| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2021 |
| 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/dx using the chain rule, showing it's always positive (simple sign analysis), and finding a tangent equation at a specific point. The algebra is clean and all steps are routine for P3 level, 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 correct chain rule or correct quotient rule to differentiate \(x\) or \(y\) | M1 | |
| Obtain \(\dfrac{dx}{dt} = \dfrac{3}{2+3t}\) or \(\dfrac{dy}{dt} = \dfrac{2}{(2+3t)^2}\) | A1 | OE |
| Use \(\dfrac{dy}{dx} = \dfrac{dy}{dt} \div \dfrac{dx}{dt}\) | M1 | |
| Obtain answer \(\dfrac{2}{3(2+3t)}\) | A1 | OE. Express as a simple fraction but not necessarily fully cancelled. |
| Explain why this is always positive | A1 | For correct gradient. e.g. \(x\) is only defined for \(2 + 3t > 0\) hence gradient \(> 0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Form equation in \(x\) and \(y\) only | M1 | |
| Obtain \(y = \dfrac{e^x - 2}{3e^x}\left(= \dfrac{1}{3} - \dfrac{2}{3}e^{-x}\right)\) | A1 | OE |
| Differentiate | M1 | |
| Obtain \(y' = \dfrac{2}{3}e^{-x}\) | A1 | OE |
| Explain why this is always positive | A1 | |
| Total | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Obtain \(y = -\frac{1}{3}\) when \(x = 0\) | B1 | |
| Use a correct method to form the given tangent | M1 | \(\frac{y + \frac{1}{3}}{x} = \frac{2}{3}\) |
| Obtain answer \(3y = 2x - 1\) | A1 | OE |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use correct chain rule **or** correct quotient rule to differentiate $x$ or $y$ | M1 | |
| Obtain $\dfrac{dx}{dt} = \dfrac{3}{2+3t}$ **or** $\dfrac{dy}{dt} = \dfrac{2}{(2+3t)^2}$ | A1 | OE |
| Use $\dfrac{dy}{dx} = \dfrac{dy}{dt} \div \dfrac{dx}{dt}$ | M1 | |
| Obtain answer $\dfrac{2}{3(2+3t)}$ | A1 | OE. Express as a simple fraction but not necessarily fully cancelled. |
| Explain why this is always positive | A1 | For correct gradient. e.g. $x$ is only defined for $2 + 3t > 0$ hence gradient $> 0$ |
**Alternative method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Form equation in $x$ and $y$ only | M1 | |
| Obtain $y = \dfrac{e^x - 2}{3e^x}\left(= \dfrac{1}{3} - \dfrac{2}{3}e^{-x}\right)$ | A1 | OE |
| Differentiate | M1 | |
| Obtain $y' = \dfrac{2}{3}e^{-x}$ | A1 | OE |
| Explain why this is always positive | A1 | |
| **Total** | **5** | |
## Question 6(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Obtain $y = -\frac{1}{3}$ when $x = 0$ | B1 | |
| Use a correct method to form the given tangent | M1 | $\frac{y + \frac{1}{3}}{x} = \frac{2}{3}$ |
| Obtain answer $3y = 2x - 1$ | A1 | OE |
---6 The parametric equations of a curve are
$$x = \ln ( 2 + 3 t ) , \quad y = \frac { t } { 2 + 3 t }$$
\begin{enumerate}[label=(\alph*)]
\item Show that the gradient of the curve is always positive.
\item Find the equation of the tangent to the curve at the point where it intersects the $y$-axis.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2021 Q6 [8]}}