| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2011 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Show gradient expression then find coordinates |
| Difficulty | Standard +0.3 This is a standard parametric differentiation question requiring the chain rule (dy/dx = dy/dt รท dx/dt), followed by straightforward applications. Part (i) involves routine differentiation and algebraic simplification using trigonometric identities. Parts (ii) and (iii) are direct substitutions and equation solving. Slightly above average due to the algebraic manipulation needed, but follows a well-practiced technique with no novel insight required. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Differentiate \(y\) to obtain \(3\sin^2 t \cos t - 3\cos^2 t \sin t\) | B1 | |
| Use \(\frac{dy}{dx} = \frac{dy}{dt} / \frac{dx}{dt}\) | M1 | |
| Obtain \(-3\sin t \cos t\) | A1cwo | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Identify parameter at origin as \(t = \frac{3}{4}\pi\) | B1 | |
| Use \(t = \frac{3}{4}\pi\) to obtain \(\frac{3}{2}\) | B1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Rewrite equation as equation in one trig variable e.g. \(\sin 2t = -\frac{2}{3}\), \(9\sin^4 x - 9\sin^2 x + 1 = 0\), \(\tan^2 x + 3\tan x + 1 = 0\) | B1 | |
| Find at least one value of \(t\) from equation of form \(\sin 2t = k\) | M1 | |
| Obtain \(1.9\) | A1 | |
| Obtain \(2.8\) and no others | A1 | [4] |
## Question 8:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Differentiate $y$ to obtain $3\sin^2 t \cos t - 3\cos^2 t \sin t$ | B1 | |
| Use $\frac{dy}{dx} = \frac{dy}{dt} / \frac{dx}{dt}$ | M1 | |
| Obtain $-3\sin t \cos t$ | A1cwo | [3] |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Identify parameter at origin as $t = \frac{3}{4}\pi$ | B1 | |
| Use $t = \frac{3}{4}\pi$ to obtain $\frac{3}{2}$ | B1 | [2] |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Rewrite equation as equation in one trig variable e.g. $\sin 2t = -\frac{2}{3}$, $9\sin^4 x - 9\sin^2 x + 1 = 0$, $\tan^2 x + 3\tan x + 1 = 0$ | B1 | |
| Find at least one value of $t$ from equation of form $\sin 2t = k$ | M1 | |
| Obtain $1.9$ | A1 | |
| Obtain $2.8$ and no others | A1 | [4] |
---8\\
\includegraphics[max width=\textwidth, alt={}, center]{6025cf1d-525e-4f12-9517-f20ef5fff2fa-3_698_1006_758_571}
The diagram shows the curve with parametric equations
$$x = \sin t + \cos t , \quad y = \sin ^ { 3 } t + \cos ^ { 3 } t$$
for $\frac { 1 } { 4 } \pi < t < \frac { 5 } { 4 } \pi$.\\
(i) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = - 3 \sin t \cos t$.\\
(ii) Find the gradient of the curve at the origin.\\
(iii) Find the values of $t$ for which the gradient of the curve is 1 , giving your answers correct to 2 significant figures.
\hfill \mbox{\textit{CAIE P3 2011 Q8 [9]}}