| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2021 |
| Session | June |
| 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 the chain rule (dy/dx = (dy/dt)/(dx/dt)), product rule for dy/dt, and solving a simple equation for the stationary point. While it involves multiple techniques, these are routine applications with no conceptual surprises, making it slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07n Stationary points: find maxima, minima using derivatives1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| State \(\frac{dx}{dt} = 1 + \frac{1}{t+2}\) | B1 | |
| Use product rule | M1 | |
| Obtain \(\frac{dy}{dt} = e^{-2t} - 2(t-1)e^{-2t}\) | A1 | OE |
| Use \(\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}\) | M1 | |
| Obtain correct answer in any simplified form, e.g. \(\frac{(3-2t)(t+2)}{t+3}e^{-2t}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Equate derivative to zero and solve for \(t\) | M1 | |
| Obtain \(t = \frac{3}{2}\) and obtain answer \(y = \frac{1}{2}e^{-3}\), or exact equivalent | A1 |
## Question 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| State $\frac{dx}{dt} = 1 + \frac{1}{t+2}$ | B1 | |
| Use product rule | M1 | |
| Obtain $\frac{dy}{dt} = e^{-2t} - 2(t-1)e^{-2t}$ | A1 | OE |
| Use $\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}$ | M1 | |
| Obtain correct answer in any simplified form, e.g. $\frac{(3-2t)(t+2)}{t+3}e^{-2t}$ | A1 | |
## Question 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Equate derivative to zero and solve for $t$ | M1 | |
| Obtain $t = \frac{3}{2}$ and obtain answer $y = \frac{1}{2}e^{-3}$, or exact equivalent | A1 | |3 The parametric equations of a curve are
$$x = t + \ln ( t + 2 ) , \quad y = ( t - 1 ) \mathrm { e } ^ { - 2 t }$$
where $t > - 2$.
\begin{enumerate}[label=(\alph*)]
\item Express $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $t$, simplifying your answer.
\item Find the exact $y$-coordinate of the stationary point of the curve.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2021 Q3 [7]}}