| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2022 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Trapezium rule applied to real-world data |
| Difficulty | Standard +0.3 This is a straightforward applied calculus question requiring standard techniques: differentiation to find maximum velocity, integration with substitution to verify an equation, and iterative calculation using a given formula. All steps are routine P3 procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.08d Evaluate definite integrals: between limits1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{dv}{dt} = -e^{t-10} + 9e^{-0.75t} = 0 \Rightarrow 9 = e^{1.75t-10}\) or \(9e^{-0.75t} = e^{t-10}\) | M1A1 | Differentiates to \(\pm Ae^{t-10} \pm Be^{-0.75t}\) and sets equal to 0 |
| \(9 = e^{1.75t-10} \Rightarrow t = \frac{10 + \ln 9}{1.75}\) or \(\ln 9 - 0.75t = t - 10 \Rightarrow t = \frac{10 + \ln 9}{1.75}\) | M1 | Proceeds from equation of form \(Ae^{Ct\pm D} - Be^{Et\pm F} = 0\) to a value for \(t\) |
| \(t =\) awrt \(6.97\) | A1 | Also accept exact forms e.g. \(\frac{40 + 4\ln 9}{7}\) or \(\frac{10+\ln 9}{1.75}\) |
| awrt \(11.9\) (ms\(^{-1}\)) | A1 | With all previous marks scored; condone lack of units |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\int 12 - 12e^{-0.75t} - e^{(t-10)}\, dt = 12t - e^{(t-10)} + 16e^{-0.75t}\ (+C)\) | M1A1 | Award for \(12t \pm Ae^{(t-10)} \pm Be^{-0.75t}\); do not accept index \((t-10)+1\) |
| \(\left[12t - e^{(t-10)} + 16e^{-0.75t}\right]_0^T = 100 \Rightarrow 12T = \ldots\) | M1 | Substitutes \(T\) and \(0\), sets equal to 100, attempts to make \(12T\) or \(T\) subject |
| \(T = \frac{1}{12}\left(116 - 16e^{-0.75T} + e^{T-10} - e^{-10}\right)\) * | A1* | No errors including bracket errors/omissions; must have \(T = \ldots\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(T_2 = \frac{1}{12}\left(116 - 16e^{-0.75\times10} + e^{10-10} - e^{-10}\right)\) | M1 | Substitutes 10 into the iterative formula |
| \(T_2 =\) awrt \(9.7493\) | A1 | May be implied by awrt 9.7492 or 9.7493, or by 9.7306 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(9.7293\) (seconds) | A1 | Cannot be awarded without the method mark; note: finding directly on calculator is not acceptable |
# Question 8:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{dv}{dt} = -e^{t-10} + 9e^{-0.75t} = 0 \Rightarrow 9 = e^{1.75t-10}$ or $9e^{-0.75t} = e^{t-10}$ | M1A1 | Differentiates to $\pm Ae^{t-10} \pm Be^{-0.75t}$ and sets equal to 0 |
| $9 = e^{1.75t-10} \Rightarrow t = \frac{10 + \ln 9}{1.75}$ or $\ln 9 - 0.75t = t - 10 \Rightarrow t = \frac{10 + \ln 9}{1.75}$ | M1 | Proceeds from equation of form $Ae^{Ct\pm D} - Be^{Et\pm F} = 0$ to a value for $t$ |
| $t =$ awrt $6.97$ | A1 | Also accept exact forms e.g. $\frac{40 + 4\ln 9}{7}$ or $\frac{10+\ln 9}{1.75}$ |
| awrt $11.9$ (ms$^{-1}$) | A1 | With all previous marks scored; condone lack of units |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\int 12 - 12e^{-0.75t} - e^{(t-10)}\, dt = 12t - e^{(t-10)} + 16e^{-0.75t}\ (+C)$ | M1A1 | Award for $12t \pm Ae^{(t-10)} \pm Be^{-0.75t}$; do not accept index $(t-10)+1$ |
| $\left[12t - e^{(t-10)} + 16e^{-0.75t}\right]_0^T = 100 \Rightarrow 12T = \ldots$ | M1 | Substitutes $T$ and $0$, sets equal to 100, attempts to make $12T$ or $T$ subject |
| $T = \frac{1}{12}\left(116 - 16e^{-0.75T} + e^{T-10} - e^{-10}\right)$ * | A1* | No errors including bracket errors/omissions; must have $T = \ldots$ |
## Part (c)(i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $T_2 = \frac{1}{12}\left(116 - 16e^{-0.75\times10} + e^{10-10} - e^{-10}\right)$ | M1 | Substitutes 10 into the iterative formula |
| $T_2 =$ awrt $9.7493$ | A1 | May be implied by awrt 9.7492 or 9.7493, or by 9.7306 |
## Part (c)(ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $9.7293$ (seconds) | A1 | Cannot be awarded without the method mark; note: finding directly on calculator is not acceptable |
---8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{44035bf8-f54c-472a-b969-b4fa4fa3d203-26_579_467_219_749}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
Figure 4 is a graph showing the velocity of a sprinter during a 100 m race.\\
The sprinter's velocity during the race, $v \mathrm {~ms} ^ { - 1 }$, is modelled by the equation
$$v = 12 - \mathrm { e } ^ { t - 10 } - 12 \mathrm { e } ^ { - 0.75 t } \quad t \geqslant 0$$
where $t$ seconds is the time after the sprinter begins to run.
According to the model,
\begin{enumerate}[label=(\alph*)]
\item find, using calculus, the sprinter's maximum velocity during the race.
Given that the sprinter runs 100 m in $T$ seconds, such that
$$\int _ { 0 } ^ { T } v \mathrm {~d} t = 100$$
\item show that $T$ is a solution of the equation
$$T = \frac { 1 } { 12 } \left( 116 - 16 \mathrm { e } ^ { - 0.75 T } + \mathrm { e } ^ { T - 10 } - \mathrm { e } ^ { - 10 } \right)$$
The iteration formula
$$T _ { n + 1 } = \frac { 1 } { 12 } \left( 116 - 16 \mathrm { e } ^ { - 0.75 T _ { n } } + \mathrm { e } ^ { T _ { n } - 10 } - \mathrm { e } ^ { - 10 } \right)$$
is used to find an approximate value for $T$
Using this iteration formula with $T _ { 1 } = 10$
\item find, to 4 decimal places,
\begin{enumerate}[label=(\roman*)]
\item the value of $T _ { 2 }$
\item the time taken by the sprinter to run the race, according to the model.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel P3 2022 Q8 [12]}}