| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Exponential model with shifted asymptote |
| Difficulty | Standard +0.3 This is a standard exponential modelling question requiring substitution of boundary conditions to find constants, differentiation for rate of change, and understanding of horizontal translation. All techniques are routine C3 material with clear scaffolding across parts (a), (b), and (c). Slightly easier than average due to the structured guidance. |
| Spec | 1.06i Exponential growth/decay: in modelling context1.07k Differentiate trig: sin(kx), cos(kx), tan(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(t = 10, T = 18 \Rightarrow 18 = 5 + Ae^{-10k}\) (1) | M1 | |
| \(t = 60, T = 12 \Rightarrow 12 = 5 + Ae^{-60k}\) (2) | M1 | |
| \(\frac{(1)}{(2)} \Rightarrow A = 13e^{10k}\) | M1 | |
| sub (2) \(\Rightarrow 7 = 13e^{10k} \times e^{-60k}\) | M1 | |
| \(e^{-50k} = \frac{7}{13}\) | A1 | |
| \(\therefore k = -\frac{1}{50}\ln\frac{13}{7} = 0.0124\) (3sf) | M1 A1 | |
| \(\therefore A = 13e^{10 \times 0.01238} = 14.7\) (3sf) | A1 | |
| (b) \(T = 5 + 14.7 \times e^{-0.01238t}\) | M1 | |
| \(\frac{dT}{dt} = -0.01238 \times 14.71 \times e^{-0.01238t} = -0.1822e^{-0.01238t}\) | M1 A1 | |
| when \(t = 20\), \(\frac{dT}{dt} = -0.1822e^{-0.01238 \times 20} = -0.142\) | M1 | |
| \(\therefore\) temperature decreasing at rate of \(0.142°\text{C}\) per minute (3sf) | A1 | |
| (c) \(T = 5 + 14.7 \times e^{-0.01238(t-60)}\) | M1 | |
| \(= 5 + 14.7 \times e^{0.7428 - 0.01238t}\) | M1 | |
| \(= 5 + 14.7 \times e^{0.7428} \times e^{-0.01238t}\) | M1 | |
| \(= 5 + 30.9 \times e^{-0.01238t}, B = 30.9\) (3sf) | A1 | (13) |
**(a)** $t = 10, T = 18 \Rightarrow 18 = 5 + Ae^{-10k}$ (1) | M1 |
$t = 60, T = 12 \Rightarrow 12 = 5 + Ae^{-60k}$ (2) | M1 |
$\frac{(1)}{(2)} \Rightarrow A = 13e^{10k}$ | M1 |
sub (2) $\Rightarrow 7 = 13e^{10k} \times e^{-60k}$ | M1 |
$e^{-50k} = \frac{7}{13}$ | A1 |
$\therefore k = -\frac{1}{50}\ln\frac{13}{7} = 0.0124$ (3sf) | M1 A1 |
$\therefore A = 13e^{10 \times 0.01238} = 14.7$ (3sf) | A1 |
**(b)** $T = 5 + 14.7 \times e^{-0.01238t}$ | M1 |
$\frac{dT}{dt} = -0.01238 \times 14.71 \times e^{-0.01238t} = -0.1822e^{-0.01238t}$ | M1 A1 |
when $t = 20$, $\frac{dT}{dt} = -0.1822e^{-0.01238 \times 20} = -0.142$ | M1 |
$\therefore$ temperature decreasing at rate of $0.142°\text{C}$ per minute (3sf) | A1 |
**(c)** $T = 5 + 14.7 \times e^{-0.01238(t-60)}$ | M1 |
$= 5 + 14.7 \times e^{0.7428 - 0.01238t}$ | M1 |
$= 5 + 14.7 \times e^{0.7428} \times e^{-0.01238t}$ | M1 |
$= 5 + 30.9 \times e^{-0.01238t}, B = 30.9$ (3sf) | A1 | (13)7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{795e472b-ad43-432a-a7cf-457b0f5e66f5-4_499_1107_242_415}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a graph of the temperature of a room, $T ^ { \circ } \mathrm { C }$, at time $t$ minutes.\\
The temperature is controlled by a thermostat such that when the temperature falls to $12 ^ { \circ } \mathrm { C }$, a heater is turned on until the temperature reaches $18 ^ { \circ } \mathrm { C }$. The room then cools until the temperature again falls to $12 ^ { \circ } \mathrm { C }$.
For $t$ in the interval $10 \leq t \leq 60$, $T$ is given by
$$T = 5 + A \mathrm { e } ^ { - k t } ,$$
where $A$ and $k$ are constants.\\
Given that $T = 18$ when $t = 10$ and that $T = 12$ when $t = 60$,
\begin{enumerate}[label=(\alph*)]
\item show that $k = 0.0124$ to 3 significant figures and find the value of $A$,
\item find the rate at which the temperature of the room is decreasing when $t = 20$.
The temperature again reaches $18 ^ { \circ } \mathrm { C }$ when $t = 70$ and the graph for $70 \leq t \leq 120$ is a translation of the graph for $10 \leq t \leq 60$.
\item Find the value of the constant $B$ such that for $70 \leq t \leq 120$
$$T = 5 + B \mathrm { e } ^ { - k t } .$$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 Q7 [13]}}