| Exam Board | OCR MEI |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2015 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Stationary points of surface (multivariable) |
| Difficulty | Challenging +1.2 This is a multi-part FP3 question on stationary points of surfaces requiring partial derivatives, second derivative test alternatives, and tangent planes. While it involves multiple techniques and is from Further Maths (inherently harder), the question is highly structured with clear guidance ('show that', 'deduce'), and each part follows standard procedures. The novel aspect is part (ii)(C) requiring investigation via cross-sections, but this is explicitly prompted. Overall, moderately above average difficulty due to Further Maths content and length, but not requiring exceptional insight. |
| Spec | 8.05a 3D surfaces: z = f(x,y) and implicit form, partial derivatives8.05b Extended surfaces: more than two variables, trig/log/exp functions8.05e 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/Working | Mark | Guidance |
| \(z=3x^2-12xy+2y^3+60\) | M1 | For finding both partials and setting \(=0\) |
| \(\dfrac{\partial z}{\partial x}=6x-12y=0\Rightarrow x=2y\) | A1 | Both |
| \(\dfrac{\partial z}{\partial y}=-12x+6y^2=0\Rightarrow y^2=2x=4y\) | ||
| \(\Rightarrow y=0\) or \(4\), \(\Rightarrow x=0\) or \(8\) | M1 | Solving simultaneously to get \(x\) or \(y\) |
| \(\Rightarrow z=60\) or \(-4\) | ||
| Stationary points at \(A(8,4,-4)\) | B1 | Also by verification |
| and \(B(0,0,60)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substitute \(x=8+h\), \(y=4+k\) | M1 | For substitution and expansion |
| \(\Rightarrow z_p=3(8+h)^2-12(8+h)(4+k)+2(4+k)^3+60\) | ||
| \(=-4+3(h^2-4hk+4k^2)+12k^2+2k^3\) | M1 | For splitting \(24k^2\) |
| \(=-4+3(h-2k)^2+2k^2(k+6)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| For all values of \(h\) and \(k\), \((h-2k)^2\geq 0\) | M1 | |
| and \(2k^2(k+6)>0\) providing \(k>-6\) | M1 | Or for small \(k\) |
| So \(z>-4\) for all values of \(x\) and \(y\) close to A | A1 | |
| So is local minimum. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| When \(x=0\), \(z=2y^3+60\) and so either side of \((0,0,60)\) the value of \(z\) will be greater or less than 60. | M1, A1 | For obtaining functions; For pt of inflexion - can be by sketch |
| When \(y=0\), \(z=3x^2+60\) and so either side of \((0,0,60)\) the value of \(z\) will always be greater than 60. | A1 | For minimum - can be by sketch |
| So B is a saddle point. | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| At \((1,1,53)\): | M1 | Finding values of derivatives |
| \(\dfrac{\partial z}{\partial x}=6x-12y=-6\) | A1 | |
| \(\dfrac{\partial z}{\partial y}=-12x+6y^2=-6\) | ||
| Equation of tangent plane: \(z-53=\dfrac{\partial z}{\partial x}(x-1)+\dfrac{\partial z}{\partial y}(y-1)\) | M1 | Eqn of plane |
| \(z-53=-6x-6y+12\) | ||
| \(\Rightarrow 6x+6y+z=65\) | A1 | ag |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\dfrac{\partial z}{\partial x}=-6\) and \(\dfrac{\partial z}{\partial y}=-6\) | M1 | Put partial derivatives \(=-6\) |
| \(\Rightarrow x=2y-1\) and \(y^2=2x-1\) | A1 | Both correct |
| \(\Rightarrow y^2-4y+3=0\) | M1 | Solve simultaneously |
| \(\Rightarrow y=1,3\) | A1 | Both values |
| \(\Rightarrow\) Coordinates of R \((5,3,9)\) | A1 |
# Question 2:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $z=3x^2-12xy+2y^3+60$ | M1 | For finding both partials and setting $=0$ |
| $\dfrac{\partial z}{\partial x}=6x-12y=0\Rightarrow x=2y$ | A1 | Both |
| $\dfrac{\partial z}{\partial y}=-12x+6y^2=0\Rightarrow y^2=2x=4y$ | | |
| $\Rightarrow y=0$ or $4$, $\Rightarrow x=0$ or $8$ | M1 | Solving simultaneously to get $x$ or $y$ |
| $\Rightarrow z=60$ or $-4$ | | |
| Stationary points at $A(8,4,-4)$ | B1 | Also by verification |
| and $B(0,0,60)$ | A1 | |
## Part (ii)(A)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitute $x=8+h$, $y=4+k$ | M1 | For substitution and expansion |
| $\Rightarrow z_p=3(8+h)^2-12(8+h)(4+k)+2(4+k)^3+60$ | | |
| $=-4+3(h^2-4hk+4k^2)+12k^2+2k^3$ | M1 | For splitting $24k^2$ |
| $=-4+3(h-2k)^2+2k^2(k+6)$ | A1 | |
## Part (ii)(B)
| Answer/Working | Mark | Guidance |
|---|---|---|
| For all values of $h$ and $k$, $(h-2k)^2\geq 0$ | M1 | |
| and $2k^2(k+6)>0$ providing $k>-6$ | M1 | Or for small $k$ |
| So $z>-4$ for all values of $x$ and $y$ close to A | A1 | |
| So is local minimum. | | |
## Part (ii)(C)
| Answer/Working | Mark | Guidance |
|---|---|---|
| When $x=0$, $z=2y^3+60$ and so either side of $(0,0,60)$ the value of $z$ will be greater or less than 60. | M1, A1 | For obtaining functions; For pt of inflexion - can be by sketch |
| When $y=0$, $z=3x^2+60$ and so either side of $(0,0,60)$ the value of $z$ will always be greater than 60. | A1 | For minimum - can be by sketch |
| So B is a saddle point. | A1 | |
## Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| At $(1,1,53)$: | M1 | Finding values of derivatives |
| $\dfrac{\partial z}{\partial x}=6x-12y=-6$ | A1 | |
| $\dfrac{\partial z}{\partial y}=-12x+6y^2=-6$ | | |
| Equation of tangent plane: $z-53=\dfrac{\partial z}{\partial x}(x-1)+\dfrac{\partial z}{\partial y}(y-1)$ | M1 | Eqn of plane | Or $-6x-6y-z=c$ |
| $z-53=-6x-6y+12$ | | |
| $\Rightarrow 6x+6y+z=65$ | A1 | ag |
## Part (iv)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{\partial z}{\partial x}=-6$ and $\dfrac{\partial z}{\partial y}=-6$ | M1 | Put partial derivatives $=-6$ |
| $\Rightarrow x=2y-1$ and $y^2=2x-1$ | A1 | Both correct |
| $\Rightarrow y^2-4y+3=0$ | M1 | Solve simultaneously |
| $\Rightarrow y=1,3$ | A1 | Both values |
| $\Rightarrow$ Coordinates of R $(5,3,9)$ | A1 | |2 A surface has equation $z = 3 x ^ { 2 } - 12 x y + 2 y ^ { 3 } + 60$.
\begin{enumerate}[label=(\roman*)]
\item Show that the point $\mathrm { A } ( 8,4 , - 4 )$ is a stationary point on the surface. Find the coordinates of the other stationary point, B , on this surface.
\item A point P with coordinates $( 8 + h , 4 + k , p )$ lies on the surface.\\
(A) Show that $p = - 4 + 3 ( h - 2 k ) ^ { 2 } + 2 k ^ { 2 } ( 6 + k )$.\\
(B) Deduce that the stationary point A is a local minimum.\\
(C) By considering sections of the surface near to B in each of the planes $x = 0$ and $y = 0$, investigate the nature of the stationary point B .
\item The point Q with coordinates $( 1,1,53 )$ lies on the surface.
Show that the equation of the tangent plane at Q is
$$6 x + 6 y + z = 65$$
\item The tangent plane at the point R has equation $6 x + 6 y + z = \lambda$ where $\lambda \neq 65$.
Find the coordinates of R .
\end{enumerate}
\hfill \mbox{\textit{OCR MEI FP3 2015 Q2 [24]}}