| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2011 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Show constant equals specific form |
| Difficulty | Moderate -0.3 This is a standard Newton's Law of Cooling application requiring substitution to find A, solving an exponential equation for k (with the answer given to verify), and differentiation for rate of change. All techniques are routine C3 material with clear signposting, making it slightly easier than average despite being multi-part. |
| Spec | 1.06g Equations with exponentials: solve a^x = b1.06i Exponential growth/decay: in modelling context1.07k Differentiate trig: sin(kx), cos(kx), tan(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substitutes \(t = 0\) and \(\theta = 90\) into \(\theta = 20 + Ae^{-kt}\): \(90 = 20 + Ae^{-k(0)}\) | M1 | Substitutes \(t=0\) and \(\theta=90\) into eqn |
| \(90 = 20 + A \Rightarrow A = 70\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(55 = 20 + 70e^{-k(5)}\), rearranges to \(\frac{35}{70} = e^{-5k}\) | M1 | Substitutes \(t=5\), \(\theta=55\) and rearranges to make \(e^{\pm 5k}\) the subject |
| \(\ln\!\left(\frac{35}{70}\right) = -5k\) | dM1 | Takes ln and proceeds to make \(\pm 5k\) the subject |
| \(-5k = \ln\!\left(\frac{1}{2}\right)\), \(-5k = \ln 1 - \ln 2 \Rightarrow -5k = -\ln 2 \Rightarrow k = \frac{1}{5}\ln 2\) | A1* | Convincing proof that \(k = \frac{1}{5}\ln 2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{d\theta}{dt} = -\frac{1}{5}\ln 2 \cdot (70)e^{-\frac{1}{5}t\ln 2}\) | M1 | \(\pm\alpha e^{-kt}\) where \(k = \frac{1}{5}\ln 2\) |
| \(-14\ln 2\, e^{-\frac{1}{5}t\ln 2}\) | A1 oe | |
| When \(t=10\): \(\frac{d\theta}{dt} = -14\ln 2\, e^{-2\ln 2} = -\frac{7}{2}\ln 2 = -2.426015...\) | ||
| Rate of decrease of \(\theta = 2.426\ ^\circ\text{C/min}\) (3 d.p.) | A1 | awrt \(\pm 2.426\) |
## Question 4:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitutes $t = 0$ and $\theta = 90$ into $\theta = 20 + Ae^{-kt}$: $90 = 20 + Ae^{-k(0)}$ | M1 | Substitutes $t=0$ and $\theta=90$ into eqn |
| $90 = 20 + A \Rightarrow A = 70$ | A1 | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $55 = 20 + 70e^{-k(5)}$, rearranges to $\frac{35}{70} = e^{-5k}$ | M1 | Substitutes $t=5$, $\theta=55$ and rearranges to make $e^{\pm 5k}$ the subject |
| $\ln\!\left(\frac{35}{70}\right) = -5k$ | dM1 | Takes ln and proceeds to make $\pm 5k$ the subject |
| $-5k = \ln\!\left(\frac{1}{2}\right)$, $-5k = \ln 1 - \ln 2 \Rightarrow -5k = -\ln 2 \Rightarrow k = \frac{1}{5}\ln 2$ | A1* | Convincing proof that $k = \frac{1}{5}\ln 2$ |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{d\theta}{dt} = -\frac{1}{5}\ln 2 \cdot (70)e^{-\frac{1}{5}t\ln 2}$ | M1 | $\pm\alpha e^{-kt}$ where $k = \frac{1}{5}\ln 2$ |
| $-14\ln 2\, e^{-\frac{1}{5}t\ln 2}$ | A1 oe | |
| When $t=10$: $\frac{d\theta}{dt} = -14\ln 2\, e^{-2\ln 2} = -\frac{7}{2}\ln 2 = -2.426015...$ | | |
| Rate of decrease of $\theta = 2.426\ ^\circ\text{C/min}$ (3 d.p.) | A1 | awrt $\pm 2.426$ |
---4. Joan brings a cup of hot tea into a room and places the cup on a table. At time $t$ minutes after Joan places the cup on the table, the temperature, $\theta ^ { \circ } \mathrm { C }$, of the tea is modelled by the equation
$$\theta = 20 + A \mathrm { e } ^ { - k t } ,$$
where $A$ and $k$ are positive constants.
Given that the initial temperature of the tea was $90 ^ { \circ } \mathrm { C }$,
\begin{enumerate}[label=(\alph*)]
\item find the value of $A$.
The tea takes 5 minutes to decrease in temperature from $90 ^ { \circ } \mathrm { C }$ to $55 ^ { \circ } \mathrm { C }$.
\item Show that $k = \frac { 1 } { 5 } \ln 2$.
\item Find the rate at which the temperature of the tea is decreasing at the instant when $t = 10$. Give your answer, in ${ } ^ { \circ } \mathrm { C }$ per minute, to 3 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2011 Q4 [8]}}