| Exam Board | OCR MEI |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2010 |
| 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 structured multi-part question combining partial differentiation, vector geometry, and linear approximation. While it involves Further Maths content (FP3), each part follows logically from the previous with clear guidance. Parts (i)-(iii) are routine applications of standard techniques; parts (iv)-(vi) require careful algebraic manipulation and understanding of linear approximation but the question scaffolds the approach extensively. The conceptual demand is moderate—higher than a typical C3 question but not requiring exceptional insight. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates4.04a Line equations: 2D and 3D, cartesian and vector forms4.04g Vector product: a x b perpendicular vector8.05a 3D surfaces: z = f(x,y) and implicit form, partial derivatives8.05d Partial differentiation: first and second order, mixed derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{\partial g}{\partial x} = (y + xy + z^2)e^{x-2y}\) | M1, A1 | Partial differentiation |
| \(\frac{\partial g}{\partial y} = (x - 2xy - 2z^2)e^{x-2y}\) | A1 | |
| \(\frac{\partial g}{\partial z} = 2ze^{x-2y}\) | A1 | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| At \((2,1,-1)\): \(\frac{\partial g}{\partial x}=4\), \(\frac{\partial g}{\partial y}=-4\), \(\frac{\partial g}{\partial z}=-2\) | M1, A1 | |
| Normal has direction \(\begin{pmatrix}4\\-4\\-2\end{pmatrix}\) | M1 | |
| \(L\) passes through \((2,1,-1)\) and has this direction | A1 (ag) | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| When \(g=0\), \(xy+z^2=0\) | ||
| \((2-2\lambda)(1+2\lambda)+(-1+\lambda)^2=0\) | M1 | |
| \(3-3\lambda^2=0\), \(\lambda=\pm 1\) | M1 | Obtaining a value of \(\lambda\) |
| \(\lambda=1\) gives \(P(0,3,0)\) | A1 (ag) | Or B1 for verifying \(g(0,3,0)=0\) and showing P is on \(L\) |
| \(\lambda=-1\) gives \(Q(4,-1,-2)\) | A1 | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| At P: \(\frac{\partial g}{\partial x}=3e^{-6}\), \(\frac{\partial g}{\partial y}=0\), \(\frac{\partial g}{\partial z}=0\) | M1 | OR give M2 A1 www for \(g(-2\mu, 3+2\mu, \mu) = (-3\mu^2-6\mu)e^{-6\mu-6} \approx -6\mu e^{-6}\) |
| \(\delta g \approx \frac{\partial g}{\partial x}\delta x + \frac{\partial g}{\partial y}\delta y + \frac{\partial g}{\partial z}\delta z\) | M1 | |
| \(= 3e^{-6}(-2\mu)+0+0 = -6\mu e^{-6}\) | A1 (ag) | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| When \(-6\mu e^{-6} \approx h\), \(\mu \approx -\frac{1}{6}e^6 h\) | M1 | |
| Point \((-2\mu, 3+2\mu, \mu)\) is approximately \((\frac{1}{3}e^6h,\ 3-\frac{1}{3}e^6h,\ -\frac{1}{6}e^6h)\) | A1 (ag) | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| At Q: \(\frac{\partial g}{\partial x}=-e^6\), \(\frac{\partial g}{\partial y}=4e^6\), \(\frac{\partial g}{\partial z}=-4e^6\) | M1 | OR give M1 M2 A1 www for \(g(4-2\mu,-1+2\mu,-2+\mu)=(-3\mu^2+6\mu)e^{-6\mu+6}\approx 6\mu e^6\) |
| When \(x=4-2\mu\), \(y=-1+2\mu\), \(z=-2+\mu\) | M1 | |
| \(\delta g \approx (-e^6)(-2\mu)+(4e^6)(2\mu)+(-4e^6)(\mu) = 6\mu e^6\) | M1A1 | |
| If \(6\mu e^6 \approx h\), then \(\mu \approx \frac{1}{6}e^{-6}h\) | M1 | |
| Point is approximately \((4-\frac{1}{3}e^{-6}h,\ -1+\frac{1}{3}e^{-6}h,\ -2+\frac{1}{6}e^{-6}h)\) | A2 | Give A1 for one coordinate correct; if partial derivatives not evaluated at Q, max mark M0M1M0M0 [7] |
# Question 2:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{\partial g}{\partial x} = (y + xy + z^2)e^{x-2y}$ | M1, A1 | Partial differentiation |
| $\frac{\partial g}{\partial y} = (x - 2xy - 2z^2)e^{x-2y}$ | A1 | |
| $\frac{\partial g}{\partial z} = 2ze^{x-2y}$ | A1 | **[4]** |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| At $(2,1,-1)$: $\frac{\partial g}{\partial x}=4$, $\frac{\partial g}{\partial y}=-4$, $\frac{\partial g}{\partial z}=-2$ | M1, A1 | |
| Normal has direction $\begin{pmatrix}4\\-4\\-2\end{pmatrix}$ | M1 | |
| $L$ passes through $(2,1,-1)$ and has this direction | A1 (ag) | **[4]** |
## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| When $g=0$, $xy+z^2=0$ | | |
| $(2-2\lambda)(1+2\lambda)+(-1+\lambda)^2=0$ | M1 | |
| $3-3\lambda^2=0$, $\lambda=\pm 1$ | M1 | Obtaining a value of $\lambda$ |
| $\lambda=1$ gives $P(0,3,0)$ | A1 (ag) | Or B1 for verifying $g(0,3,0)=0$ and showing P is on $L$ |
| $\lambda=-1$ gives $Q(4,-1,-2)$ | A1 | **[4]** |
## Part (iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| At P: $\frac{\partial g}{\partial x}=3e^{-6}$, $\frac{\partial g}{\partial y}=0$, $\frac{\partial g}{\partial z}=0$ | M1 | OR give M2 A1 www for $g(-2\mu, 3+2\mu, \mu) = (-3\mu^2-6\mu)e^{-6\mu-6} \approx -6\mu e^{-6}$ |
| $\delta g \approx \frac{\partial g}{\partial x}\delta x + \frac{\partial g}{\partial y}\delta y + \frac{\partial g}{\partial z}\delta z$ | M1 | |
| $= 3e^{-6}(-2\mu)+0+0 = -6\mu e^{-6}$ | A1 (ag) | **[3]** |
## Part (v):
| Answer | Marks | Guidance |
|--------|-------|----------|
| When $-6\mu e^{-6} \approx h$, $\mu \approx -\frac{1}{6}e^6 h$ | M1 | |
| Point $(-2\mu, 3+2\mu, \mu)$ is approximately $(\frac{1}{3}e^6h,\ 3-\frac{1}{3}e^6h,\ -\frac{1}{6}e^6h)$ | A1 (ag) | **[2]** |
## Part (vi):
| Answer | Marks | Guidance |
|--------|-------|----------|
| At Q: $\frac{\partial g}{\partial x}=-e^6$, $\frac{\partial g}{\partial y}=4e^6$, $\frac{\partial g}{\partial z}=-4e^6$ | M1 | OR give M1 M2 A1 www for $g(4-2\mu,-1+2\mu,-2+\mu)=(-3\mu^2+6\mu)e^{-6\mu+6}\approx 6\mu e^6$ |
| When $x=4-2\mu$, $y=-1+2\mu$, $z=-2+\mu$ | M1 | |
| $\delta g \approx (-e^6)(-2\mu)+(4e^6)(2\mu)+(-4e^6)(\mu) = 6\mu e^6$ | M1A1 | |
| If $6\mu e^6 \approx h$, then $\mu \approx \frac{1}{6}e^{-6}h$ | M1 | |
| Point is approximately $(4-\frac{1}{3}e^{-6}h,\ -1+\frac{1}{3}e^{-6}h,\ -2+\frac{1}{6}e^{-6}h)$ | A2 | Give A1 for one coordinate correct; if partial derivatives not evaluated at Q, max mark M0M1M0M0 **[7]** |
---2 In this question, $L$ is the straight line with equation $\mathbf { r } = \left( \begin{array} { r } 2 \\ 1 \\ - 1 \end{array} \right) + \lambda \left( \begin{array} { r } - 2 \\ 2 \\ 1 \end{array} \right)$, and $\mathrm { g } ( x , y , z ) = \left( x y + z ^ { 2 } \right) \mathrm { e } ^ { x - 2 y }$.\\
(i) Find $\frac { \partial \mathrm { g } } { \partial x } , \frac { \partial \mathrm {~g} } { \partial y }$ and $\frac { \partial \mathrm { g } } { \partial z }$.\\
(ii) Show that the normal to the surface $\mathrm { g } ( x , y , z ) = 3$ at the point $( 2,1 , - 1 )$ is the line $L$.
On the line $L$, there are two points at which $\mathrm { g } ( x , y , z ) = 0$.\\
(iii) Show that one of these points is $\mathrm { P } ( 0,3,0 )$, and find the coordinates of the other point Q .\\
(iv) Show that, if $x = - 2 \mu , y = 3 + 2 \mu , z = \mu$, and $\mu$ is small, then
$$\mathrm { g } ( x , y , z ) \approx - 6 \mu \mathrm { e } ^ { - 6 }$$
You are given that $h$ is a small number.\\
(v) There is a point on $L$, close to P , at which $\mathrm { g } ( x , y , z ) = h$. Show that this point is approximately
$$\left( \frac { 1 } { 3 } \mathrm { e } ^ { 6 } h , 3 - \frac { 1 } { 3 } \mathrm { e } ^ { 6 } h , - \frac { 1 } { 6 } \mathrm { e } ^ { 6 } h \right)$$
(vi) Find the approximate coordinates of the point on $L$, close to Q , at which $\mathrm { g } ( x , y , z ) = h$.
\hfill \mbox{\textit{OCR MEI FP3 2010 Q2 [24]}}