CAIE P2 2011 June — Question 2 4 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2011
SessionJune
Marks4
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Mark schemeDownload PDF ↗
TopicParametric differentiation
TypeFind gradient at given parameter
DifficultyModerate -0.5 This is a straightforward application of the parametric differentiation formula dy/dx = (dy/dt)/(dx/dt), requiring basic differentiation of trigonometric functions and evaluation at a specific value. The arithmetic is clean with t = π/6 giving exact values, making this slightly easier than average but still requiring correct technique.
Spec1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation
2 A curve has parametric equations $$x = 3 t + \sin 2 t , \quad y = 4 + 2 \cos 2 t$$ Find the exact gradient of the curve at the point for which \(t = \frac { 1 } { 6 } \pi\).
AnswerMarks Guidance
State \(\frac{dx}{dt} = 3 + 2\cos 2t\) or \(\frac{dy}{dt} = -4\sin 2t\) (or both)B1
Use \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\)M1
Obtain or imply \(\frac{-4\sin 2t}{3 + 2\cos 2t}\)A1
Substitute \(\frac{1}{6}\pi\) to obtain \(-\frac{1}{2}\sqrt{3}\) or exact equivalentA1 [4] 
State $\frac{dx}{dt} = 3 + 2\cos 2t$ or $\frac{dy}{dt} = -4\sin 2t$ (or both) | B1 |
Use $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ | M1 |
Obtain or imply $\frac{-4\sin 2t}{3 + 2\cos 2t}$ | A1 |
Substitute $\frac{1}{6}\pi$ to obtain $-\frac{1}{2}\sqrt{3}$ or exact equivalent | A1 | [4]
2 A curve has parametric equations

$$x = 3 t + \sin 2 t , \quad y = 4 + 2 \cos 2 t$$

Find the exact gradient of the curve at the point for which $t = \frac { 1 } { 6 } \pi$.

\hfill \mbox{\textit{CAIE P2 2011 Q2 [4]}}
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