| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Show gradient expression then find coordinates |
| Difficulty | Standard +0.8 This parametric question requires the chain rule (dy/dx = (dy/dt)/(dx/dt)), product rule for differentiating y = 5e^(-t)cos(2t), then solving dy/dx = 0 which involves tan(2t) = -1/2. The multi-step differentiation and trigonometric equation solving elevate this above standard parametric exercises, though it follows a familiar stationary point framework. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.06a Exponential function: a^x and e^x graphs and properties1.07n Stationary points: find maxima, minima using derivatives1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use product rule to find \(\frac{dy}{dt}\) | M1 | |
| Obtain \(\frac{dy}{dt} = -5e^{-t}\cos 2t - 10e^{-t}\sin 2t\) | A1 | Or unsimplified equivalent (do not condone poor use of brackets) |
| Obtain \(\frac{dy}{dx} = \frac{-5e^{-t}\cos 2t - 10e^{-t}\sin 2t}{8e^{2t}}\) | A1 | OE following *their* expression for \(\frac{dy}{dt}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Equate \(\frac{dy}{dx}\) to zero and simplify at least as far as \(\tan 2t = \ldots\) | M1* | Now condoning any error with \(\frac{dx}{dt}\) |
| Obtain \(\tan 2t = -\frac{1}{2}\) | A1 | |
| Obtain \(t = -0.231\ldots\) | A1 | Allow \(t = -0.232\) |
| Substitute negative value of \(t\) in expressions for \(x\) and \(y\) | DM1 | |
| Obtain \(x = 2.52\) and \(y = 5.64\) | A1 | Or greater accuracy |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use product rule to find $\frac{dy}{dt}$ | M1 | |
| Obtain $\frac{dy}{dt} = -5e^{-t}\cos 2t - 10e^{-t}\sin 2t$ | A1 | Or unsimplified equivalent (do not condone poor use of brackets) |
| Obtain $\frac{dy}{dx} = \frac{-5e^{-t}\cos 2t - 10e^{-t}\sin 2t}{8e^{2t}}$ | A1 | OE following *their* expression for $\frac{dy}{dt}$ |
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Equate $\frac{dy}{dx}$ to zero and simplify at least as far as $\tan 2t = \ldots$ | M1* | Now condoning any error with $\frac{dx}{dt}$ |
| Obtain $\tan 2t = -\frac{1}{2}$ | A1 | |
| Obtain $t = -0.231\ldots$ | A1 | Allow $t = -0.232$ |
| Substitute negative value of $t$ in expressions for $x$ and $y$ | DM1 | |
| Obtain $x = 2.52$ and $y = 5.64$ | A1 | Or greater accuracy |5\\
\includegraphics[max width=\textwidth, alt={}, center]{3966e088-0a2f-434a-94fc-40765cd157a7-06_376_848_269_644}
The diagram shows the curve with parametric equations
$$x = 4 \mathrm { e } ^ { 2 t } , \quad y = 5 \mathrm { e } ^ { - t } \cos 2 t$$
for $- \frac { 1 } { 4 } \pi \leqslant t \leqslant \frac { 1 } { 4 } \pi$. The curve has a maximum point $M$.
\begin{enumerate}[label=(\alph*)]
\item Find an expression for $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $t$.
\item Find the coordinates of $M$, giving each coordinate correct to 3 significant figures.
\end{enumerate}
\hfill \mbox{\textit{CAIE P2 2023 Q5 [8]}}