Question 2 - A-Level Maths

OCR MEI FP3 2011 June — Question 2 24 marks

Exam Board	OCR MEI
Module	FP3 (Further Pure Mathematics 3)
Year	2011
Session	June
Marks	24
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Vector Product and Surfaces
Type	Tangent plane with given equation
Difficulty	Challenging +1.8 This is a substantial Further Maths question requiring partial differentiation, stationary point analysis in 3D, and finding tangent planes with a given normal direction. Part (iv) is particularly demanding as it requires setting up the tangent plane condition ∂z/∂x = 120, ∂z/∂y = 0 simultaneously, solving the resulting system, and verifying exactly two solutions exist. The multi-step reasoning and algebraic manipulation place this well above average difficulty.
Spec	8.05a 3D surfaces: z = f(x,y) and implicit form, partial derivatives 8.05d Partial differentiation: first and second order, mixed derivatives 8.05e Stationary points: where partial derivatives are zero 8.05f Nature of stationary points: classify using Hessian matrix 8.05g Tangent planes: equation at a given point on surface

2 A surface $S$ has equation $z = 8 y ^ { 3 } - 6 x ^ { 2 } y - 15 x ^ { 2 } + 36 x$.

Sketch the section of $S$ given by $y = - 3$, and sketch the section of $S$ given by $x = - 6$. Your sketches should include the coordinates of any stationary points but need not include the coordinates of the points where the sections cross the axes.
From your sketches in part (i), deduce that $( - 6 , - 3 , - 324 )$ is a stationary point on $S$, and state the nature of this stationary point.
Find $\frac { \partial z } { \partial x }$ and $\frac { \partial z } { \partial y }$, and hence find the coordinates of the other three stationary points on $S$.
Show that there are exactly two values of $k$ for which the plane with equation $$120 x - z = k$$ is a tangent plane to $S$, and find these values of $k$.

Show mark scheme Show mark scheme source

2(i)

Answer	Marks	Guidance
When $y = -3$, $z = 3x^2 + 36x - 216$	B1
Graph 1: Parabola, vertex at $(-6, -324)$	B1 B1	Correct shape (parabola) and position; For $(-6, -324)$
When $x = -6$, $z = 8y^3 - 216y - 756$	B1
Graph 2: Correct cubic shape and position	B1 B1	Correct shape and position; For $(-3, -324)$; For $(3, -1188)$; If BOBO then give B1 for $x = \pm 3$ (7 marks)

2(ii)

Answer	Marks	Guidance
$(-6, -3, -324)$ is a SP on both sections; hence it is a SP on $S$	B1 B1	Saddle point (2 marks)

2(iii)

Answer	Marks
$\frac{\partial z}{\partial x} = -12xy - 30x + 36$, $\frac{\partial z}{\partial y} = 24y^2 - 6x^2$	B1B1 M1
At a SP, $\frac{\partial z}{\partial x} = \frac{\partial z}{\partial y} = 0$	M1
$24y^2 - 6x^2 = 0 \Rightarrow y = \pm \frac{1}{4}x$	M1
$y = \frac{1}{4}x \Rightarrow -6x^2 - 30x + 36 = 0 \Rightarrow x = -6, 1$; SPs are $(-6, -3, -324)$, $(1, 0.5, 19)$	M1 A1
$y = -\frac{1}{4}x \Rightarrow 6x^2 - 30x + 36 = 0 \Rightarrow x = 2, 3$; SPs are $(2, -1, 28)$, $(3, -1.5, 27)$	M1 A1 (8 marks)

2(iv)

Answer	Marks	Guidance
$\frac{\partial z}{\partial x} = -120$ and $\frac{\partial z}{\partial y} = 0$	M1	(Allow M1 for $\frac{\partial z}{\partial x} = -120$)
$y = \frac{1}{4}x \Rightarrow -6x^2 - 30x - 84 = 0$; $D = 30^2 - 4 \times 6 \times 84 = -1116 < 0$; so this has no roots	M1 A1
$y = -\frac{1}{4}x \Rightarrow 6x^2 - 30x - 84 = 0 \Rightarrow x = 7, -2$	M1 M1	Obtaining at least one value of $x$; Obtaining a value of $k$
When $x = 7$, $y = -3.5$, $z = 203$; so $k = 637$	A1
When $x = -2$, $y = 1$, $z = -148$; so $k = -92$	A1 (7 marks)

### 2(i)
When $y = -3$, $z = 3x^2 + 36x - 216$ | B1 |

**Graph 1:** Parabola, vertex at $(-6, -324)$ | B1 B1 | Correct shape (parabola) and position; For $(-6, -324)$

When $x = -6$, $z = 8y^3 - 216y - 756$ | B1 |

**Graph 2:** Correct cubic shape and position | B1 B1 | Correct shape and position; For $(-3, -324)$; For $(3, -1188)$; If BOBO then give B1 for $x = \pm 3$ (7 marks)

### 2(ii)
$(-6, -3, -324)$ is a SP on both sections; hence it is a SP on $S$ | B1 B1 | Saddle point (2 marks)

### 2(iii)
$\frac{\partial z}{\partial x} = -12xy - 30x + 36$, $\frac{\partial z}{\partial y} = 24y^2 - 6x^2$ | B1B1 M1 |

At a SP, $\frac{\partial z}{\partial x} = \frac{\partial z}{\partial y} = 0$ | M1 |

$24y^2 - 6x^2 = 0 \Rightarrow y = \pm \frac{1}{4}x$ | M1 |

$y = \frac{1}{4}x \Rightarrow -6x^2 - 30x + 36 = 0 \Rightarrow x = -6, 1$; SPs are $(-6, -3, -324)$, $(1, 0.5, 19)$ | M1 A1 |

$y = -\frac{1}{4}x \Rightarrow 6x^2 - 30x + 36 = 0 \Rightarrow x = 2, 3$; SPs are $(2, -1, 28)$, $(3, -1.5, 27)$ | M1 A1 (8 marks)

### 2(iv)
$\frac{\partial z}{\partial x} = -120$ and $\frac{\partial z}{\partial y} = 0$ | M1 | (Allow M1 for $\frac{\partial z}{\partial x} = -120$)

$y = \frac{1}{4}x \Rightarrow -6x^2 - 30x - 84 = 0$; $D = 30^2 - 4 \times 6 \times 84 = -1116 < 0$; so this has no roots | M1 A1 |

$y = -\frac{1}{4}x \Rightarrow 6x^2 - 30x - 84 = 0 \Rightarrow x = 7, -2$ | M1 M1 | Obtaining at least one value of $x$; Obtaining a value of $k$

When $x = 7$, $y = -3.5$, $z = 203$; so $k = 637$ | A1 |

When $x = -2$, $y = 1$, $z = -148$; so $k = -92$ | A1 (7 marks)

Show LaTeX source

2 A surface $S$ has equation $z = 8 y ^ { 3 } - 6 x ^ { 2 } y - 15 x ^ { 2 } + 36 x$.\\
(i) Sketch the section of $S$ given by $y = - 3$, and sketch the section of $S$ given by $x = - 6$. Your sketches should include the coordinates of any stationary points but need not include the coordinates of the points where the sections cross the axes.\\
(ii) From your sketches in part (i), deduce that $( - 6 , - 3 , - 324 )$ is a stationary point on $S$, and state the nature of this stationary point.\\
(iii) Find $\frac { \partial z } { \partial x }$ and $\frac { \partial z } { \partial y }$, and hence find the coordinates of the other three stationary points on $S$.\\
(iv) Show that there are exactly two values of $k$ for which the plane with equation

$$120 x - z = k$$

is a tangent plane to $S$, and find these values of $k$.

\hfill \mbox{\textit{OCR MEI FP3 2011 Q2 [24]}}

This paper (5 questions)

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Q1 24 Q2 24 Q3 24 Q4 24 Q5 24

Answer	Marks	Guidance
When \(y = -3\), \(z = 3x^2 + 36x - 216\)	B1
Graph 1: Parabola, vertex at \((-6, -324)\)	B1 B1	Correct shape (parabola) and position; For \((-6, -324)\)
When \(x = -6\), \(z = 8y^3 - 216y - 756\)	B1
Graph 2: Correct cubic shape and position	B1 B1	Correct shape and position; For \((-3, -324)\); For \((3, -1188)\); If BOBO then give B1 for \(x = \pm 3\) (7 marks)

Answer	Marks
\(\frac{\partial z}{\partial x} = -12xy - 30x + 36\), \(\frac{\partial z}{\partial y} = 24y^2 - 6x^2\)	B1B1 M1
At a SP, \(\frac{\partial z}{\partial x} = \frac{\partial z}{\partial y} = 0\)	M1
\(24y^2 - 6x^2 = 0 \Rightarrow y = \pm \frac{1}{4}x\)	M1
\(y = \frac{1}{4}x \Rightarrow -6x^2 - 30x + 36 = 0 \Rightarrow x = -6, 1\); SPs are \((-6, -3, -324)\), \((1, 0.5, 19)\)	M1 A1
\(y = -\frac{1}{4}x \Rightarrow 6x^2 - 30x + 36 = 0 \Rightarrow x = 2, 3\); SPs are \((2, -1, 28)\), \((3, -1.5, 27)\)	M1 A1 (8 marks)

Answer	Marks	Guidance
\(\frac{\partial z}{\partial x} = -120\) and \(\frac{\partial z}{\partial y} = 0\)	M1	(Allow M1 for \(\frac{\partial z}{\partial x} = -120\))
\(y = \frac{1}{4}x \Rightarrow -6x^2 - 30x - 84 = 0\); \(D = 30^2 - 4 \times 6 \times 84 = -1116 < 0\); so this has no roots	M1 A1
\(y = -\frac{1}{4}x \Rightarrow 6x^2 - 30x - 84 = 0 \Rightarrow x = 7, -2\)	M1 M1	Obtaining at least one value of \(x\); Obtaining a value of \(k\)
When \(x = 7\), \(y = -3.5\), \(z = 203\); so \(k = 637\)	A1
When \(x = -2\), \(y = 1\), \(z = -148\); so \(k = -92\)	A1 (7 marks)