Question 4 - A-Level Maths

CAIE FP1 2009 November — Question 4 8 marks

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2009
Session	November
Marks	8
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Parametric differentiation
Type	Find second derivative d²y/dx²
Difficulty	Standard +0.8 This is a Further Maths parametric differentiation question requiring the chain rule for dy/dx, then the quotient rule for d²y/dx², followed by showing a derivative is positive. While the techniques are standard for FP1, the algebraic manipulation to reach the given form of d²y/dx² is non-trivial, and proving monotonicity requires careful sign analysis of trigonometric expressions over the given interval. This is moderately challenging even for Further Maths students.
Spec	1.07o Increasing/decreasing: functions using sign of dy/dx 1.07s Parametric and implicit differentiation

4 It is given that $$x = t + \sin t , \quad y = t ^ { 2 } + 2 \cos t$$ where $- \pi < t < \pi$. Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $t$. Show that $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \frac { 2 t \sin t } { ( 1 + \cos t ) ^ { 3 } }$$ Show that $\frac { \mathrm { d } y } { \mathrm {~d} x }$ increases with $x$ over the given interval of $t$.

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Answer	Marks
$dx/dt = 1 + \cos t, dy/dt = 2t - 2\sin t$	M1A1
$dy/dx = (2t - 2\sin t)/(1 + \cos t)$	M1A1
Any correct result for $d(dy/dx)/dt$ in terms of $t$	B1
$d^2y/dx^2 = [(2 - 2\cos t)(1 + \cos t) + \sin(2t - 2\sin t)]/[(1 + \cos t)^3]$ (AEF)	M1A1
$d^2y/dx^2 = 2\sin t/(1 + \cos t)^3$ (AG)	A1
Consideration of sign of $d^2y/dx^2$	M1
$-\pi < t < 0$: $(-)(-)/ +> 0$; $0 < t < \pi$: $(+)(+)/+> 0 \Rightarrow d^2y/dx^2 > 0, \forall$ non-zero $t \in (-\pi,\pi)$	A1

$dx/dt = 1 + \cos t, dy/dt = 2t - 2\sin t$ | M1A1 |
$dy/dx = (2t - 2\sin t)/(1 + \cos t)$ | M1A1 |
Any correct result for $d(dy/dx)/dt$ in terms of $t$ | B1 |
$d^2y/dx^2 = [(2 - 2\cos t)(1 + \cos t) + \sin(2t - 2\sin t)]/[(1 + \cos t)^3]$ (AEF) | M1A1 |
$d^2y/dx^2 = 2\sin t/(1 + \cos t)^3$ (AG) | A1 |
Consideration of sign of $d^2y/dx^2$ | M1 |
$-\pi < t < 0$: $(-)(-)/ +> 0$; $0 < t < \pi$: $(+)(+)/+> 0 \Rightarrow d^2y/dx^2 > 0, \forall$ non-zero $t \in (-\pi,\pi)$ | A1 |

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4 It is given that

$$x = t + \sin t , \quad y = t ^ { 2 } + 2 \cos t$$

where $- \pi < t < \pi$. Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $t$.

Show that

$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \frac { 2 t \sin t } { ( 1 + \cos t ) ^ { 3 } }$$

Show that $\frac { \mathrm { d } y } { \mathrm {~d} x }$ increases with $x$ over the given interval of $t$.

\hfill \mbox{\textit{CAIE FP1 2009 Q4 [8]}}

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Q1 4 Q2 6 Q3 8 Q4 8 Q5 9 Q6 9 Q7 9 Q8 10 Q9 11 Q10 12 Q11 EITHER Q11 OR

Answer	Marks
\(dx/dt = 1 + \cos t, dy/dt = 2t - 2\sin t\)	M1A1
\(dy/dx = (2t - 2\sin t)/(1 + \cos t)\)	M1A1
Any correct result for \(d(dy/dx)/dt\) in terms of \(t\)	B1
\(d^2y/dx^2 = [(2 - 2\cos t)(1 + \cos t) + \sin(2t - 2\sin t)]/[(1 + \cos t)^3]\) (AEF)	M1A1
\(d^2y/dx^2 = 2\sin t/(1 + \cos t)^3\) (AG)	A1
Consideration of sign of \(d^2y/dx^2\)	M1
\(-\pi < t < 0\): \((-)(-)/ +> 0\); \(0 < t < \pi\): \((+)(+)/+> 0 \Rightarrow d^2y/dx^2 > 0, \forall\) non-zero \(t \in (-\pi,\pi)\)	A1