| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2018 |
| Session | Specimen |
| 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 standard separation of variables technique. Part (a) requires routine integration and applying initial conditions to find the constant, while part (b) is simple substitution and logarithm manipulation. The question is slightly easier than average as it's a textbook application of Newton's law of cooling with clear guidance ('show that') and no conceptual surprises. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y) |
| VIIIV SIHI NI JIIYM ION OC | VIIV SIHI NI JIIIAM ION OO | VI4V SIHII NI JIIYM IONOO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int\frac{1}{120-\theta}\,d\theta = \int\lambda\,dt\) | B1 | Separates variables correctly |
| \(-\ln(120-\theta) = \lambda t + c\) | M1A1; M1A1 | M1A1 for integrating lhs; M1A1 for integrating rhs |
| \(t=0,\ \theta=20 \Rightarrow -\ln(100) = \lambda(0) + c\) | M1 | Substitutes \(t=0\) AND \(\theta=20\) in integrated equation leading to \(\pm\lambda t = \ln(f(\theta))\) |
| \(\Rightarrow -\lambda t = \ln\left(\frac{120-\theta}{100}\right)\) | — | — |
| \(e^{-\lambda t} = \frac{120-\theta}{100}\) | dddM1 | Fully correct method to eliminate logarithms with evaluated constant of integration |
| \(\theta = 120 - 100e^{-\lambda t}\) | A1* | Correct answer with no errors (given answer) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\lambda = 0.01,\ \theta = 100 \Rightarrow 100 = 120 - 100e^{-0.01t}\) | M1 | Substitutes \(\lambda = 0.01\), \(\theta = 100\) into given equation |
| \(-0.01t = \ln\left(\frac{120-100}{100}\right) \Rightarrow t = \frac{1}{-0.01}\ln\left(\frac{1}{5}\right) = 100\ln 5\) | dM1 | Correct order of operations moving from \(100 = 120 - 100e^{-0.01t}\) to give \(t = ...\) and \(t = A\ln B\), \(B > 0\) |
| \(t = 160.9437...\approx 161\) (s) (nearest second) | A1 | Awrt 161 seconds |
## Question 8(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int\frac{1}{120-\theta}\,d\theta = \int\lambda\,dt$ | B1 | Separates variables correctly |
| $-\ln(120-\theta) = \lambda t + c$ | M1A1; M1A1 | M1A1 for integrating lhs; M1A1 for integrating rhs |
| $t=0,\ \theta=20 \Rightarrow -\ln(100) = \lambda(0) + c$ | M1 | Substitutes $t=0$ AND $\theta=20$ in integrated equation leading to $\pm\lambda t = \ln(f(\theta))$ |
| $\Rightarrow -\lambda t = \ln\left(\frac{120-\theta}{100}\right)$ | — | — |
| $e^{-\lambda t} = \frac{120-\theta}{100}$ | dddM1 | Fully correct method to eliminate logarithms with evaluated constant of integration |
| $\theta = 120 - 100e^{-\lambda t}$ | A1* | Correct answer with no errors (given answer) |
## Question 8(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\lambda = 0.01,\ \theta = 100 \Rightarrow 100 = 120 - 100e^{-0.01t}$ | M1 | Substitutes $\lambda = 0.01$, $\theta = 100$ into given equation |
| $-0.01t = \ln\left(\frac{120-100}{100}\right) \Rightarrow t = \frac{1}{-0.01}\ln\left(\frac{1}{5}\right) = 100\ln 5$ | dM1 | Correct order of operations moving from $100 = 120 - 100e^{-0.01t}$ to give $t = ...$ and $t = A\ln B$, $B > 0$ |
| $t = 160.9437...\approx 161$ (s) (nearest second) | A1 | Awrt 161 seconds |8. 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 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.
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VIIIV SIHI NI JIIYM ION OC & VIIV SIHI NI JIIIAM ION OO & VI4V SIHII NI JIIYM IONOO \\
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\hfill \mbox{\textit{Edexcel P4 2018 Q8 [11]}}