| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find tangent equation at parameter |
| Difficulty | Standard +0.3 This is a straightforward parametric differentiation question requiring standard techniques: finding dy/dt and dx/dt, then dividing them. Part (a) involves routine differentiation of sec(t) and ln(tan(t)), then simplification using trig identities. Part (b) requires finding the parameter value where y=0 (tan t = 1, so t = π/4), then applying the tangent line formula. While it requires multiple steps and some trig manipulation, these are all standard A-level techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use chain rule at least once | M1 | Needs \(\frac{dy}{dt} = \frac{1}{\tan t}\frac{d}{dt}(\tan t)\) or \(\frac{dx}{dt} = (-1)(\cos^{-2}t)\frac{d}{dt}(\cos t)\). BOD if \(+\) and \((-1)(-1)\) not seen. \(\frac{dx}{dt} = \sec t \tan t\) (from MF19) M1 A1. If \(\frac{dx}{dt} = -\sec t \tan t\) M1 A0. |
| Obtain \(\frac{dx}{dt} = \sec t \tan t\) | A1 | OE e.g. \(\sin t(\cos t)^{-2}\). If e.g. \(\frac{dx}{dt} = \sec x \tan x\) or \(\sec\theta\tan\theta\) or \(\sec t\tan x\), condone recovery on next line. |
| Obtain \(\frac{dy}{dt} = \frac{\sec^2 t}{\tan t}\) | A1 | OE e.g. \(\frac{1}{\sin t \cos t}\). If e.g. \(\frac{dy}{dt} = \frac{\sec^2 x}{\tan x}\) or \(\frac{\sec^2\theta}{\tan\theta}\), condone recovery on next line. Only penalise notation errors once in \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) if no recovery. |
| Use \(\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}\) | M1 | Allow even if previous M0 scored, but must be using derivatives. |
| Obtain given answer \(\frac{\cos t}{\sin^2 t}\) | A1 | AG. After \(\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}\) used, any notation error A0. Must cancel \(\cos t\) correctly. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| State or imply \(t = \frac{1}{4}\pi\) when \(y = 0\) | B1 | |
| Form the equation of the tangent at \(y = 0\) or find \(c\) | M1 | \(x = \sqrt{2}\), \(\frac{dy}{dx} = \sqrt{2}\) and \(y = 0\), *their* coordinates and gradient used in \(y = mx + c\). |
| Obtain answer \(y = \sqrt{2}x - 2\) | A1 | OE e.g. \(y = \sqrt{2}(x - \sqrt{2})\). Allow \(y = 1.41x - 2[.00]\) or \(1.41(x-1.41)\). |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use chain rule at least once | M1 | Needs $\frac{dy}{dt} = \frac{1}{\tan t}\frac{d}{dt}(\tan t)$ or $\frac{dx}{dt} = (-1)(\cos^{-2}t)\frac{d}{dt}(\cos t)$. BOD if $+$ and $(-1)(-1)$ not seen. $\frac{dx}{dt} = \sec t \tan t$ (from MF19) M1 A1. If $\frac{dx}{dt} = -\sec t \tan t$ M1 A0. |
| Obtain $\frac{dx}{dt} = \sec t \tan t$ | A1 | OE e.g. $\sin t(\cos t)^{-2}$. If e.g. $\frac{dx}{dt} = \sec x \tan x$ or $\sec\theta\tan\theta$ or $\sec t\tan x$, condone recovery on next line. |
| Obtain $\frac{dy}{dt} = \frac{\sec^2 t}{\tan t}$ | A1 | OE e.g. $\frac{1}{\sin t \cos t}$. If e.g. $\frac{dy}{dt} = \frac{\sec^2 x}{\tan x}$ or $\frac{\sec^2\theta}{\tan\theta}$, condone recovery on next line. Only penalise notation errors **once** in $\frac{dx}{dt}$ and $\frac{dy}{dt}$ if no recovery. |
| Use $\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}$ | M1 | Allow even if previous M0 scored, but must be using derivatives. |
| Obtain given answer $\frac{\cos t}{\sin^2 t}$ | A1 | AG. After $\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}$ used, any notation error A0. Must cancel $\cos t$ correctly. |
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| State or imply $t = \frac{1}{4}\pi$ when $y = 0$ | B1 | |
| Form the equation of the tangent at $y = 0$ or find $c$ | M1 | $x = \sqrt{2}$, $\frac{dy}{dx} = \sqrt{2}$ and $y = 0$, *their* coordinates and gradient used in $y = mx + c$. |
| Obtain answer $y = \sqrt{2}x - 2$ | A1 | OE e.g. $y = \sqrt{2}(x - \sqrt{2})$. Allow $y = 1.41x - 2[.00]$ or $1.41(x-1.41)$. |6 The parametric equations of a curve are $x = \frac { 1 } { \cos t } , y = \ln \tan t$, where $0 < t < \frac { 1 } { 2 } \pi$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { \cos t } { \sin ^ { 2 } t }$.
\item Find the equation of the tangent to the curve at the point where $y = 0$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2022 Q6 [8]}}