| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure (Further Additional Pure) |
| Year | 2018 |
| Session | December |
| Marks | 7 |
| Topic | Vector Product and Surfaces |
| Type | Tangent plane equation at a point |
| Difficulty | Standard +0.8 This is a Further Maths question on partial derivatives and tangent planes to surfaces. Part (a) requires straightforward partial differentiation (one trivial, one using chain rule with tan). Part (b) requires applying the tangent plane formula and substituting specific values. While the topic is beyond standard A-level, the execution is methodical with no novel insight required—standard Further Maths fare but elevated above typical single maths questions. |
| Spec | 8.05a 3D surfaces: z = f(x,y) and implicit form, partial derivatives8.05g Tangent planes: equation at a given point on surface |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{\partial z}{\partial x} = \tan y\) and \(\frac{\partial z}{\partial y} = x \sec^2 y\) | B1, B1 | Both required |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = 1, y = \frac{1}{4}\pi\) substituted into the two partial derivatives | M1 | 1.1a |
| \(\frac{\partial z}{\partial x} = 1\) and \(\frac{\partial z}{\partial y} = 2\) | A1 | 1.1 |
| \(z = 1\) | B1 | 1.1 |
| Equation of tangent plane is \(z - 1 = 1(x - 1) + 2(y - \frac{1}{4}\pi)\) | M1FT | 1.1a |
| i.e. \(x + 2y - z = \frac{1}{2}\pi\) oe | A1 | 1.1 |
| Total: [5] | FT three previous values |
**Part (a)**
| $\frac{\partial z}{\partial x} = \tan y$ and $\frac{\partial z}{\partial y} = x \sec^2 y$ | B1, B1 | Both required |
**Part (b)**
| $x = 1, y = \frac{1}{4}\pi$ substituted into the two partial derivatives | M1 | 1.1a |
| $\frac{\partial z}{\partial x} = 1$ and $\frac{\partial z}{\partial y} = 2$ | A1 | 1.1 |
| $z = 1$ | B1 | 1.1 |
| Equation of tangent plane is $z - 1 = 1(x - 1) + 2(y - \frac{1}{4}\pi)$ | M1FT | 1.1a |
| i.e. $x + 2y - z = \frac{1}{2}\pi$ oe | A1 | 1.1 |
| Total: [5] | FT three previous values | |1 A surface has equation $z = x \tan y$ for $- \frac { 1 } { 2 } \pi < y < \frac { 1 } { 2 } \pi$.
\begin{enumerate}[label=(\alph*)]
\item Find
\begin{itemize}
\item $\frac { \partial z } { \partial x }$,
\item $\frac { \partial z } { \partial y }$.
\item Find in cartesian form, the equation of the tangent plane to the surface at the point where $x = 1$ and $y = \frac { 1 } { 4 } \pi$.
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure 2018 Q1 [7]}}