| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2021 |
| Session | October |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Find dy/dx at a point |
| Difficulty | Standard +0.3 This is a straightforward parametric differentiation question requiring standard techniques: finding dy/dx using the chain rule, eliminating the parameter using the identity sec²t = 1 + tan²t, and completing the square. All steps are routine for P4 students with no novel insight required, making it slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{16\sec^2 t \tan t}{2\sec^2 t} = 8\tan t\) | M1 A1 | M1: Attempts to use the rule \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\). Condone incorrect attempts on \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\) |
| At \(x=3\), \(\tan t = -1 \Rightarrow\) Gradient \(= -8\) | dM1 A1 | dM1: Dependent on previous M. Either substituting \(\tan t = -1\) into \(\frac{dy}{dx} = g(t)\), or substituting \(t = -\frac{\pi}{4}\). A1: CSO Gradient \(= -8\). Cannot be awarded from differentiation of \(y = 2(x-5)^2 + 8\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Attempts to use \(1 + \tan^2 t = \sec^2 t \Rightarrow 1 + \frac{(x-5)^2}{4} = \frac{y}{8}\) | M1 A1 | M1: Attempts to use \(\pm 1 \pm \tan^2 t = \pm \sec^2 t\) with \(\tan t\) replaced by expression in \(x\) and \(\sec^2 t\) replaced by expression in \(y\) |
| \(y = 2(x-5)^2 + 8\) | A1 | Also fine as \(f(x) = 2(x-5)^2 + 8\). NB1: Possible to use part (a), find \(\frac{dy}{dx} = 4(x-5)\), then integrate using point \((3,16)\). NB2: Can use points to generate \(f(x) = ax^2 + bx + c\) with 3 simultaneous equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(8\text{ , } f\text{ , } 32\) | M1 A1 | M1: One correct end found, condoning strict inequalities. Look for lower value \(= 8\) or higher value \(= 32\). A1: \(8_n\ f_n\ 32\) or equivalent such as \([8, 32]\) |
## Question 5:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{16\sec^2 t \tan t}{2\sec^2 t} = 8\tan t$ | M1 A1 | M1: Attempts to use the rule $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$. Condone incorrect attempts on $\frac{dy}{dt}$ and $\frac{dx}{dt}$ |
| At $x=3$, $\tan t = -1 \Rightarrow$ Gradient $= -8$ | dM1 A1 | dM1: Dependent on previous M. Either substituting $\tan t = -1$ into $\frac{dy}{dx} = g(t)$, or substituting $t = -\frac{\pi}{4}$. A1: CSO Gradient $= -8$. Cannot be awarded from differentiation of $y = 2(x-5)^2 + 8$ |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Attempts to use $1 + \tan^2 t = \sec^2 t \Rightarrow 1 + \frac{(x-5)^2}{4} = \frac{y}{8}$ | M1 A1 | M1: Attempts to use $\pm 1 \pm \tan^2 t = \pm \sec^2 t$ with $\tan t$ replaced by expression in $x$ and $\sec^2 t$ replaced by expression in $y$ |
| $y = 2(x-5)^2 + 8$ | A1 | Also fine as $f(x) = 2(x-5)^2 + 8$. NB1: Possible to use part (a), find $\frac{dy}{dx} = 4(x-5)$, then integrate using point $(3,16)$. NB2: Can use points to generate $f(x) = ax^2 + bx + c$ with 3 simultaneous equations |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $8\text{ , } f\text{ , } 32$ | M1 A1 | M1: One correct end found, condoning strict inequalities. Look for lower value $= 8$ or higher value $= 32$. A1: $8_n\ f_n\ 32$ or equivalent such as $[8, 32]$ |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{08756c4b-6619-42da-ac8a-2bf065c01de8-14_787_638_251_653}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of the curve $C$ with parametric equations
$$x = 5 + 2 \tan t \quad y = 8 \sec ^ { 2 } t \quad - \frac { \pi } { 3 } \leqslant t \leqslant \frac { \pi } { 4 }$$
\begin{enumerate}[label=(\alph*)]
\item Use parametric differentiation to find the gradient of $C$ at $x = 3$
The curve $C$ has equation $y = \mathrm { f } ( x )$, where f is a quadratic function.
\item Find $\mathrm { f } ( x )$ in the form $a ( x + b ) ^ { 2 } + c$, where $a$, $b$ and $c$ are constants to be found.
\item Find the range of f.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P4 2021 Q5 [9]}}