CAIE P3 2016 June — Question 4 8 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2016
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicParametric differentiation
TypeFind parameter value given gradient condition
DifficultyModerate -0.3 Part (i) is a straightforward application of the chain rule for parametric differentiation (dy/dx = (dy/dt)/(dx/dt)) with standard derivatives of ln, sin, and cos. Part (ii) requires solving sec t = 3, which is routine trigonometry. This is slightly easier than average as it involves direct application of well-practiced techniques with no conceptual challenges or multi-step problem-solving.
Spec1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation

4 The parametric equations of a curve are $$x = t + \cos t , \quad y = \ln ( 1 + \sin t )$$ where \(- \frac { 1 } { 2 } \pi < t < \frac { 1 } { 2 } \pi\).
  1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \sec t\).
  2. Hence find the \(x\)-coordinates of the points on the curve at which the gradient is equal to 3 . Give your answers correct to 3 significant figures.

AnswerMarks Guidance
(i) State \(\frac{dx}{dt} = 1 - \sin t\)B1
Use chain rule to find the derivative of \(y\)M1
Obtain \(\frac{dy}{dt} = \frac{-\cos t}{1 + \sin t}\), or equivalentA1
Use \(\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dx}{dt}\)M1
Obtain the given answer correctlyA1 [5]
(ii) State or imply \(t = \cos^{-1}\left(\frac{1}{x}\right)\)B1
Obtain answers \(x = 1.56\) and \(x = -0.898\)B1 + B1 [3]
**(i)** State $\frac{dx}{dt} = 1 - \sin t$ | B1 | 

Use chain rule to find the derivative of $y$ | M1 | 

Obtain $\frac{dy}{dt} = \frac{-\cos t}{1 + \sin t}$, or equivalent | A1 | 

Use $\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dx}{dt}$ | M1 | 

Obtain the given answer correctly | A1 | [5]

**(ii)** State or imply $t = \cos^{-1}\left(\frac{1}{x}\right)$ | B1 | 

Obtain answers $x = 1.56$ and $x = -0.898$ | B1 + B1 | [3]

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4 The parametric equations of a curve are

$$x = t + \cos t , \quad y = \ln ( 1 + \sin t )$$

where $- \frac { 1 } { 2 } \pi < t < \frac { 1 } { 2 } \pi$.\\
(i) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \sec t$.\\
(ii) Hence find the $x$-coordinates of the points on the curve at which the gradient is equal to 3 . Give your answers correct to 3 significant figures.

\hfill \mbox{\textit{CAIE P3 2016 Q4 [8]}}