# Extension 2 Mathematics

## Notes about these Notes

These notes are from the 2008 NSW HSC course so some parts may not be relevant to the new courseTitle: 16 October 2007 Ext 2 Maths

Date: 16 October 2007 7:14 PM

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Mod-Arg Form.

Modulus = r

Arguement = θ

The Cartesian plane for Complex numbers follows a square grid. There is a polar grid for complex numbers

- cisθ = cosθ + i sinθ
- On the polar grid the first number is the radius then the turn (Modulus and Argument)

Title: 22 October 2007 Ext 2 Maths

Date: 22 October 2007 8:30 PM

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Converting Cartesian to Mod-Arg and back again.

__z, z bar, cz, iz In mod-arg form__

z = a complex number.

z bar (line on top of the z meaning conjugate) = z but mirrored in the Real axis.

cz (constant x z) = an extension of r, the point has a bigger radius but the same turn.

iz = z rotated 90˚ anticlockwise

__Multiplication in mod-arg form__

Let z1 = rcisθ

Let z2 = rcisØ

Find z1z2

…r1r2cis(θ+Ø)

We add arguments and multiply moduli

(rcisθ)n = rncis(nθ)

Title: 23 October 2007 Ext 2 Maths

Date: 23 October 2007 6:14 PM

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Consider

z = cisθ

then z3 = cis3θ

But z3 = (cosθ + isinθ)3

*expand using binomial theorem*

Turn the expression you get into a complex number (Cartesian form) and substitute where the cos and sin were.

__Vectors__

A complex number in mod-arg form is like a physics vector. You can add them like physics. To subtract one you flip it about the x axis then add.

Title: 30 October 2007 Ext 2 Maths

Date: 30 October 2007 6:11 PM

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__Triangle Inequality__

|z1 + z2| ≤ |z1| + |z2|

This proves the shortest way to anything is a straight line.

More work on vectors.

In a square where x is a side and y is the diagonal:

x = _

__1____

√2 y

y = √2 x

IN CASE YOU MISSED IT:

|z| = the modulus/radius of the complex

arg(z) = the argument/angle of the complex number

Title: 13 November 2007 Ext 2 Maths

Date: 13 November 2007 7:18 PM

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This is very long.

An ellipse: The locus of a point where the sum of the distances to two fixed points is constant.

Sketch locus of

|z-2-2i| + |z-5-2i| = 5

Therefore the two foci are at (2+2i) and (5 + 2i) and the distance to the points is added up to be 5.

____________________________________________________________________

|z1|/|z2| = |z1/z2|

|z1z2| = |z1| |z2|

|a+ib| = √a2+b2

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__Regions__

Sketch the region on the Argand Diagram where |z| ≥ 1 & 0 ≤ argz ≤ π/3

__De Moivre's Theorem__

zn = (rcosθ + risinθ)n

= cis(nθ) for r=1

n can be real

The theorem holds for n<0

__Roots of unity__

z = cosø + isinø

= cos (ø + 2kπ) + isin (ø + 2kπ)

z(1/n) = cos ((ø+2kπ)/(n)) + isin ((ø+2kπ)/(n))

Now you can find all the n roots of 1 including complex ones.

Roots of the cube root of 1 are 1, ω, ω2

All roots of ±1 lie on a circle with radius 1 and they form an n sided polygon.

When finding the roots of +1 it will always have a root lying on the real axis and the roots of -1 will never be on the real axis.

Title: 5 February 2008 Ext 2 Maths

Date: 5 February 2008 3:06 PM

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y = b + a sin (nx + ø)

b moves the graph vertically

a changes the amplitude of the graph

The period of the graph is:

__2π__

n

(this squishes or lengthens the graph)

ø is the phase shift of the graph by:

__ø__

n

if - then moves y axis left

is +then moves y axis right.

Graphs of the form y=f(x)+c

c moves the x axis up or down.

When adding two functions you simply add all the y values together to get a new graph.

Title: 8 February 2008 Ext 2 Math

Date: 8 February 2008 3:09 PM

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y=c.f(x)

changes amplitude

y=|f(x)|

anything bellow the x axis is reflected up in the x axis

y=f(|x|)

CLONES and reflects the 1 and 4 quadrants into the 3 and 2 quadrants. i.e. the 2 quadrant will look the same as the 1st but reflected in the y axis. Makes an even function. The 2 and 3 quadrents are overwritten or ignored.

y= -f(x)

reflection of entire function in x-axis

y=f(-x)

reflection of entire function in y-axis

Title: 11 February 2008 Ext 2 Maths

Date: 11 February 2008 5:58 PM

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f(x).g(x)

y=xsinx

It looks like sin x but the crest and trough always hits the line y=x

1/f(x)

zeros of f(x) will give an asymptote.

If f(x) approaches infintiy

function approaches 0

Guess what happens when f(x) approaches zero.

f(x)/g(x)

Title: 13 February 2008 Ext 2 Maths

Date: 13 February 2008 6:01 PM

Category: School

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y = g(x)/f(x)

Horizontal asymptotes when degree of g(x) is equal to the degree of f(x) (deg[g(x)] = deg[f(x)])

Oblique asymptotes if deg[g(x)] > deg[f(x)] Try a division to see why

y = sqroot(f(x))

Exists only for f(x) ≥ 0

y2 = f(x)

The graph of y = root(f(x)) but also reflected in the x axis

y = [f(x)]n

The zeros of f(x) are stationary points of [f(x)]n

You can show this by diffrentiating

__Implicit differentiation__

Differentiate xy = 1

(Use the chain rule ie. vu’ + uv’ rule)

u’ = 1 v’ = y’

xy’ + y = 0

therefore y’ = -y/x

Implicit differentiation is like normal differentiation but for non-functions.

y = ef(x)

GASP!

Where f(x) = 0 then ef(x) = 1

Title: 21 February 2008 Ext 2 Maths

Date: 21 February 2008 6:50 PM

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__POLYNOMIALS__

1. If a polynomial is of degree n it will have at most n real roots and exactly n complex roots (remember complex roots can also be real) Complex roots occur in conjugate pairs if all coefficients are real (unconfirmed).

2. If the degree of P(x) is odd then P(x) has at least one real root. This can be factored into one linear factor and some quadratic factors.

3. If a polynomial P(x) degree is even then it can be factored into a series of quadratics.

4. If a polynomial P(x) has a root α of multiplicity n then (x - α)n is a factor and (x - α)n-1 is a factor of the derivative. You can prove this through the derivative of P(x) = (x - α)n . Q(x)

Title: 22 February 2008 Ext 2 Maths

Date: 22 February 2008 7:13 PM

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You will be asked this question:

Polynomial P(x)=...

It has roots α, β, γ

Find the equation that has the opposite OR reciprocal OR squares of these roots.

To do this: say m is the roots of the new equation. If they ask you to find the equation whose roots are the squares of the original roots then

m=x2

therefore √m=x

now substitute √m into the original polynomial equal to zero. Solve. Present as a nice polynomial.

Title: 27 February 2008 Ext 2 Maths

Date: 27 February 2008 2:53 PM

Category: School

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Use De Moivre’s Theorem to express cos4ø in terms of cosø

Use the result to solve

8x4 = 8x2 + 1 = 0

Hence show that cosπ/8 + cos3π/8 + cos 5π/8 + cos7π/8 = 0

De Moivre’s Theorem is that one that says (rcisø)n = rncisnø

So to do this question:

(cosø + isinø)4 = *expand with binomial thorem*

You will get a bunch of crap that you can separate into real part and imaginary part. Completely ignore the imaginary part and just use the real part to equal cos4ø

Change the sin2ø into 1-cos2ø and rearrange to make the entire thing look like 8x4 = 8x2 + 1 = 0 but where x = cosø

so cos4ø = 0

4ø = π/2, 3π2 etc.

ø = π/8, 3π/8, 5π/5, 7π/8

so the roots of the equation 8x4 = 8x2 + 1 = 0 are cosπ/8, cos3π/8, cos5π/5, cos7π/8

so all roots added together = -b/a = 0

__Partial Fractions__

Express (5x+1)/(x-1)(x+2) in the form A/(x-1) + B/(x+2)

If you get that second bit of the question and try to turn it into the first bit of the question you get A(x+2) + B(x-1) on top. That should equal 5x+1 right? Now simultaneous equations.

There is the possibility that A and B could be polynomials so sometimes you will have to write them as Ax + C for example.

If the degree of the top polynomial is larger than the bottom polynomial then do a polynomial division before splitting into fractions.

Title: 12 March 2008 Ext 2 Maths

Date: 12 March 2008 3:30 PM

Category: School

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__Ellipse__

Definition 1

The locus of a point P such that the sum of the distances from two other fixed points is always the same. The fixed points are called foci (singular: focus)

Definition 2

The ratio of distance from P to a focus to distance from P to the directrix is always the same. And less than 1.

Say the foci are named F and F’ and R and R’ is a point on the directrix such that PR is perpendicular to the directrix then

Then

__PF__=

__PF’__= eccentricity (e)

PR PR’

e is a measure of the deviation from a circle. (e = 0)

The equation of an ellipse is

x2/a2 + y2/b2 = 1

Chords are lines that touch two points on the ellipse.

Focal chords are chords through one of the foci.

A diameter chord runs through the center of the ellipse.

The semi-major axis is the axis that runs through the long side of the ellipse, usually the x-axis

The semi-minor axis is the axis that runs through the short side.

The coordinates of the foci are ae and -ae

The ellipse cuts the x-axis at a and -a and the y-axis at b and -b

The equation of the directrices is a/e and -a/e

b2/a2 = 1 - e2

Title: 13 March 2008 Ext 2 Maths

Date: 13 March 2008 12:16 PM

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The equation of a tangent to a ellipse is given by

__xx__1 +

__yy__1 = 1

a2 b2

Use implicit differentiation to do this.

Relationships:

b2x2 + a2y2 = a2b2

a2 - b2 = a2e2

The equation of a normal to a point on an ellipse is:

__a__2

__x__-

__b__2

__y__= a2 - b2

x1 y1

__Parametric Definition__

__Reflective property of the ellipse__

If you stood on the focus in an elliptical room the a person could hear everything you say very clearly if they are standing on the other focus.

Title: 30 April 2008 Ext 2 Maths

Date: 30 April 2008 6:05 PM

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The Chord of Contact for an ellipse from an external point:

__xx__1 +

__yy__1 = 1

a2 b2

The equation for a normal tangent is the same.

The equation for the tangent and chord of contact to a hyperbola is the same but with a minus sign instead of a plus.

__Hyperbola__

Locus: The difference of the distances to two fixed points is equal

The fixed points are the foci.

The ratio of the distance of a point P to the focus to the distance to the directrix is > 1

This ratio is the eccentricity (e)

Foci: (±ae, 0)

Directrices: x = ±a/e

Asymptotes: y = (±b/a)x

b2/a2 = e2 - 1

Title: 1 May 2008 Ext 2 Maths

Date: 1 May 2008 8:22 PM

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A rectangular hyperbola is one where the asymptotes are perpendicular. y = 1/x is one of them.

conjugate axis is the one with the b’s on it while the transverse axis has the a’s on it. Diameter goes through the origin

A conjugate hyperbola is one that occurs on the other sides of the asymptotes and just means the x2/a2 is swapped with the y2/b2

Parametric definition of a hyperbola: x = asecø y = btanø

Title: 16 May 2008 Ext 2 Maths

Date: 16 May 2008 1:38 PM

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INTEGRATION

You can split the fraction into multiple fractions. Try to manipulate the fraction to make the bottom resemble a standard integral or the top the derivative of the bottom.

Title: 28 May 2008 Ext 2 Maths

Date: 28 May 2008 3:11 PM

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INTEGRATION BY PARTS

∫u dv = uv - ∫v du

It’s a very useful formula

Title: 29 May 2008 Ext 2 Maths

Date: 29 May 2008 3:13 PM

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Recurrence Integrals

e.g. ∫cosnx dx

Split into cosx . cosn-1x

Use integration by parts

Why not use the t results when integrating trigonometric functions! Hours of fun!

Title: 5 June 2008 Ext 2 Maths

Date: 5 June 2008 3:10 PM

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VOLUMES!

I have a circle. It is of radius a around the origin.

This isn’t going to work.

Title: 12 June 2008 Ext 2 Maths

Date: 12 June 2008 8:23 PM

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Volumes by circular cross-sections

Want to spin the curve y = 2√x around y=4? Here’s how.

Taking vertical slices:

radius of slices: 4 - x

area: π(4-x)2

volume: π(4-x)2∆y (∆y being a tiny height of each slice)

Integrate!

When integrating, make sure you either change the x into a y or dy into dx somehow.

Title: 13 June 2008 Ext 2 Maths

Date: 13 June 2008 11:53 AM

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The curve of y=cosx is rotated around the line y =1 from x = 0 to π/2

Find the volume generated.

Answer is 3π2/4 - 2π

y = √(x-1) rotated around x=0 from x=0 to 5

You have to find the inner/outer radius/area/volume then take one away from the other.

Washers:

y=(1-x)x

Volume of the washer is π x22∆y = π x12∆y

V = ∫x22 - x12 dy (from 0 to 1/4 (I cant write it properly))

but y = x-x2

x2-x+y=0

use x1 and x2 as the roots to this with the whole alpha beta = c/a

Answer is π/6

Title: 25 June 2008 Ext 2 Maths

Date: 25 June 2008 11:59 AM

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Theorem of Pappus.

If a symmetrical area is rotated about a line outside the area then the volume generated is given by

Area times distance traversed by the axis of symmetry.

METHOD OF CYLINDRICAL SHELLS

General formula:

2π∫xy dx

Don’t be fooled. x is actually the radius of each shell. So if you are rotating around something other than x=0 then x will not be x but the radius (eg maybe 1-x)

Title: 30 July 2008 Ext 2 Maths

Date: 30 July 2008 12:09 PM

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Banked tracks.

N is sideways, mg is straight down and sideways is provided by mv2/r

Vertical component of N counterbalances mg therefore is the track is banked at ø then Nsinø = mv2/r and Ncosø = mg

Title: 31 July 2008 Ext 2 Maths

Date: 31 July 2008 12:11 PM

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Resolving forces on banked tracks could involve multiplying by sin or cos or squaring everything.

Then add the two equations you have together and solve.

Title: 6 August 2008 Ext 2 Maths

Date: 6 August 2008 12:59 PM

Category: School

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Resisted motion!

Upwards:

When shot out of a canon upwards: forces acting on a particle are -mg and -mkv

So:

(F=ma)

a = - g - kv

Change the a to a v(dv/dx) and integrate to get displacement. Remember to find the constant of integration.

Title: 7 August 2008 Ext 2 Maths

Date: 7 August 2008 1:01 PM

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Motion downwards (resisted):

A particle is dropped from a height.

Forces acting with: g

Forces acting against: -kv

So:

a = g - kv

You can find terminal velocity (which is g/k) by integrating the above equation if a = dv/dt

Once you have t (with the constant figured out) you could say as t approaches infinity v approaches ...

Or isolate the v and make t = infinity.