| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2022 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Tangent/normal meets curve again |
| Difficulty | Standard +0.8 This is a multi-part parametric question requiring differentiation using the chain rule, finding tangent equations, solving simultaneous equations by substitution (leading to a trigonometric equation), and using exact trigonometric values. The algebraic manipulation in part (c) to derive the given equation is non-trivial, and part (d) requires careful handling of exact values. While the techniques are standard P4 content, the question requires sustained reasoning across multiple steps with potential for algebraic errors, placing it moderately above average difficulty. |
| Spec | 1.02q Use intersection points: of graphs to solve equations1.03g Parametric equations: of curves and conversion to cartesian1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dx}{dt} = \cos t + 6\cos t \sin t\) | B1 | Or \(\cos t + 3\sin 2t\) |
| \(\frac{dy}{dt} = 3\cos t - 2\sin t\) | B1 | |
| \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3\cos t - 2\sin t}{\cos t + 6\cos t\sin t} = \frac{3\cos\pi - 2\sin\pi}{\cos\pi + 6\cos\pi\sin\pi} = 3\) | M1A1* | M1: attempts \(\frac{dy}{dx}=\frac{dy/dt}{dx/dt}\) substituting \(t=\pi\); A1*: achieves \(\frac{dy}{dx}=3\) with full working and no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| When \(t=\pi\), \(x=-3\), \(y=-2\) | B1 | May be seen within working |
| \(y - (-2) = 3(x-(-3))\) | M1 | Attempts tangent with gradient 3; if using \(y=mx+c\) must proceed to find \(c\) |
| \(y = 3x+7\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y=3x+7 \Rightarrow 3\sin t + 2\cos t = 3(\sin t - 3\cos^2 t)+7\) or \(y=3(x+3\cos^2 t)+2\cos t \Rightarrow 3x+7=3x+9\cos^2 t+2\cos t\) | M1 | Full attempt to solve parametric equations simultaneously with tangent equation |
| \(\Rightarrow 9\cos^2 t + 2\cos t - 7 = 0\) | A1* | Achieves given answer with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\cos t = \frac{7}{9}\) | B1 | Seen or implied; allow if seen in (c) |
| \(y = 3\times\frac{\sqrt{32}}{9}+2\times\frac{7}{9} = \frac{4\sqrt{2}}{3}+\frac{14}{9}\) | M1A1 | M1: attempts to find \(y\)-coordinate, must evaluate trig terms; A1: \(\frac{4\sqrt{2}}{3}+\frac{14}{9}\) or exact equivalent; withhold if additional answers given |
# Question 7:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dx}{dt} = \cos t + 6\cos t \sin t$ | B1 | Or $\cos t + 3\sin 2t$ |
| $\frac{dy}{dt} = 3\cos t - 2\sin t$ | B1 | |
| $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3\cos t - 2\sin t}{\cos t + 6\cos t\sin t} = \frac{3\cos\pi - 2\sin\pi}{\cos\pi + 6\cos\pi\sin\pi} = 3$ | M1A1* | M1: attempts $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$ substituting $t=\pi$; A1*: achieves $\frac{dy}{dx}=3$ with full working and no errors |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| When $t=\pi$, $x=-3$, $y=-2$ | B1 | May be seen within working |
| $y - (-2) = 3(x-(-3))$ | M1 | Attempts tangent with gradient 3; if using $y=mx+c$ must proceed to find $c$ |
| $y = 3x+7$ | A1 | |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y=3x+7 \Rightarrow 3\sin t + 2\cos t = 3(\sin t - 3\cos^2 t)+7$ or $y=3(x+3\cos^2 t)+2\cos t \Rightarrow 3x+7=3x+9\cos^2 t+2\cos t$ | M1 | Full attempt to solve parametric equations simultaneously with tangent equation |
| $\Rightarrow 9\cos^2 t + 2\cos t - 7 = 0$ | A1* | Achieves given answer with no errors |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\cos t = \frac{7}{9}$ | B1 | Seen or implied; allow if seen in (c) |
| $y = 3\times\frac{\sqrt{32}}{9}+2\times\frac{7}{9} = \frac{4\sqrt{2}}{3}+\frac{14}{9}$ | M1A1 | M1: attempts to find $y$-coordinate, must evaluate trig terms; A1: $\frac{4\sqrt{2}}{3}+\frac{14}{9}$ or exact equivalent; withhold if additional answers given |
---\begin{enumerate}
\item In this question you must show all stages of your working.
\end{enumerate}
\section*{Solutions relying entirely on calculator technology are not acceptable.}
The curve $C$ has parametric equations
$$x = \sin t - 3 \cos ^ { 2 } t \quad y = 3 \sin t + 2 \cos t \quad 0 \leqslant t \leqslant 5$$
(a) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = 3$ where $t = \pi$
The point $P$ lies on $C$ where $t = \pi$\\
(b) Find the equation of the tangent to the curve at $P$ in the form $y = m x + c$ where $m$ and $c$ are constants to be found.
Given that the tangent to the curve at $P$ cuts $C$ at the point $Q$\\
(c) show that the value of $t$ at point $Q$ satisfies the equation
$$9 \cos ^ { 2 } t + 2 \cos t - 7 = 0$$
(d) Hence find the exact value of the $y$ coordinate of $Q$
\hfill \mbox{\textit{Edexcel P4 2022 Q7 [12]}}