| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2007 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Exponential model with shifted asymptote |
| Difficulty | Moderate -0.3 This is a standard exponential modelling question with a shifted asymptote (Newton's Law of Cooling). Students substitute given values to find constants a and k, then evaluate the model. The structure is straightforward with clear guidance, requiring only algebraic manipulation and understanding that the asymptote represents room temperature. Slightly easier than average due to explicit model form and routine calculations. |
| Spec | 1.06i Exponential growth/decay: in modelling context |
| Answer | Marks | Guidance |
|---|---|---|
| \(t=0, T=100\): \(100 = 25 + a \Rightarrow a = 75\) | B1 | Correct value of \(a\) |
| \(t=3, T=80\): \(80 = 25 + 75e^{-3k}\) | M1 | Substituting second condition |
| \(e^{-3k} = \frac{55}{75} = \frac{11}{15}\) | A1 | Correct equation |
| \(k = -\frac{1}{3}\ln\frac{11}{15} = \frac{1}{3}\ln\frac{15}{11}\) | M1 | Taking logarithms |
| \(k \approx 0.1034\) | A1 | [5] Correct value of \(k\) |
| Answer | Marks | Guidance |
|---|---|---|
| \((A)\) \(T = 25 + 75e^{-5k} \approx 68.0°C\) | B1 | Correct temperature at 5 mins (follow through) |
| \((B)\) As \(t \to \infty\), \(e^{-kt} \to 0\), so \(T \to 25°C\) | B1 | \(25°C\) stated |
| — | B1 | [3] Correct reasoning for long term |
# Question 4:
**Part (i):**
| $t=0, T=100$: $100 = 25 + a \Rightarrow a = 75$ | B1 | Correct value of $a$ |
| $t=3, T=80$: $80 = 25 + 75e^{-3k}$ | M1 | Substituting second condition |
| $e^{-3k} = \frac{55}{75} = \frac{11}{15}$ | A1 | Correct equation |
| $k = -\frac{1}{3}\ln\frac{11}{15} = \frac{1}{3}\ln\frac{15}{11}$ | M1 | Taking logarithms |
| $k \approx 0.1034$ | A1 | [5] Correct value of $k$ |
**Part (ii):**
| $(A)$ $T = 25 + 75e^{-5k} \approx 68.0°C$ | B1 | Correct temperature at 5 mins (follow through) |
| $(B)$ As $t \to \infty$, $e^{-kt} \to 0$, so $T \to 25°C$ | B1 | $25°C$ stated |
| — | B1 | [3] Correct reasoning for long term |
---4 A cup of water is cooling. Its initial temperature is $100 ^ { \circ } \mathrm { C }$. After 3 minutes, its temperature is $80 ^ { \circ } \mathrm { C }$.
\begin{enumerate}[label=(\roman*)]
\item Given that $T = 25 + a \mathrm { e } ^ { - k t }$, where $T$ is the temperature in ${ } ^ { \circ } \mathrm { C } , t$ is the time in minutes and $a$ and $k$ are constants, find the values of $a$ and $k$.
\item What is the temperature of the water\\
(A) after 5 minutes,\\
(B) in the long term?
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C3 2007 Q4 [8]}}