| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2013 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | Finding x from given y value |
| Difficulty | Standard +0.3 This is a standard exponential modelling question requiring substitution of given conditions to find constants, then solving a logarithmic equation. The structure is formulaic (find constants, then use them) with straightforward algebra throughout. It's slightly easier than average because the steps are clearly signposted and the techniques (substituting boundary conditions, solving for exponentials using logarithms) are routine C3 material with no conceptual surprises. |
| Spec | 1.06i Exponential growth/decay: in modelling context |
| Answer | Marks | Guidance |
|---|---|---|
| \(\theta = a - be^{-kt}\) | M1, B1, A1cao [6] | must have \(e^0 = 1\); \(15 = a - b\); \(a = 100\) |
| When \(t = 0, \theta = 15 \Rightarrow 15 = a - b\) | ||
| When \(t = \infty, \theta = 100 \Rightarrow 100 = a\) | ||
| \(\Rightarrow b = 85\) | ||
| When \(t = 1, \theta = 30 \Rightarrow 30 = 100 - 85e^{-k}\) | ||
| \(\Rightarrow e^{-k} = \frac{70}{85}\) | ||
| \(\Rightarrow -k = \ln(70/85) = -0.194(156\ldots)\) | M1, M1, A1 | (need not substitute for \(a\) and \(b\)); Re-arranging and taking lns; allow \(-k = \ln[(a-30)/b]\) ft on \(a, b\); mark final ans; \(0.19\) or better, or \(-\ln(70/85)\) o.e. |
| \(\Rightarrow k = 0.194\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(80 = 100 - 85e^{-0.194t}\) | M1, A1 [2] | ft their values for \(a, b\) and \(k\); but must substitute values; art 7.5 or 7 min 30 s or better |
| \(\Rightarrow e^{-0.194t} = \frac{20}{85}\) | ||
| \(\Rightarrow t = -\ln(4/17) / 0.194 = 7.45\) (min) |
### (i)
$\theta = a - be^{-kt}$ | M1, B1, A1cao [6] | must have $e^0 = 1$; $15 = a - b$; $a = 100$
When $t = 0, \theta = 15 \Rightarrow 15 = a - b$ | |
When $t = \infty, \theta = 100 \Rightarrow 100 = a$ | |
$\Rightarrow b = 85$ | |
When $t = 1, \theta = 30 \Rightarrow 30 = 100 - 85e^{-k}$ | |
$\Rightarrow e^{-k} = \frac{70}{85}$ | |
$\Rightarrow -k = \ln(70/85) = -0.194(156\ldots)$ | M1, M1, A1 | (need not substitute for $a$ and $b$); Re-arranging and taking lns; allow $-k = \ln[(a-30)/b]$ ft on $a, b$; mark final ans; $0.19$ or better, or $-\ln(70/85)$ o.e.
$\Rightarrow k = 0.194$ | |
### (ii)
$80 = 100 - 85e^{-0.194t}$ | M1, A1 [2] | ft their values for $a, b$ and $k$; but must substitute values; art 7.5 or 7 min 30 s or better
$\Rightarrow e^{-0.194t} = \frac{20}{85}$ | |
$\Rightarrow t = -\ln(4/17) / 0.194 = 7.45$ (min) | |
---The temperature $\theta$ °C of water in a container after $t$ minutes is modelled by the equation
$$\theta = a - be^{-kt},$$
where $a$, $b$ and $k$ are positive constants.
The initial and long-term temperatures of the water are 15°C and 100°C respectively. After 1 minute, the temperature is 30°C.
\begin{enumerate}[label=(\roman*)]
\item Find $a$, $b$ and $k$. [6]
\item Find how long it takes for the temperature to reach 80°C. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C3 2013 Q4 [8]}}