MEI Conference 2017

MEI Extra Pure:

Multivariable Calculus

Claire Baldwin claire.baldwin@mei.org.uk

Surfaces of the form 𝑧 = 𝑓(𝑥, 𝑦)

Can you visualise these surfaces?

𝑧 = 𝑥2

𝑧 = 𝑦3

𝑧 = cos 𝑥 + sin 𝑦

Use GeoGebra to illustrate each of the surfaces.

HOW TO GUIDE

Open a new window / GeoGebra file

Click View → 3D graphics (close the Graphics window)

Type the function into the input bar in the form 𝑓(𝑥, 𝑦) = 𝑥2

For powers use ^

Click the toggle next to ‘3D graphics’ to get the toolbar of options underneath.

Click on the arrow next to the last small icon on the toolbar and select

‘projection for glasses’ to view the surface in 3D

Click on the ‘rotate 3D graphics’ button on the main toolbar then click in

the 3D graphics window and rotate the window to see the surface from

different perspectives.

Click on the down arrow on the same button and click ‘zoom out’. Then click

in the middle of the screen to zoom out – this allows you to see more of the

surface.

Note: In the latest version of GeoGebra instead of being along the top of the screen

the menus are accessible by clicking on the icon at the top right of the screen.

Select the 3D graphics windows and the input bar and deselect the 2D graphics

window.

To see the view in 3D click on at the top right of the screen and then to find

the ‘glasses’ tool.

Contours and sections

Create the section 𝑥 = 0 for the surface 𝑧 = cos 𝑥 + sin 𝑦. Vary this to sections of the

form 𝑥 = 𝑘 and 𝑦 = 𝑙.

HOW TO GUIDE

Type 𝑥 = 0 into the input bar

Right click on the plane and select ‘object properties’. Under colour drag the

‘opacity’ to 100 (you can also change the default colour if you wish). Close

the window.

Right click on the surface and select ‘object properties’. Under colour drag

the opacity to around 25 (again, change the colour if you wish).

Rotate the graphics window as before to see the image from different

perspectives. You should be able to clearly see the section formed where the

plane cuts the surface.

Click View and select Graphics. Resize this window so it is the same size as

the 3D graphics window.

In the graphics window select the slider tool and then click anywhere in

the Graphics window. In the box name the slider 𝑘 and ensure the option

‘number’ is selected. Change the interval to be between (say) -8 and 9 and

click OK. (You can move a slider by right clicking on it and dragging it to

where you want it to be)

In the algebra pane right click on 𝑥 = 0 and in the Object Properties, under

basic, change the equation to 𝑥 = 𝑘.

Drag the slider from left to right to see how the section changes for different

values of 𝑥.

Repeat the process for 𝑦 = 𝑙

What would the contours look like for this surface? Contours have the form 𝑧 = 𝑘

and indicate points that have the same height (𝑧-coordinate).

HOW TO GUIDE

Open a new window (this will be 2D graphics)

Input curves of the form cos 𝑥 + sin 𝑦 = 𝑘 (you will automatically be asked if

you want to create a slider for 𝑘).

Right click on the slider and select ‘animation on’.

Choose a sensible range of values for 𝑘

Finding stationary points and their nature

Question 1 (adapted from legacy FP3 June 2011)

A surface S has equation 𝑧 = 8𝑦3 − 6𝑥2𝑦 − 15𝑥2 + 36𝑥

Sketch the sections of S given by 𝑦 = −3 and 𝑥 = −6. From your sketches deduce

that (-6, -3, -324) is a stationary point on S and state the nature of the stationary

point.

Find the coordinates of the other three stationary points on S

Question 2 (adapted from legacy FP3 June 2015)

A surface has equation 𝑧 = 3𝑥2 − 12𝑥𝑦 + 2𝑦3 + 60

Show that the point A(8,4, −4) is a stationary point on the surface and find the

coordinates of the other stationary point B.

A point P with coordinates (8 + ℎ, 4 + 𝑘, 𝑝) lies on the surface. Show that

𝑝 = −4 + 3(ℎ − 2𝑘)2 + 2𝑘2(6 + 𝑘)

and deduce that the stationary point A is a local minimum.

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.

Normal line and tangent plane

PREREQUISITE KNOWLEDGE

The equation of a plane can be written in the vector form 𝒓 = 𝒂 + 𝜇𝒃 + 𝜆𝒄 where 𝜆

and 𝜇 are constants, 𝒂 is the position vector of a point on the plane and 𝒃 and 𝒄 are

two non-parallel vectors in the plane. .

A plane can also be written in the cartesian form 𝑎𝑥 + 𝑏𝑦 + 𝑐𝑥 + 𝑑 = 0.

The vector equation of a line can be written in the form 𝒓 = 𝒂 + 𝜆𝒅 where 𝜆 is a

constant, 𝒂 is the position vector of a point on the line and 𝒅 is the direction vector of

the line.

The vector product finds a vector which is perpendicular to two given vectors (

𝑎1

𝑎2

𝑎3

)

and (

𝑏1

𝑏2

𝑏3

) and is given by (

𝑎1

𝑎2

𝑎3

) × (

𝑏1

𝑏2

𝑏3

) = (

𝑎2𝑏3 − 𝑎3𝑏2

𝑎3𝑏1 − 𝑎1𝑏3

𝑎1𝑏2 − 𝑎2𝑏1

)

Use GeoGebra to construct a 3D graphic of the surface 𝑧 = 𝑥2 + 𝑦2 showing the

tangent plane and normal plane at the point A(1,2,5) and the direction vectors

parallel to the x-axis and the y-axis at that point.

HOW TO GUIDE

Click View – 3D graphics to open a 3D graphic window and input the function

𝑧 = 𝑥2 + 𝑦2 by typing 𝑓(𝑥, 𝑦) = 𝑥2 + 𝑦2

In the Graphics window click on the second button from the left (point) and

click anywhere on the 𝑥 axis to create the point A. Click anywhere on the 𝑦

axis to create a point B.

In the input bar enter 𝑎 = x(A) then enter 𝑏 = y(B). This creates two numerical

quantities 𝑎 and 𝑏 associated with the 𝑥 and 𝑦 coordinates of A and B

respectively.

Input the point (𝑎, 𝑏) – this creates point C in the 𝑥𝑦 plane

Input the point (a, b, 𝑓(𝑎, 𝑏)) – this creates the D point in 3D

You should now be able to drag the points A and B in the Graphics window

and see the effect in 3D in the 3D graphics window

Input 𝑓_𝑥(𝑥, 𝑦) = derivative [𝑓(𝑥, 𝑦), 𝑥] and 𝑓_𝑦(𝑥, 𝑦) = derivative [𝑓(𝑥, 𝑦), 𝑦] –

this will produce graphs which show the value of the partial derivatives 𝜕𝑓

𝜕𝑥 and

𝜕𝑓

𝜕𝑦. Hide these graphs by clicking on the circular button next to their equations

in the algebra pane. (continued)

Input vector[D, D+(1, 0, 𝑓_𝑥(𝑎, 𝑏))]. This will default to have the name 𝑢 and

represents a vector in the direction of the section parallel to the 𝑥 axis.

Input vector[D, D+(0, 1, 𝑓_𝑦(𝑎, 𝑏))]. This will default to have the name 𝑣 and

represents a vector in the direction of the section parallel to the 𝑦 axis.

If you wish, right click on the vectors 𝑢 and 𝑣 and go to object properties to

change the colour / thickness of each of the vectors. Move points A and B in

the graphics window to see how 𝑢 and 𝑣 behave.

Input 𝑛=vector[D, D+cross(𝑢, 𝑣)]. This creates a vector perpendicular to 𝑢

and 𝑣 which is the normal vector to the tangent plane at that point. You can

leave this visible or hide it, whichever you prefer.

Input line[D,𝑛] – this creates the normal line at the point D.

Input perpendicular plane[D,𝑛] to create the tangent plane at D.

Again move points A and B in the graphics window to see how the vectors

and plane behave. You can use object properties to change the colour or

opacity of the surface and the plane if this helps to visualise things more

clearly.

Mark Schemes for stationary points exam questions

Question 1 (adapted from legacy FP3 June 2011)

Question 2 (adapted from legacy FP3 June 2015)

A little trip ‘off spec’

The specification does not require students to know any rules for determining the

nature of stationary points, but they might be interested to explore a rule and learn

about second partial derivatives at the same time.

A simple test involves finding 𝐷 =𝜕2𝑓

𝜕𝑥2×

𝜕2𝑓

𝜕𝑦2− (

𝜕2𝑓

𝜕𝑥𝜕𝑦)

2

Step 1: At each stationary point work out the values of the second partial derivatives

Step 2: Calculate D at each stationary point

Step 3: If D < 0 the stationary point is a saddle point

If D > 0 and 𝜕2𝑓

𝜕𝑥2 > 0 the stationary point is a (local) minimum

If D > 0 and 𝜕2𝑓

𝜕𝑥2< 0 the stationary point is a (local) maximum

If D = 0 an alternative method is needed

Explore the stationary points on the surfaces:

𝑧 = 8𝑥2 + 6𝑦2 − 2𝑦3 + 5

𝑧 = 2𝑥2 + 𝑦2 + 3𝑥𝑦 − 3𝑦 − 5𝑥 + 8

𝑧 = 𝑥3 − 3𝑥2𝑦

How is this guy involved with one of them?

MEI Extra Pure:

Multivariable

calculus

Claire Baldwin

FMSP Central Coordinator

claire.baldwin@mei.org.uk

MEI Further Mathematics A levelAssessment Overview

Mandatory unit:

Core Pure

144 raw marks

2 hrs 40 mins50% of A level

Major options:

Mechanics Major

Statistics Major

120 raw marks

2 hrs 15 mins33⅓% of A level

Minor options:

Mechanics Minor

Statistics Minor

Modelling with Algorithms

Numerical Methods

Extra Pure

Further Pure with Technology

60 raw marks

1 hr 15 mins

(1 hr 45 mins for FPT)

16⅔% of A level

Extra Pure is not one of the units that is designated as being co-teachable

with AS Further Mathematics or AS Mathematics

Modular MEI specification• Multivariable calculus is currently on the Further

Applications of Advanced Mathematics (FP3) module

• This is a 1½ hour examination where candidates choose

3 questions from 5, worth 24 marks each.

Option 1: Vectors

Option 2: Multi-variable calculus

Option 3: Differential Geometry

Option 4: Groups

Option 5: Markov Chains

• The content of the multivariable calculus section is

essentially the same in the new specification

Linear MEI specification• Multivariable calculus is in the Extra Pure minor option

along with Recurrence Relations, Matrices and Groups.

• There are no optional questions – candidates must

answer all the questions in the printed answer booklet.

• The four topics may not be evenly weighted in the

assessment e.g. on the sample assessment materials:

Q1 (10 marks) and Q2 (4 marks) – Groups

Q3 (12 marks) – Recurrence Relations

Q4 (16 marks) – Multivariable calculus

Q5 (18 marks) – Matrices

Total: 60 marks, 1 hour 15 mins

Key ideas

normal line

and tangent

plane

partial

derivatives

contours

and

sections

Any hikers in the room??

But first…… a quiz!!

http://www.rgs.org/webcasts/activities/contours/co

ntours.html

or

http://bit.ly/2tcN0yc

Applications of contour diagrams

Partial differentiation

Partial differentiation

Partial differentiation

Partial differentiation

Partial differentiation

Partial differentiation

Partial differentiation

𝜕

𝜕𝑦𝑦𝑒𝑥𝑦 = 𝑒𝑥𝑦 + 𝑥𝑦𝑒𝑥𝑦

Stationary points on surfacesIn two dimensions a stationary point can be:

a maximum a minimum a point of inflexion

What might a stationary point look like on a three

dimensional surface?

Stationary points on surfaces

Maximum

point

Minimum

point

Two types of

saddle point

How to find stationary points on surfaces

The method will be given if investigation of the nature of

the stationary points is required.

Explore the location and nature of the stationary points on

the surfaces 𝑧 = 𝑒−(𝑥2+𝑦2) and 𝑧 = 𝑥𝑦 𝑥 + 𝑦 − 1

Check your answers using GeoGebra

Explore the location and nature of the stationary points on the

surfaces𝑧 = 𝑒−(𝑥2+𝑦2) and 𝑧 = 𝑥𝑦 𝑥 + 𝑦 − 1

Check your answers using GeoGebra

Acknowledgements• Royal Geographical Society quiz

http://www.rgs.org/webcasts/activities/contours/contours.html

• Seabed image

http://www.infomar.ie/surveying/Bays/BayProfileImages/Dingle/BaysDingleBY1200.jpg

• Isobars image http://www.metoffice.gov.uk/public/weather/surface-

pressure/#?tab=surfacePressureColour&fcTime=1494849600

• Rainfall map

http://www.lordgrey.org.uk/~f014/usefulresources/aric/Resources/Teaching_Packs/Key_Sta

ge_4/Weather_Climate/11.html

• EEG image http://www.neurotherapytacoma.com/resources/brain-at-work

• Corneal topography image https://entokey.com/wp-content/uploads/2017/01/gr2-691.jpg

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