| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2015 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find tangent equation at parameter |
| Difficulty | Standard +0.3 This is a straightforward parametric differentiation question requiring the chain rule (dy/dx = (dy/dt)/(dx/dt)), trigonometric differentiation, and application of a compound angle formula. Part (i) is routine algebraic manipulation to reach the given form, and part (ii) requires finding the parameter value where x=0 and substituting. Slightly above average due to the compound angle work, but still a standard textbook exercise with no novel problem-solving required. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.05l Double angle formulae: and compound angle formulae1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Either: Obtain \(\frac{dy}{dt} = -3\sin t\) | B1 | |
| Obtain \(\frac{dy}{dt} = -2\sin(t - \frac{1}{6}\pi)\) | B1 | |
| Use \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\) | M1 | |
| Expand \(-2\sin(t - \frac{1}{6}\pi)\) to obtain \(k_1 \sin t + k_2 \cos t\) | M1 | |
| Confirm given result \(\frac{1}{3}(\sqrt{3} - \cot t)\) correctly | A1 | |
| Or: Obtain \(\frac{dy}{dt} = -3\sin t\) | B1 | |
| Expand \(y\) to obtain \(k_3 \cos t + k_4 \sin t\) | M1 | |
| Obtain \(\frac{dy}{dt} = -\sqrt{3} \sin t + \cos t\) or equivalent | A1 | |
| Use \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\) | M1 | |
| Confirm given result \(\frac{1}{3}(\sqrt{3} - \cot t)\) correctly | A1 | [5] |
| (ii) Identify value of \(t\) as \(\frac{1}{3}\pi\) only | B1 | |
| Obtain gradient at relevant point as \(\frac{1}{3}\sqrt{3}\) or \(0.577\) or better | B1 | |
| Form equation of tangent through \((0, 1)\), using their gradient | M1 | |
| Obtain \(y = \frac{1}{3}\sqrt{3}x + 1\) or equivalent | A1 | [4] |
(i) Either: Obtain $\frac{dy}{dt} = -3\sin t$ | B1 |
Obtain $\frac{dy}{dt} = -2\sin(t - \frac{1}{6}\pi)$ | B1 |
Use $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ | M1 |
Expand $-2\sin(t - \frac{1}{6}\pi)$ to obtain $k_1 \sin t + k_2 \cos t$ | M1 |
Confirm given result $\frac{1}{3}(\sqrt{3} - \cot t)$ correctly | A1 |
Or: Obtain $\frac{dy}{dt} = -3\sin t$ | B1 |
Expand $y$ to obtain $k_3 \cos t + k_4 \sin t$ | M1 |
Obtain $\frac{dy}{dt} = -\sqrt{3} \sin t + \cos t$ or equivalent | A1 |
Use $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ | M1 |
Confirm given result $\frac{1}{3}(\sqrt{3} - \cot t)$ correctly | A1 | [5]
(ii) Identify value of $t$ as $\frac{1}{3}\pi$ only | B1 |
Obtain gradient at relevant point as $\frac{1}{3}\sqrt{3}$ or $0.577$ or better | B1 |
Form equation of tangent through $(0, 1)$, using their gradient | M1 |
Obtain $y = \frac{1}{3}\sqrt{3}x + 1$ or equivalent | A1 | [4]
---6\\
\includegraphics[max width=\textwidth, alt={}, center]{7e100be2-9768-4fcd-b516-c714e53b0665-3_453_650_258_744}
The diagram shows the curve with parametric equations
$$x = 3 \cos t , \quad y = 2 \cos \left( t - \frac { 1 } { 6 } \pi \right)$$
for $0 \leqslant t < 2 \pi$.\\
(i) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 3 } ( \sqrt { } 3 - \cot t )$.\\
(ii) Find the equation of the tangent to the curve at the point where the curve crosses the positive $y$-axis. Give the answer in the form $y = m x + c$.
\hfill \mbox{\textit{CAIE P2 2015 Q6 [9]}}