| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure AS (Further Additional Pure AS) |
| Year | 2018 |
| Session | March |
| Marks | 13 |
| Topic | Chain Rule |
| Type | Find stationary points and nature |
| Difficulty | Standard +0.8 This is a multivariable calculus question requiring partial differentiation, finding stationary points by solving simultaneous equations, and computing second partial derivatives. While the differentiation itself is routine, solving the system of equations from setting both partials to zero requires algebraic manipulation and is non-trivial. This is Further Maths content (AS level Further Additional Pure), which places it above standard A-level, but the techniques are fairly standard for this syllabus. |
| Spec | 8.05a 3D surfaces: z = f(x,y) and implicit form, partial derivatives8.05d Partial differentiation: first and second order, mixed derivatives8.05e Stationary points: where partial derivatives are zero |
| Answer | Marks | Guidance |
|---|---|---|
| M1, A1, A1 | 1.1a, 1.1, 1.1 | More than 2 terms correct in either derivative; First correct; Second correct |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| M1 | 1.1a | Setting \(\frac{\partial z}{\partial x} = 0\) and getting \(x = \ldots\) |
| M1 | 1.1a | Setting \(\frac{\partial z}{\partial y} = 0\) and getting \(x = \ldots\) |
| A1 | 1.1 | Both values for \(x\) noted |
| B1 | 1.1 | |
| M1, A1 | 1.1a, 1.1 | Eliminating \(x\) and solving for \(y\) (e.g. BC); All three coordinates correct |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| B1 | 1.1 | |
| B1 | 1.1 | |
| B1 | 1.1 | |
| B1 | 1.1 | Must give both or note that they are equal |
| [4] |
## Part (i)(a)
**Answer:** $\frac{\partial z}{\partial x} = 2xy - 8y^2 + \frac{1}{y}$
$\frac{\partial z}{\partial y} = x^2 - 16xy - \frac{x}{y^2}$
| **M1, A1, A1** | **1.1a, 1.1, 1.1** | More than 2 terms correct in either derivative; First correct; Second correct |
| **[3]** | | |
## Part (i)(b)
**Answer:** $\frac{\partial z}{\partial x} = 0 \Rightarrow 2xy - 8y^3 + 1 = 0$ i.e. $x = \frac{8y^3 - 1}{2y^2}$ or $x = 4y - \frac{1}{2y^2}$
$\frac{\partial z}{\partial y} = 0 \Rightarrow x = 0$ or $x = 16y + \frac{1}{y^2}$ or $x = \frac{16y^3 + 1}{y^2}$
If $x = 0, y = \frac{1}{4}, z = 0$
If $x = \frac{16y^3 + 1}{y^2} = \frac{8y^3 - 1}{2y^2}$ then $32y^3 + 2 = 8y^3 - 1 \Rightarrow y = -\frac{1}{3}, x = -4, z = 8$
| **M1** | **1.1a** | Setting $\frac{\partial z}{\partial x} = 0$ and getting $x = \ldots$ |
| **M1** | **1.1a** | Setting $\frac{\partial z}{\partial y} = 0$ and getting $x = \ldots$ |
| **A1** | **1.1** | Both values for $x$ noted |
| **B1** | **1.1** | |
| **M1, A1** | **1.1a, 1.1** | Eliminating $x$ and solving for $y$ (e.g. BC); All three coordinates correct |
| **[6]** | | |
## Part (ii)
**Answer:** $\frac{\partial^2 z}{\partial x^2} = 2y$
$\frac{\partial^2 z}{\partial y^2} = -16x + \frac{2x}{y^3}$
$\frac{\partial^2 z}{\partial x \partial y} = 2x - 16y - \frac{1}{y^2}$
$\frac{\partial^2 z}{\partial y \partial x} = 2x - 16y - \frac{1}{y^2}$
| **B1** | **1.1** | |
| **B1** | **1.1** | |
| **B1** | **1.1** | |
| **B1** | **1.1** | Must give both or note that they are equal |
| **[4]** | | |
---2 The surface $S$ has equation $z = x ^ { 2 } y - 8 x y ^ { 2 } + \frac { x } { y }$ for $y \neq 0$.
\begin{enumerate}[label=(\roman*)]
\item (a) Find the following.
\begin{itemize}
\item $\frac { \partial z } { \partial x }$
\item $\frac { \partial z } { \partial y }$\\
(b) Find the coordinates of all stationary points of $S$.
\item Find all four second partial derivatives of $z$ with respect to $x$ and/or $y$.
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure AS 2018 Q2 [13]}}