| Exam Board | OCR MEI |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2012 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vector Product and Surfaces |
| Type | Normal vector or normal line to surface |
| Difficulty | Challenging +1.2 This is a Further Maths question on partial derivatives and surfaces, requiring multiple techniques (finding partial derivatives, normal lines, tangent planes) but following standard procedures. Parts (i)-(iii) are routine applications of formulas, part (iv) requires simple algebraic reasoning, and part (v) involves solving a system but is still methodical. The multi-part structure and Further Maths content place it above average difficulty, but no part requires significant novel 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 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{\partial g}{\partial x} = 2x + 2z\) | B1 | |
| \(\dfrac{\partial g}{\partial y} = 4y + 2z\) | B1 | |
| \(\dfrac{\partial g}{\partial z} = -2z + 2x + 2y + 4\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| At P: \(\dfrac{\partial g}{\partial x} = -2,\quad \dfrac{\partial g}{\partial y} = -2,\quad \dfrac{\partial g}{\partial z} = -4\) | B1 | |
| Normal line is \(\mathbf{r} = \begin{pmatrix}-2\\-1\\1\end{pmatrix} + \lambda\begin{pmatrix}1\\1\\2\end{pmatrix}\) | M1 | For direction of normal line |
| A1 | FT. Condone omission of \(\mathbf{r} =\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| If Q is \((-2+\lambda,\ -1+\lambda,\ 1+2\lambda)\) | ||
| \(h = \delta g \approx \dfrac{\partial g}{\partial x}\delta x + \dfrac{\partial g}{\partial y}\delta y + \dfrac{\partial g}{\partial z}\delta z\) | M1 | Requires some substitution in RHS or \(h\) on LHS. Allow M1 for \(\delta x = -2+\lambda\) etc |
| \(= (-2)(\lambda) + (-2)(\lambda) + (-4)(2\lambda)\quad [= -12\lambda]\) | A1 | FT |
| \(PQ = \sqrt{(\lambda)^2 + (\lambda)^2 + (2\lambda)^2} = \sqrt{6} | \lambda | \) |
| \(c = \dfrac{\sqrt{6}}{12}\) | A1 | A0 for \(c = -\dfrac{\sqrt{6}}{12}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Require \(\dfrac{\partial g}{\partial x} = \dfrac{\partial g}{\partial y} = 0\) | M1 | |
| \(x = -z,\quad y = -\tfrac{1}{2}z\) | ||
| \(z^2 + \tfrac{1}{2}z^2 - z^2 - 2z^2 - z^2 + 4z - 3 = 0\) | M1 | Obtaining equation in one variable |
| \(5z^2 - 8z + 6 = 0\) | A1 | Or \(5x^2+8x+6=0\) or \(10y^2+8y+3=0\) or \(\lambda^2+14=0\) |
| Discriminant is \(64 - 120 = -56 < 0\) | M1 | Dependent on quadratic with negative discriminant |
| Hence there are no such points | E1 | Correctly shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Require \(\dfrac{\partial g}{\partial x} = \dfrac{\partial g}{\partial y} = \dfrac{\partial g}{\partial z}\ (= \lambda)\) | M1 | Allow M1 if \(\lambda = 1\) |
| \(2x+2z = 4y+2z = -2z+2x+2y+4\) | A1 | FT |
| \(x = 2y,\quad y = 2z-2\) | ||
| \((4z-4)^2 + 2(2z-2)^2 - z^2 + \ldots + 4z - 3 = 0\) | M1 | Obtaining equation in one variable |
| \(5z^2 - 8z + 3 = 0\) | A1 | Or \(5x^2+8x=0\) or \(5y^2+4y=0\). Or \(\lambda^2-4=0\) |
| Points \((0,\ 0,\ 1)\) and \((-1.6,\ -0.8,\ 0.6)\) | M1 | Obtaining at least one point. Implies previous M1 if values of \(x, y, z\) not seen |
| \(k = 0+0+1\) or \(k = -1.6-0.8+0.6\) | M1 | Obtaining a value of \(k\) |
| \(k = 1,\quad -1.8\) | A1A1 |
# Question 2:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{\partial g}{\partial x} = 2x + 2z$ | B1 | |
| $\dfrac{\partial g}{\partial y} = 4y + 2z$ | B1 | |
| $\dfrac{\partial g}{\partial z} = -2z + 2x + 2y + 4$ | B1 | |
**[3 marks]**
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| At P: $\dfrac{\partial g}{\partial x} = -2,\quad \dfrac{\partial g}{\partial y} = -2,\quad \dfrac{\partial g}{\partial z} = -4$ | B1 | |
| Normal line is $\mathbf{r} = \begin{pmatrix}-2\\-1\\1\end{pmatrix} + \lambda\begin{pmatrix}1\\1\\2\end{pmatrix}$ | M1 | For direction of normal line |
| | A1 | FT. Condone omission of $\mathbf{r} =$ |
**[3 marks]**
## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| If Q is $(-2+\lambda,\ -1+\lambda,\ 1+2\lambda)$ | | |
| $h = \delta g \approx \dfrac{\partial g}{\partial x}\delta x + \dfrac{\partial g}{\partial y}\delta y + \dfrac{\partial g}{\partial z}\delta z$ | M1 | Requires some substitution in RHS or $h$ on LHS. Allow M1 for $\delta x = -2+\lambda$ etc |
| $= (-2)(\lambda) + (-2)(\lambda) + (-4)(2\lambda)\quad [= -12\lambda]$ | A1 | FT |
| $PQ = \sqrt{(\lambda)^2 + (\lambda)^2 + (2\lambda)^2} = \sqrt{6}|\lambda|$ | M1, A1 | FT. Allow $\sqrt{6}\,\lambda$ |
| $c = \dfrac{\sqrt{6}}{12}$ | A1 | A0 for $c = -\dfrac{\sqrt{6}}{12}$ |
**[5 marks]**
## Part (iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Require $\dfrac{\partial g}{\partial x} = \dfrac{\partial g}{\partial y} = 0$ | M1 | |
| $x = -z,\quad y = -\tfrac{1}{2}z$ | | |
| $z^2 + \tfrac{1}{2}z^2 - z^2 - 2z^2 - z^2 + 4z - 3 = 0$ | M1 | Obtaining equation in one variable |
| $5z^2 - 8z + 6 = 0$ | A1 | Or $5x^2+8x+6=0$ or $10y^2+8y+3=0$ or $\lambda^2+14=0$ |
| Discriminant is $64 - 120 = -56 < 0$ | M1 | Dependent on quadratic with negative discriminant |
| Hence there are no such points | E1 | Correctly shown |
**[5 marks]**
## Part (v):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Require $\dfrac{\partial g}{\partial x} = \dfrac{\partial g}{\partial y} = \dfrac{\partial g}{\partial z}\ (= \lambda)$ | M1 | Allow M1 if $\lambda = 1$ |
| $2x+2z = 4y+2z = -2z+2x+2y+4$ | A1 | FT |
| $x = 2y,\quad y = 2z-2$ | | |
| $(4z-4)^2 + 2(2z-2)^2 - z^2 + \ldots + 4z - 3 = 0$ | M1 | Obtaining equation in one variable |
| $5z^2 - 8z + 3 = 0$ | A1 | Or $5x^2+8x=0$ or $5y^2+4y=0$. Or $\lambda^2-4=0$ |
| Points $(0,\ 0,\ 1)$ and $(-1.6,\ -0.8,\ 0.6)$ | M1 | Obtaining at least one point. Implies previous M1 if values of $x, y, z$ not seen |
| $k = 0+0+1$ or $k = -1.6-0.8+0.6$ | M1 | Obtaining a value of $k$ |
| $k = 1,\quad -1.8$ | A1A1 | |
**[8 marks]**
---2 You are given that $\mathrm { g } ( x , y , z ) = x ^ { 2 } + 2 y ^ { 2 } - z ^ { 2 } + 2 x z + 2 y z + 4 z - 3$.\\
(i) Find $\frac { \partial \mathrm { g } } { \partial x } , \frac { \partial \mathrm {~g} } { \partial y }$ and $\frac { \partial \mathrm { g } } { \partial z }$.
The surface $S$ has equation $\mathrm { g } ( x , y , z ) = 0$, and $\mathrm { P } ( - 2 , - 1,1 )$ is a point on $S$.\\
(ii) Find an equation for the normal line to the surface $S$ at the point P .\\
(iii) A point Q lies on this normal line and is close to P . At $\mathrm { Q } , \mathrm { g } ( x , y , z ) = h$, where $h$ is small. Find the constant $c$ such that $\mathrm { PQ } \approx c | h |$.\\
(iv) Show that there is no point on $S$ at which the normal line is parallel to the $z$-axis.\\
(v) Given that $x + y + z = k$ is a tangent plane to the surface $S$, find the two possible values of $k$.
\hfill \mbox{\textit{OCR MEI FP3 2012 Q2 [24]}}