| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure (Further Additional Pure) |
| Year | 2018 |
| Session | March |
| Marks | 10 |
| Topic | Vector Product and Surfaces |
| Type | Tangent plane equation at a point |
| Difficulty | Challenging +1.2 This is a Further Maths question on partial derivatives and tangent planes, requiring quotient/product rule differentiation and standard tangent plane formula application. While the algebraic manipulation is moderately involved and it's a Further Maths topic, the techniques are routine and methodical with no novel problem-solving required. |
| Spec | 8.05d Partial differentiation: first and second order, mixed derivatives8.05g Tangent planes: equation at a given point on surface |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{\partial z}{\partial x} = \frac{1}{y}\sin y - \frac{1}{x}\sin x - \frac{y}{x^2}\cos x\) oe | M1, A1 | Partial differentiation w.r.t \(x\) (including use of the *Product* or *Quotient Rule*) |
| \(\frac{\partial z}{\partial y} = \frac{x}{y}\cos x - \frac{x}{y^2}\sin y + \frac{1}{x}\cos x\) oe | M1, A1 [4] | Partial differentiation w.r.t \(y\) (including use of the *Product* or *Quotient Rule*) |
| Answer | Marks |
|---|---|
| When \(x = y = \frac{1}{4}\pi, z = \sqrt{2}\) | B1 |
| \(\frac{\partial z}{\partial x} = \frac{-1}{\sqrt{2}}\) and \(\frac{\partial z}{\partial y} = \frac{1}{\sqrt{2}}\) oe | B1, B1, B1 |
| Eqn. of tangent-plane is \(z = \frac{-1}{\sqrt{2}}(x - \frac{1}{4}\pi) + \frac{1}{\sqrt{2}}(y - \frac{1}{4}\pi) + \sqrt{2}\) \(\Rightarrow x - y + z\sqrt{2} = 2\) oe | M1, A1 [5] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\begin{pmatrix} 1 \\ -1 \\ \frac{1}{\sqrt{2}} \end{pmatrix}\) or any suitable multiple | B1 [1] | FT from their answer to (ii) |
## (i)
| $\frac{\partial z}{\partial x} = \frac{1}{y}\sin y - \frac{1}{x}\sin x - \frac{y}{x^2}\cos x$ **oe** | M1, A1 | Partial differentiation w.r.t $x$ (including use of the *Product* or *Quotient Rule*) |
| $\frac{\partial z}{\partial y} = \frac{x}{y}\cos x - \frac{x}{y^2}\sin y + \frac{1}{x}\cos x$ **oe** | M1, A1 [4] | Partial differentiation w.r.t $y$ (including use of the *Product* or *Quotient Rule*) |
## (ii)
| When $x = y = \frac{1}{4}\pi, z = \sqrt{2}$ | B1 | |
| $\frac{\partial z}{\partial x} = \frac{-1}{\sqrt{2}}$ and $\frac{\partial z}{\partial y} = \frac{1}{\sqrt{2}}$ **oe** | B1, B1, B1 | |
| Eqn. of tangent-plane is $z = \frac{-1}{\sqrt{2}}(x - \frac{1}{4}\pi) + \frac{1}{\sqrt{2}}(y - \frac{1}{4}\pi) + \sqrt{2}$ $\Rightarrow x - y + z\sqrt{2} = 2$ **oe** | M1, A1 [5] | |
## (iii)
| $\begin{pmatrix} 1 \\ -1 \\ \frac{1}{\sqrt{2}} \end{pmatrix}$ or any suitable multiple | B1 [1] | FT from their answer to (ii) |
---3 The surface $S$ has equation $z = \frac { x } { y } \sin y + \frac { y } { x } \cos x$ where $0 < x \leqslant \pi$ and $0 < y \leqslant \pi$.\\
(i) Find
\begin{itemize}
\item $\frac { \partial z } { \partial x }$,
\item $\frac { \partial z } { \partial y }$.\\
(ii) Determine the equation of the tangent plane to $S$ at the point $A$ where $x = y = \frac { 1 } { 4 } \pi$. Give your answer in the form $a x + b y + c z = d$ where $a , b , c$ and $d$ are exact constants.\\
(iii) Write down a normal vector to $S$ at $A$.
\end{itemize}
\hfill \mbox{\textit{OCR Further Additional Pure 2018 Q3 [10]}}