| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | October |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Parametric differentiation |
| Difficulty | Standard +0.3 This is a standard parametric differentiation question requiring chain rule application (dx/dy then reciprocal for dy/dx), followed by algebraic manipulation using trigonometric identities. Part (a) involves solving sin²(4y) = 1/4, parts (b-c) require routine differentiation and identity application (sin(2θ) = 2sinθcosθ), and part (d) is straightforward interpretation. While multi-step, each component is a textbook technique with no novel insight required, making it slightly easier than average. |
| Spec | 1.05a Sine, cosine, tangent: definitions for all arguments1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{1}{4} = \sin^2 4y \Rightarrow y = \frac{\pi}{24}\) | M1A1 | M1: Substitute \(x = \frac{1}{4}\), rearrange to find exact value for \(y\). Look for \(y = M\arcsin\left(\pm\frac{1}{2}\right) \rightarrow k\pi\) or \(k(30°)\), allowing errors in dividing by 4, via double angle identity, correct up to sign error. A1: \(\frac{\pi}{24}\). Ignore extra solutions outside domain. |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{dx}{dy} = 8\sin 4y \cos 4y\) | M1A1 | M1: Differentiate achieving form \(\frac{dx}{dy} = A\sin 4y\cos 4y\) or \(A\sin 8y\). Also accept via implicit differentiation \(1 = A\sin 4y\cos 4y\frac{dy}{dx}\). The \(\frac{dx}{dy}\) may be missing or labelled \(\frac{dy}{dx}\). A1: \(\frac{dx}{dy} = 8\sin 4y\cos 4y\) (or e.g. \(\frac{dx}{dy} = 4\sin 8y\)). Coefficients must be simplified. Must include \(\frac{dx}{dy}\). |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{dx}{dy} = 8\sin 4y\cos 4y \rightarrow \frac{dy}{dx} = \frac{1}{8\sin 4y\cos 4y}\) | M1 | Uses \(\frac{dy}{dx} = 1 \div \frac{dx}{dy}\). Allow reciprocal of answer to (b) if LHS omitted. Variables must be consistent. |
| \(\frac{dy}{dx} = \frac{1}{8\sqrt{x}(1-x)}\) | M1 | Attempts to use \(\sin 4y = \pm\sqrt{x}\) and \(\cos 4y = \pm\sqrt{1-x}\) to write \(\frac{dy}{dx}\) or \(\frac{dx}{dy}\) in terms of \(x\) only, with no trig terms. |
| \(\frac{dy}{dx} = \dfrac{1}{\sqrt{16-64\left(x-\frac{1}{2}\right)^2}}\) | A1 | Accept isw after correct answer. Allow with terms under square root reversed. |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(x = \frac{1}{2}\) | B1ft | ft their \(-s\) provided their \(q > 0\), \(r < 0\) and \(x\) is in range \(0 \leq x \leq 1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{dy}{dx} = \frac{1}{4}\) | B1ft | ft their \(q\) provided it is positive and their \(r\) was negative. |
# Question 10:
## Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{1}{4} = \sin^2 4y \Rightarrow y = \frac{\pi}{24}$ | M1A1 | M1: Substitute $x = \frac{1}{4}$, rearrange to find exact value for $y$. Look for $y = M\arcsin\left(\pm\frac{1}{2}\right) \rightarrow k\pi$ or $k(30°)$, allowing errors in dividing by 4, via double angle identity, correct up to sign error. A1: $\frac{\pi}{24}$. Ignore extra solutions outside domain. |
**Total: 2 marks**
---
## Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{dx}{dy} = 8\sin 4y \cos 4y$ | M1A1 | M1: Differentiate achieving form $\frac{dx}{dy} = A\sin 4y\cos 4y$ or $A\sin 8y$. Also accept via implicit differentiation $1 = A\sin 4y\cos 4y\frac{dy}{dx}$. The $\frac{dx}{dy}$ may be missing or labelled $\frac{dy}{dx}$. A1: $\frac{dx}{dy} = 8\sin 4y\cos 4y$ (or e.g. $\frac{dx}{dy} = 4\sin 8y$). Coefficients must be simplified. Must include $\frac{dx}{dy}$. |
**Total: 2 marks**
---
## Part (c):
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{dx}{dy} = 8\sin 4y\cos 4y \rightarrow \frac{dy}{dx} = \frac{1}{8\sin 4y\cos 4y}$ | M1 | Uses $\frac{dy}{dx} = 1 \div \frac{dx}{dy}$. Allow reciprocal of answer to (b) if LHS omitted. Variables must be consistent. |
| $\frac{dy}{dx} = \frac{1}{8\sqrt{x}(1-x)}$ | M1 | Attempts to use $\sin 4y = \pm\sqrt{x}$ and $\cos 4y = \pm\sqrt{1-x}$ to write $\frac{dy}{dx}$ or $\frac{dx}{dy}$ in terms of $x$ only, with no trig terms. |
| $\frac{dy}{dx} = \dfrac{1}{\sqrt{16-64\left(x-\frac{1}{2}\right)^2}}$ | A1 | Accept isw after correct answer. Allow with terms under square root reversed. |
**Total: 3 marks**
---
## Part (d)(i):
| Working | Mark | Guidance |
|---------|------|----------|
| $x = \frac{1}{2}$ | B1ft | ft their $-s$ provided their $q > 0$, $r < 0$ and $x$ is in range $0 \leq x \leq 1$ |
---
## Part (d)(ii):
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{dy}{dx} = \frac{1}{4}$ | B1ft | ft their $q$ provided it is positive and their $r$ was negative. |
**Total: 2 marks**
**Question 10 Total: 9 marks**\begin{enumerate}
\item In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable.
\end{enumerate}
A curve $C$ has equation
$$x = \sin ^ { 2 } 4 y \quad 0 \leqslant y \leqslant \frac { \pi } { 8 } \quad 0 \leqslant x \leqslant 1$$
The point $P$ with $x$ coordinate $\frac { 1 } { 4 }$ lies on $C$\\
(a) Find the exact $y$ coordinate of $P$\\
(b) Find $\frac { \mathrm { d } x } { \mathrm {~d} y }$\\
(c) Hence show that $\frac { \mathrm { d } y } { \mathrm {~d} x }$ can be written in the form
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { \sqrt { q + r ( x + s ) ^ { 2 } } }$$
where $q , r$ and $s$ are constants to be found.
Using the answer to part (c),\\
(d) (i) state the $x$ coordinate of the point where the value of $\frac { \mathrm { d } y } { \mathrm {~d} x }$ is a minimum,\\
(ii) state the value of $\frac { \mathrm { d } y } { \mathrm {~d} x }$ at this point.
\hfill \mbox{\textit{Edexcel P3 2023 Q10 [9]}}