| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2024 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find normal equation at parameter |
| Difficulty | Standard +0.3 This is a straightforward parametric differentiation question requiring standard techniques: differentiate both x and y with respect to t, apply the chain rule dy/dx = (dy/dt)/(dx/dt), and simplify using trigonometric identities. Part (b) adds a routine normal line calculation. While it requires careful algebraic manipulation and knowledge of trig identities (particularly for simplifying tan t + cot t), it follows a completely standard template with no novel problem-solving required. Slightly easier than average due to its predictable structure. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Obtain \(\dfrac{dx}{dt} = 6\cos 2t\) | B1 | Allow \(3.2\cos 2t\). \(dx = 6\cos 2t\) is B0. |
| Obtain \(\dfrac{dy}{dt} = \sec^2 t - \text{cosec}^2 t\) | B1 | Any equivalent form. \(dy = \sec^2 t - \text{cosec}^2 t\) is B0, but \(\dfrac{dy}{dx} = \dfrac{\sec^2 t - \text{cosec}^2 t}{6\cos 2t}\) can go on to gain M1M1A1, so 3/5 possible. |
| Use \(\dfrac{dy}{dx} = \dfrac{dy}{dt} \div \dfrac{dx}{dt}\) | *M1 | \(\dfrac{dy}{dx} = \dfrac{\sec^2 t - \text{cosec}^2 t}{6\cos 2t}\) |
| Express as a single fraction with \(\dfrac{dy}{dt}\) correctly simplified as a single fraction in terms of \(\sin t\) and \(\cos t\). Allow with \(6\cos 2t\) expressed as \(\dfrac{1}{6\cos 2t}\) outside bracket | DM1 | Allow \(\dfrac{\sin^2 t - \cos^2 t}{\cos^2 t \sin^2 t} \times \dfrac{1}{6\cos 2t}\) or \(\dfrac{\sin^2 t - \cos^2 t}{\cos^2 t \sin^2 t} \div 6\cos 2t\) or \(\dfrac{\sin^2 t - \cos^2 t}{6\cos 2t \cdot \cos^2 t \sin^2 t}\) |
| Obtain \(\left(\dfrac{-4\cos 2t}{6\cos 2t \times \sin^2 2t}\right) = \dfrac{-2}{3\sin^2 2t}\) from full and correct working. Numerator and denominator must have identical terms before cancelling. | A1 | AG. Allow slips in \(\theta\) and \(x\) for \(t\) to recover earlier marks, provided corrected before final line. Do not allow serious errors in working for the final mark. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = \frac{6}{x}\) | B2 | Using \(y = \frac{\tan^2 t + 1}{\tan t} = \frac{\sec^2 t}{\frac{\sin t}{\cos t}} = \frac{1}{\sin t \cos t}\) |
| \(\frac{\mathrm{d}y}{\mathrm{d}x} = -6x^{-2}\) | B1 | |
| \(\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{6}{9\sin^2 2t}\) | M1 | |
| Obtain \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{-2}{3\sin^2 2t}\) from full and correct working | A1 | |
| Total | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Gradient of normal \(= \frac{3}{2}\) | B1 | |
| Use correct method to find the equation of the normal | M1 | E.g. \((y-2) = \frac{3}{2}(x-3)\) or find \(c\) in \(y = \frac{3}{2}x + c\). Allow a wrong value for \(x\) or \(y\) but not both, with *their* normal gradient. |
| Obtain \(2y - 3x + 5 = 0\) | A1 | Or \(k(2y - 3x + 5) = 0\), where \(k\) is an integer. |
| Total | 3 |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Obtain $\dfrac{dx}{dt} = 6\cos 2t$ | B1 | Allow $3.2\cos 2t$. $dx = 6\cos 2t$ is B0. |
| Obtain $\dfrac{dy}{dt} = \sec^2 t - \text{cosec}^2 t$ | B1 | Any equivalent form. $dy = \sec^2 t - \text{cosec}^2 t$ is B0, but $\dfrac{dy}{dx} = \dfrac{\sec^2 t - \text{cosec}^2 t}{6\cos 2t}$ can go on to gain M1M1A1, so 3/5 possible. |
| Use $\dfrac{dy}{dx} = \dfrac{dy}{dt} \div \dfrac{dx}{dt}$ | *M1 | $\dfrac{dy}{dx} = \dfrac{\sec^2 t - \text{cosec}^2 t}{6\cos 2t}$ |
| Express as a single fraction with $\dfrac{dy}{dt}$ correctly simplified as a single fraction in terms of $\sin t$ and $\cos t$. Allow with $6\cos 2t$ expressed as $\dfrac{1}{6\cos 2t}$ outside bracket | DM1 | Allow $\dfrac{\sin^2 t - \cos^2 t}{\cos^2 t \sin^2 t} \times \dfrac{1}{6\cos 2t}$ or $\dfrac{\sin^2 t - \cos^2 t}{\cos^2 t \sin^2 t} \div 6\cos 2t$ or $\dfrac{\sin^2 t - \cos^2 t}{6\cos 2t \cdot \cos^2 t \sin^2 t}$ |
| Obtain $\left(\dfrac{-4\cos 2t}{6\cos 2t \times \sin^2 2t}\right) = \dfrac{-2}{3\sin^2 2t}$ from full and correct working. Numerator and denominator must have identical terms before cancelling. | A1 | AG. Allow slips in $\theta$ and $x$ for $t$ to recover earlier marks, provided corrected before final line. Do not allow serious errors in working for the final mark. |
## Question 7(a) [Alternative Method]:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = \frac{6}{x}$ | B2 | Using $y = \frac{\tan^2 t + 1}{\tan t} = \frac{\sec^2 t}{\frac{\sin t}{\cos t}} = \frac{1}{\sin t \cos t}$ |
| $\frac{\mathrm{d}y}{\mathrm{d}x} = -6x^{-2}$ | B1 | |
| $\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{6}{9\sin^2 2t}$ | M1 | |
| Obtain $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{-2}{3\sin^2 2t}$ from full and correct working | A1 | |
| **Total** | **5** | |
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Gradient of normal $= \frac{3}{2}$ | B1 | |
| Use correct method to find the equation of the normal | M1 | E.g. $(y-2) = \frac{3}{2}(x-3)$ or find $c$ in $y = \frac{3}{2}x + c$. Allow a wrong value for $x$ or $y$ but not both, with *their* normal gradient. |
| Obtain $2y - 3x + 5 = 0$ | A1 | Or $k(2y - 3x + 5) = 0$, where $k$ is an integer. |
| **Total** | **3** | |
---7 The parametric equations of a curve are
$$x = 3 \sin 2 t , \quad y = \tan t + \cot t$$
for $0 < t < \frac { 1 } { 2 } \pi$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { - 2 } { 3 \sin ^ { 2 } 2 t }$.\\
\includegraphics[max width=\textwidth, alt={}, center]{6280ab81-0bdb-47b4-8651-bff1261a0adf-10_2716_40_109_2009}\\
\includegraphics[max width=\textwidth, alt={}, center]{6280ab81-0bdb-47b4-8651-bff1261a0adf-11_2723_33_99_22}
\item Find the equation of the normal to the curve at the point where $t = \frac { 1 } { 4 } \pi$. Give your answer in the form $p y + q x + r = 0$, where $p , q$ and $r$ are integers.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2024 Q7 [8]}}