lambda$ is a positive constant.
Given that $\theta = 20$ when $t = 0$,
\begin{enumerate}[label=(\alph*)]
\item solve this differential equation to show that
$$\theta = 120 - 100 \mathrm { e } ^ { - \lambda t }$$
When the temperature of the water reaches $100 ^ { \circ } \mathrm { C }$, the kettle switches off.
\item Given that $\lambda = 0.01$, find the time, to the nearest second, when the kettle switches off.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2013 Q6 [11]}}