| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2018 |
| Session | Specimen |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find tangent equation at point |
| Difficulty | Standard +0.3 This is a straightforward implicit differentiation problem requiring finding a point coordinate, then using dy/dx to find a tangent equation. The differentiation involves chain rule with standard trig functions (sin 2y), and the algebra is routine. Slightly above average difficulty due to the implicit form and exact value work with π, but follows standard P3 techniques without requiring novel insight. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(p = 4\pi^2\) or \((2\pi)^2\) | B1 | Also allow \(x = 4\pi^2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = (4y - \sin 2y)^2 \Rightarrow \frac{dx}{dy} = 2(4y - \sin 2y)(4 - 2\cos 2y)\) | M1 A1 | Uses chain rule to get form \(A(4y - \sin 2y)(B \pm C\cos 2y)\), \(A,B,C \neq 0\) |
| Sub \(y = \frac{\pi}{2} \Rightarrow \frac{dx}{dy} = 24\pi\) (= 75.4) OR \(\frac{dy}{dx} = \frac{1}{24\pi}\) (= 0.013) | M1 | Sub \(y = \frac{\pi}{2}\) into \(\frac{dx}{dy}\) or inverted. Evidence can be minimal |
| Equation of tangent \(y - \frac{\pi}{2} = \frac{1}{24\pi}\left(x - 4\pi^2\right)\) | M1 | Score for correct method finding tangent at \(\left(4\pi^2, \frac{\pi}{2}\right)\) |
| Using \(y - \frac{\pi}{2} = \frac{1}{24\pi}\left(x - 4\pi^2\right)\) with \(x = 0 \Rightarrow y = \frac{\pi}{3}\) | M1 A1 | cso |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = (4y - \sin 2y)^2 \Rightarrow x^{0.5} = 4y - \sin 2y \Rightarrow 0.5x^{-0.5}\frac{dx}{dy} = 4 - 2\cos 2y\) | M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = (16y^2 - 8y\sin 2y + \sin^2 2y) \Rightarrow 1 = 32y\frac{dy}{dx} - 8\sin 2y\frac{dy}{dx} - 16y\cos 2y\frac{dy}{dx} + 4\sin 2y\cos 2y\frac{dy}{dx}\) | M1A1 |
## Question 7:
### Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $p = 4\pi^2$ or $(2\pi)^2$ | B1 | Also allow $x = 4\pi^2$ |
### Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = (4y - \sin 2y)^2 \Rightarrow \frac{dx}{dy} = 2(4y - \sin 2y)(4 - 2\cos 2y)$ | M1 A1 | Uses chain rule to get form $A(4y - \sin 2y)(B \pm C\cos 2y)$, $A,B,C \neq 0$ |
| Sub $y = \frac{\pi}{2} \Rightarrow \frac{dx}{dy} = 24\pi$ (= 75.4) OR $\frac{dy}{dx} = \frac{1}{24\pi}$ (= 0.013) | M1 | Sub $y = \frac{\pi}{2}$ into $\frac{dx}{dy}$ or inverted. Evidence can be minimal |
| Equation of tangent $y - \frac{\pi}{2} = \frac{1}{24\pi}\left(x - 4\pi^2\right)$ | M1 | Score for correct method finding tangent at $\left(4\pi^2, \frac{\pi}{2}\right)$ |
| Using $y - \frac{\pi}{2} = \frac{1}{24\pi}\left(x - 4\pi^2\right)$ with $x = 0 \Rightarrow y = \frac{\pi}{3}$ | M1 A1 | **cso** |
**Alternative I:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = (4y - \sin 2y)^2 \Rightarrow x^{0.5} = 4y - \sin 2y \Rightarrow 0.5x^{-0.5}\frac{dx}{dy} = 4 - 2\cos 2y$ | M1A1 | |
**Alternative II:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = (16y^2 - 8y\sin 2y + \sin^2 2y) \Rightarrow 1 = 32y\frac{dy}{dx} - 8\sin 2y\frac{dy}{dx} - 16y\cos 2y\frac{dy}{dx} + 4\sin 2y\cos 2y\frac{dy}{dx}$ | M1A1 | |
---\begin{enumerate}
\item The point $P$ lies on the curve with equation
\end{enumerate}
$$x = ( 4 y - \sin 2 y ) ^ { 2 }$$
Given that $P$ has $( x , y )$ coordinates $\left( p , \frac { \pi } { 2 } \right)$, where $p$ is a constant,\\
(a) find the exact value of $p$
The tangent to the curve at $P$ cuts the $y$-axis at the point $A$.\\
(b) Use calculus to find the coordinates of $A$.
\hfill \mbox{\textit{Edexcel P3 2018 Q7 [7]}}