| Exam Board | OCR MEI |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2009 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vector Product and Surfaces |
| Type | Finding stationary points on surfaces |
| Difficulty | Challenging +1.8 This is a substantial Further Maths question requiring partial differentiation, finding stationary points by solving simultaneous equations, determining normal lines in 3D, linear approximation using the tangent plane, and solving a constraint problem. While each technique is standard for FP3, the multi-part nature, algebraic complexity (especially solving the cubic system for stationary points), and the conceptual demand of part (iv) requiring understanding of tangent plane approximation elevate this significantly above average A-level difficulty. |
| Spec | 8.05a 3D surfaces: z = f(x,y) and implicit form, partial derivatives8.05d Partial differentiation: first and second order, mixed derivatives8.05e Stationary points: where partial derivatives are zero8.05g Tangent planes: equation at a given point on surface |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{\partial z}{\partial x} = 3(x+y)^3 + 9x(x+y)^2 - 6x^2 + 24\) | M1, A2 | Partial differentiation; give A1 if just one minor error |
| \(\frac{\partial z}{\partial y} = 9x(x+y)^2\) | A1 | 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| At stationary points, \(\frac{\partial z}{\partial x}=0\) and \(\frac{\partial z}{\partial y}=0\) | M1 | |
| \(9x(x+y)^2=0 \Rightarrow x=0\) or \(y=-x\) | M1 | |
| If \(x=0\) then \(3y^3+24=0\), \(y=-2\); stationary point is \((0,-2,0)\) | A1A1 | |
| If \(y=-x\) then \(-6x^2+24=0\), \(x=\pm 2\); stationary points are \((2,-2,32)\) and \((-2,2,-32)\) | M1, A1, A1 | If A0A0, give A1 for \(x=\pm2\); 7 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| At P\((1,-2,19)\): \(\frac{\partial z}{\partial x}=24\), \(\frac{\partial z}{\partial y}=9\) | B1 | |
| Normal line is \(\mathbf{r} = \begin{pmatrix}1\\-2\\19\end{pmatrix}+\lambda\begin{pmatrix}24\\9\\-1\end{pmatrix}\) | M1, A1 ft | For normal vector (allow sign error); condone omission of '\(\mathbf{r}=\)'; 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\delta z \approx \frac{\partial z}{\partial x}\delta x + \frac{\partial z}{\partial y}\delta y = 24\,\delta x + 9\,\delta y\) | M1, A1 ft | |
| \(3h \approx 24k + 9h\) | M1 | |
| \(k = -\frac{1}{4}h\) | A1 | 4 marks |
| OR: Tangent plane is \(24x+9y-z=-13\); \(24(1+k)+9(-2+h)-(19+3h)\approx-13\); \(k\approx-\frac{1}{4}h\) | M2A1 ft, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{\partial z}{\partial x}=27\) and \(\frac{\partial z}{\partial y}=0\) | M1 | Allow M1 for \(\frac{\partial z}{\partial x}=-27\) |
| \(9x(x+y)^2=0 \Rightarrow x=0\) or \(y=-x\) | ||
| If \(x=0\) then \(3y^3+24=27\), \(y=1\), \(z=0\); point is \((0,1,0)\), \(d=0\) | M1, A1, A1 | |
| If \(y=-x\) then \(-6x^2+24=27\), \(x^2=-\frac{1}{2}\); there are no other points | M1, A1 | 6 marks |
# Question 2:
## Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{\partial z}{\partial x} = 3(x+y)^3 + 9x(x+y)^2 - 6x^2 + 24$ | M1, A2 | Partial differentiation; give A1 if just one minor error |
| $\frac{\partial z}{\partial y} = 9x(x+y)^2$ | A1 | **4 marks** |
## Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| At stationary points, $\frac{\partial z}{\partial x}=0$ and $\frac{\partial z}{\partial y}=0$ | M1 | |
| $9x(x+y)^2=0 \Rightarrow x=0$ or $y=-x$ | M1 | |
| If $x=0$ then $3y^3+24=0$, $y=-2$; stationary point is $(0,-2,0)$ | A1A1 | |
| If $y=-x$ then $-6x^2+24=0$, $x=\pm 2$; stationary points are $(2,-2,32)$ and $(-2,2,-32)$ | M1, A1, A1 | If A0A0, give A1 for $x=\pm2$; **7 marks** |
## Part (iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| At P$(1,-2,19)$: $\frac{\partial z}{\partial x}=24$, $\frac{\partial z}{\partial y}=9$ | B1 | |
| Normal line is $\mathbf{r} = \begin{pmatrix}1\\-2\\19\end{pmatrix}+\lambda\begin{pmatrix}24\\9\\-1\end{pmatrix}$ | M1, A1 ft | For normal vector (allow sign error); condone omission of '$\mathbf{r}=$'; **3 marks** |
## Part (iv)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\delta z \approx \frac{\partial z}{\partial x}\delta x + \frac{\partial z}{\partial y}\delta y = 24\,\delta x + 9\,\delta y$ | M1, A1 ft | |
| $3h \approx 24k + 9h$ | M1 | |
| $k = -\frac{1}{4}h$ | A1 | **4 marks** |
| OR: Tangent plane is $24x+9y-z=-13$; $24(1+k)+9(-2+h)-(19+3h)\approx-13$; $k\approx-\frac{1}{4}h$ | M2A1 ft, A1 | |
## Part (v)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{\partial z}{\partial x}=27$ and $\frac{\partial z}{\partial y}=0$ | M1 | Allow M1 for $\frac{\partial z}{\partial x}=-27$ |
| $9x(x+y)^2=0 \Rightarrow x=0$ or $y=-x$ | | |
| If $x=0$ then $3y^3+24=27$, $y=1$, $z=0$; point is $(0,1,0)$, $d=0$ | M1, A1, A1 | |
| If $y=-x$ then $-6x^2+24=27$, $x^2=-\frac{1}{2}$; there are no other points | M1, A1 | **6 marks** |
---2 A surface has equation $z = 3 x ( x + y ) ^ { 3 } - 2 x ^ { 3 } + 24 x$.\\
(i) Find $\frac { \partial z } { \partial x }$ and $\frac { \partial z } { \partial y }$.\\
(ii) Find the coordinates of the three stationary points on the surface.\\
(iii) Find the equation of the normal line at the point $\mathrm { P } ( 1 , - 2,19 )$ on the surface.\\
(iv) The point $\mathrm { Q } ( 1 + k , - 2 + h , 19 + 3 h )$ is on the surface and is close to P . Find an approximate expression for $k$ in terms of $h$.\\
(v) Show that there is only one point on the surface at which the tangent plane has an equation of the form $27 x - z = d$. Find the coordinates of this point and the corresponding value of $d$.
\hfill \mbox{\textit{OCR MEI FP3 2009 Q2 [24]}}