Question 10 - A-Level Maths

Edexcel P3 2021 January — Question 10 10 marks

Exam Board	Edexcel
Module	P3 (Pure Mathematics 3)
Year	2021
Session	January
Marks	10
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Chain Rule
Type	Equation of normal line
Difficulty	Standard +0.3 This is a straightforward P3 chain rule question with parametric differentiation. Part (a) requires basic differentiation of sec²(2y), part (b) involves reciprocal and algebraic manipulation using trig identities (tan²θ = sec²θ - 1), and part (c) is standard tangent/normal application. While it requires multiple techniques, each step follows standard procedures with no novel insight needed, making it slightly easier than average.
Spec	1.07m Tangents and normals: gradient and equations 1.07s Parametric and implicit differentiation

10. The curve $C$ has equation $$x = 3 \sec ^ { 2 } 2 y \quad x > 3 \quad 0 < y < \frac { \pi } { 4 }$$

Find $\frac { \mathrm { d } x } { \mathrm {~d} y }$ in terms of $y$.
Hence show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { p } { q x \sqrt { x - 3 } }$$ where $p$ is irrational and $q$ is an integer, stating the values of $p$ and $q$.
Find the equation of the normal to $C$ at the point where $y = \frac { \pi } { 12 }$, giving your answer in the form $y = m x + c$, giving $m$ and $c$ as exact irrational numbers.

END

VI4V SIHI NI JIIIM IONOO VIAV SIHI NI JIIIM ION OO VI4V SIHI NI IIIYM ION OC

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Question 10:

Part (a):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$\frac{dx}{dy} = 12\sec^2 2y\tan 2y$	M1, A1	M1: differentiates to form $\alpha\sec^2 2y\tan 2y$. Note scheme also applies if student adapts $x=3\sec^2 2y$ to $x=\pm3\tan^2 2y\pm3$. A1: $\frac{dx}{dy}=12\sec^2 2y\tan 2y$; if LHS included it must be correct. Condone unsimplified $6\sec 2y\times 2\sec 2y\tan 2y$. ISW after correct answer

Part (b):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$\frac{dx}{dy}=12\left(\frac{x}{3}\right)\sqrt{\sec^2 2y-1} \Rightarrow \frac{dx}{dy}=12\left(\frac{x}{3}\right)\sqrt{\frac{x}{3}-1}$	M1, A1ft	M1: replace $\sec^2 2y$ with $\alpha x$; use identity $\pm1\pm\tan^2 2y=\pm\sec^2 2y$ and replace $\tan 2y = b\sqrt{\pm1\pm dx}$. A1ft: requires substitution of both $\sec^2 2y$ with $\frac{x}{3}$ and $\tan 2y=\sqrt{\frac{x}{3}-1}$ following through on their $\frac{dx}{dy}=\alpha\sec^2 2y\tan 2y$
$\frac{dy}{dx}=\frac{\sqrt{3}}{4x\sqrt{x-3}}$	A1

Part (c):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$y=\frac{\pi}{12}\Rightarrow x=4$	B1
$\frac{dx}{dy}=12\times\frac{4}{3}\times\frac{1}{\sqrt{3}}$ or $\frac{dy}{dx}=\frac{1}{12\left(\frac{4}{3}\right)\sqrt{\frac{4}{3}-1}}$	M1
Correct $m_N = -\frac{16}{\sqrt{3}}$ o.e.	A1
$y-\frac{\pi}{12}=-\frac{16}{\sqrt{3}}(x-4)$	dM1
$y=-\frac{16\sqrt{3}}{3}x+\frac{64\sqrt{3}}{3}+\frac{\pi}{12}$	A1

Question (a) - Alt Method:

Answer	Marks
$x = 3(\cos 2y)^{-2} \Rightarrow \frac{dx}{dy} = 12(\cos 2y)^{-3} \sin 2y$	M1 A1

Question (b) - Alt Method:

Answer	Marks	Guidance
If $x = 3(\cos 2y)^{-2} \Rightarrow \cos 2y = \frac{\sqrt{3}}{\sqrt{x}}$	stated or implied
$\frac{dx}{dy} = \frac{12\sin 2y}{(\cos 2y)^3} = \frac{12\sqrt{\frac{x-3}{x}}}{\left(\sqrt{\frac{3}{x}}\right)^3}$	M1 A1	Score in similar way to main scheme

Alt (b) via arccos:

Answer	Marks	Guidance
$y = \frac{1}{2}\arccos\sqrt{\frac{3}{x}} \Rightarrow \frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{\sqrt{1-\left(\sqrt{\frac{3}{x}}\right)^2}} \times \frac{\sqrt{3}}{2} x^{-\frac{3}{2}}$
For $\frac{dy}{dx} = \lambda \frac{1}{\sqrt{1-\left(\sqrt{\frac{\alpha}{x}}\right)^2}} \times -x^{-\frac{3}{2}}$	M1	Correct unsimplified
Correct and in required form	A1

Question (c):

Answer	Marks	Guidance
Correct value for $x$	B1
Attempts to find value of $\frac{dx}{dy}$ using part (a) with $y = \frac{\pi}{12}$	M1	May be called $m$ or $f'$, not identified as $\frac{dx}{dy}$ or $\frac{dy}{dx}$
the value of $\frac{dy}{dx}$ from an inverted $\frac{dx}{dy}$ using part (a) with $y = \frac{\pi}{12}$, or the value of $\frac{dy}{dx}$ using part (b) with value of $x$ found using $y = \frac{\pi}{12}$	M1
Correct normal gradient	A1
Attempt at equation of normal at $y = \frac{\pi}{12}$	dM1	Gradient should be either $-\frac{dx}{dy}$ at $y=\frac{\pi}{12}$ for their $\frac{dx}{dy}$, or negative reciprocal of their $\frac{dy}{dx}$ at their "4" found from $y = \frac{\pi}{12}$
Fully correct equation in required form	A1	ISW after correct answer. Allow equivalents e.g. $y = -\frac{16}{\sqrt{3}}x + \frac{64}{\sqrt{3}} + \frac{\pi}{12}$, $y = -\frac{16}{\sqrt{3}}x + \frac{256\sqrt{3}+\pi}{12}$

# Question 10:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dx}{dy} = 12\sec^2 2y\tan 2y$ | M1, A1 | M1: differentiates to form $\alpha\sec^2 2y\tan 2y$. Note scheme also applies if student adapts $x=3\sec^2 2y$ to $x=\pm3\tan^2 2y\pm3$. A1: $\frac{dx}{dy}=12\sec^2 2y\tan 2y$; if LHS included it must be correct. Condone unsimplified $6\sec 2y\times 2\sec 2y\tan 2y$. ISW after correct answer |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dx}{dy}=12\left(\frac{x}{3}\right)\sqrt{\sec^2 2y-1} \Rightarrow \frac{dx}{dy}=12\left(\frac{x}{3}\right)\sqrt{\frac{x}{3}-1}$ | M1, A1ft | M1: replace $\sec^2 2y$ with $\alpha x$; use identity $\pm1\pm\tan^2 2y=\pm\sec^2 2y$ and replace $\tan 2y = b\sqrt{\pm1\pm dx}$. A1ft: requires substitution of both $\sec^2 2y$ with $\frac{x}{3}$ and $\tan 2y=\sqrt{\frac{x}{3}-1}$ following through on their $\frac{dx}{dy}=\alpha\sec^2 2y\tan 2y$ |
| $\frac{dy}{dx}=\frac{\sqrt{3}}{4x\sqrt{x-3}}$ | A1 | |

## Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $y=\frac{\pi}{12}\Rightarrow x=4$ | B1 | |
| $\frac{dx}{dy}=12\times\frac{4}{3}\times\frac{1}{\sqrt{3}}$ or $\frac{dy}{dx}=\frac{1}{12\left(\frac{4}{3}\right)\sqrt{\frac{4}{3}-1}}$ | M1 | |
| Correct $m_N = -\frac{16}{\sqrt{3}}$ o.e. | A1 | |
| $y-\frac{\pi}{12}=-\frac{16}{\sqrt{3}}(x-4)$ | dM1 | |
| $y=-\frac{16\sqrt{3}}{3}x+\frac{64\sqrt{3}}{3}+\frac{\pi}{12}$ | A1 | |

## Question (a) - Alt Method:

$x = 3(\cos 2y)^{-2} \Rightarrow \frac{dx}{dy} = 12(\cos 2y)^{-3} \sin 2y$ | M1 A1 |

---

## Question (b) - Alt Method:

If $x = 3(\cos 2y)^{-2} \Rightarrow \cos 2y = \frac{\sqrt{3}}{\sqrt{x}}$ | stated or implied |

$\frac{dx}{dy} = \frac{12\sin 2y}{(\cos 2y)^3} = \frac{12\sqrt{\frac{x-3}{x}}}{\left(\sqrt{\frac{3}{x}}\right)^3}$ | M1 A1 | Score in similar way to main scheme |

---

## Alt (b) via arccos:

$y = \frac{1}{2}\arccos\sqrt{\frac{3}{x}} \Rightarrow \frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{\sqrt{1-\left(\sqrt{\frac{3}{x}}\right)^2}} \times \frac{\sqrt{3}}{2} x^{-\frac{3}{2}}$ | |

For $\frac{dy}{dx} = \lambda \frac{1}{\sqrt{1-\left(\sqrt{\frac{\alpha}{x}}\right)^2}} \times -x^{-\frac{3}{2}}$ | M1 | Correct unsimplified |

Correct and in required form | A1 | |

---

## Question (c):

Correct value for $x$ | B1 | |

Attempts to find value of $\frac{dx}{dy}$ using part (a) with $y = \frac{\pi}{12}$ | M1 | May be called $m$ or $f'$, not identified as $\frac{dx}{dy}$ or $\frac{dy}{dx}$ |

the value of $\frac{dy}{dx}$ from an inverted $\frac{dx}{dy}$ using part (a) with $y = \frac{\pi}{12}$, or the value of $\frac{dy}{dx}$ using part (b) with value of $x$ found using $y = \frac{\pi}{12}$ | M1 | |

Correct normal gradient | A1 | |

Attempt at equation of normal at $y = \frac{\pi}{12}$ | dM1 | Gradient should be either $-\frac{dx}{dy}$ at $y=\frac{\pi}{12}$ for their $\frac{dx}{dy}$, or negative reciprocal of their $\frac{dy}{dx}$ at their "4" found from $y = \frac{\pi}{12}$ |

Fully correct equation in required form | A1 | ISW after correct answer. Allow equivalents e.g. $y = -\frac{16}{\sqrt{3}}x + \frac{64}{\sqrt{3}} + \frac{\pi}{12}$, $y = -\frac{16}{\sqrt{3}}x + \frac{256\sqrt{3}+\pi}{12}$ |

Show LaTeX source

10. The curve $C$ has equation

$$x = 3 \sec ^ { 2 } 2 y \quad x > 3 \quad 0 < y < \frac { \pi } { 4 }$$
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } x } { \mathrm {~d} y }$ in terms of $y$.
\item Hence show that

$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { p } { q x \sqrt { x - 3 } }$$

where $p$ is irrational and $q$ is an integer, stating the values of $p$ and $q$.
\item Find the equation of the normal to $C$ at the point where $y = \frac { \pi } { 12 }$, giving your answer in the form $y = m x + c$, giving $m$ and $c$ as exact irrational numbers.\\

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END &  \\
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VI4V SIHI NI JIIIM IONOO & VIAV SIHI NI JIIIM ION OO & VI4V SIHI NI IIIYM ION OC \\
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\hfill \mbox{\textit{Edexcel P3 2021 Q10 [10]}}

This paper (10 questions)

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Q1 3 Q2 6 Q3 8 Q4 9 Q5 11 Q6 8 Q7 9 Q8 7 Q9 4 Q10 10

Answer	Marks	Guidance
If \(x = 3(\cos 2y)^{-2} \Rightarrow \cos 2y = \frac{\sqrt{3}}{\sqrt{x}}\)	stated or implied
\(\frac{dx}{dy} = \frac{12\sin 2y}{(\cos 2y)^3} = \frac{12\sqrt{\frac{x-3}{x}}}{\left(\sqrt{\frac{3}{x}}\right)^3}\)	M1 A1	Score in similar way to main scheme

Answer	Marks	Guidance
\(y = \frac{1}{2}\arccos\sqrt{\frac{3}{x}} \Rightarrow \frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{\sqrt{1-\left(\sqrt{\frac{3}{x}}\right)^2}} \times \frac{\sqrt{3}}{2} x^{-\frac{3}{2}}\)
For \(\frac{dy}{dx} = \lambda \frac{1}{\sqrt{1-\left(\sqrt{\frac{\alpha}{x}}\right)^2}} \times -x^{-\frac{3}{2}}\)	M1	Correct unsimplified
Correct and in required form	A1

Answer	Marks	Guidance
Correct value for \(x\)	B1
Attempts to find value of \(\frac{dx}{dy}\) using part (a) with \(y = \frac{\pi}{12}\)	M1	May be called \(m\) or \(f'\), not identified as \(\frac{dx}{dy}\) or \(\frac{dy}{dx}\)
the value of \(\frac{dy}{dx}\) from an inverted \(\frac{dx}{dy}\) using part (a) with \(y = \frac{\pi}{12}\), or the value of \(\frac{dy}{dx}\) using part (b) with value of \(x\) found using \(y = \frac{\pi}{12}\)	M1
Correct normal gradient	A1
Attempt at equation of normal at \(y = \frac{\pi}{12}\)	dM1	Gradient should be either \(-\frac{dx}{dy}\) at \(y=\frac{\pi}{12}\) for their \(\frac{dx}{dy}\), or negative reciprocal of their \(\frac{dy}{dx}\) at their "4" found from \(y = \frac{\pi}{12}\)
Fully correct equation in required form	A1	ISW after correct answer. Allow equivalents e.g. \(y = -\frac{16}{\sqrt{3}}x + \frac{64}{\sqrt{3}} + \frac{\pi}{12}\), \(y = -\frac{16}{\sqrt{3}}x + \frac{256\sqrt{3}+\pi}{12}\)