CAIE P3 2024 November — Question 8 8 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2024
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicParametric differentiation
TypeFind normal equation at parameter
DifficultyStandard +0.3 This is a straightforward parametric differentiation question requiring standard application of dy/dx = (dy/dt)/(dx/dt), followed by routine normal line calculation. The algebra involves basic trigonometric identities and substitution at a specific parameter value. While it requires multiple steps (8 marks total), each step follows a well-practiced procedure with no novel insight needed, making it slightly easier than average.
Spec1.03g Parametric equations: of curves and conversion to cartesian1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation
The parametric equations of a curve are $$x = \tan^2 2t, \quad y = \cos 2t,$$ for \(0 < t < \frac{1}{4}\pi\).
  1. Show that \(\frac{dy}{dx} = -\frac{1}{2}\cos^3 2t\). [4]
  2. Hence find the equation of the normal to the curve at the point where \(t = \frac{1}{8}\pi\). Give your answer in the form \(y = mx + c\). [4]
Question 8:
AnswerMarks Guidance
8(a)Use correct product rule or chain rule to find derivative of x with respect to t M1
dx
Obtain =4tan2tsec22t oe
AnswerMarks
dtA1
dy
=−2sin2t
AnswerMarks
dtB1
dy dy dt dy 1
Use =  to obtain given answer =− cos32t
AnswerMarks Guidance
dx dt dx dx 2B1 dy
Condone if missing.
dx
4
AnswerMarks
8(b)2
Obtain x = 1 and y =
AnswerMarks Guidance
2B1 Accept y = 0.707…
− 2
State or imply gradient of tangent is or gradient of normal is 4 2
AnswerMarks Guidance
8B1 5
Any equivalent form, e.g. 22.
Accept – 0.177 or 5.66.
AnswerMarks Guidance
Use correct method to find equation of normal using their valuesM1 Need a fully substituted equation for the normal
(in any form) or to get at least as far as finding
value for m and expression for c.
7 2
Obtain equation of normal is y=4 2x− or equivalent 3 term equation
AnswerMarks Guidance
2A1 E.g., y = 5.66x – 4.95.
Must be y = …
4
AnswerMarks Guidance
QuestionAnswer Marks 
Question 8:
--- 8(a) ---
8(a) | Use correct product rule or chain rule to find derivative of x with respect to t | M1 | Obtain ktan2tsec22t.
dx
Obtain =4tan2tsec22t oe
dt | A1
dy
=−2sin2t
dt | B1
dy dy dt dy 1
Use =  to obtain given answer =− cos32t
dx dt dx dx 2 | B1 | dy
Condone if missing.
dx
4
--- 8(b) ---
8(b) | 2
Obtain x = 1 and y =
2 | B1 | Accept y = 0.707…
− 2
State or imply gradient of tangent is or gradient of normal is 4 2
8 | B1 | 5
Any equivalent form, e.g. 22.
Accept – 0.177 or 5.66.
Use correct method to find equation of normal using their values | M1 | Need a fully substituted equation for the normal
(in any form) or to get at least as far as finding
value for m and expression for c.
7 2
Obtain equation of normal is y=4 2x− or equivalent 3 term equation
2 | A1 | E.g., y = 5.66x – 4.95.
Must be y = …
4
Question | Answer | Marks | Guidance
The parametric equations of a curve are
$$x = \tan^2 2t, \quad y = \cos 2t,$$
for $0 < t < \frac{1}{4}\pi$.

\begin{enumerate}[label=(\alph*)]
\item Show that $\frac{dy}{dx} = -\frac{1}{2}\cos^3 2t$. [4]

\item Hence find the equation of the normal to the curve at the point where $t = \frac{1}{8}\pi$. Give your answer in the form $y = mx + c$. [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE P3 2024 Q8 [8]}}
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