| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2020 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find stationary points of parametric curve |
| Difficulty | Standard +0.3 This is a standard parametric differentiation question requiring dy/dx = (dy/dθ)/(dx/dθ), followed by finding the maximum of the gradient function using calculus. Part (a) is routine manipulation, and part (b) requires differentiating the gradient and solving a trigonometric equation, but all steps follow standard procedures with no novel insight required. Slightly easier than average due to straightforward calculus techniques. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| State \(\dfrac{dx}{d\theta} = \sec^2\theta\) or \(\dfrac{dy}{d\theta} = -2\sin\theta\cos\theta\) | B1 | CWO, AEF. |
| Use \(\dfrac{dy}{dx} = \dfrac{dy}{d\theta} \div \dfrac{dx}{d\theta}\) | M1 | |
| Obtain \(\dfrac{dy}{dx} = -2\sin\theta\cos^3\theta\) from correct working | A1 | AG |
| Alternative method: Convert to Cartesian form and differentiate | M1 | \(y = \dfrac{1}{1+x^2}\) |
| \(\dfrac{dy}{dx} = \dfrac{-2x}{(1+x^2)^2}\) | A1 | OE |
| Obtain \(\dfrac{dy}{dx} = -2\sin\theta\cos^3\theta\) from correct working | A1 | AG |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use correct product rule to obtain \(\dfrac{d}{d\theta}(\pm 2\cos^3\theta\sin\theta)\) | M1 | Condone incorrect naming of the derivative. For work done in correct context |
| Obtain correct derivative in any form | A1 | e.g. \(\pm(-2\cos^4\theta + 6\sin^2\theta\cos^2\theta)\) |
| Equate derivative to zero and obtain an equation in one trig ratio | A1 | e.g. \(3\tan^2\theta = 1\), or \(4\sin^2\theta = 1\) or \(4\cos^2\theta = 3\) |
| Obtain answer \(x = -\dfrac{1}{\sqrt{3}}\) | A1 | Or \(-\dfrac{\sqrt{3}}{3}\) |
| Alternative method: Use correct quotient rule to obtain \(\dfrac{d^2y}{dx^2}\) | M1 | |
| Obtain correct derivative in any form | A1 | \(\dfrac{-2(1+x^2)^2 + 2\times 2x\times 2x(1+x^2)}{(1+x^2)^4}\) |
| Equate derivative to zero and obtain an equation in \(x^2\) | A1 | e.g. \(6x^2 = 2\) |
| Obtain answer \(x = -\dfrac{1}{\sqrt{3}}\) | A1 | |
| 4 |
## Question 5:
### Part 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| State $\dfrac{dx}{d\theta} = \sec^2\theta$ or $\dfrac{dy}{d\theta} = -2\sin\theta\cos\theta$ | B1 | CWO, AEF. |
| Use $\dfrac{dy}{dx} = \dfrac{dy}{d\theta} \div \dfrac{dx}{d\theta}$ | M1 | |
| Obtain $\dfrac{dy}{dx} = -2\sin\theta\cos^3\theta$ from correct working | A1 | AG |
| **Alternative method:** Convert to Cartesian form and differentiate | M1 | $y = \dfrac{1}{1+x^2}$ |
| $\dfrac{dy}{dx} = \dfrac{-2x}{(1+x^2)^2}$ | A1 | OE |
| Obtain $\dfrac{dy}{dx} = -2\sin\theta\cos^3\theta$ from correct working | A1 | AG |
| | **3** | |
### Part 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use correct product rule to obtain $\dfrac{d}{d\theta}(\pm 2\cos^3\theta\sin\theta)$ | M1 | Condone incorrect naming of the derivative. For work done in correct context |
| Obtain correct derivative in any form | A1 | e.g. $\pm(-2\cos^4\theta + 6\sin^2\theta\cos^2\theta)$ |
| Equate derivative to zero and obtain an equation in one trig ratio | A1 | e.g. $3\tan^2\theta = 1$, or $4\sin^2\theta = 1$ or $4\cos^2\theta = 3$ |
| Obtain answer $x = -\dfrac{1}{\sqrt{3}}$ | A1 | Or $-\dfrac{\sqrt{3}}{3}$ |
| **Alternative method:** Use correct quotient rule to obtain $\dfrac{d^2y}{dx^2}$ | M1 | |
| Obtain correct derivative in any form | A1 | $\dfrac{-2(1+x^2)^2 + 2\times 2x\times 2x(1+x^2)}{(1+x^2)^4}$ |
| Equate derivative to zero and obtain an equation in $x^2$ | A1 | e.g. $6x^2 = 2$ |
| Obtain answer $x = -\dfrac{1}{\sqrt{3}}$ | A1 | |
| | **4** | |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{77a45360-8e1d-4f4f-9830-075d832a14cf-08_334_895_258_625}
The diagram shows the curve with parametric equations
$$x = \tan \theta , \quad y = \cos ^ { 2 } \theta$$
for $- \frac { 1 } { 2 } \pi < \theta < \frac { 1 } { 2 } \pi$.
\begin{enumerate}[label=(\alph*)]
\item Show that the gradient of the curve at the point with parameter $\theta$ is $- 2 \sin \theta \cos ^ { 3 } \theta$.\\
The gradient of the curve has its maximum value at the point $P$.
\item Find the exact value of the $x$-coordinate of $P$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2020 Q5 [7]}}