| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2015 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Iterative method for parameter value |
| Difficulty | Standard +0.3 This is a straightforward parametric differentiation question with standard iterative method application. Part (i) requires finding dy/dx at the origin (routine). Part (ii)(a) involves algebraic manipulation to rearrange an equation (standard). Part (ii)(b) applies a given iterative formula with clear instructions—purely procedural calculation requiring no problem-solving insight. Slightly above average only due to the multi-step nature and iteration component. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| (i) | ||
| Obtain \(\frac{dx}{dt} = \frac{2}{t+2}\) and \(\frac{dy}{dt} = 3t^2 + 2\) | B1 | |
| Use \(\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}\) | M1 | |
| Obtain \(\frac{dy}{dx} = \frac{1}{2}(3t^2+2)(t+2)\) | A1 | |
| Identify value of \(t\) at the origin as \(-1\) | B1 | |
| Substitute to obtain \(\frac{5}{2}\) as gradient at the origin | A1 | [5] |
| (ii)(a) | ||
| Equate derivative to \(\frac{1}{2}\) and confirm \(p = \frac{1}{3p^2+2} - 2\) | B1 | [1] |
| (ii)(b) | ||
| Use the iterative formula correctly at least once | M1 | |
| Obtain value \(p = -1.924\) or better (\(-1.92367\ldots\)) | A1 | |
| Show sufficient iterations to justify accuracy or show a sign change in appropriate interval | A1 | |
| Obtain coordinates \((-5.15, -7.97)\) | A1 | [4] |
**(i)** |
Obtain $\frac{dx}{dt} = \frac{2}{t+2}$ and $\frac{dy}{dt} = 3t^2 + 2$ | B1 |
Use $\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}$ | M1 |
Obtain $\frac{dy}{dx} = \frac{1}{2}(3t^2+2)(t+2)$ | A1 |
Identify value of $t$ at the origin as $-1$ | B1 |
Substitute to obtain $\frac{5}{2}$ as gradient at the origin | A1 | [5]
**(ii)(a)** |
Equate derivative to $\frac{1}{2}$ and confirm $p = \frac{1}{3p^2+2} - 2$ | B1 | [1]
**(ii)(b)** |
Use the iterative formula correctly at least once | M1 |
Obtain value $p = -1.924$ or better ($-1.92367\ldots$) | A1 |
Show sufficient iterations to justify accuracy or show a sign change in appropriate interval | A1 |
Obtain coordinates $(-5.15, -7.97)$ | A1 | [4]10\\
\includegraphics[max width=\textwidth, alt={}, center]{3eefd6c1-924c-4b7e-8d17-a2942fb48234-3_515_508_1105_815}
The diagram shows part of the curve with parametric equations
$$x = 2 \ln ( t + 2 ) , \quad y = t ^ { 3 } + 2 t + 3$$
(i) Find the gradient of the curve at the origin.\\
(ii) At the point $P$ on the curve, the value of the parameter is $p$. It is given that the gradient of the curve at $P$ is $\frac { 1 } { 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that $p = \frac { 1 } { 3 p ^ { 2 } + 2 } - 2$.
\item By first using an iterative formula based on the equation in part (a), determine the coordinates of the point $P$. Give the result of each iteration to 5 decimal places and each coordinate of $P$ correct to 2 decimal places.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2015 Q10 [10]}}