| Exam Board | OCR MEI |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2008 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vector Product and Surfaces |
| Type | Finding partial derivatives |
| Difficulty | Challenging +1.2 This is a Further Maths question on partial derivatives and surfaces, requiring multiple techniques (partial differentiation, gradient vectors, normal lines, tangent planes). While it involves several steps and FM content, each part follows standard procedures: computing partials is routine, finding normals uses ∇g directly, and the tangent plane condition is a straightforward system. The conceptual demand is moderate—higher than typical A-level but not requiring deep insight. |
| Spec | 8.05a 3D surfaces: z = f(x,y) and implicit form, partial derivatives8.05d Partial differentiation: first and second order, mixed derivatives8.05g Tangent planes: equation at a given point on surface |
## 2(i)
$$\frac{\partial g}{\partial x} = 6z - 2(x + 2y + 3z) = -2x - 4y$$
$$\frac{\partial g}{\partial y} = -4(x + 2y + 3z)$$
$$\frac{\partial g}{\partial z} = 6x - 6(x + 2y + 3z) = -12y - 18z$$
**M1**
**A1** - Partial differentiation. Any correct form, ISW
**A1**
**A1**
**Mark: 4**
## 2(ii)
At P, $\frac{\partial g}{\partial x} = 16$, $\frac{\partial g}{\partial y} = -4$, $\frac{\partial g}{\partial z} = 36$
**M1** - Evaluating partial derivatives at P
**A1** - All correct
Normal line is $\mathbf{r} = \begin{pmatrix} 7 \\ -7.5 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 4 \\ -1 \\ 9 \end{pmatrix}$
**A1 ft**
**3** - Condone omission of '$\mathbf{r} =$'
**Mark: 3**
## 2(iii)
$\delta g = 16\delta x - 4\delta y + 36\delta z$
**M1** - Alternative: M3 for substituting $x = 7 + 4\lambda, \ldots$ into $g = 125 + h$ and neglecting $\lambda^2$
If $PQ = \lambda \begin{pmatrix} 4 \\ -1 \\ 9 \end{pmatrix}$
$\delta g = 16(4\lambda) - 4(-\lambda) + 36(9\lambda) = 392\lambda$
**M1**
**A1 ft** - $h = 392\lambda$
**A1 ft** - For linear equation in $\lambda$ and $h$
**A1** - For n correct
**Mark: 5**
## 2(iv)
Require $\frac{\partial g}{\partial x} = \frac{\partial g}{\partial y} = 0$
**M1**
$-2x - 4y = 0$ and $x + 2y + 3z = 0$
$x + 2y = 0$ and $z = 0$
$g(x, y, z) = 0^2 = 0 \neq 125$
Hence there is no such point on $S$
**M1** - Useful manipulation using both eqns
**M1**
**A1** - Showing there is no such point on $S$. Fully correct proof
**Mark: 4**
## 2(v)
Require $\frac{\partial g}{\partial z} = 0$
and $\frac{\partial g}{\partial y} = \lambda \frac{\partial g}{\partial x}$
$-4x - 8y - 12z = 5(2x - 4y)$
**M1**
**M1**
**M1** - Implied by $\frac{\partial g}{\partial x} = \lambda$, $\frac{\partial g}{\partial y} = 5\lambda$
This M1 can be awarded for $-2x - 4y = 1$ and $-4x - 8y - 12z = 5$
**Mark: [continued from previous]**
---2 You are given $\mathrm { g } ( x , y , z ) = 6 x z - ( x + 2 y + 3 z ) ^ { 2 }$.\\
(i) Find $\frac { \partial \mathrm { g } } { \partial x } , \frac { \partial \mathrm {~g} } { \partial y }$ and $\frac { \partial \mathrm { g } } { \partial z }$.
A surface $S$ has equation $\mathrm { g } ( x , y , z ) = 125$.\\
(ii) Find the equation of the normal line to $S$ at the point $\mathrm { P } ( 7 , - 7.5,3 )$.\\
(iii) The point Q is on this normal line and is close to P . At $\mathrm { Q } , \mathrm { g } ( x , y , z ) = 125 + h$, where $h$ is small. Find the vector $\mathbf { n }$ such that $\overrightarrow { \mathrm { PQ } } = h \mathbf { n }$ approximately.\\
(iv) Show that there is no point on $S$ at which the normal line is parallel to the $z$-axis.\\
(v) Find the two points on $S$ at which the tangent plane is parallel to $x + 5 y = 0$.
\hfill \mbox{\textit{OCR MEI FP3 2008 Q2 [24]}}