| Exam Board | OCR MEI |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2016 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Standard non-homogeneous with polynomial RHS |
| Difficulty | Challenging +1.2 This is a multi-part question on 3D surfaces and stationary points requiring partial derivatives, completing the square, and tangent plane calculations. While it involves several techniques and extended working across 5 parts, each individual part follows standard Further Maths procedures without requiring novel insight. The Taylor expansion approach in parts (iii)-(iv) is a textbook method for classifying stationary points. Slightly above average difficulty due to the length and coordination required, but well within expected FP3 scope. |
| 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 zero8.05f Nature of stationary points: classify using Hessian matrix8.05g Tangent planes: equation at a given point on surface |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| [Graph with correct shape] | B1 | Correct shape |
| [Minimum in third quadrant and positive intercept on \(z\)-axis] | B1 | Minimum in third quadrant and positive intercept on \(z\)-axis |
| \(y=1 \Rightarrow z=3x^2+6x+1\ [=3(x+1)^2-2]\) | B1 | |
| Line of symmetry is \(\mathbf{r} = \begin{pmatrix}-1\\1\\-2\end{pmatrix} + \lambda\begin{pmatrix}0\\0\\1\end{pmatrix}\) | B1B1 | For \(\begin{pmatrix}-1\\1\\\cdot\end{pmatrix}\) and \(\lambda\begin{pmatrix}0\\0\\1\end{pmatrix}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| We require \(\dfrac{\partial z}{\partial x} = \dfrac{\partial z}{\partial y} = 0\) | M1 | |
| \(\dfrac{\partial z}{\partial x} = 6x+6y=0\ [\Rightarrow y=-x]\) | A1 | For either equation |
| \(\dfrac{\partial z}{\partial y} = 6x+3y^2=0\) and hence \(y^2=2y\) | A1 | Correct equation in \(y\) or \(x\); Or \(6x+3x^2=0\) |
| \(y=0,2;\ x=0,-2;\ z=0,-4\); Stationary points are \((0,0,0)\) and \((-2,2,-4)\) | E1 | Some working required for \(-4\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| For \(x = -2+h\), \(y = 2+k\), \(z = \lambda\) | ||
| \(\lambda = 3(-2+h)^2 + 6(-2+h)(2+k) + (2+k)^3\) | M1, A1 | Substitution |
| \(= 12 - 12h + 3h^2 + 6hk + 12h - 12k - 24 + 8 + 12k + 6k^2 + k^3\) | ||
| \(= -4 + 3h^2 + 6hk + 6k^2 + k^3\) | ||
| \(= -4 + 3(h+k)^2 + k^3 + 3k^2\) | ||
| \(= -4 + 3(h+k)^2 + k^2(k+3)\) | E1 | |
| Since \(3(h+k)^2 > 0\) and \(k^2(k+3) > 0\) for small \(k\) | M1 | M0 for numerical work |
| \(\lambda > -4\) for all small values of \(h\) and \(k\), so P is a minimum | E1 | Must mention small \(k\) or \(k > -3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| For small \(x\) and \(y\), \(z\) can be positive or negative | M1 | Numerical demonstration can earn M1A0E0 |
| If \(x = 0\) and \(y > 0\), then \(z > 0\) | A1 | Correct argument which applies arbitrarily close to O; When \(x=0\), \(z=y^3\) which has a point of inflection |
| If \(x = 0\) and \(y < 0\), then \(z < 0\) | ||
| Hence O is neither a maximum nor a minimum | E1 | Correctly shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| We require \(\frac{\partial z}{\partial x} = 18\), \(\frac{\partial z}{\partial y} = 18\) | M1 | Allow \(-18\) for M1 |
| \(6x + 6y = 18\), \(6x + 3y^2 = 18\) | ||
| \(2(3-y) + y^2 = 6\) | M1 | Obtaining equation for \(y\) or \(x\) |
| A1 | or \(2x + (3-x)^2 = 6\) | |
| Points are \((3, 0, 27)\) and \((1, 2, 23)\) | M1, A1 | Obtaining values of \(x\), \(y\), \(z\) |
| \(18x + 18y - z = d\) | M1 | |
| So \(d = 27, 31\) | A1 |
# Question 2:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| [Graph with correct shape] | B1 | Correct shape |
| [Minimum in third quadrant and positive intercept on $z$-axis] | B1 | Minimum in third quadrant and positive intercept on $z$-axis |
| $y=1 \Rightarrow z=3x^2+6x+1\ [=3(x+1)^2-2]$ | B1 | |
| Line of symmetry is $\mathbf{r} = \begin{pmatrix}-1\\1\\-2\end{pmatrix} + \lambda\begin{pmatrix}0\\0\\1\end{pmatrix}$ | B1B1 | For $\begin{pmatrix}-1\\1\\\cdot\end{pmatrix}$ and $\lambda\begin{pmatrix}0\\0\\1\end{pmatrix}$ |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| We require $\dfrac{\partial z}{\partial x} = \dfrac{\partial z}{\partial y} = 0$ | M1 | |
| $\dfrac{\partial z}{\partial x} = 6x+6y=0\ [\Rightarrow y=-x]$ | A1 | For either equation |
| $\dfrac{\partial z}{\partial y} = 6x+3y^2=0$ and hence $y^2=2y$ | A1 | Correct equation in $y$ or $x$; Or $6x+3x^2=0$ |
| $y=0,2;\ x=0,-2;\ z=0,-4$; Stationary points are $(0,0,0)$ and $(-2,2,-4)$ | E1 | Some working required for $-4$ |
# Question 2 (part iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| For $x = -2+h$, $y = 2+k$, $z = \lambda$ | | |
| $\lambda = 3(-2+h)^2 + 6(-2+h)(2+k) + (2+k)^3$ | M1, A1 | Substitution |
| $= 12 - 12h + 3h^2 + 6hk + 12h - 12k - 24 + 8 + 12k + 6k^2 + k^3$ | | |
| $= -4 + 3h^2 + 6hk + 6k^2 + k^3$ | | |
| $= -4 + 3(h+k)^2 + k^3 + 3k^2$ | | |
| $= -4 + 3(h+k)^2 + k^2(k+3)$ | E1 | |
| Since $3(h+k)^2 > 0$ and $k^2(k+3) > 0$ for small $k$ | M1 | M0 for numerical work |
| $\lambda > -4$ for all small values of $h$ and $k$, so P is a minimum | E1 | Must mention small $k$ or $k > -3$ |
---
# Question 2 (part iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| For small $x$ and $y$, $z$ can be positive or negative | M1 | Numerical demonstration can earn M1A0E0 |
| If $x = 0$ and $y > 0$, then $z > 0$ | A1 | Correct argument which applies arbitrarily close to O; When $x=0$, $z=y^3$ which has a point of inflection |
| If $x = 0$ and $y < 0$, then $z < 0$ | | |
| Hence O is neither a maximum nor a minimum | E1 | Correctly shown |
---
# Question 2 (part v):
| Answer | Marks | Guidance |
|--------|-------|----------|
| We require $\frac{\partial z}{\partial x} = 18$, $\frac{\partial z}{\partial y} = 18$ | M1 | Allow $-18$ for M1 |
| $6x + 6y = 18$, $6x + 3y^2 = 18$ | | |
| $2(3-y) + y^2 = 6$ | M1 | Obtaining equation for $y$ or $x$ |
| | A1 | or $2x + (3-x)^2 = 6$ |
| Points are $(3, 0, 27)$ and $(1, 2, 23)$ | M1, A1 | Obtaining values of $x$, $y$, $z$ |
| $18x + 18y - z = d$ | M1 | |
| So $d = 27, 31$ | A1 | |
---2 A surface, S , has equation $z = 3 x ^ { 2 } + 6 x y + y ^ { 3 }$.\\
(i) Find the equation of the section where $y = 1$ in the form $z = \mathrm { f } ( x )$. Sketch this section.
Find in three-dimensional vector form the equation of the line of symmetry of this section.\\
(ii) Show that there are two stationary points on S , at $\mathrm { O } ( 0,0,0 )$ and at $\mathrm { P } ( - 2,2 , - 4 )$.\\
(iii) Given that the point ( $- 2 + h , 2 + k , \lambda$ ) lies on the surface, show that
$$\lambda = - 4 + 3 ( h + k ) ^ { 2 } + k ^ { 2 } ( k + 3 ) .$$
By considering small values of $h$ and $k$, deduce that there is a local minimum at P .\\
(iv) By considering small values of $x$ and $y$, show that the stationary point at O is neither a maximum nor a minimum.\\
(v) Given that $18 x + 18 y - z = d$ is a tangent plane to S , find the two possible values of $d$.
\hfill \mbox{\textit{OCR MEI FP3 2016 Q2 [24]}}