| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2016 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Newton's law of cooling |
| Difficulty | Moderate -0.8 This is a straightforward application of exponential models with routine substitution (part a), logarithm manipulation (part b), and basic differentiation (part c). All parts follow standard textbook procedures with no problem-solving insight required, making it easier than average but not trivial due to the algebraic manipulation needed in parts (b) and (c). |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties1.06g Equations with exponentials: solve a^x = b1.07j Differentiate exponentials: e^(kx) and a^(kx) |
| VIII SIHI NI I IVM I I ON OC | VIIV SIHI NI JIIIM IONOO | VI4V SIHI NI BIIIM ION OO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(320\ (°C)\) | B1 | cao, do not need °C |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T = 180 \Rightarrow 300e^{-0.04t} = 160 \Rightarrow e^{-0.04t} = \frac{160}{300}\) (awrt 0.53) | M1, A1 | Substitutes \(T=180\), proceeds to \(Ae^{-0.04t} = B\) or \(Ce^{0.04t} = D\) |
| \(t = \frac{1}{-0.04}\ln\left(\frac{160}{300}\right)\) or \(\frac{1}{0.04}\ln\left(\frac{300}{160}\right)\) | dM1 | Dependent on first M1; moving from \(e^{kt}=c, c>0 \Rightarrow t = \frac{\ln c}{k}\) |
| \(15.7\) (minutes) cao | A1cso | Correct answer and correct solution only, do not accept awrt |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(180 = 300e^{-0.04t} + 20\) and \(300e^{-0.04t} = 160\) | M1 | |
| \(\ln 300 - 0.04t = \ln 160 \Rightarrow t = \frac{\ln 300 - \ln 160}{0.04}\) | dM1, A1 | |
| \(15.7\) (minutes) cao | A1cso |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dT}{dt} = (-0.04) \times 300e^{-0.04t} = (-0.04) \times (T-20)\) | M1 A1 | Differentiates to give \(\frac{dT}{dt} = ke^{-0.04t}\); correct derivative eliminating \(t\) |
| \(= \frac{20-T}{25}\ *\) | A1* | Obtains printed answer correctly — no errors |
## Question 6(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $320\ (°C)$ | B1 | cao, do not need °C |
## Question 6(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T = 180 \Rightarrow 300e^{-0.04t} = 160 \Rightarrow e^{-0.04t} = \frac{160}{300}$ (awrt 0.53) | M1, A1 | Substitutes $T=180$, proceeds to $Ae^{-0.04t} = B$ or $Ce^{0.04t} = D$ |
| $t = \frac{1}{-0.04}\ln\left(\frac{160}{300}\right)$ or $\frac{1}{0.04}\ln\left(\frac{300}{160}\right)$ | dM1 | Dependent on first M1; moving from $e^{kt}=c, c>0 \Rightarrow t = \frac{\ln c}{k}$ |
| $15.7$ (minutes) cao | A1cso | Correct answer and correct solution only, do not accept awrt |
## Question 6(b) Alternative:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $180 = 300e^{-0.04t} + 20$ and $300e^{-0.04t} = 160$ | M1 | |
| $\ln 300 - 0.04t = \ln 160 \Rightarrow t = \frac{\ln 300 - \ln 160}{0.04}$ | dM1, A1 | |
| $15.7$ (minutes) cao | A1cso | |
## Question 6(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dT}{dt} = (-0.04) \times 300e^{-0.04t} = (-0.04) \times (T-20)$ | M1 A1 | Differentiates to give $\frac{dT}{dt} = ke^{-0.04t}$; correct derivative eliminating $t$ |
| $= \frac{20-T}{25}\ *$ | A1* | Obtains printed answer correctly — no errors |
---6. A hot piece of metal is dropped into a cool liquid. As the metal cools, its temperature $T$ degrees Celsius, $t$ minutes after it enters the liquid, is modelled by
$$T = 300 \mathrm { e } ^ { - 0.04 t } + 20 , \quad t \geqslant 0$$
\begin{enumerate}[label=(\alph*)]
\item Find the temperature of the piece of metal as it enters the liquid.
\item Find the value of $t$ for which $T = 180$, giving your answer to 3 significant figures. (Solutions based entirely on graphical or numerical methods are not acceptable.)
\item Show, by differentiation, that the rate, in degrees Celsius per minute, at which the temperature of the metal is changing, is given by the expression
$$\frac { 20 - T } { 25 }$$
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VIII SIHI NI I IVM I I ON OC & VIIV SIHI NI JIIIM IONOO & VI4V SIHI NI BIIIM ION OO \\
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\hfill \mbox{\textit{Edexcel C34 2016 Q6 [8]}}