4u Maths Summary Page 1 of 117
4u Maths Summary
Contents: Graphs 3
Basic Graphs 3 Trigonometric Graphs 7 Addition & Subtraction of Ordinates 11 Reflection in Axis 13 Rules for Graphing 15 Multiplication of Ordinates 20 Division of Ordinates 21 Behaviours at Critical Points 23 Index Rules 24 Implicit Differentiation 26 Inequalities 30
Complex Numbers 32 Arithmetic of Complex Numbers 32 Geometrical Representation 35 Basic Proofs 38 Vectors 41 De Moivre’s Theorem 44 nth Roots 46 Loci of Complex Numbers 47 Regions 49
Conics 51 Ellipse 51 Tangents & Normals 56 Equation of a Chord 56 Chord of Contact 56 Basic Proofs 57 Hyperbola 63 Tangents & Normals 65 Equation of a Chord 68 Chord of Contact 68 Basic Proofs 68 Rectangular Hyperbola 69 Equation of a Chord 71 Tangent 72 Normal 73 Chord of Contact 74 Basic Proofs 74 Loci 76 Conics in Cones 78 Eccentricity Range 78
4u Maths Summary Page 2 of 117
Integration 80 Algebraic Substitutions 80 Trigonometric Integrals 80 Trigonometric Substitutions 81 Integration by Parts 82 Recurrence Formulas 82 Partial Fractions 83
Volumes 84
Volumes Using Limits 85 Volumes by slicing 86 Volumes by shells 87 Volumes with non-circular slices 89
Mechanics 90
Projectile Motion Simple Harmonic Motion Mathematical Descriptions of Motion Upwards Motion Downwards Motion Circular Motion Uniform Circular Motion Conical Pendulum Banked Circular Track
91 93 95 96 97 99
100 101 104
Polynomials 107
Integer Roots of Polynomials Factoring Polynomials Roots & Coefficients of Polynomials Partial Fractions
107 109 111 114
Harder 3 Unit 117
Circle Geometry Induction Inequalities
117 117 117
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SYLLABUS
REFERENCE RELATED INFORMATION
Topic 1 Graphs
Basic Curves
• Graph a linear equation. ax + by + c = 0
y = mx + b
ax + by + c = 0 Gradient = – a
b
y-intercept: y = – cb
y = mx + b Gradient = m y-intercept: y = c
x-10 -5 5 10
y
-10
-5
5
10
• Graph a quadratic
Function. y = ax2 + bx + c
Axis of symmetry: x = – b2a
Roots: ax2 + bx + c = 0 If a > 0, then concave up. If a < 0, then concave down.
x-10 -5 5 10
y
-10
-5
5
10
• Graph a cubic
function. Steps:
1. Find factor of whole equation.
(0, -3)
m
4u Maths Summary Page 4 of 117
y = ax3 + bx2 + cx + d 2. Use polynomial division. 3. Find roots of remaining equation.
Steps: 1. Find stationary points and their nature. 2. Find points of inflexion. 3. Find intercepts. 4. Use table of values.
Properties: • Has at most three roots.
x-10 -5 5 10
y
-10
-5
5
10
y = x
3 + 4x
2 – x – 3
• Graph a quartic function.
y = ax4 + bx3 + cx2 + dx + e
Steps: 1. Find factor of whole equation. 2. Use polynomial division. 3. Find roots of remaining equation.
Steps: 1. Find stationary points and their nature. 2. Find points of inflexion. 3. Find intercepts. 4. Use table of values.
Properties: o Has at most four roots.
4u Maths Summary Page 5 of 117
x-4 -2 2 4
y
-10
-5
5
10
y = x
4 – 2x
3 – 2x
2 + 4x – 1
• Graph a rectangular hyperbola.
xy = k y = k
x
Properties: • Asymptote: x ? 0 • Use limit to find horizontal asymptotes.
x-10 -5 5 10
y
-10
-5
5
10
xy = 4
• Graph a circle. x2 + y2 + 2gx + 2fy + c
= 0
(x – h)2 + (y – k)
2 = r
2
x2 + y2 + 2gx + 2fy + c = 0 1. Complete the square to get equation into the form
below.
(x – h)2 + (y – k)
2 = r
2
• Centre of circle: (h, k) • Radius of circle = r
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x-10 -5 5 10
y
-10
-5
5
10
(x – 3)
2+ (y + 2)
2= 9
• Graph an exponential function.
y = ax
For both cases: a > 1 & 0 < a < 1
a > 1 Properties: • y-intercept: y = 1 • Horizontal asymptote: y = 0 • The large the value for a, the steeper the curve. • The graph is always increasing.
x-4 -2 2 4
y
-10
-5
5
10
y = 4
x
0 < a < 1 Properties: • y-intercept: y = 1 • Horizontal Asymptote: y = 0 • The graph is always decreasing.
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x-4 -2 2 4
y
-10
-5
5
10
y = 0.2
x
• Graph a logarithmic function.
y = logax
Function can be re-written as: x = a y
Properties: • Root: x = 1 • Vertical Asymptote: x = 0
• Graph trigonometric functions.
y = a Sin bx y = a Cos bx
y = k + a Sin (bx + c)
Basic Sine Curve:
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x-360 -180 180 360
y
-1.5
-1
-0.5
0.5
1
1.5
y = Sin x
Basic Cosine Curve:
x-360 -180 180 360
y
-1.5
-1
-0.5
0.5
1
1.5
y = Cos x
Trigonometric Graph Transformations: y = k + a Sin (bx + c)
4u Maths Summary Page 9 of 117
x-360 -180 180 360
y
-4
-2
2
4
6
y = 2 + 3sin (2x + 10)
Properties: • k moves the graph up or down by a value of k. • b increases the number of cycles per 360°. Number of
normal cycles in 360° is b. • c moves the graph left or right by c. • a increases the amplitude. The maximum displacement
on either side of an equilibrium is a. The same properties apply for cosine curves.
• Graph inverse trigonometric functions.
(Eg: y = a Sin-1bx)
Basic Sine Curve:
x-360 -180 180 360
y
-1.5
-1
-0.5
0.5
1
1.5
y = Sin x
Inverse Sine Curve:
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x-1.5 -1 -0.5 0.5 1 1.5
y
-360
-180
180
360
y = sin
-1 x
Properties: • Domain: -1 = x = 1 • Range: -90 = y = 90
• Odd Function • Passes through origin. • All movements are the same as for normal sine curve. Basically, an inverse trig graph is the equiva lent of switching the values on the x & y axis.
• Graph the functions:
y = x½ &
y = x13
y = x½
x-10 -5 5 10
y
-10
-5
5
10
y = x
12
Properties:
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• Domain: x = 0 • Range: y = 0
• Always concave down • Always increasing.
y = x13
x-4 -2 2 4
y
-4
-2
2
4
y = x
13
Properties:
• Odd Function • Inflexion point at (0,0) • Can be moved by constants.
Drawing graphs by addition and
subtraction of ordinates
Addition and Subtraction of Ordinates:
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• Graph a function y = f(x) ± c
by initially graphing y = f(x)
x-10 -5 5 10
y
-10
-5
5
10
y = (x – 2)
2, y = (x – 2)
2 + 4 & y = (x – 2)
2 – 4
y = f(x) ± c Rules:
• If c is positive, the function is moved up by c. • If c is negative, the function is shifted down by c.
• Graph a function y = f(x) ± g(x)
By initially graphing: y = f(x) & y = g(x)
Addition & Subtraction of Ordinates:
x-360 -180 180 360
y
-6
-4
-2
2
4
6
y = 2cosx y = 3sinx y = 2cosx + 3sinx
Rules: • In essence, the y coordinates at each x value are either
added to, or subtracted from each other. Addition Hints:
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• Where f(x) & g(x) intersect, the combined graph is double the value at the intersection.
• If either f(x) or g(x) are equal to zero at a point, the combined graph is equal to the value of the graph not equalling zero.
Subtraction Hints: • Where the two graphs intersect, the combined graph is
equal to zero. Drawing graphs by reflecting functions in coordinate axes.
• Graph y = -f(x) by initially graphing y = f(x).
y = -f(x)
x-10 -5 5 10
y
-10
-5
5
10
f(x) = (x – 2)
2 – 2 y = – f(x)
Rules: • Function is reflected about the x-axis. • All positive values become negative • All negative values become positive.
• Graph y = |f(x)| by initially graphing y = f(x)
y = | f(x) |
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x-10 -5 5 10
y
-10
-5
5
10
f(x) = (x – 2)
2 – 2 y = |f(x)|
Rules: • All negative y-values become the positive equivalent. • All positive values remain the same.
• Graph y = f(-x) by initially graphing y = f(x)
y = f(-x)
x-10 -5 5 10
y
-10
-5
5
10
f(x) = (x – 2)
2 – 2 y = f( – x)
Rules: • Reflected about the y-axis.
• The initial emphasis in this topic is operating on graphs of these basic
Ways to Graph complex graphs: • Break graph into two or more parts. (Eg: y = logex can be
broken into y = x and y = logex) • Multiply the y-values of each of these.
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functions in order to produce a graph of a more complex function (eg: the graph of y = x logex will be developed by considering properties of the graphs of y = x and y = logex). x-10 -5 5 10
y
-10
-5
5
10
y = ln x , y = x , y = x ln x.
• The notations logex and ln x are used to denote the natural logarithm of x and students should be familiar with both notations.
Basic Fact: logex = ln x
• Students will need to be able to produce quickly a neat sketch of these basic functions in order to use them in the sketching of further functions.
Rules for graphing: 1. Look for families or recognise graph type. Mark original
function and y = ± 1.
• y • y2 • | y |
• 1y
2. Find x & y intercepts. 3. Find asymptotes:
• Vertical • Horizontal • Oblique
4. Find Stationary points • y’ = 0
5. Find inflexion points • y’’ = 0
6. Find critical points • y’ = 8
7. Find domain & range 8. Use symmetry.
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• Odd or even functions 9. Examine behaviour at extremities.
• Right beside asymptotes. • As x à ±8
10. Table of values • Students need to
examine the behaviour of the derivatives of y = x½ and y = x1/3
near x = 0 and investigate the behaviour of these functions at x = 0. They must be familiar with the term ‘critical point’ and with the possibility of curves having vertical tangent lines at points on them.
y = x½
y' = x– 1
2
2
y' = 12 x
x-4 -2 2 4
y
-4
-2
2
4
y = x12
y = 1
2 x Properties of the derivative:
• Vertical Asymptote: x = 0 • Horizontal asymptote: y = 0
y = x13
y' =
1
3 3 x2
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x-4 -2 2 4
y
-4
-2
2
4
y = x13
y' =
1
3 3 x2
Properties of the derivative:
• Vertical Asymptote: x = 0 • Horizontal asymptote: y = 0 • Symmetrical about the y-axis.
Critical Point: A critical point is where a tangent drawn to a curve is exactly vertical. That is, where the gradient of a graph is equal to infinity or where the first derivative returns a math error. • m = 8 • y’ = ERROR (Usually division by zero or square root of
negative) • Tangent to curve is vertical.
• Typical functions involving addition of ordinates could include y = 1 + 3 sin 2x for –2p = x = 2p and y = cos–
1x – p. Students should realise that the graph of 3 sin 2x can be transformed to the graph of 1 + 3 sin 2x by either translating the
y = 1 + 3sin 2x for – 2π ≤ x ≤ 2π
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graph one unit upwards or translating the x-axis one unit in the opposite direction.
x-360 -180 180 360
y
-3
-2
-1
1
2
3
4
The graph has been shifted one unit up, or the x-axis has been moved one unit down.
• Other types could include graphing functions such as f (x) = 3 sin x + x for 0 < x < 4. This may be developed from the graphs of y = x and y = 3 sin x. The points where y = 3 sin x cuts the x-axis correspond to the points where y = 3 sin x + x cuts y = x. Once the shape of the curve has been roughed out using addition of ordinates the position of stationary points and points of inflexion may be obtained when appropriate.
f(x) = 3sinx + x
x-10 -5 5 10
y
-10
-5
5
10
y = x
4u Maths Summary Page 19 of 117
x-360 -180 180 360
y
-4
-3
-2
-1
1
2
3
4
y = 3sinx
Combined Graph: • y= 3sin x cuts the x-axis at 0, 180 & 360. • y = 3sin x + x cuts y = x at the same points. • Basically, y = 3sin x is transposed over y = x.
• A function such as y = – logex may be graphed by reflecting the graph of y = logex in the x-axis. The graph of y = 2 – logex may then be obtained by a suitable translation.
Reflection of Graphs:
x-1 1 2 3 4 5 6
y
-4
-2
2
4
y = ln x y = – ln x y = 2 – ln x
• The graph is reflected in the x-axis. • The translation is achieved by moving the reflection up
two units. • The relationship
between the graphs of y = f (x) and of y = f (x – a)
Relationship between y = f(x) & y = f(x-a) • If a is positive, the graph is moved by a units to the right. • If a is negative, the graph is moved by a units to the left.
4u Maths Summary Page 20 of 117
should be discussed and used also in examples involving the reflection properties, such as, for example, the graph of |1 – sin (x – 2)|. x-360 -180 180 360
y
-2
-1
1
2
y = | 1 – sin(x – 2) |
Sketching functions by multiplication of
ordinates.
• Graph a function y = c.f(x) by initially graphing y = f(x).
y = C . f(x)
x-10 -5 5 10
y
-10
-5
5
10
y = (x – 2)
2 y = 3(x – 2)
2 y = 1
3(x – 2)
2 y = – (x – 2)
2
Rules: • If C > 1, then the function becomes steeper. All y-values
are multiplied by C. • If 0 < C < 1, then the function becomes shallower. All y-
values are multiplied by C. • If C < 0, then the function is reflected in the x-axis. Then
apply the rules for a positive value of C in terms of
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‘steepness’ and ‘shallowness.’ • Graph a function
y = f(x).g(x) by initially graphing y = f(x) & y = g(x).
y = f(x).g(x)
x-360 -180 180 360
y
-10
-5
5
10
y = 4sinx y = x
100 y = x100
(4sin x)
Rules: • Y-coordinates are multiplied together.
Sketching functions by division of
ordinates.
• Graph a function
y = 1f(x)
by
initially graphing y = f(x).
y = 1f(x)
x-2 2 4 6
y
-4
-2
2
4
y = (x – 2)2 – 3
y = 1
(x – 2)2 – 3
Rules:
4u Maths Summary Page 22 of 117
• Vertical asymptotes of inverse function where original
function’s roots lie. That is, where f(x) = 0, y = 1f(x)
has
an asymptote. • Where the original function is larger than 1, the inverse
function becomes smaller. • Where the original function is less than 1, the inverse
function becomes larger. • Same rules apply for when original function is less than
zero, however the values are negative. • Points at y = ±1 do not move. • Where the original is less than zero, the inverse is also
less than zero. • Where y = f(x) is increasing then the inverse graph is
decreasing and vice versa. • Graph a function
y = f(x)g(x) by
initially graphing y = f(x) and y = g(x).
y = f(x)g(x)
x5 10 15
y
-2
2
4
6
8
y = (x – 3)
2 y = x
y = (x – 3)2
x Rules: • Where the denominator is equal to 0, (where the
denominator graph, g(x), cuts the x-axis), a vertical asymptote occurs at that point.
• Can be graphed using reciprocal rule . Eg: graph y = f(x)
and y = 1g(x) . Then use multiplication of ordinates.
• Where f (x) = 0, f(x)g(x) = 0.
• Where g(x) = 0, f(x)g(x) is undefined and a discontinuity
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exists. • A good initial idea
of the behaviour of functions of the form y = f(x)·g(x) may be obtained by examining the graphs of y = f(x) and y = g(x) independently. To graph y = xe–x, the functions y = x and y = e–x may be graphed on the same set of axes. Important features of the graph of y = xe–x can be obtained.
Behaviours:
x-3 -2 -1 1 2 3
y
-3
-2
-1
1
2
3
y = x y = e
– x y = xe
– x
Properties: • For x < 0, xe–x < 0 • For x = 0, xe–x = 0 • For x > 0, xe–x > 0 • As x à -8, xe–x à -8 • As x à 8, xe–x à 0. This enables a rough shape to be quickly sketched. The exact positions of the stationary points and points of inflexion may be determined by calculus.
• To graph
y = x (x + 1)x – 2
, the
graphs of y = x, y = x + 1 and y = x – 2 can be first sketched. The regions in the number plane, in which the graph exists, can be then shaded, discontinuities determined, points of intersection with coordinate axes marked on and the behaviour of the function for x à ±
y = x (x + 1)x – 2
4u Maths Summary Page 24 of 117
8 investigated. Exact positions of stationary points and points of inflexion could lastly be determined if required.
x-2 2 4 6
y
-5
5
10
15
y = x y = x + 1 y = 1
x – 2 y = x (x + 1)x – 2
Properties: • Asymptotes of the inverse function of the denominator
are also asymptotes for the overall graph. • Roots occur where the numerator functions’ roots occur.
• To sketch y = x (x + 1)x – 2
, a rough sketch of
y = x (x + 1)x – 2
can first be drawn and then square root
rules (see below) can be used. Drawing graphs of the form y = [f(x)]n
• Graph a function
y = [ f(x)] n by first graphing y = f(x).
y = [ f(x)] n
4u Maths Summary Page 25 of 117
x-2 2 4 6
y
-5
5
10
x-1 -0.5 0.5 1 1.5
y
-2
-1
1
2
3
y = (x – 2)
2 – 3 y = ((x – 2)
2 – 3)
2 y = ((x – 2)
2 – 3)
3
Rules:
• When f(x) = 0 or 1, y remains at 0 or 1.
• When f(x) > 1, y becomes larger
• When 0 > f(x) > 1, y becomes smaller.
• If n is even, y is always = 0.
• If n is odd, y has the same sign as f(x). • Its derivative is n [f (x)]n–1 f '(x), then all stationary points
and intercepts on the x-axis of y = f (x) are stationary points of [f(x)]n.
• Graph a function y = f(x) by first graphing y = f(x).
y = ± f(x)
4u Maths Summary Page 26 of 117
x-10 -5 5 10
y
-10
-5
5
10
y = (x – 2)
2 – 3 y = (x – 2)
2 – 3 y = – (x – 2)
2 – 3
Properties: • For a positive square root, only values above the x-axis
are shown. • For a negative square root, only values below the x-axis
are shown. • For the graph of y2 = f(x), both values above and below
the x-axis are graphed. • Where f(x) < 0, the square root graph does not exist. • Where f(x) > 1, the values of y are decreased. • Where 0 < f(x) < 1, the values of y are increased. • Where f(x) = 0 or 1, the values of y do not change.
• If y = (f(x) then y' = f'(x)2 f(x)
.This leads to the
position of stationary points. General approach to
curve sketching.
• Use implicit differentiation to
compute dydx for
curves given in implicit form.
Implicit Differentiation:
• Differentiate each part separately.
• Any part containing a y, multiply that derivative by dydx
• Make dydx the subject.
• Make y the subject in the original equation.
• Sub the value for y into the derivative function. Sample Problem:
Differentiate x12 + y
12 = 4
4u Maths Summary Page 27 of 117
Solution:
x12 + y
12 = 4
x– 1
2
2 + y
– 12
2× dy
dx = 0
y
– 12
2× dy
dx = – x
– 12
2 dydx
= – x– 1
2
2× 2
y– 1
2
dydx
= – y
12
x12
Rearrange original to find expression for y½
y12 = 4 – x
12
Sub into derivative.
dydx
= – 4 – x
12
x12
• Curves graphed
could include:
y = x4
x2
– 1
y = x2e
– x
y = x ln (x2
– 1) y
2 = x
2 – 9x
x2 + 2y
2 = 4
y = sinxx
y = x cosx
y = x4
x2
– 1
x-3 -2 -1 1 2 3
y
-2
2
4
6
y = x2e
– x
4u Maths Summary Page 28 of 117
x-2 2 4 6 8
y
-2
2
4
6
y = x ln (x2
– 1)
x-10 -5 5 10
y
-10
-5
5
10
y2 = x
2 – 9x
4u Maths Summary Page 29 of 117
x-5 5 10
y
-10
-5
5
10
x2 + 2y
2 = 4
x-3 -2 -1 1 2 3
y
-3
-2
-1
1
2
3
y = sinxx
4u Maths Summary Page 30 of 117
x-360 -180 180 360
y
-0.04
-0.02
0.02
0.04
y = x cosx
x-360 -180 180 360
y
-360
-180
180
360
• Solve inequalities by sketching an appropriate graph.
Inequalities:
• For graphs where a function is expressed as being larger or smaller than a fixed value or other graph, graph them as separate functions.
Eg: Graph:
x
2– 3 ≥ 2x
4u Maths Summary Page 31 of 117
x-10 -5 5 10
y
-10
-5
5
10
Rules: • Graph each function separately. • Where the blue function has a higher y-value than the
red graph, the inequality is fulfilled. • Find the number
of solutions of an equation by graphical considerations.
Graph Solutions: For the graph: x
2– 3 = 2x
The number of solutions can be determined by the number of times the graphs cross. The x-values at which they cross are the solutions to the equation.
4u Maths Summary Page 32 of 117
Topic 2
Complex Numbers Arithmetic of
complex numbers and solving quadratic
equations.
• Appreciate the necessity of introducing the symbol i, where i2 = -1, in order to solve quadratic equations.
The complex number, i, is used when a quadratic equation has no real roots. For instance the equation x2 + 2x + 3 = 0 has no real roots and therefore i must be introduced in order to solve it. This occurs when the quadratic formula:
x = – b ± b2
– 4ac2a returns a math error.
• Write down the real part Re(z) and the imaginary part Im(z) of a complex number z = x + iy.
For the complex number z = x + iy: Real Part = Re(z) = x Imaginary Part = Im(z) = y
• Add, subtract and multiply complex numbers written in the form x + iy.
Addition of Complex Numbers: The real parts are added together and the imaginary parts are added together.
(a + ib) + (c + id) = (a + c) + (b + d)i Subtraction of Complex Numbers: The real parts are subtracted and the imaginary parts are subtracted. (a + ib) – (c + id) = (a – c) + (b – d)i Multiplication of Complex Numbers: Use the same rule as for two pairs of brackets. (Eg: first term by first term, first term by second, etc) (a + ib)(c + id) = ac + iad + ibc + bdi
2
= ac + iad + ibc – bd
= (ac – bd) + (ad + bc)i • Find the complex
conjugate z of the number x + iy.
Complex Conjugate: For the equation: z = x + iy The conjugate is: z = x – iy
• Divide a complex number a + ib by a
Complex Number Division: Multiply both the numerator and the denominator by the
4u Maths Summary Page 33 of 117
complex number c + id.
conjugate of the denominator. a + ibc + id
= a + ibc + id
× c – idc – id
= ac – iad + ibc + bd
c2 – i
2d
2
= (ac + bd)
c2 + d
2 + i bc – ad
c2 + d
2
• Write down a
condition for a + ib to be equal to c + id.
Complex Number Equality: Two complex numbers are equal is their real parts are equal and their imaginary parts are equal. ∴ a + ib = c + id IF a = c & b = d
• Prove that there are always two square roots of a non-zero complex number.
Square Roots of Complex Numbers: Find z given z2 = a + ib. Let z = x + iy
∴ (x + iy)2 = a + ib
x2 + 2xyi + i
2y
2 = a + ib
(x2 – y
2) + (2xy)i = a + ib
Equating Real & Imaginary Parts:
x2 – y
2 = a
2xy = b
Solving these two equations simultaneously gives a quartic equation whereby an expression for x2 can be derived. One of these expressions will be negative and therefore has no solutions. The other remaining expression will be positive and therefore has two solutions, one positive and one negative. Therefore, a non-zero complex number has two square roots.
• Find the square Square Roots of Complex Numbers:
4u Maths Summary Page 34 of 117
roots of a complex number a + ib.
Example: Find the square root of 3 + 4i. Solution:
Let z2 = 3 + 4i
Let z = x + iy
(x + iy)2 = 3 + 4i
x
2 – y
2 + (2xy) i = 3 + 4i
x2 – y
2 = 3 & 2xy = 4
Multiply first equation by x2
x4 – x
2y
2 = 3x
2
Square the second equation
x2y
2 = 4
Solve Simultaneously
x4 – 3x
2 – 4 = 0
x
2 – 4
x
2 + 1
= 0
x2 = 4, – 1
∴ x = ± 2∴ x = 2, y = 1 & x = -2, y = -1 i is a device by which quadratic equations with real coefficients could be always solvable. It could be shown that there exist 2 complex roots for a complex number. This then leads to the discovery that a quadratic equation with complex coefficients will have 2 complex roots. In finding the square roots of a + ib, the statement a + ib = x + iy, where a, b, x, & y are real, leads to the
need to solve the equations x2 – y2 = a & 2xy = b. Examining graphs of these curves for various values of a and b will lead to the conclusion that two roots will always exist for a complex number.
4u Maths Summary Page 35 of 117
x-10 -5 5 10
y
-10
-5
5
10
x2 – y
2 = 5 2xy = 9
• Solve quadratic equations in the form: ax
2 + bx + c = 0 ,
where a, b & c are complex.
Expand
Geometrical Representation
• Appreciate that there exists a one to one correspondence between the complex number a + ib and the ordered pair (a, b).
The complex number a + ib , represents an ordered pair of (a, b) on the argand diagram.
x0.5 1 1.5 2 2.5 3
y
1
2
3
4
5
6
1 + 4i
• Plot the point corresponding to a + ib on the Argand diagram.
For the complex number a + ib. The x-value = a The y-value = b
4u Maths Summary Page 36 of 117
x0.5 1 1.5 2 2.5 3
y
1
2
3
4
5
6
1 + 4i
• Define the modulus ( |z| ) and argument (arg z) of a complex number z.
The modulus of z is the distance from the origin to the complex number z. It can be referred to |z|. The argument of z is the angle that a line drawn from the origin to a complex number, z, makes with the x-axis in the positive direction. It can be referred to as arg (z).
• Find the modulus and argument of a complex number.
The modulus of the complex number z = x + iy is:
|z| = x2
+ y2
The argument of the complex number z = x + iy is:
Tan θ = yx
• Write a + ib in Modulus-argument form.
Modulus-argument form is also known as: • Mod-Arg form • Polar form Mod-Arg Form of a complex number: z = r (cos θ + i sin θ)z = r cis θ Where:
r = x2 + y
2
Tan θ = yx
Remember:
Mod z = |z| = |x + iy| = x2 + y
2 = zz = r
Arg z = θ = ph z
4u Maths Summary Page 37 of 117
Arg z is any value of θ for which x = |z| cos θ and y = |z| sin θ .
• Prove basic relations involving modulus and argument.
|z1z2| = |z1||z2|
z1
z2 = |z1|
|z2|
Arg (z1z2) = Arg z1 + Arg z2 ± 2π
Argz1
z2 = Arg z1 – Arg z2 ± 2π
|z| = |z| = x2 + y
2
zz = |z|2 = |z|
2 = x
2 + y
2
z + z = 2x
(z1z2) = z1 × z2
z – z = 2yi
Arg z = – Arg z
z-1
= 1z = z
|z|2
• Use modulus-
argument relations to do calculations involving complex numbers.
Multiplication in Mod-Arg Form: Let z1 = r1 cis θ1 & z2 = r2 cis θ2
z1z2 = r1r2 cis (θ1 + θ2) Division in Mod-Arg Form:
z1
z2
= r1
r2
cis (θ1 – θ2)
y
x
r y
x
x + iy (r, θ )
θ
4u Maths Summary Page 38 of 117
• Recognize the geometrical relationships between the point representing z and points representing z , cz (c is real) & iz.
x-10 -5 5 10
y
-10
-5
5
10
z = 4 + 3i z = 4 – 3i iz = -3 + 4i
cz = 8 + 6i where c = 2 Rules:
• i2 = -1
• i3 = – i
• i4 = 1
• Multiplying a complex number by i is equal to a quarter turn about the origin.
• z is a reflection of z in the x-axis. • Multiplying a complex number by a real value, c, results
in both the real and imaginary part being multiplied by that number.
• Students should be able to prove these relations.
Proof #1: |z1z2| = |z1| × |z2| & arg (z1z2) = arg z1 + arg z2 Let z1 = r1cisθ1
z2 = r2cisθ2
z1z2 = (r1cisθ1)(r2cisθ2)= r1r2 (cosθ1 + isinθ1)(cosθ2 + isinθ2)
= r1r2 (cosθ1cosθ2 + isinθ2cosθ1 + isinθ1cosθ2 + i2sinθ1sinθ2 )
= r1r2 [(cosθ1cosθ2 – sinθ1sinθ2) + i (sinθ1cosθ2 + cosθ1sinθ2)]= r1r2 [cos(θ1 + θ2) + isin(θ1 + θ2)]= r1r2 cis(θ1 + θ2)
|z1z2| = r1r2
∴ |z1z2| = r1r2 = |z1||z2|arg (z1z2) = θ1 + θ2
∴ arg (z1z2) = θ1 + θ2 = arg z1 + arg z2
4u Maths Summary Page 39 of 117
Proof #2:
z1
z2
= |z1|
|z2| & arg
z1
z2
= arg z1 – arg z2
Let z1 = r1 cis θ1
z2 = r2 cis θ2
z1
z2
= r1 cis θ1
r2 cis θ2
= r1 cis θ1
r2 cis θ2
× r2 cis( – θ2)r2 cis( – θ2)
= r1r2 cis(θ1 – θ2)
r22 cis 0°
However, since cis 0° = 1
= r1
r2
cis(θ1 – θ2)
∴ z1
z2
= r1
r2
= |z1||z2|
∴ arg z1
z2
= θ1 – θ2 = arg z1 – arg z2
Proof #3:
Let z1 = a + ib
z1 = a – ibz2 = c + id
z2 = c – id
z1 + z2 = (a – ib) + (c – id)= (a + c) – ib – id= (a + c) – i (b + d)
z1 + z2 = a + ib + c + id= (a + c) + i (b + d)
= (a + c) – i (b + d) = LHS Proof #4:
4u Maths Summary Page 40 of 117
Let z1 = a + ib
z1 = a – ibz2 = c + id
z2 = c – id
z1z2 = (a – ib)(c – id)
= ac – adi – bci + bdi2
= ac – bd – adi – bci= (ac – bd) – i (ad + bc)
z1z2 = (a + ib)(c + id)
= ac + adi + bci + bdi2
= ac – bd + adi + bci= (ac – bd) + i (ad + bc)
LHS = (ac – bd) – i (ad + bc)= RHS
Proof #5: |z
n| = |z|
n
Let z = rcis θ
zn = r
n cis nθ
|zn| = r
n
z = rcis θ|z| = r
|z|n
= rn
∴ |zn| = |z|
n
Proof #6: arg(z
n) = n argz
Let z = rcis θ
zn
= rn cis nθ
arg(zn) = nθ
z = rcis θargz = θn argz = nθ
∴ arg(zn) = n argz
Sample Question Q4 (1987) a. Let OABC be a
square on an argand diagram where O is the origin. The points
Answer: a. Let z = 3 + 2i ∴ iz = -2 + 3i
4u Maths Summary Page 41 of 117
A and C represent the complex numbers z and iz respectively. Find the complex number represented by B.
b. The square is now rotated about O through 45° in an anticlockwise direction to OA’B’C’. Find the complex numbers represented by the points A’, B’ and C’.
By graphical solution, B = 1 + 5i 1 + 5i = z + iz ∴ B = z + iz b. To rotate a complex number about the origin by 45° is the equivalent of multiplying a complex number by:
|z|
|z|2 × z
∴ A' = z
|z|
B' = z + iz|z|
C' = iz|z|
• Appreciate that a complex number can be represented as a vector on the Argand diagram.
The complex number z = x + iy can be represented as a vector OZ .
• Appreciate the geometrical
For the addition of two complex numbers. Use vector addition to find the resulting vector.
O
Z
iz
z
O
?
4u Maths Summary Page 42 of 117
significance of the addition of two complex numbers.
• Given the points representing z1 and z2, find the position of the point representing z, where z = z1 + z2.
For z1 + z2
Where z1 = 5 + 2i & z2 = 1 + 3i
Familiarity with the vector representation of a complex number is extremely useful when work on curves and loci is encountered.
• Appreciate that the vector z = z1 + z2, corresponds to the diagonal of a parallelogram with vectors representing z1 and z2 as adjacent sides.
That is: • The gradient of the vector joining z1 to the origin is equal
to the gradient of the vector joining (z1 + z2) to z2. • The gradient of the vector joining z2 to the origin is equal
to the gradient of the vector joining (z1 + z2) to the point z1.
• Given vectors z1 and z2, construct vectors z1 – z2 and z2 – z1.
Subtract the appropriate real parts and separately subtract the appropriate imaginary parts.
z1 = 1 + 4i z2 = 3 + 6i z2 – z1 z1 – z2
• We need to be able to interpret the expression | z – (a +
ib) | as the magnitude of a vector joining (a, b) to the point representing z. Basically it is the same graph only it is based at (a, b) instead of the origin.
• Students need to recognise that the expression arg(z – z1) refers to the angle, which a vector joining the point representing z1 to the point representing z, makes with
z1
z2
z2 - z1
z1 – z2
4u Maths Summary Page 43 of 117
the positive direction of the real axis. Once again, basically the same graph but with its base at z1 instead of the origin.
• Given z1 and z2, construct the vector z1z2.
z1 = 1 + 3i z2 = 3 + 2i z1 × z2
• Prove geometrically that
|z1 + z2| ≤ |z1| + |z2| .
z1
z2
z1z2
z1
z2
z1 + z2
4u Maths Summary Page 44 of 117
Using z1 = 2 + 2i = A z2 = 1 + 3i = B z1 + z2 = 3 + 5i = C
Let |z1| = r1
|z2| = r2
Let AD be perpindicular to OC
∴ ∆ADO is a right ∠∆ where OA is the hypotenuse
∆ACD is also a right ∠∆ where AC is the hypotenuse
∴ OD ≤ r1 (Pythagoras' theorem)
DC ≤ r2 (Pythagoras' theorem)
∴ |OD| + |DC| ≤ r1 + r2
|z1 + z2| ≤ |z1| + |z2| • Prove, by
induction, that (cos θ + isin θ)
n =
cos nθ + isin nθ for positive integers n.
Proof:
C
A
r1
r2
O
D
4u Maths Summary Page 45 of 117
(cos θ + isin θ)n = cos nθ + isin nθ
Proof by Mathematical Induction:Let n = 1
LHS = (cos θ + isin θ)1
= cos θ + isin θ
RHS = cos(1)θ + isin(1)θ= cos θ + isin θ
∴ True for n = 1
Assume true for n = k
(cos θ + isin θ)k = cos kθ + isin kθ
Let n = k + 1RHS = cos(k + 1)θ + isin(k + 1)θ
LHS = (cos θ + isin θ)k + 1
= (cos θ + isin θ)1(cos θ + isin θ)k
= (cos θ + isin θ)(cos kθ + isin kθ)= cosθcoskθ – sinθsinkθ + icosθsinkθ + isinθcosθ
= (cosθcoskθ – sinθsinkθ) + i (cosθsinkθ + sinθcosθ) = cos (kθ + θ) + isin (kθ + θ)= cos(k + 1)θ + isin(k + 1)θ= RHS
∴ True for n = k + 1∴ True for all positive integer values Be able to reproduce this.
• Prove that (cos θ + isin θ)
n =
cos nθ + isin nθ for negative integers n.
(cos θ + isin θ)n = cos nθ + isin nθ
Let n = – m, where m is a positive integer.
LHS = (cos θ + isin θ)– m
= 1
(cos θ + isin θ)m
= 1cos mθ + isin mθ
= 1cos mθ + isin mθ
× cos mθ – isin mθcos mθ – isin mθ
= cos mθ – isin mθ
cos2 mθ + sin
2 mθ
= cos mθ – isin mθ1
= cos( – mθ) + isin( – mθ)= cos nθ + isin nθ= RHS
• Find any integer This is known a De Moivre’s theorem:
4u Maths Summary Page 46 of 117
power of a given complex number.
(r cis θ)
n = r
n cis nθ
= rn (cos nθ + isin nθ)
• Find the complex nth roots of ±1 in modulus-argument form.
To find the nth roots of any complex number we use the fact: If R
n (cos φ + isin φ)
n = r(cos θ + isin θ) then:
R = r
1n and φ = θ + 2kπ
n , where k = 0, 1, 2,...(n-1)
Therefore the nth roots of the complex number
r (cos θ + isin θ) have the modulus r1n and arguments
given by φ = θ + 2kπn
, where k = 0, 1, 2,…(n – 1).
Finding nth roots of ±1: z
n = 1
Let n = 5
z5 = 1
Using z = r
1n cis
θ + 2kπ
n where k = 0,1,2,3,4
z1 = cis 0° = 1
z2 = cis 2π
5
z3 = cis 4π
5
z4 = cis 6π
5
z5 = cis 8π
5
Use the same method for zn = -1.
4u Maths Summary Page 47 of 117
• Sketch the nth roots ±1 on an Argand diagram.
• Illustrate the geometrical relationship connecting the nth roots of ±1.
• The roots represent evenly spaced points on the unit
circle. • They are also the vertices of an nth sided polygon.
• Given equations Re(z) = c, Im(z) = k, sketch lines parallel to the appropriate axis.
x-4 -2 2 4
y
-4
-2
2
4
Re(z) = 3 Im(z) = 2
Rules: • For Re(z) = c, the real part of z becomes the graph.
Therefore the locus of the graph is x = c. • For Im(z) = k, the imaginary part of z becomes the
graph. Therefore the locus of the graph is y = k . • Given an
equation, |z – z1| = |z – z2| ,
For this type of locus, let z1 and z2 represent points on the argand diagram. If the distance from P to z1 is equal to the distance from P to z2, then we know from plane geometry
1
z2
z3
z4
z5
P(z)
4u Maths Summary Page 48 of 117
sketch the corresponding line.
that the locus is the perpendicular bisector of the line joining z1 and z2.
z1 = 2 + 0i z2 = -1 + i
Rules: • Let z1 and z2 be normal Cartesian co-ordinates. • Find the midpoint of z1 and z2. • Find the gradient of the line joining z1 to z2. • Inverse the gradient to find the gradient of the line
perpendicular to it. • Use point-gradient formula to find equation of the line
representing the point P. • Given equations
|z| = R and |z – z1| = R , sketch the corresponding circles.
These two equations represent the locus of a circle. |z| = R Is the locus of a circle with a centre at the origin and radius R units.
O
z2
z1
4u Maths Summary Page 49 of 117
x-4 -2 2 4
y
-3
-2
-1
1
2
3
|z| = 2 |z – z1| = 2 where z1 = 2 – i
Rules: • The value of R becomes the length of the radius. • The point z1 becomes the centre of the circle.
• Given equations arg z = θ and arg(z – z1) = θ , sketch the corresponding rays.
The graph is a ray which originates from either the origin or z1, and goes off at angle θ in the positive direction.
arg z = π
4 arg (z – z1) = π
3 where z1 = -3 + 2i
Rules: • The graph is just a ray that starts at z1 • The ray makes an angle θ with the positive direction.
• Sketch regions associated with any of the above curves.
The graphs are sketched exactly the same way but the area in which the equation is fulfilled must be determined. Remember:
z1 θ
θ
O
4u Maths Summary Page 50 of 117
• Dotted lines for < or >. • Solid lines for = or =.
• Give a geometrical description of any such curves or regions.
Use basic names to describe the locus of a point or the region in which an equation is fulfilled. Eg: The area inside a circle with radius 3 units and centre at (3, 2). Be able to graph and describe these:
• Be able to understand the intersection of more than one
region or graph. Examples need only involve replacing z by z = x + iy in relations such as:
4u Maths Summary Page 51 of 117
Topic 3 Conics
• Write down the defining equation of an ellipse with centre at the origin.
Equation of an ellipse with centre at the origin:
x2
a2 + y
2
b2 = 1
This is regarded as the defining equation.
• Sketch the ellipse showing points of intersection with the axes of symmetry.
x
y
x
2
a2 + y
2
b2 = 1
• Find the lengths of the major and minor axes and semi-major and semi-minor axes of an ellipse.
x
y
Rules: • Major axis length = 2a
(0, b)
(0, -b)
(a, 0) (-a, 0)
2a
2b
4u Maths Summary Page 52 of 117
• Minor axis length = 2b • Semi-major axis length = a • Semi-minor axis length = b
• Write down the parametric coordinates of a point on an ellipse.
Parametric Coordinates: x = a cos θy = b sin θ
• Sketch an ellipse using its auxiliary circle.
x
y
x2 + y
2 = a
2
x2
a2 + y
2
b2 = 1
Equation of the Ellipse: x = a cos θy = b sin θ Equation of the Auxiliary Circle: x = a cos θy = a sin θ The parametric representation x = a cos θ , y = b sin θ is useful in graphing the ellipse from an auxiliary circle. The shape of an ellipse should be examined as the ratio b
a
varies. • Find the equation
of an ellipse from its focus-directrix definition.
The definition of a conic is:
PS
PM = e
Where P – Is any point on the conic S – The focus of the conic M – A point on the directrix
(acos θ , bsin θ )
(acos θ , asin θ )
4u Maths Summary Page 53 of 117
e – The degree of eccentricity For an ellipse, 0 < e < 1 Rearranging the definition of a conic gives: PS = ePM Therefore, the equation of an ellipse can be calculated by substituting in appropriate values for the focus, directrix and eccentricity. The focus-directrix definition should be used whenever a focal distance is to be calculated.
• Find the eccentricity from the defining equation of an ellipse.
Using the definition of a conic: PS = ePM
(x – ae)2
+ (y – 0)2 = e
x – a
e 2 + (y – y)
2
(x – ae)2 + y
2 = e
2
x – a
e 2
x2 – 2aex + a
2e
2 + y
2 = e
2x
2 – 2aex + a
2
x2 – e
2x
2 + y
2 = a
2 – a
2e
2
1 – e
2x
2 + y
2 = a
21 – e
2
x2
a2 +
y2
a2
1 – e
2 = 1
∴ b2 = a
21 – e
2
• Given the equation of an ellipse, find the co-ordinates of the foci and equations of the directrices.
For the ellipse with equation: x
2
a2 + y
2
b2 = 1
Foci: (±ae, 0) Directrices:
x = ± ae
4u Maths Summary Page 54 of 117
• Sketch an ellipse, marking on it the positions of its foci and directrices.
x
y
x2
a2 + y
2
b2 = 1
The major properties of an ellipse are to be proven for both a general ellipse with centre O and for ellipses with given values of a and b.
• Use implicit differentiation to find the equations of the tangent and the normal at P(x1, y1) on an ellipse.
x2
a2 + y
2
b2 = 1
Implicitly Differentiating Gives:2x
a2 + 2y
b2 × dy
dx = 0
2y
b2 × dy
dx = – 2x
a2
dydx
= – 2xb2
2a2y
dydx
= – b2x
a2y
Let P = (x1, y1)
∴ Gradient = – b2x1
a2y1
(a, 0)
(ae, 0) (-ae, 0)
(-a, 0)
(0, b)
(0, -b)
O
D’
D
D’
D
x = ae
x = – a
e
4u Maths Summary Page 55 of 117
∴ Equation of the tangent at P is:
y – y1 = – b2x1
a2y1
(x – x1)
a2y1(y – y1) = – b
2x1(x – x1)
a2y1y – a
2y1
2 = – b
2x1x + b
2x1
2
a2y1y + b
2x1x = b
2x1
2 + a
2y1
2
Dividing everything by a2b
2 gives
y y1
b2 + x x1
a2 = x
2
a2 + y
2
b2
Since x2
a2 + y
2
b2 = 1
x x1
a2 + y y1
b2 = 1
Gradient of the Normal = a
2y1
b2x1
∴ Equation of the Normal:
y = y1 = a2y1
b2x1
(x – x1)
b2x1 (y – y1) = a
2y1 (x – x1)
b2x1y – b
2x1y1 = a
2y1x – a
2x1y1
Dividing everything by x1y1 gives:
b2y
y1
– b2 = a
2x
y1
– a2
a2x
x1
– b2y
y1
= a2 – b
2
4u Maths Summary Page 56 of 117
• Find the equations of the tangent and the normal at P (acos θ, bsin θ) on an ellipse.
x = acos θdxdθ
= – asin θ
y = bsin θdydθ
= bcos θ
∴ dydx
= – bcos θasin θ
∴ Equation of the tangent is:
y – bsin θ = – bcosθasin θ
(x – acos θ)
aysin θ – absin2θ = – bxcos θ + abcos
2θ
aysin θ + bxcos θ = abcos2θ + absin
2θDividing Everything by abysin θ
b + xcos θ
a = cos
2θ + sin2θ
∴ Tangent is:xcos θ
a + ysin θ
b = 1
Gradient of the Normal = asin θ
bcos θ
y – bsin θ = asin θbcos θ
(x – acos θ)
bycos θ – b2sinθcosθ = axsin θ – a
2sinθcosθ
Dividing everything by sinθcosθby
sin θ – b
2 = ax
cos θ – a
2
axcos θ
– bysin θ
= a2 – b
2
• Find the equation of a chord of an ellipse.
The equation for the chord of an ellipse is:
xa cos
θ + φ
2 + y
b sin
θ + φ
2 = cos
θ – φ
2
• Find the equation of a chord of contact.
Let PT and PQ be tangents of an ellipse.
4u Maths Summary Page 57 of 117
Let P = (x1, y1)
Q = (x2, y2)T = (x0, y0)
Equation of tangent PT:x0 x1
a2 + y0 y1
b2 = 1
Equation of tangent PQ:x0 x2
a2 + y0 y2
b2 = 1
Hence both P & Q satisfy:x x0
a2 + y y0
b2 = 1
The chord of contact is useful as a tool in the proof of a number of properties of an ellipse.
• Prove that the sum of the focal lengths is constant.
Let P = (acos θ, bsin θ)
S = (ae, 0)S' = ( – ae, 0)
PS = (acos θ – ae)2 + (bsin θ – 0)
2
PS2 = a
2cos
2θ – 2a
2ecosθ + a
2e
2 + b
2sin
2θ
Using b2 = a
21 – e
2
b2 = a
2 – a
2e
2
a2e
2 = a
2 – b
2
PS2 = a
2cos
2θ – 2a
2ecos θ + a
2 – b
2 + b
2sin
2θ
PS2 = a
2cos
2θ – 2a
2ecosθ + a
2 – b
21 – sin
2θ
PS2 = a
2cos
2θ – 2a
2ecosθ + a
2 – b
2cos
2θ
PS2 =
a
2– b
2cos
2θ – 2a
2ecosθ + a
2
PS2 = a
2e
2cos
2θ – 2a
2ecosθ + a
2
PS2 = a
2(1 – ecosθ)
2
PS = a (1 – ecosθ)
PS' = a (1 + ecosθ)
PS + PS' = 2a (A constant) The focus-directrix definition leads to a simple proof that the sum of the focal lengths is constant (2a).
4u Maths Summary Page 58 of 117
• Prove the reflection property, namely that the tangent to an ellipse at a point P on it is equally inclined to the focal chords through P.
4u Maths Summary Page 59 of 117
4u Maths Summary Page 60 of 117
tan φ =
b
aesin θ
4u Maths Summary Page 61 of 117
tan θ =
– b
aesin θ
∴ tan θ = tan φ θ = φ∴ ∠RPS' = ∠TPS • The reflection property of the ellipse may be approached
by using the result that the bisector of an angle of a triangle divides the opposite side into two intervals, whose lengths are in the same ratio as the lengths of the other two sides.
4u Maths Summary Page 62 of 117
• Prove that the chord of contact from a point on a directrix is a focal chord.
∴ QR is a focal chord
• Prove that part of the tangent between the point of contact and the directrix subtends a right angle at the
4u Maths Summary Page 63 of 117
corresponding focus.
• Prove simple properties for both the general
ellipse x2
a2 + y
2
b2 = 1
and for ellipses with given values of a and b.
Students are not expected to do proofs, under examination conditions, which are more difficult than those involved in the Contents and Skills objectives. Locus problems on the ellipse are not included.
• Write down the defining equation of a hyperbola with centre at the origin.
General Equation of a Hyperbola:
x2
a2 – y
2
b2 = 1
• Sketch the hyperbola
x2
a2 – y
2
b2 = 1
, showing points of intersection with axes of symmetry and positions of asymptotes.
Rules:
• Cuts the x-axis at (-a, 0) & (a, 0)
• The asymptotes are: y = ± bxa
• The shape of the hyperbola should be examined as ba
varies.
• Find the length of the major and minor axes and semi-major and semi-minor axes of a hyperbola.
Rules: • Length of major axis = 2a • Length of minor axis = 2b • Length of semi-major axis = a • Length of semi-minor axis = b
• Write down the Parametric Coordinates:
(-a, 0) (a, 0)
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parametric coordinates of a point on the hyperbola.
x = a sec θy = b tan θ
• Find the equation of a hyperbola from its focus-directrix definition.
The definition of a conic is:
PS
PM = e
Where P – Is any point on the conic S – The focus of the conic M – A point on the directrix e – The degree of eccentricity For a hyperbola , e > 1 Rearranging the definition of a conic gives: PS = ePM Therefore, the equation of a hyperbola can be calculated by substituting in appropriate values for the focus, directrix and eccentricity. The focus-directrix definition should be used whenever a focal distance is to be calculated.
• Find the eccentricity from the defining equation of a hyperbola.
Using the definition of a conic:
PS = ePM
(x – ae)2
+ (y – 0)2 = e
x – a
e 2 + (y – y)
2
(x – ae)2 + y
2 = e
2
x – a
e 2
x2 – 2aex + a
2e
2 + y
2 = e
2x
2 – 2aex + a
2
x2 – e
2x
2 + y
2 = a
2 – a
2e
2
1 – e
2x
2 + y
2 = a
21 – e
2
x2
a2 +
y2
a2
1 – e
2 = 1
From before b2 = a
21 – e
2
Since e > 1, 1 – e2 < 0
∴ b2 = a
2e
2– 1
• Given the equation of the hyperbola, find the coordinates of its foci and equations of its directrices.
For the hyperbola with equation: x
2
a2 – y
2
b2 = 1
Foci: (±ae, 0) Directrices:
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x = ± ae
• Sketch a hyperbola, marking on it the positions of its foci and directrices.
• Use implicit
differentiation to find the equations of the tangent and normal at P(x1, y1) on a hyperbola.
(-ae, 0) (ae, 0)
y = ae
y = – ae
Gradient of Tangent
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Equation of tangent to hyperbola:x x1
a2 – y y1
b2 = 1
Equation of normal to hyperbola:
a2x
x1 + b
2y
y1 = a
2 + b
2
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• Find the equations of the tangent and normal at P (asec θ, btan θ) on the hyperbola.
Equation of tangent to hyperbolaxsec θ
a – ytan θ
b = 1
Equation of normal to hyperbola
bytan θ
+ axsec θ
= a2 + b
2
Gradient of Tangent
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• Find the equation of a chord of a hyperbola.
The equation of a chord from P (asec θ, btan θ) to Q(asec φ, btan φ) is:
xa cos
θ – φ
2 – y
b sin
θ + φ
2 = cos
θ + φ
2
• Find the equation of a chord of contact.
Let the point on directrix = T(x0, y0) Let P(x1, y1) & Q(x2, y2) Therefore the chord formulas for both are:
x0 x1
a2 – y0 y1
b2 = 1 & x0 x2
a2 – y0 y2
b2 = 1
Both equations satisfy the relation:
x x0
a2
– y y0
b2
= 1
The chord of contact is useful in proving some properties.
• Prove that the difference of the focal lengths is a constant.
Proving PS – PS’ = k
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• Prove the
reflection property for a hyperbola.
The same geometry theorem, as used in the case of the ellipse, is useful in proving the reflection property of the hyperbola.
• Prove that the chord of contact from a point on the directrix is a focal chord.
This proof is exactly the same for the hyperbola as it is with the ellipse.
• Prove simple properties for the general hyperbola and also hyperbolae with given values of a and b.
The major properties of the hyperbola are to be proven for both the general hyperbola with centre O and for hyperbolae with given values of a and b. Students are not expected to do proofs, under examination conditions, which are more difficult than those involved in the skills objectives. Locus problems, on a hyperbola with the general equation, are not in the course.
The Rectangular Hyperbola
• Prove that the hyperbola with equation
xy = 12 a
2 is the
hyperbola
x2 – y
2 = a
2
referred to different axes.
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A definition needs to be given for a rectangular hyperbola. It quickly follows, from seeing the connection between x2 – y2 = a2 and xy = ½a2, that the eccentricity is 2 .
• Write down the eccentricity, coordinates of foci and vertices, equations of directrices and equations of asymptotes.
Eccentricity = 2Foci = ± (a, a)Directrices = x + y = ±a
Vertices = ±
a 2
, a 2
or ± (c, c)
Asymptotes = x = 0, y = 0
• Sketch the hyperbola
xy = 12 a
2
, for varying values of a, marking on vertices, foci, directrices and asymptotes.
S
S’
D
D’
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• Asymptotes are the x and y axes. • Write down the
parametric coordinates for the rectangular hyperbola
xy = c2 , for
varying values of c.
When the value of c changes, substitute the new value into the following formulas: x = ct
y = ct
∴ P ct, c
t
• Find the equation of the chord
joining P cp, c
p
to Q cq, c
q .
Gradient of PQ
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• Find the equation
of the tangent at P.
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• Find the equation of the normal at P.
• Find the equation
of the chord joining P(x1, y1) to Q(x2, y2).
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Therefore, the equation of the chord from P to Q is:
• Find the equation
of the chord of contact from T(x0, y0).
• Find the point of
intersection of tangents and of normals.
Use the equations for tangents and normals. Solve these simultaneously to find the points of intersection.
• Prove simple geometrical properties of the rectangular hyperbola including: o The area of the
triangle bounded by a tangent and the asymptotes is a constant.
o The length of the intercept, cut off a tangent by the asymptotes, equals twice
Area of the Triangle:
P
R
T
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the distance of the point of contact from the intersection of the asymptotes.
Length of Intercept:
P
T
R O
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• Find loci of points
including: o Loci
determined by intersection points of tangents.
o Loci determined by intersection points of normals
o Loci determined by midpoints of intervals.
Intersection of tangents:
This example is for a problem whereby the points P & Q must join to form a chord that passes through a given point, in this case (0, 4). Also, for this example, c = 3.
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Intersection of Normals: Midpoints of intervals:
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The locus is therefore a hyperbola that shares the same asymptotes as the original equation. It is not intended that locus problems should include sophisticated techniques for elimination of parameters. Students are expected to be able to proceed from a pair of parametric equations to obtain a locus expressible by a linear equation (perhaps with constraints on x or y). In cases where the resulting locus is not expressible in terms of a linear equation, it will be given in algebraic o r geometric form and students will verify that this form is satisfied (perhaps with additional constraints).
• Appreciate that the various conic sections (circle, ellipse, parabola, hyperbola and pairs of intersecting lines) are indeed the curves obtained when a plane intersects a (double) cone
• Relate the various ranges of values of the eccentricity e to the appropriate conic and to understand how the shape of a conic varies as its eccentricity varies.
For e = 1: • The locus is a parabola • PS = PM For e < 1: • The locus in an ellipse. • PS < PM For e > 1: • The locus is a hyperbola • PS > PM
• Appreciate that the equations of all conic sections
All conic sections deal with just equations in x and y.
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involve only quadratic expressions in x and y.
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Topic 4
Integration
• Use a table of standard integrals.
The table of standard integrals is supplied in an exam and can be referred to for integration.
• Change an integrand into an appropriate form by use of algebra.
Look for constant values that can be removed, trig substitutions, etc.
• Evaluate integrals using algebraic substitutions.
Let u equal some portion of the integrand so that the integrand can be rearranged to give an integrand in terms of u and du. Eg:
⌡⌠ u du
• Only simple substitutions are needed, eg u = 1 + x2,
v2 = 1 – x in ⌡⌠ x(1 + x
2)4 dx ,
⌡⌠ x
1 – x dx .
The effect on limits of integration is required, and definite integrals are to be treated.
• Evaluate simple trigonometric integrals.
Trigonometric Identities:
sin2θ + cos
2θ = 1
1 + cot2θ = cosec
2θ
tan2θ + 1 = sec
2θ Sums & Differences: sin(θ + φ) = sinθcosφ + sinφcosθ
sin(θ – φ) = sinθcosφ – sinφcosθ
cos(θ + φ) = cosθcosφ – sinθsinφ
cos(θ – φ) = cosθcosφ + sinθsinφ
tan(θ + φ) = tanθ + tanφ1 – tanθtanφ
tan(θ – φ) = tanθ – tanφ1 + tanθtanφ
Double Angles: sin 2θ = 2sinθcosθ
cos 2θ = cos2θ – sin
2θ
= 2cos2θ – 1
= 1 – 2sin2θ
tan 2θ = 2tanθ
1 – tan2θ
Transformations:
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asinθ + bcosθ = rsin(θ + α)asinθ – bcosθ = rsin(θ – α)acosθ + bsinθ = rcos(θ – α)acosθ – bsinθ = rcos(θ + α)
where r = a2 + b2
tanα = ba
Integration of Trigonometric Functions: Basic Trigonometric Integration: ⌡⌠ sinax dx = – 1
a cosax + c
⌡⌠ cosax dx = 1
a sinax + c
⌡⌠ sec
2ax dx = 1
a tanax + c
Integration of Squared Trig Functions: ⌡⌠ sin
2ax = 1
2 x – 1
4a sin 2ax + c
⌡⌠
cos2ax = 1
2 x + 1
4a sin 2ax + c
Inverse Trigonometric Functions:
⌡⌠
1
a2
– x2
dx = sin-1
xa + c
⌡⌠ – 1
a2
– x2
dx = cos-1
xa
+ c
⌡⌠ a
a2 + x
2 dx = tan-1
xa
+ c
• Evaluate
Integrals using trigonometric substitutions.
The Substitution Method:
θ2
1 + t2
t
1
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Let t = tan θ2
θ2
Ratios:
tan
θ2
= t
sin θ2
= t
1 + t2
cos θ2
= 1
1 + t2
Basic Ratios:
sinθ = 2t
1 + t2
cosθ = 1 – t2
1 + t2
tanθ = 2t
1 – t2
Derivative:
dθdt
= 2
1 + t2
• Evaluate
integrals using integration by parts.
This works for an integrand whereby one part is also to be differentiated and the other part capable of being integrated. ⌡⌠ u dv
dx dx = uv –
⌡⌠ v du
dx dx
or
⌡⌠ uv' dx = uv –
⌡⌠ vu' dx
• Derive and use
recurrence relations.
Recurrence Formula:
⌡⌠ cos
nx dx
In = ⌡⌠
cosnx dx
= ⌡⌠ cosn – 1x . cosx dx
n – 1
⌡
Let u = cosn – 1
xdudx
= – sinx (n – 1) cosn – 2
x
dvdx
= cosx
v = sinx
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In = sinxcosn – 1
x + ⌡⌠ (n – 1)sinxcos
n – 2x . sinx dx
= sinxcosn – 1
x + (n – 1)⌡⌠
sin2x cos
n – 2x dx
= sinxcosn – 1
x + (n – 1)⌡⌠ (1 – cos
2x)cos
n – 2x dx
= sinxcosn – 1
x + (n – 1)⌡⌠ cos
n – 2x – cos
nx dx
= sinxcosn – 1
x + (n – 1)⌡⌠ cos
n – 2x dx – (n – 1)
⌡⌠ cos
nx dx
In = sinxcosn – 1
x + (n – 1)In – 2 – (n – 1)In
In + (n – 1)In = sinxcosn – 1
x + (n – 1)In – 2
nIn = sinxcosn – 1
x + (n – 1)In – 2
In = 1n sinxcos
n – 1x + n – 1
n In – 2
Then use this new-found equation to solve an integral such as:
⌡⌠ cos
6x dx
• Integrate rational
functions by completing the square in a quadratic denominator.
Completing a square will often result in an inverse tan result.
• Integrate rational functions whose denominators have simple linear or quadratic factors.
Partial Fractions: Used in situations whereby the denominator is broken into parts so it can be integrated. Example:
⌡⌠ 5x + 1
(x – 1)(x + 2)
Let 5x + 1(x – 1)(x + 2)
= ax – 1
+ bx + 2
∴ 5x + 1 = a(x + 2) + b(x – 1)Then solve for x = -2 & 1 to find values of a & bThen integrate:
⌡⌠ a
x – 1 +
⌡⌠ b
x + 2
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Integrals involving the Log Function: ⌡⌠ 1
x dx = lnx + c
⌡⌠
h'(x)h(x)
dx = ln h(x) + c
• If one of the factors in the denominator is a quadratic, then
instead of b being part of the equivalence, use a basic linear function.
Example:
⌡⌠ 5x + 1
(x2 + 3x + 2)(x – 2)
= ax – 2
+ bx + c
x2
+ 3x + 2 • Note that sometimes polynomial division is needed to
solve an integrand.
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Topic 5
Volumes
• Appreciate that, by dividing a solid into a number of slices or shells, whose volumes can be simply estimated, the volume of the solid is the value of the definite integral obtained as the limit of the corresponding approximating sums.
A solid’s volume can be calculated by dividing it into slices or shells. Let us say ∆V = A(x).∆x By summing these together in a series and taking the limit as x à 0.
V = lim∆x → 0
Σa
b
A(x).∆x
V = ⌡⌠
a
b
A(x) dx
This can then be adjusted so a definite integral is reached. • The purpose of this topic is to provide practical examples
of the use of a definite integral to represent a quantity (in this case, a volume) whose value can be regarded as the limit of an appropriate approximating sum. Emphasis is to be placed on understanding the various approximation methods given, deriving the relevant approximate expression for the corresponding element of volume and proceeding from this to expressing the volume as a definite integral. The evaluations of infinite series by a definite integral, or of integrals by summation of series, are not included in this topic.
• Volumes of revolution could lead, from questions involving rotation about a coordinate axis, to rotation about a line parallel to a coordinate axis, eg find the volume of the solid formed when the region bounded by y = 2 x , the x-axis and x = 4 is rotated about the line x = 4.
• Always draw a sketch
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• Find the volume of a solid of revolution by summing the volumes of slices with circular cross-sections.
• Find the volume of a solid of revolution by summing the volumes of slices with annular cross-sections.
x = y
2
From this revolution we take a typical slice, which in this case is a spherical slice.
So, therefore the area of the circle can be expressed as:
A(x) = π y2
Meaning the volume of this slice is:
∆V = π y2 ∆x
By summing together and taking the limit, the volume of the whole solid is:
V = lim∆x → 0
Σ0
5
A(x) ∆x
V =
⌡⌠
0
5
A(x) dx
V = ⌡⌠
0
5
(π y2) dx
V = π ⌡⌠
0
5
y2 dx
Since y2 = x
∆x
radius = y
A(x)
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V = π ⌡⌠
0
5
x dx
V = π x
2
2
5
0
V = 25π2
units3
This method can also be applied to rotations around the y-axis. Notes: • Examples involving annular shells should include
questions as difficult as the following. The region R, bounded by: 0 ≤ x ≤ 2 , 0 ≤ y ≤ 4x
2– x
4, is rotated
about the y-axis. The solid so formed is sliced by planes perpendicular to the y-axis. Express the areas of the cross-sections so formed as a function of y, the distance of the plane from the origin. Use this result to calculate the volume of the solid.
• Find the volume of a solid of revolution by summing the volumes of cylindrical shells.
This is used when a graph is being rotated about the y-axis between x = a & x = b. Take for instance, this example, where we rotate the area enclosed between the line y = -x + 2 and the x & y-axes.
We firstly take our typical slice. This is a cylinder with: Radius = x Height = y Width = ?x This is then unfolded to give a rectangle. ?x
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Therefore: ∆V = Volume of a cylindrical shell . ∆V = [π (x + ∆x)
2 – π x
2] y
∆V = [π (x2
+ 2x∆x + ∆x2) – π x
2] y
∆V = π y (2x∆x + ∆x2)
However, ∆x2 is negligible, so:
∆V = 2π x y.∆x Taking the limit and integrating, gives:
V = lim
∆x → 0 Σ
0
2
∆V
V = 2π ⌡⌠
0
2
xy dx
Since y = – x + 2
V = 2π ⌡⌠
0
2
x ( – x + 2) dx
V = 2π ⌡⌠
0
2
( – x2 + 2x ) dx
Finding definite integral gives: V = 2π
– x
3
3 + x
2
2
0
V = 8π3
units3
Notes: • A formula for summing by cylindrical shells should not be
learnt. Each problem should rather be developed from first principles.
y
2πx
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• Find the volume of a solid which has parallel cross-sections of similar shape.
We are given the dimensions of the base and told the height is 4m. That is, 4m right through the middle of the pyramid from top to bottom. For this our typical slice is:
Since the sizes of a, b and c will vary proportional to how far along the pyramid we go, we break them down into three single triangles. This way, each of the values for a, b and c can be calculated in terms of x. Since each of these vary proportionally with x:
a5
= x4
b4
= x4
c3
= x4
a = 5x4
b = x
c = 3x4
4
3
5 4
a
c
b
∆x
5 4 3
a b c
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We now have values for our typical slice.
We then use the area of a triangle. A(x) = 1
2 bh
A(x) = 12
× 3x4
× x
Meaning the volume of the typical slice is:
∆V = 3x2
8 ∆x
By summing and taking the limit:
V = lim
∆x → 0 Σ
0
4
3x2
8 ∆x
V = ⌡⌠
0
4
3x2
8 dx
V = 3x
3
24
4
0
V = 8 units3
Notes: • The process of writing the limiting sum as an integral
should be extended to cases where cross-sections are other than circular. These cases should only involve problems in which the geometrical shape is able to be visualised, eg prove that the volume of a pyramid of height
h on a square base of side a is 13 a
2h
5x4
3x4
x
∆x
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Topic 6
Mechanics
Projectile Motion • Derive the
equations of motion of a projectile
Horizontally:
∴ x = vt cosθ
.x = v cosθ..x = 0
Vertically:
∴ y = vt sinθ – gt
2
2.y = v sinθ – gt..y = – g
Cartesian Equation of Motion:
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y = – gx2
2V2 (1 + tan
2θ) + x tanθ
Maximum Height: Max height occurs when
.y = 0
h = V
2sin
2θ
2g Range: Since projectile motion is a parabola, the time taken for the entire journey will be double that taken to reach the maximum height.
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R = V2sin 2θg
Maximum Range: Max range occurs when the projectile is fired at 45°.
R = V2
g Time of Flight: Since projectile motion is a parabola, the time taken for the entire journey will be double that taken to reach the maximum height.
T = 2Vsinθg
• Use equations for horizontal and vertical components of velocity and displacement to answer harder problems on projectiles.
Use the above equations and substitute in values that are known in order to find those that aren’t.
Simple Harmonic Motion
• Write down equations for displacement, velocity and acceleration given that a motion is simple harmonic.
Simple Harmonic Motion:
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x = a cos(nt + ∈).x = d
dx (a cos(nt + ∈)
.x = – an sin(nt + ∈)..x = d
dx ( – an sin(nt + ∈)
..x = – an
2cos(nt + ∈)
..x = – n
2[acos(nt + ∈)]
..x = – n
2x
• Use relevant
formulae and graphs to solve harder problems on simple harmonic motion.
Substitute in known values to the formulas above to find unknown values.
• Use Newton’s laws to obtain equations of motion of a particle in situations other than projectile motion and simple harmonic motion.
Newton’s Laws: • Every object in a state of uniform motion tends to
remain in that state of motion unless an external force is applied to it.
• The relationship between an object's mass m, its acceleration a, and the applied force F is F = ma. Acceleration and force are vectors (as indicated by their symbols being displayed in slant bold font); in this law the direction of the force vector is the same as the direction of the acceleration vector.
• For every action there is an equal and opposite reaction.
This translates to a few necessary formulas: F = mawhere F - Force in Newtons
m - Mass in kg
a - Acceleration in ms– 2
p = mvwhere p - Momentum
m - Mass
v - Velocity in ms– 1
1 kg wt = 9.8 N • The classical statement of Newton’s first and second laws
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of motion should be given as an illustration of the application of calculus to the physical world. Resolution of forces, accelerations and velocities in horizontal and vertical directions is to be used to obtain the appropriate equations of motion in two dimensions.
• Describe mathematically the motion of particles in situations other than projectile motion and simple harmonic motion.
Use the above formulas and represent situations using them. • Students should be able to represent mathematically,
motions described in physical terms. They should be able to explain, in physical terms, features given by mathematical descriptions of motion in one or two dimensions.
Mathematical Descriptions of
Motion
• Given ..x = f(x)
and initial conditions
derive v2 = g(x)
and describe the resultant motion.
If ..x = f(x), use d
dx 1
2 v
2
f(x) = ddx
1
2 v
2
12 v
2 =
⌡⌠
f(x) dx
• Recognise that a motion is simple harmonic given an equation for either acceleration, velocity or displacement, and describe the resultant motion.
Look for equations that look like the ones above. Also, if the motion needs to be proven, integrate or differentiate to show that:
..x = – n
2x
Resisted Motion along a horizontal
line
• Derive, from Newton’s laws of motion, the equation of motion of a particle moving in a single direction under a resistance proportional to a power of the speed.
Remembering F = ma, If resistance is – kv
n
∴ a = – kvn
F = – mkvn
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• Derive an expression for velocity as a function of time
If ..x = f(t), use dv
dtdvdt
= f(t)
v = ⌡⌠ f(t) dt
If ..x = f(v), use dv
dt if initial conditions are (t,v)
dvdt
= f(v)
dtdv
= 1f(v)
t = ⌡⌠ 1
f(v) dv
Then rearrange • Derive an
expression for velocity as a function of displacement.
If ..x = f(x), use d
dx 1
2 v
2
ddx
1
2 v
2
= f(x)
12 v
2 =
⌡⌠
f(x) dx
• Derive an expression for displacement as a function of time.
If
..x = f(t), use d
2x
dt2
d2x
dt2 = f(t)
Then intergrate twice Motion Vertically
Upwards
• Derive, from Newton’s laws of motion, the equation of motion of a particle, moving vertically upwards in a medium, with a resistance R proportional to the first or second power of its speed.
Remember F = ma, ..x = – (g + kv
n)
F = – m (g + kvn)
• Typical cases to consider include those in which the
resistance is proportional to the speed and to the square of the speed.
• Analysis of the motion of a particle should include consideration of the behaviour of the particle as t becomes large. Graphs can offer assistance in understanding the behaviour of the particle.
• The origin should be placed at the point of projection. • The maximum height reached by the particle can be
obtained from the expression relating speed and displacement.
• The time taken to reach this maximum height can be
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obtained from the expression relating speed and displacement.
• The time taken to reach this maximum height can be obtained from the expression relating speed and time.
• Problems should include cases where the magnitude of
the resistance is given. (eg: R = 110
v2
)
• Derive an expression for velocity as a function of time
If ..x = f(t), use dv
dtdvdt
= f(t)
v = ⌡⌠ f(t) dt
If ..x = f(v), use dv
dt if initial conditions are (t,v)
dvdt
= f(v)
dtdv
= 1f(v)
t = ⌡⌠ 1
f(v) dv
Then rearrange • Derive an
expression for displacement as a function of time.
If
..x = f(t), use d
2x
dt2
d2x
dt2 = f(t)
Then intergrate twice • Solve problems
by using the expressions derived for acceleration, velocity and displacement.
Use the above equations and substitute in values.
Motion of a Particle Falling Downwards
• Derive, from Newton’s laws of motion, the equation of motion of a particle falling in a medium with a resistance R proportional to the first or
Remember, F = ma a = g – kv
n
F = ma
F = m (g – kvn)
• Cases, other than where the resistance is proportional to
the first or second power of the speed, are not required to be investigated.
• Students should place the origin at the point from which
4u Maths Summary Page 98 of 117
second power of its speed.
the particle initially falls. If the motion of a particle both upwards and then downwards is considered then the position of the origin should be changed as soon as the particle reaches its maximum height. Care must then be taken in determining the correct initial conditions for the downward motion.
• The time taken for the particle to reach the ground should be found.
• Problems should include a study of the complete motion of a particle, projected vertically upwards, which then returns to its starting point. For specific resistance functions, comparisons should be made between the times required for its upward and downward journeys and between the speed of projection and the speed of its return.
• Determine the terminal velocity of a falling particle, from its equation of motion.
Terminal velocity occurs when acceleration has ceased. g – kvn
= 0
g = kvn
vn = g
k
v = n gk
• Derive expressions for velocity as a function of time and for velocity as a function of displacement.
Time:
If ..x = f(t), use dv
dtdvdt
= f(t)
v = ⌡⌠ f(t) dt
If ..x = f(v), use dv
dt if initial conditions are (t,v)
dvdt
= f(v)
dtdv
= 1f(v)
t = ⌡⌠ 1
f(v) dv
Then rearrange
Displacement:
If ..x = f(x), use d
dx 1
2 v
2
ddx
1
2 v
2
= f(x)
12 v
2 =
⌡⌠
f(x) dx
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• Derive an expression for displacement as a function of time.
If
..x = f(t), use d
2x
dt2
d2x
dt2 = f(t)
Then intergrate twice • Solve problems
by using the expressions derived for acceleration, velocity and displacement.
Use above equations and substitute in known values.
Circular Motion • Define angular
velocity of a point moving about a fixed point.
Angular Velocity:
ω = dθdt
= .θ
• Deduce, from
this definition of angular velocity, expressions for angular acceleration of a point around a fixed point.
Through differentiation ω = dθ
dt.ω = d
2θ
dt2 =
..θ
• Prove that the instantaneous velocity of a particle moving in a circle of radius R, with angular velocity ω , is Rω .
Velocity: v = Change in arc AB
Change in time
v = ddt
Rθ
v = R dθdt
v = Rω
• Prove that the
tangential and normal components of the force acting on a particle
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moving in a circle of radius R, with angular velocity ω , need to be Mr
.ω and
– mRω2
respectively. Uniform Circular
Motion
• Write down the formula appropriate for a particle moving around a circle with uniform angular velocity.
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• Apply these
formulae to the solution of simple problems
Use above equations, substitute in known values.
Conical Pendulum • Use Newton’s
law to analyse the forces acting on the bob of a conical
Conical Pendulum:
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pendulum.
Vertically: T cosθ – mg = 0 Radially:
T sinθ = mv2
r = mrω
2
• Derive results
Tension = 4π2
mn2l
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h = g
ω2
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tanθ = v2
rg • Discuss the
behaviour of the pendulum as its features vary.
• The vertical depth of the bob below the pivot point is independent of the length of the string and the mass of the bob.
• As the speed of the particle increases, it rises upwards. • Apply derived
formulae to the solution of simple problems.
Use above formulas and substitute in known values.
Banked Circular Track
• Use Newton’s laws to analyse the forces acting on a body, represented by a particle, moving at constant speed around a banked circular track.
Vertically: N cosθ – F sinθ – mg = 0 Radially:
N sinθ + F sinθ = mv2
r • Derive results
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h = v2d
Rg
tanθ = v2
Rg • Calculate the
optimum speed around a banked track given the construction specifications.
v = Rg tanθ
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• Calculate the forces acting on a body, travelling around a banked track, at a speed other than the optimum speed.
F = mv2
r cosθ – mg sinθ
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Topic 7
Polynomials
Integer roots of polynomials with
integer coefficients.
• Prove that, if a polynomial has integer coefficients and if a is an integer root, then a is a divisor of the constant term.
The general form of a polynomial is: anx
n + an – 1x
n – 1 + an – 2x
n – 2 + ..... + a1x + a0
a0 is the constant term If a is an integer root, then a is a factor of a0 .
• Test a given polynomial with integer coefficients for possible integer roots.
This is known as the remainder theorem. If P(a) = 0 then: • (x – a) is a factor of P(x) • a is a root of P(x) Example: P(x) = x
2– 6x + 5 = 0
Testing x = 5
P(5) = 52 – 6 × 5 + 5 = 0
Since P(5) = 0 , (x – 5) is a factor of P(x) • All possible integer roots of polynomials lie among the
positive and negative integer divisors of its constant term. However, not all polynomials contain integer coefficients.
If P ba
= 0 then:
• (ax – b) is a factor of P(x) • b
a is a factor of P(x)
Where b is a factor of the constant term and a is a factor of the leading term. Example: P(z) = 2z
3 – 3z
2 + 2z – 3 = 0
z could potentially be any of the following:
±3 , ±1 , ±32 , ±1
2
In this case, P
32
= 0.
∴ (2z – 3) is a factor of P(z)
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Multiple Roots • Define a multiple
root A polynomial of degree n has n roots, but they are not necessarily all different. We say that ‘a’ is a root of multiplicity ‘r’ when the factor (z – a) occurs ‘r’ times. Therefore: P(x) = (z – a)
r.Q(x)
• Write down the order (multiplicity) of a root.
The order of a root is the number of times it appears as a factor.
• Prove that if P(x) = (x – a)r.S(x), where r > 1 and S(a) ≠ 0 , then P'(x) has a root a of multiplicity (r – 1).
P(x) = (x – a)
r.S(x) where r > 0 and S(a) ≠ 0
u = (x – a)r
dudx
= r (x – a)r – 1
v = S(x)
dvdx
= S'(x)
P'(x) = (x – a)r.S'(x) + r (x – a)
r – 1.S(x)
= (x – a)r – 1
[(x – a)S'(x) + r S(x)]
= (x – a)r – 1
Q(x) • Solve simple
problems involving multiple roots of a polynomial.
These can include examples whereby the derivative of a function must first be established.
• State the fundamental theorem of algebra.
The Fundamental Theorem of Algebra asserts that every polynomial P(x) of degree ‘n’ over the complex plane has at least one complex root.
• Deduce that a polynomial of degree n > 0, with real or complex coefficients, has exactly n complex roots, allowing for multiplicities.
Let P(z) = anzn
+ an – 1zn – 1
+ .... + a1z + a0 = 0
Let z1 be a root. So P(z1) = 0
∴ P(z) = (z – z1)Qn – 1(z)
where Qn – 1(z) is a polynomial of degree (n – 1)
By continuing this pattern we get:
P(z) = an(z – z1)(z – z2)....(z – zn) where an ≠ 0 Using this result, the factor theorem should now be used to prove (by induction on the degree) that a polynomial of degree n > 0 with real or complex coefficients has exactly n complex roots.
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Factoring Polynomials
• Recognise that a real polynomial of degree n can be written as a product of real linear and real quadratic factors.
• Factor a real polynomial into a product of real linear and real quadratic factors.
A real polynomial can be written as a product of real linear and quadratic factors. Example: P(z) = 2z
3– 3z
2+ 2z – 3
Using the remainder theorem:
P 3
2 = 0
∴ 2z – 3 is a factor of P(z) z
2+ 0z + 1
2z – 3 |2z3
– 3z2
+ 2z – 3
2z3
– 3z2
0z2
+ 2z
0z2
+ 0z 2z – 3 2z – 3
0 P(z) = (2z – 3)(z
2+ 1)
• Recognise that a complex polynomial of degree n can be written as a product of n complex linear factors.
• Factor a polynomial into a product of complex linear factors.
A complex polynomial of degree n can be written as a product of n complex linear factors. Using the example from above: P(z) = (2z – 3)(z
2+ 1)
= (2z – 3)(z2
– i2)
= (2z – 3)(z – i)(z + i) • As you can see above, the complex roots of real
polynomials occur in conjugate pairs. That means that if (z – i) is a factor of P(x), (z + i) is also a factor. • Students should be able to factor cubic and quartic
polynomials over both the real and complex planes. There are instances where a polynomial cannot be factorised using the remainder theorem. There are a number of alternatives that need to be familiarised: • Difference of cubes:
z3
– 1 = 0
(z – 1)(z2
+ z + 1) = 0
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• Difference of two squares: z
4= 1
z4
– 1 = 0
(z2
– 1)(z2
+ 1) = 0z = ±1, ±i
• Completing the square: z
4= -1
z4
+ 1 = 0
z4
+ 2z2
+ 1 – 2z2
= 0
(z2
+ 1)2
– 2z2
= 0
(z2
+ 1 – 2 z)(z2
+ 1 + 2 z) = 0
(z2
– 2 z + 1)(z2
+ 2 z + 1) = 0 • Any combination of two or more of the above methods.
• Write down a polynomial given a set of properties sufficient to define it.
If given the roots, say a, b & c. P(x) = (x – a)(x – b)(x – c)
• Solve polynomial equations over the real and complex planes.
Solve as shown above. There is, however, another method that can be employed. De Moivre’s Theorem: If z = r(cosθ + isinθ) then
zn = r
n(cosθ + isinθ)
n
zn = r
n(cosnθ + isinnθ)
∴ rn(cosθ + isinθ)
n = r
n(cosnθ + isinnθ)
Example: Use De Moivre's theorem to express cos 3θ in terms of cosθ and sin 2θ in terms of sinθ. cos 3θ + isin 3θ
= (cosθ + isinθ)3
= cos3θ + 3icos
2θsinθ – 3cosθsin
2θ – isin
3θ
Equating Real & Imaginary parts:
cos 3θ = cos3θ – 3cosθsin
2θ
= cos3θ – 3cosθ(1 – cos
2θ)
= 4cos3θ – 3cosθ
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sin 3θ = 3cos
2θsinθ – sin
3θ
= 3(1 – sin2θ)sinθ – sin
3θ
= 3sinθ – 4sin3θ
Roots &
Coefficients of Polynomials
• Write down the relationships between the roots and coefficients of polynomial equations of degrees 2, 3 & 4.
Quadratic Equations: If α & β are roots of ax
2+ bx + c = 0 then:
α + β = – ba
αβ = ca
Cubic Equations: If α , β & γ are roots of ax
3+ bx
2+ cx + d = 0 then:
α + β + γ = – ba
αβ + βγ + αγ = ca
αβγ = – da
Quartic Equations: If α , β, γ & δ are roots of ax
4+ bx
3+ cx
2+ dx + e = 0 then:
α + β + γ + δ = – ba
αβ + αγ + αδ + βγ + βδ + γδ = ca
αβγ + αβδ + βγδ + αγδ = – da
αβγδ = ea
• Use these relationships to form a polynomial equation given its roots.
Quadratic: P(x) = x
2 – (sum of roots )x + (Product of roots )
Cubic:
P(x) = x
3 – (Sum of roots )x
2 + (Sum of roots 2 at a time )x
– (Product of roots ) Quartic: P(x) = x
4 – (Sum of roots )x
3 + (Sum of roots 2 at a time )x
2
– (Sum of roots 3 at a time )x + (Product of roots )
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• Form an equation, whose roots are a multiple of the roots of a given equation.
Roots are: mα, mβ & mδ
m(α + β + δ) = – mba
m2(αβ + αδ + βδ) = m
2c
a
m3αβδ = – m
3d
a
∴ x3
+ mba
x2
+ m2c
a x + m
3d
a = 0
ax3
+ mbx2
+ m2cx + m
3d = 0
• Form an equation, whose roots are the reciprocals of the roots of a given equation.
Roots are: 1α
, 1β
& 1γ
1α
+ 1β
+ 1γ
= βγ + αγ + αβαβγ
= ca
× – ad
∴ 1α
+ 1β
+ 1γ = – c
d 1
αβ+ 1
αγ+ 1
βγ
= α + β + γαβγ
= – ba
× – ad
∴ 1αβ
+ 1αγ
+ 1βγ
= bd
1αβγ
= – ad
∴ x3 + cd x2 + b
d x + a
d = 0
dx3 + cx
2 + bx + a = 0
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• Form an equation, whose roots differ by a constant from the roots of a given equation.
Roots are: α + k , β + k & γ + k
α + β + γ + 3k = – ba
+ 3k = – b + 3aka
(α + k)(β + k) + (α + k)(γ + k) + (β + k)(γ + k)
= αβ + αk + βk + k2+ αγ + αk + γk + k2
+ βγ + βk + γk + k2
= αβ + αγ + βγ + 2k(α + β + γ) + 3k2
= ca
+ 2k – b
a + 3k
2
= ca
– 2bka
+ 3k2 = c – 2bk + 3ak
2
a
∴ Sum of Roots Two at a time = c – 2bk + 3ak2
a (α + k)(β + k)(γ + k)
= αβγ + αβk + αγk + αk2+ βγk + βk2
+ γk2+ k3
= αβγ + k(αβ + αγ + βγ) + k2(α + β + γ) + k
3
= – da + kc
a +
– k
2b
a + k
3
= – d + kc – k2b + ak
3
a
= – d – kc + k
2b – ak
3
a
∴ x
3 + b – 3ak
a x
2 + c – 2bk + 3ak
2
a x + d – kc + k
2b – ak
3
a = 0
ax3 + (b – 3ak)x
2 + (c – 2bk + 3ak
2)x + (d – kc + k
2b – ak
3) = 0
• Form an
equation, whose roots are the squares of the roots of a given equation.
Roots are: α
2 & β
2
α2
+ β2
= (α + β)2 – 2αβ
= – b
a 2
– 2ca
= b2
a2 – 2c
a
= b2
– 2ac
a2
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α2β2 =
ca 2
= c2
a2
∴ x2 + 2ac – b
2
a2 x + c
2
a2 = 0
a2x
2 + (2ac – b
2)x + c
2 = 0
Partial Fractions
• Write a fraction in terms of quotient and remainder.
f(x) = 2x2
+ 3x + 8x + 2 2x – 1
x + 2 2x2
+ 3x + 8
2x2
+ 4x – x + 8 – x – 2 10
f(x) = Q(x) + R(x)B(x)
f(x) = (2x – 1) + 10x + 2
• Write a fraction
in terms of distinct linear factors.
f(x) = x + 3
x2
– 6x + 5x + 3
x2
– 6x + 5 ≡ a
x – 5 + b
x – 1
x + 3 = a(x – 1) + b(x – 5)Let x = 14 = -4bb = -1Let x = 58 = 4aa = 2
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∴ x + 3
x2
– 6x + 5 = 2
x – 5 – 1
x – 1
• Write a fraction in terms of distinct linear factors and a simple quadratic factor.
6x2
– 53x + 42
(2x – 3)(2x2
– 5x – 3) ≡ a
2x – 3 + bx + c
2x2
– 5x – 3
6x2
– 53x + 42 = a(2x2
– 5x – 3) + (bx + c)(2x – 3)Let x = 1.5-24 = -6aa = 4Let x = 042 = -12 – 3cc = -18Let x = 1-5 = (-6 × 4) + (b – 18)(-1) -19 = b – 18b = -1
f(x) = 42x – 3
– x + 18
2x2
– 5x – 3
• Write a fraction in terms of the product of two different simple quadratic factors.
f(x) = 2x
3+ 7x
2– 4x + 3
(x2
– 1)(x2
+ 1)
2x3
+ 7x2
– 4x + 3
(x2
– 1)(x2
+ 1) ≡ ax + b
x2
– 1 + cx + d
x2
+ 1
2x3
+ 7x2
– 4x + 3 = (ax + b)(x2
+ 1) + (cx + d)(x2
– 1)Let x = 18 = 2(a + b)a + b = 4 (1)Let x = 03 = b – d (2)
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Let x = 239 = 5(2a + b) + 3(2c + d) (3)Let x = -112 = 2(b – a)b – a = 6 (4)(1) into (4)4 – a – a = 6-2a = 2a = -1b = 5d = 2Sub a, b & d into (3) 39 = 15 + 6c + 66c = 18c = 3
f(x) = 3x + 2
x2
+ 1 – x – 5
x2
– 1
• Apply these partial fraction decompositions to the integration of corresponding functions.
Refer to integration summaries.
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Topic 8
Harder 3 Unit
Circle Geometry • Solve more
difficult problems in geometry.
See attached sheet.
Induction • Carry out proofs
by mathematical induction in which S(1), S(2)…S(k) are assumed to be true in order to prove S(k+1) is true.
• Use mathematical induction to prove results in topics which include geometry, inequalities, sequences and series, calculus and algebra.
Mathematical induction occurs in four steps. STEP 1) Prove the statement is true for the lowest possible integer value, usually n = 1. STEP 2) Assume the result is true for n = k. STEP 3) Use algebraic manipulation to prove the result is true for n = k + 1. STEP 4) Have a concluding statement.
Since the formulas was proven true for n=1, it was assumed true for n=k. It was then proven true for n=k+1, meaning it is true for all n = 1.
Inequalities • Prove simple
inequalities by use of the definition of a > b for real a and b.
• Prove further results involving inequalities by logical use of previously obtained inequalities.
If a > b then:
(a – b)2 > 0
a2
– 2ab + b2 > 0
a2
+ b2 > 2ab
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