| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2017 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find parameter value given gradient condition |
| Difficulty | Standard +0.3 Part (i) is a standard parametric differentiation exercise requiring chain rule and basic trigonometric derivatives. Part (ii) requires finding where dy/dx = -1, then solving a trigonometric equation and substituting back—routine multi-step work but no novel insight needed. Slightly easier than average A-level due to straightforward calculus application. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use chain rule to differentiate \(x\) \(\left(\dfrac{dx}{d\theta} = -\dfrac{\sin\theta}{\cos\theta}\right)\) | M1 | |
| State \(\dfrac{dy}{d\theta} = 3 - \sec^2\theta\) | B1 | |
| Use \(\dfrac{dy}{dx} = \dfrac{dy}{d\theta} \div \dfrac{dx}{d\theta}\) | M1 | |
| Obtain correct \(\dfrac{dy}{dx}\) in any form e.g. \(\dfrac{3 - \sec^2\theta}{-\tan\theta}\) | A1 | |
| Obtain \(\dfrac{dy}{dx} = \dfrac{\tan^2\theta - 2}{\tan\theta}\), or equivalent | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Equate gradient to \(-1\) and obtain an equation in \(\tan\theta\) | M1 | |
| Solve a 3-term quadratic \(\left(\tan^2\theta + \tan\theta - 2 = 0\right)\) in \(\tan\theta\) | M1 | |
| Obtain \(\theta = \dfrac{\pi}{4}\) and \(y = \dfrac{3\pi}{4} - 1\) only | A1 |
## Question 4(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use chain rule to differentiate $x$ $\left(\dfrac{dx}{d\theta} = -\dfrac{\sin\theta}{\cos\theta}\right)$ | M1 | |
| State $\dfrac{dy}{d\theta} = 3 - \sec^2\theta$ | B1 | |
| Use $\dfrac{dy}{dx} = \dfrac{dy}{d\theta} \div \dfrac{dx}{d\theta}$ | M1 | |
| Obtain correct $\dfrac{dy}{dx}$ in any form e.g. $\dfrac{3 - \sec^2\theta}{-\tan\theta}$ | A1 | |
| Obtain $\dfrac{dy}{dx} = \dfrac{\tan^2\theta - 2}{\tan\theta}$, or equivalent | A1 | |
**Total: 5**
---
## Question 4(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Equate gradient to $-1$ and obtain an equation in $\tan\theta$ | M1 | |
| Solve a 3-term quadratic $\left(\tan^2\theta + \tan\theta - 2 = 0\right)$ in $\tan\theta$ | M1 | |
| Obtain $\theta = \dfrac{\pi}{4}$ and $y = \dfrac{3\pi}{4} - 1$ only | A1 | |
**Total: 3**
---4 The parametric equations of a curve are
$$x = \ln \cos \theta , \quad y = 3 \theta - \tan \theta ,$$
where $0 \leqslant \theta < \frac { 1 } { 2 } \pi$.\\
(i) Express $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $\tan \theta$.\\
(ii) Find the exact $y$-coordinate of the point on the curve at which the gradient of the normal is equal to 1 .\\
\includegraphics[max width=\textwidth, alt={}, center]{b00cefad-7c3c-4672-b309-f19aafab8b01-08_378_689_260_726}
The diagram shows a semicircle with centre $O$, radius $r$ and diameter $A B$. The point $P$ on its circumference is such that the area of the minor segment on $A P$ is equal to half the area of the minor segment on $B P$. The angle $A O P$ is $x$ radians.\\
\hfill \mbox{\textit{CAIE P3 2017 Q4 [8]}}