Question 7 - A-Level Maths

CAIE FP1 2014 June — Question 7 10 marks

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2014
Session	June
Marks	10
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Parametric differentiation
Type	Find second derivative d²y/dx²
Difficulty	Standard +0.3 This is a standard Further Maths parametric differentiation question requiring the chain rule formula for d²y/dx² and finding stationary points. While it involves multiple steps (finding dy/dt, dx/dt, then d²y/dx², setting dy/dx=0, and using the second derivative test), these are routine techniques for FP1 students with no novel insight required. The trigonometric identities involved (sin 2t = 2sin t cos t) are straightforward, making this slightly easier than average overall.
Spec	1.03g Parametric equations: of curves and conversion to cartesian 1.07n Stationary points: find maxima, minima using derivatives 1.07s Parametric and implicit differentiation

7 The curve $C$ has parametric equations $$x = \sin t , \quad y = \sin 2 t , \quad \text { for } 0 \leqslant t \leqslant \pi .$$ Find $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$ in terms of $t$. Hence, or otherwise, find the coordinates of the stationary points on $C$ and determine their nature.

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Question 7:

Answer	Marks	Guidance
Working/Answer	Mark	Guidance
$\dot{x} = \cos t$, $\dot{y} = 2\cos 2t \Rightarrow y' = \frac{2\cos 2t}{\cos t}$	M1A1
$y'' = \frac{-4\cos t\sin 2t + 2\cos 2t\sin t}{\cos^2 t} \times \frac{1}{\cos t} = -\frac{4\sin 2t}{\cos^2 t} + \frac{2\cos 2t\sin t}{\cos^3 t}$	M1A1	OE
e.g. $y'' = \frac{4\sin^3 t - 6\sin t}{\cos^3 t}$	A1
$y' = 0$ or $\dot{y} = 0 \Rightarrow \cos 2t = 0$	M1
$2t = \frac{\pi}{2}, \frac{3\pi}{2} \Rightarrow t = \frac{\pi}{4}, \frac{3\pi}{4}$	A1
Stationary points are $\left(\frac{1}{\sqrt{2}}, 1\right)$, $\left(\frac{1}{\sqrt{2}}, -1\right)$	A1	S.C. if only one value of $t$ given with correct coordinates — A1
$y''\!\left(\frac{\pi}{4}\right) = -8 \Rightarrow$ max	B1	CWO
$y''\!\left(\frac{3\pi}{4}\right) = +8 \Rightarrow$ min	B1	CWO
Total	[5]

# Question 7:

| Working/Answer | Mark | Guidance |
|---|---|---|
| $\dot{x} = \cos t$, $\dot{y} = 2\cos 2t \Rightarrow y' = \frac{2\cos 2t}{\cos t}$ | M1A1 | |
| $y'' = \frac{-4\cos t\sin 2t + 2\cos 2t\sin t}{\cos^2 t} \times \frac{1}{\cos t} = -\frac{4\sin 2t}{\cos^2 t} + \frac{2\cos 2t\sin t}{\cos^3 t}$ | M1A1 | OE |
| e.g. $y'' = \frac{4\sin^3 t - 6\sin t}{\cos^3 t}$ | A1 | |
| $y' = 0$ or $\dot{y} = 0 \Rightarrow \cos 2t = 0$ | M1 | |
| $2t = \frac{\pi}{2}, \frac{3\pi}{2} \Rightarrow t = \frac{\pi}{4}, \frac{3\pi}{4}$ | A1 | |
| Stationary points are $\left(\frac{1}{\sqrt{2}}, 1\right)$, $\left(\frac{1}{\sqrt{2}}, -1\right)$ | A1 | S.C. if only one value of $t$ given with correct coordinates — A1 |
| $y''\!\left(\frac{\pi}{4}\right) = -8 \Rightarrow$ max | B1 | CWO |
| $y''\!\left(\frac{3\pi}{4}\right) = +8 \Rightarrow$ min | B1 | CWO |
| Total | [5] | |

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7 The curve $C$ has parametric equations

$$x = \sin t , \quad y = \sin 2 t , \quad \text { for } 0 \leqslant t \leqslant \pi .$$

Find $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$ in terms of $t$.

Hence, or otherwise, find the coordinates of the stationary points on $C$ and determine their nature.

\hfill \mbox{\textit{CAIE FP1 2014 Q7 [10]}}

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Q1 5 Q2 6 Q3 7 Q4 7 Q5 8 Q6 10 Q7 10 Q8 11 Q9 10 Q10 12 Q11 EITHER Q11 OR