| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2015 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Properties of specific curves |
| Difficulty | Standard +0.8 This is a multi-part parametric question requiring chain rule differentiation, tangent equation derivation, and a geometric proof involving intercepts. While the differentiation is standard, part (iii) requires algebraic manipulation and insight to show the constant sum property, which goes beyond routine exercises. The question demands careful algebraic work across multiple steps but uses familiar techniques. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (i) State \(\frac{dx}{dt} = -4a\cos^3 t \sin t\), or \(\frac{dy}{dt} = 4a\sin^3 t \cos t\) | B1 | |
| Use \(\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}\) | M1 | |
| Obtain correct expression for \(\frac{dy}{dx}\) in a simplified form | A1 | 3 |
| (ii) Form the equation of the tangent | M1 | |
| Obtain a correct equation in any form | A1 | |
| Obtain the given answer | A1 | 3 |
| (iii) State the x-coordinate of \(P\) or the y-coordinate of \(Q\) in any form | B1 | |
| Obtain the given result correctly | B1 | 2 |
**(i)** State $\frac{dx}{dt} = -4a\cos^3 t \sin t$, or $\frac{dy}{dt} = 4a\sin^3 t \cos t$ | B1 |
Use $\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}$ | M1 |
Obtain correct expression for $\frac{dy}{dx}$ in a simplified form | A1 | 3 |
**(ii)** Form the equation of the tangent | M1 |
Obtain a correct equation in any form | A1 |
Obtain the given answer | A1 | 3 |
**(iii)** State the x-coordinate of $P$ or the y-coordinate of $Q$ in any form | B1 |
Obtain the given result correctly | B1 | 2 |5 The parametric equations of a curve are
$$x = a \cos ^ { 4 } t , \quad y = a \sin ^ { 4 } t$$
where $a$ is a positive constant.\\
(i) Express $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $t$.\\
(ii) Show that the equation of the tangent to the curve at the point with parameter $t$ is
$$x \sin ^ { 2 } t + y \cos ^ { 2 } t = a \sin ^ { 2 } t \cos ^ { 2 } t$$
(iii) Hence show that if the tangent meets the $x$-axis at $P$ and the $y$-axis at $Q$, then
$$O P + O Q = a$$
where $O$ is the origin.
\hfill \mbox{\textit{CAIE P3 2015 Q5 [8]}}