based on 1990 syllabus -...

HSC Mathematics Extension 2

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Mathematics Extension 2 Based on 1990 Syllabus

This summary has been written to be as short as possible, to cover everything if you already

have a good idea of the course contents, only need a quick revision and does not want to print

too many pages. The depth of coverage of a particular topic does not necessarily correspond to

its significance. Proofs have been given when they can be easily derived, in case you forget the

formulae. Important results have been boxed for convenience.

The majority of this work is based on the Cambridge Mathematics textbook. Sections are

numbered according to the syllabus. This document was last updated on 3 September 2010.

1. Graphs Critical point

It is a point at which the derivative if not defined (remember that derivative is a two-sided

limit). This happens:

at any point of discontinuity

at an endpoint of a finite domain, eg point (0, 0) in xy

at sharp edges, eg point (0, 0) in xy

if the tangent is vertical

Addition and substraction of functions, xgxfy

Sketch the two functions and add or subtract their heights.

Translations

kxFhy is y = F(x) translated h upwards and k to the right.

Reflection in coordinate axes

F(x) reflected in y-axis becomes F(-x). F(x) reflected in x-axis becomes -F(x).

Reflection in lines ax or by : Replace x by xa 2 or y by yb 2 , respectively. Note

that successive reflections is the same as rotation of 180 about ba, .

Functions involving absolute values, xFy , xFy and xFy

Inspect the sign of the thing inside |…| to break the functions.

For example, 13 xxy .

For x = -1 and x = 3, we can use more than 1 sub-

function, since the thing inside |…| equals zero.

Multiplication of functions, xgxfy

Draw the two functions, then examine them for things such as:

When f(x) lies below or above g(x), and when y lies below or above them, and when they

intersect. Particularly useful is the conditions when 11 x , 1x and 1x . Try to

multiply the heights of the two curves in your mind, eg when f(x) = 0.5, y will be under

g(x) as y = 0.5 * g(x).

What happens as x approaches positive and negative infinity (asymptotes) and as x

approaches zero for some curves. Which sub-function, f(x) or g(x), is dominant? Also, if

both y and f(x) has the same horizontal (or vertical) asymptote, which curve is below which

(which reaches the asymptote faster)?

3,13

31,13

1,13

xxx

xxx

xxx

y


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domain and vertical asymptotes

odd, even and other symmetries. ODD x ODD = EVEN. ODD x EVEN = ODD, etc. The

rule when multiplying (and dividing) functions is the same as that of positive and negative

numbers (treat odd like negative number). Other symmetry can include symmetry in the

line y = x (the function won‟t be affected by interchanging y and x in its equation) and

symmetry in x = a or y = a.

turning points and inflexion points: use calculus if they can‟t be easily determined

These steps apply to all sketches in general.

Division of functions xg

xfy

same principles as above

x-intercepts of g(x) correspond to vertical asymptotes

Reciprocal Functions xg

y1

g(x) and y have the same sign, intersect where 1xg

y decreases when g(x) increases, maximum point in g(x) becomes minimum point in y

Rational Functions xQ

xPy (capital letters denote polynomials)

if deg P(x) is less than deg Q(x), divide top and bottom by the highest power of x. eg,

21

1

2

2

2 11x

x

x

x

x

xy

if deg P(x) is equal to deg Q(x), get y to the form xQ

cky , by rewriting P(x) so that a

part of it is divisible by Q(x). eg,

1

32

1

312

1

1222

2

2

2

xx

x

x

xy

if deg P(x) exceeds deg Q(x) by 1, get y to the form xQ

cxFy , F(x) linear

Doing these transformation will enable us to look for asymptotes (as x gets large).

Graphs of nxfy ( ,1n n integral) -- xfxfndx

dy n'

1

the x-intercepts and x-coordinates of stationary points of f(x) give the stationary points of y.

Their nature will depend on the signs of f and f‟ and on whether n is odd or even.

Remember that when n is even, 0y .

Graphs of xfy --

xf

xf

dx

dy

2

'

x-intercepts of f(x) give critical points of y, and the x-coordinates of stationary points of

f(x) give stationary points of y, provided 0xf

pay attention to domain, and where y lies below or above f(x)

Graphs of composite functions, xguwhereufy ,

sketch g(x) and f(x), and use the graphs to guess what y looks like

pay attention to things like domain and asymptotes and where the curve is increasing and

decreasing. Find the nature of f(x) then translates the conditions to g(x). For example,

consider the function xy cosln . f(x) = ln x increases as x increases. This means y


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increases when g(x) = cos x increases. Also, f(x) = ln x has vertical asymptote at x = 0, so y

has vertical asymptote(s) whenever cos x = 0, that is when 21 nx , n integers. Note

the domain of y, too. This is where 0cos x .

Implicit Differentiation

This technique is especially for relations and not functions. Take the derivatives of both sides

with respect to x, treating y is a function of x, and using the chain rule.

2. Complex Numbers About the Number System

Real numbers(R)Rational numbers(Q)Integers(Z or J)Cardinal numbers(N)

Imaginary numbers (M)

Both R and M are part of the complex number system (C). Note that complex numbers are not

ordered. In particular, non-real numbers cannot be compared using the symbols > or < and

can‟t be designated positive or negative.

Notations (Algebraic and Geometrical)

Any complex number z can be represented as 1,,, iRyxiyxz . We should

write “ iba ” (i before pronumeral b) but “ i32 ” (i after actual number).

x is called the real part of z, or Re z. y is called the imaginary part, or Im z.

In an Argand diagram, iyxz is represented by the point P (x, y).

This position can also be specified by polar coordinates (r, θ) where r is the length of OP

( 0r ) and θ is the angle from the positive direction of the real axis (the “x-axis”) to the

ray OP (θ in radians, , positive θ is anticlockwise).

r is called the modulus of z, mod z OR z . 22 yxzzz .

θ is the principal argument of z, . The argument α of z, arg z, is technically

Znn ,2 , although the term usually refers to θ. To find θ, find its related angle,

x

y1tan , then just adjust the quadrant.

Drawing point P, we notice sin,cos ryrx . Hence iyxz can also be written

as 0,sincos rrcisirz . This is called the modulus-argument form. Note

that 000 iz can‟t be written in this form.

Conjugates and Reciprocals and Divisions

The conjugate of iyxz is iyx , written as z (a dash denotes conjugate). It can be proven

that 21212121 zzzzandzzzz .

The product zz is real. Hence, the reciprocal and quotient (result of division) of complex

numbers can be found by realising the denominator, for example:

formyixiniiii

i

i

25

3

25

434

25

1

3434

341

34

1.

The conjugate of rcisz is rcisz .

Quadratic Equations with Real Coefficients and Negative Discriminants

Note that every negative real number has two complex square roots. For example, ,1616 2i

-16 has square roots 4i and -4i. Similarly, this type of quadratics has solutions a

ibx

2

(notice, they are conjugates).


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Quadratics with Complex Coefficients and Square Roots of Complex Numbers

Quadratics with complex coefficients often have complex discriminants. Hence we must be

able to find the square roots of a complex number a + ib.

Let the roots be iyxz . ibaz 2 . ibaixyyxiyx 2222, then by equating

real and imaginary parts, bxyandayx 222 . Both equations represent rectangular

hyperbolas. By examining their graphs for various values of a and b, we know that there are

always 2 values for z, 11 zandz [can be assumed I think].

Products and Quotients in Modulus-Argument Form

21212121 argargarg zzzzandzzzz 21

2

1

2

1

2

1 argargarg zzz

zand

z

z

z

z

Geometrical Relationships

Using the products result above, we can see that:

z is the reflection of z in the real axis on an Argand diagram.

iz is z rotated anticlockwise about O through 2

.

cz, where c is a positive real number, is an enlargement (or reduction) of z about the origin

O by a factor of c.

–cz, where c is a positive real number, is z enlarged and then rotated anticlockwise by π, or

equivalently, reflected in the origin. iczicz .

Geometrical Representation

iyxz represents the point P (x, y) on an Argand diagram, which can also be represented by

the „position vector‟ OP . Its magnitude is z and its direction is given by the principal arg z.

We can translate this vector. Hence we have infinitely many vectors representing z, and they

are called „free vectors‟.

Addition and Substraction of Vectors

To add vectors, qp

, on an Argand diagram, translate q

so that its tail is at the tip of p

(tip

= the arrow end). Then join the tail of p

with the tip of the translated q

. The tip of qp

is

the tip of the translated q

.

To subtract vectors, qp

, translate q

as before, but reverse the arrow of q

. Join the two

vectors as before, the tip of qp

is the tip of the translated (and reversed) q

.

Addition and Substraction Using Parallelograms

If we let OAz 1 and OBz 2 and construct a parallelogram with the other vertex C, 21 zz

will be the diagonal OC . Diagonal BA will be a translation of 21 zz (you can think of

12 zz which starts at B and ends at A).

Using That Parallelogram to Show 212121 zzzzzz (Triangle Inequality)

ACOAOC , with equality holding if O, A and C are collinear.

OBOAOC , since opposite sides of parallelogram are equal.

2121 zzzz , with equality if 1z and 2z are parallel, i.e. Rcczz ,21 and 21 zz

has the greatest value. Also, 2121 zzzz , with equality if 1z and 2z are parallel in

opposite directions, i.e. 21 zz has the least value, Rcczz ,21 .

De Moivre’s Theorem


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This theorem says that Znnciscisn

, . This can be seen from the result that

2121 argargarg zzzz , or can be proven by mathematical induction for positive n (with

the help of compound-angle identity). To prove for negative integers –n,

nciscisciscisnnn

1

. Also holds for n = 0.

Further, if ncisrzrcisz nn , .

De Moive’s Theorem to Express Powers of cos and sin in Other Notations

Using this theorem when cisz gives nzz nn cos2 and nizz nn sin2

. This

enables us to express powers of cos θ and sin θ in terms of the cosine and sine ratios of

multiples of θ. Sample question: Show that 3sinsin34

1sin3 . Let cisz .

1sin2 zzi and sin63sin23sin8 1333133 iizzzzzzi .

De Moivre’s Theorem to Find Roots of ±1 and Other Numbers

Let z be the nth complex roots of unity (meaning, roots of 1). Then 01 cisz n . Let

ncisrzrcisz nn , . Hence, 1r and 0cisncis . Zkkn ,20 .

1...,,2,1,0,2

nkkn

. Similar method can be used to find roots of -1 and of any

complex number. Some properties of the roots of nz [ nz is any complex number with modulus

1]:

They are equally spaced around the unit circle (circle with a radius of 1) with centre O and

so form the vertices of a regular n-sided polygon

The non-real roots occur in conjugate pairs [not true if 0Im nz ]

All roots add up to zero

All roots of +1 can be expressed as powers of the „primitive non-real root‟ ω (the root

when k = 1), i.e. as 110 ,...,, n

Curves and Regions on Argand Diagram

Firstly, notice that 2

1

2

1

2

111 ImRemod zzzzzzzzzz and that

it represents the distance from z to the point 1z . Also, 1arg zz refers to the angle which a

vector joining point 1z to point z makes with the positive direction of the real axis. Below are

some loci of P representing z you should remember.

RkkzzORkz ,ReRe 1 vertical line

RkkzzORkz ,ImIm 1 horizontal line

21 zzzz line, perpendicularly bisects the line connecting 1z and 2z

RkkzzORkz ,1 circle with radius k, centre O or point 1z

1argarg zzORz ray originating from O or 1z (which is not part of locus)

with gradient tan

Other loci: often the best approach is to substitute iyxz into the locus condition to get

the Cartesian equation. Some loci which may appear: aziaz Im : parabola

ayx 42 ; azaz Re : parabola axy 42 ; azzzz 221 : ellipse, major axis

2a, foci 1z and 2z ; azzzz 221 : hyperbola, transverse axis 2a, foci 1z and 2z ,


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only the branch in 2z side; azz 422 : hyperbola axy ;

2

1argzz

zz: arc of a

circle; kzz

zz

2

1: circle.

You should be able to find maximum and minimum z and arg z in some loci, eg circles.

You should be able to sketch and describe regions involving inequalities of the above

curves and combination of inequalities ( = union / “or”; = “and”).

Questions

1. Use De Moivre‟s theorem to find, in mod-arg form, the cube roots of i3232 .

SOLN: Znnciscisi

,2

4

38

4

383232

. Using De Moivre‟s theorem,

ncisncis 2

4

3

3

182

4

38 3

13

1

1,0,1,3

2

4

12

nncis .

2. A complex number z is such that when it is divided by i57 , the real part is twice the

imaginary part. Find all possible values of z. SOLN: 74

7557

57

57

57

iyixyx

i

i

i

iyx

.

xyyxyx

9

17

74

752

74

57

. kikiyxz

9

17 or ic 179 , k and c real.

3. By mathematical induction, prove that nn zz and znz n argarg , where rcisz and

Zn . Extend this to prove for all integers n.

4. Sketch the locus 11 iz and find the maximum and minimum values of z .

5. Find the locus of 23

3

z

z. SOLN: Let iyxz . 22

3323 yxzz

22222234332 yxyxyx which gives 165 22

yx .

6. iyxz is such that 1

z

izI is purely imaginary. Find the equation of the locus of point

P representing z and show this locus on an Argand diagram. If it‟s imaginary, the real part

is zero. iyx

iyx

iyx

iiyx

z

iz

1

1

11. Real part will be

22

1

11

yx

yyxx

. To be zero,

5.05.05.001122 yxyyxx which is a circle. The locus excludes

iz (at which 0I ) and 1z (at which I is undefined).

Questions Requiring Algebraic Manipulation

7. If w is a complex cube root of unity, evaluate 2

2

awcwb

cwbwa

where a, b and c are real

numbers. SOLN:

wacwbw

cwbwaw

awcwbw

cwbwaw

w

w

awcwb

cwbwa

2

2

32

2

2

2

(note:

13 w ).

8. 2,,1 wandw are the three cube roots of unity. State the values of 3w and 21 ww .

Hence simplify each of the expressions 2231 ww and 2231 ww and show that

their sum is 4 and their product is 16 . SOLN: 13 w ; 2,,1 wandw are the roots to

0100 23 zzz . Sum of roots = 21 ww = 0 (using a theorem from the


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Polynomials topic). Now, 222222 4202131 wwwwwww and

wwwwwwwww 4442131 3422222 . Sum = 144 2 ww

since 01 2 ww . Product = 1616 3w .

9. Show that 2

sin2

cos2

cos2sincos1 nini nnn , Zn . SOLN: since

the RHS have 2

n as its argument, change the LHS to have

2

. In LHS, sincos1 i

2sin

2cos

2cos2

2cos

2sin21

2cos21 2

ii . Now, De Moivre‟s

theorem.

10. Show that the roots to 01166 zz are given by

1212cot

kiz ,

6,5,4,3,2,1k . SOLN: The equation can be re-written as

11

16

6

z

z. Let

11

1 6

w

z

zw . w

ciskcis

36.

sin1cos

sin1cos

1

111

1

1

i

i

cis

ciszciszzcis

z

z

1cos

sin

cos12

sin2

sin1cos

sin1cos

i

i

i

i

2cot

2cot

2sin2

2cos

2sin2

2

iii ,

6

1236

kk . 6,5,4,3,2,1,

1212cot

1212cot rrikiz

.

Questions Requiring Understanding of Vectors

11. Use the vector representation of 1z and 2z on an Argand diagram to show that if 21 zz ,

21

21

zz

zz

is purely imaginary. SOLN: A number is purely imaginary if its argument is

2 . 21 zz and 21 zz are diagonals of a rhombus and bisect at right angles. Hence

2

argargarg 2121

21

21

zzzz

zz

zz.

12. Show geometrically that all unique complex roots of any complex number add up to zero,

by using their representation on the Argand diagram. SOLN: In short, because those roots

are evenly spaced on the Argand diagram, they can be translated to form a regular polygon

with a vertex at the origin.

13. Prove that 4123

4321

zzzz

zzzzK

is real if 4321 ,,, zzzz

are concyclic. SOLN:

greenzzzzzz

zz

2321

23

21 argargarg ,

pinkzz

zz

41

43arg , Karg green + pink =

(property of cyclic quadrilateral), hence K is real.

14. Sketch the curve in the Argand diagram determined by

41

1arg

z

z and find its Cartesian equation. SOLN:


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Draw P(z) on the Argand diagram such that 4

1arg1arg

pinkgreenzz . This

will be an arc of a circle with 4

APB as drawn in the diagram. We can also find that

the equation is 0,2122 yyx . Point A is excluded since at A we have

undefinedz

z

0arg

1

1arg . Point B is excluded because at B we have

0

2arg

1

1arg

z

z.

15. Sketch the curve determined by 41

1arg

z

z, then sketch

41

1arg

z

z. SOLN: The

two curves are identical, and will be similar to the previous question but flipped upside

down.

16. Sketch the curve in the Argand diagram determined by 4

3

1

1arg

z

z. SOLN: It will be

like in 41

1arg

z

z but it‟s the minor arc instead of the major arc.

3. Conics A „conic section‟ or simply „conic‟ is the locus of point P such that the

ratio of the distances from P to a fixed point S (the focus) and to a fixed

straight line m (the directrix) is a constant e (the eccentricity). In fact it

could have two pairs of focus and directrix. Three possible cases are

shown on the right.

Some terms to remember:

extremities = farthest points, eg extremities of a chord is where it cuts the conic

diameter = chord through the centre

latus rectum = focal chord perpendicular to the major axis

Ellipse

Major axis is an axis of symmetry, perpendicular to the directrix (and hence passes the two

foci). Minor axis is also an axis of symmetry, but parallel to the directrix.

Cartesian Equation: 12

2

2

2

b

y

a

x Parametric Equation:

sin

cos

by

ax

20

a = half the length of the axis parallel to x-axis [Using the definition above, the major axis is

the longest axis of symmetry and is vertical when ab ]

b = half the length of the axis parallel to the y-axis

θ = the angle made by the line from O to P’, a point on the auxiliary circle 222 ayx with

the same x-coordinate and in the same quadrant as P

Notice that an ellipse is formed by dilation of its auxiliary circle by a factor of ab

perpendicular to its major axis.

Translation:

12

2

2

2

b

ky

a

hx or

kby

hax

sin

cos has centre (h, k).

Key Features

Comparison between the ab and ab type ellipses shows that the roles of x and y are

interchanged, as are the roles of a and b in determining e, S and m.

Ellipse Eccentricity Foci Directrices

hyperbolae

ellipsee

parabolae

1

10

1


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typeab

222 1 eab 2

2

1a

be kaeh , e

ahx

typeab

222 1 eba 2

2

1b

ae bekh , e

bky

Proofs of Key Features

Let P represents any point on the ab ellipse with centre at origin which cuts the x-axis at A

and B. Let one focus be S and its corresponding directrix intersects the x-axis at N. From

definition, eBNBSandeANAS .

Directrices: BNANeBSAS . Notice that ONBNANandaBSAS 22 , since

the ellipse is symmetrical about y-axis. Finally, e

axsdirectricee

aON : .

Foci: 0,... aefociaeOAONeaeANaASOAOS

Relations between a, b and e: Draw a line from P perpendicular to the directrix, and call the

point of intersection M. By definition, ePMPS then by substitution of values, we get a

Cartesian equation in which 222 1 eab .

Relationships Between a

b , e and Shape

ab : as 0:1a

be more elongated (flat); as 1:0a

be near-circle.

ab : as

ab

bae 0:1 more vertically elongated (thin).

The following results are intended for ellipses with centre at origin. Simply replacing x with x-

h and y with y-k will not work. It‟s unlikely that you have a question with the centre at (h, k).

Tangent at sin,cos, 11 bayxP : 12

1

2

1 b

yy

a

xx OR 1sincos

b

y

a

x [Hint: the

Cartesian form is similar to the ellipse equation, and we can put sin,cos 11 byax to get

the parametric form]

Proof: differentiate ellipse equation: implicitly (Cartesian), by chain rule (parametric).

Chord of Contact of Tangents from an External Point 00 , yxP : 12

0

2

0 b

yy

a

xx [Hint: again,

this is similar to the ellipse equation 12

2

2

2

b

y

a

x]

Equation of Any Chord PQ, sin,cos,sin,cos baQbaP :

2cos

2sin

2cos

b

y

a

x [Proving hint: when finding the gradient of PQ, use

the „sums to products‟ formulae; in finding the equation of PQ, use the compound-angle

formula]

Hyperbola, Oblique Asymptotes

In a hyperbola, major axis is more correctly called transverse axis and minor axis should be

called conjugate axis. Their definitions are similar to those of the ellipse.


2

2

2

b

y

a

x Parametric Equation:

tan

sec

by

ax

2,

2


2

2

2

a

x

b

y Parametric Equation: [Not Part Of This Course]


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a = half the length of the axis parallel to x-axis [for the first type of hyperbola, it is the length

of the semi-transverse axis]

b = half the length of the axis parallel to the y-axis

θ = the angle made by the line from O to P’ on the auxiliary circle 222 ayx , whose

tangent at P’ cuts the x-axis and this becomes the x-coordinate of P. Right-hand branch: P and

P‟ are in the same quadrant; left-hand branch: different quadrants.

Translation:

12

2

2

2

b

ky

a

hx or

kby

hax

tan

sec has centre (h, k).

Key Features

Comparison between the hyperbolas 12

2

2

2

b

y

a

x and 1

2

2

2

2

a

x

b

y shows that the roles of x

and y are interchanged, as are the roles of a and b in determining e, S and m.

Hyperbola Eccentricity Foci Directrices Asymptotes

typeyx 22

1222 eab 2

2

1a

be kaeh , e

ahx khxa

by

which is the same as

hkyb

ax typexy 22

1222 eba 2

2

1b

ae bekh , e

bky

Relationships Between a

b , e and Shape ( 22 yx and 22 xy types)

Notice that e affects the asymptotes since they are xa

by .

As typeyxa

be 220:1 more curvature and foci move closer typesboth

As typeyxa

be 22: less curvature and foci move away typesboth

When typesbotha

be 1:2 rectangular hyperbola typesboth

The following results are intended for the hyperbola 12

2

2

2

b

y

a

x.

Tangent at tan,sec, 11 bayxP : 12

1

2

1 b

yy

a

xx OR 1tansec

b

y

a

x

Chord of Contact of Tangents from an External Point 00 , yxP : 12

0

2

0 b

yy

a

xx

Equation of Any Chord PQ, tan,sec,tan,sec baQbaP :

2cos

2sin

2cos

b

y

a

x

Proof: gradient =

coscos

sin

coscos

coscos

secsec

tantan

a

b

a

b

by double angle formula (for the numerator) and by sums to

products formula (for the denominator). Then we find the

equation of PQ. Manipulate the equation so that we can use a

“rearranged” compound-angle formula, eg

coscossinsincos .

Rectangular Hyperbola, Asymptotes Coordinate Axes

2sin

2sin2

2cos

2sin2

a

b


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Rectangular means that its two asymptotes are at right angles. Hence, ab , 222 ayx ,

2e and asymptotes xy . Construct 0,aA .

If this hyperbola if rotated through 4

anticlockwise about the origin (or alternatively, rotate

the coordinate axes 4

clockwise), its equation will have the form 0,2 ccxy . A will have

coordinate (c, c) and 2cOA . Note that rotation preserves distances.

The rotated hyperbola is then 2

21 axy . Asymptotes: the x-axis and y-axis. Foci: aa , .

Directrices: ayx . Vertices:

2,

2

aa.

Cartesian Equation: 0,2 ccxy Parametric Equation: 0,, tt

cyctx

Let 002211 ,,,,,, yxTandqccqyxQpccpyxP :

Tangent at P: 2

11 2cyxxy OR cpypx 22

Chord of Contact from an External Point T: 2

00 2cyxxy

Any Chord PQ: 21

2

21

2 xxcyxxxc OR qpcpqyx

[Hint for derivations: remember 2

11 cyx when finding equations of the lines]

[Hint for memorising: look for patterns, eg Cartesian tangent looks like 222 cxy ]

Geometrical Properties of Ellipse and Hyperbola

The chord of contact of tangents from a point on a directrix is a focal chord through the

corresponding focus.

Proof: Consider an ellipse with centre at origin, ab . Suppose 00 , yxT lies on directrix

eax . Then chord of contact PQ has equation 1

2

0 b

yy

ae

x. But 0,aeS satisfies this

equation so PQ is a focal chord. Similarly, if T lies on the other directrix it will pass the

other focus. Since it is a purely geometric property, it is also true if the ellipse is translated /

rotated (true for any ellipse). The proof for hyperbola is similar.

The segment of the tangent between the point of contact and the directrix subtends a right

angle at the corresponding focus.

Proof: Consider an ellipse with centre at origin, ab . Let the tangent at P meet the

directrices at T and T‟. Firstly, prove for segment PT by noting that T lies on tangent PT

and on the directrix (i.e. constant x-coordinate), hence we can find T. Finally, use

222 1 eab to simplify product of gradients of PS and ST to -1. The proof for segment

PT‟ is similar. It is also true for any ellipse. Proof for hyperbola is similar.

The tangent at a point P on the locus is equally inclined to the focal chords through P [the

reflection property].

Proof: Use the ellipse in the previous point. Let the feet of perpendiculars from P to the

directrices be M and M‟. Hence we have 2 similar right-angle triangles at M and M‟. Using

'

'

PM

PS

PM

PSe and the previous property (right angles at foci), prove that the triangles

PST and PS‟T‟ are similar by the RHS similarity theorem.

The sum of focal lengths is constant.

Proof: Use the ellipse in the previous point. By focus-directrix definition,

ae

aePMPMePSSP 2

2'' .


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Properties Unique to Rectangular Hyperbola ( 2cxy and thus its transformations)

The area of the triangle bounded by a tangent and the asymptotes is a constant.

Proof: use the equation of tangent to find x and y intercepts, then find area.

The length of the intercept cut off a tangent by the asymptotes is twice the distance from

the point of intersection of the asymptotes (i.e. the origin) to the point of contact of the

tangent.

Proof: as before, find x and y intercepts, then find the two lengths asked.

Physical Understanding of Conics

A conic section is the set of points on the intersection of a (double) cone and a plane. Any line

from the vertex to any point on the circumference of the base is called a „generator‟ (the

generators form the non-base part of a cone). A double cone is two identical cones placed on

top of each other with their vertices touching.

Plane intersects all generators of one cone (hence won‟t touch the other cone): an ellipse. If

the plane is parallel to the base, a circle is formed.

Plane is parallel to a generator (hence won‟t touch the other cone): a parabola.

Neither of those conditions (hence the plane intersects both cones): a hyperbola

How the plane intersects is determined by its angle of intersection relative to the axis of the

cone and by the angle between (any) generator and that axis (in other words, the slope of the

cone).

4. Integration Using the Reverse Chain Rule

duudxdx

duu , the objective is to have u and its derivative

Examples: xuletdxx

e x

tan,cos 2

tan

; dxx

xxdx

cos

sintan

We also look for similarity to the standard integrals: x

x

x

euletdxe

e

,

1 2

Also pay attention to the natural domain: cxdxx

x

xx

dxlnln

ln

1

ln

The domain of xx

dx

ln includes 10 x and so allows the “ xln ” to be negative. Hence the

integral “ cx lnln ” must be written as “ cx lnln ”. Sometimes the |…| is not necessary

because there never will be negative values to be ln-ed. Remember that |…| actually means

modulus (in case the thing inside is a complex number).

Reminder: for indefinite integrals, don‟t forget to convert u back to its x expression.

Algebraic Manipulation to Reduce Fractional Integrand to Standard Form

Expand and simplify:

dx

xxdx

x

x 121

2

;

dx

x

x2

1

Note: (1) each term in numerator is divided by denominator; (2) coefficients can be

manipulated:

dx

xx

xdx

xx

xdx

x

x

1

1

1

2

2

3

1

1

1

3

1

1322222

Rewrite as partial fractions: this will often result in a series of ln functions, so revise the

log rules to combine them into one. Some special cases:


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Both functions linear:

dx

xdx

x

xdx

x

x

12

1

2

7

2

3

12

2712

23

12

23

Not linear but can be rewritten:

dx

xdx

x

xdx

x

x

1

11

1

11

1 22

2

2

2

Denominator is an irreducible quadratic ( 0 ) or a quadratic in a square root, numerator

is linear or constant: complete the square of the quadratic, then you may need to

substitute u = (the thing in brackets) [if the quadratic is reducible, use partial fraction];

Example:

1,11

12

22

1222

xuletdx

x

xdx

xx

x

Using More Sophisticated Substitution to Reduce the Integrand

Algebraic substitutions, often of squares:

xuletdxxx

,4

1 ;

11,1

222

2

3

xuorxuletdxx

x (then do implicit differentiation)

Trigonometric substitutions tansec,sin aoraax : especially when the integrand

contains 22 xa , 22 ax or even 22 xa [a can be 1] but is not in the table of

standard integrals; then use Pythagorean identity to write it differently

[There‟s a restriction on the values of θ, as implied by the inverse trig functions.]

[The primitive may have “ cx 1sin ” part; this is the same as “ dx 1cos ”.]

An example illustrating algebraic manipulation OR trigonometric substitution [with recurrence

formula]: :12

dxx

xI

n

let tanx OR write as

dxx

xxx nn

1

12

222

Integration of Trigonometric Functions

Rewrite the integrand so it can be easily integrated (for example, it‟s in the table of standard

integrals or it has u and its derivative). Do this by trigonometric identities.

Integrals of the form dORd mm cossin [OR damsin etc.], m +ve even

Write msin as 22sinm

etc. and 2cos12

1cos;2cos12

1sin 22

Integrals of the form dORd mm cossin [OR damsin etc.], m +ve odd

Use Pythagorean identity and substitution, for example:

duuddd 2223 1cossin1coscoscos , xu sin

Pythagorean identities: 1cossin 22 , etc. [dividing 1cossin 22 by 2cos gives

22 sec1tan ]

Integrals of the form dnm cossin , where at least one of m and n is +ve odd

Suppose n is odd [so 1n is even]. Note that the derivative of sin is cos. Rewrite the integral

as

ddn

mnm cossin1sincoscossin 2

121 .

Integrals of the form dnmdnmdnm sinsin,coscos,cossin

Use product to sum formulae: ,sinsincossin2 qpqpqp

qpqpqpqpqpqp coscossinsin2,coscoscoscos2 . Perhaps it‟s

enough knowing any 1 formula because we can use the identity xx cos2

sin .


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The substitution 2

tan xt : will convert trigonometric functions into an algebraic function,

used e.g. in dx

xsin1

1 ;

22

2

2 1

2tan,

1

1cos,

1

2sin

t

tx

t

tx

t

tx

Integration by Parts

Formula: dxvdx

duvudx

dx

dvu

In words: int. of existing parts = parts without dx – int. of the new parts

We want the new parts to be easily integrated. This is generally done by: choosing dv/dx as the

function which can be integrated, and often without raising the power; choosing u as the

function which can be easily differentiated, and hopefully looks simpler after it‟s differentiated.

Some functions, such as e and sin, display a nice pattern when differentiated or integrated. If

your choice doesn‟t work, you can usually swap them.

LIATE (Log, Inverse trig, Algebra, Trig, Exponential): is a useful acronym for determining

which part should be u and which should be dv/dx. The function whose letter comes first in

LIATE is u. Example: xdxx 1sin , x is algebra so it‟s dv/dx.

Sometimes we need to introduce “1”. Example: xudxxdxx ln;ln1ln .

Sometimes integration by parts must be carried out more than once, the second of which

brings us back to the original integral. Example: let dxxeI x sin Ixe x cos ;

moving I to LHS gives xeI x cos2 .

You may need to use simple substitutions to find out the integral of the new parts.

Note on Definite Integral

k

c

k

c

k

cdxv

dx

dudx

dx

dvu vu

Recurrence or Reduction Formula

Sometimes the integral is so naughty that the integral of the new parts looks similar to the

original integral, except that it has a different power. The notation nI is used to denote the

original integral with power n, 1nI denotes the same integral but with power 1n , etc. A

question usually gives the formula which you have to show.

Application: suppose we have shown that 1 n

xnxn

n nIexdxexI , find 3I .

0

23

1

23

1

23

2

3

3 16363233 IxeexxIexxIexexIexI xxxxxx But we

can find 0I , which is Cedxex xx0 .

Definite Integral: we can let k

cn dxxfI . A nice thing with this is that the recurrence

formula won‟t contain any x. Example: show 1

1

0 32

21

n

n

n In

ndxxxI ; find 3I .

01235

2

7

4

9

6

7

4

9

6

9

6IIII , and

3

20 I .

We can also find expression for nI in factorial notation (without recurrence relation).

Example: referring to the previous question, show that

14

!32

!1!

n

nn

nnI . To do this, spell out

132

2

nn I

n

nI 0..

3

2

5

2

7

4

9

6

12

42

12

22

32

2Iei

n

n

n

n

n

n

then proceed with your

skills…


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Recurrence Formula Involving Single Trigonometric Function Raised to a Power

Example: 2

0cos

xdxI n

n . Show that 2

1

nn I

n

nI . Convert this expression into factorial

notations.

2

0

12

0coscoscos

xdxxxdxI nn

n , we choose xdx

dvxu n cos,cos 1 because there‟s

2nI in the question so we want 2cos n in the new parts.

Because the expression 2

1

nn I

n

nI relates nI to 2nI , we need to consider separate cases

for odd and even n when finding the factorial notations.

Further Properties of Definite Integrals

Assumption: b

a

b

aduufdxxf [proof: kinda common sense]

This idea can be used, for example, to evaluate

2

2

2

1dx

e

xx

using the substitution xu . It is

also used to prove the following theorems.

Theorem #1:

aa

adxxfxfdxxf

0 [proof: let xu ]

Theorem #2: aa

dxxafdxxf00

[proof: xau ]

I think questions should hint you if you need to use these theorems.

5. Volumes Definite Integrals as Limiting Sums

Suppose that we have a curve xfy . The area under the curve from ax to bx can be

approximated by making thin rectangular strips of width x .

b

ax

xxfA

b

ax

xxfx

A 0

lim and by the thing called the fundamental theorem of calculus,

b

adxxfA

Volumes of Solids of Revolutions

Any region bounded by curve(s) rotated about an axis will form a solid. Its cross-sections

(found by slicing it perpendicular to the axis of rotation) will look like a circular disc [how?

every point on the curve sweeps out a circle centred on the axis of rotation]. The disc may have

a hole in the middle or a hole around the circular track (but this course only concerns the hole

in the middle, and this kind of disc is called annulus or washer). There are two methods:

Slices: Take strips perpendicular to the axis of rotation, rotate them to form circular slices

and sum their volumes

Cylindrical Shells: Take strips parallel to the axis of rotation, rotate them to form

cylindrical shells and sum their volumes

The second method is often easier, especially for annulus; and some volumes can‟t be found by

the first method.

With this topic what you need is the ability to visualise and to regard curves as physical objects

(to find their dimensions).


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Circle Slices

Question #1: The region bounded by the curve xxy 4 and the x-axis is rotated through

180o about the line 2x . Find the volume of solid of revolution.

The typical slice is shown on the left. Since the parabola

is symmetrical about the axis of rotation, its volume is

the same as if only the shaded region is rotated. The

volume of a typical slice is denoted by V . A is the

area of either the upper or lower surface of the slice. 2rA 2

2 x

yAV yx 2

2

Since we have y , we will be integrating with dy, so we

want to replace x with something in y, using the equation

of the curve, xxy 4 . We could do that but a more efficient way would be to note that

xxy 42 so completing the square in the RHS would give

42422

24 xyxy , which is exactly what we need. Continuing,

yyV 4 .

4

0

4

0

8440

lim

dyyyy

yV

y

cubic units.

Annulus Slices

Any annulus has area rRrRrRA 22 . When you can‟t actually find R and r,

such as in question #2 below, you need the second expression.

Question #2: The region bounded by the curve xxy 4 and the x-axis is rotated about the

y-axis. Find the volume of the solid generated.

The annulus has radii 21, xx , where 21, xx are the roots of xxy 4

considered as a quadratic equation.

1212 xxxxA

044 2 yxxxxy (this is the quadratic equation)

Sum of roots: 412 xx Product of roots: yxx 21

Finding 12 xx : 21

2

12

2

12 4 xxxxxx

yyxx 4241612

yyV 48 …… ANS: 3128 units3.

Question #3: Find the volume of solid obtained by rotating the region

4260,60:, xxyxyx about the y-axis.

From 426 xxy , we can form a quadratic in 2x : 06 222 yxx .

Using the quadratic formula we can find 2

1x and 2

2x . ANS: 36 units3.

Question #4: The region bounded by the curve 21 xy and the x- and y-axes is rotated

about the line 5.0y . Find the volume of the solid. ANS: 158 units3.

Question #5: The area bounded by 22 xy and 52 xy is rotated about the x-axis. Find

the volume of the solid so formed.

Notice one curve is always above the other (in the region we are concerned with).

52 xR and 22 xr …… ANS: 2.115 units3.


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Question #6: A hole of radius a is bored through the centre of a sphere of radius 2a. Find the

volume of the remaining solid by vertical slices.

The solid has circular cross-sections so it can be formed by rotating

curve(s). Because it asks to take vertical slices, you are rotating,

about the x-axis, 222 4ayx and ay .

Notice that the solid begins from 1x .

yR and ar and ax 31

a

dxxaV3

0

2232

334 a units3.

Cylindrical Shells

The following picture illustrates the process when the area under xf between ax and

bx is rotated about the y-axis, where 0xf and is continuous for bxa .

You can quote the result dxhrV 2 provided that you draw the rectangular solid like

in the diagram before. A more formal treatment is given in the following example.

Question #7: The region bounded by the curve xy 1sin , the x-axis and the ordinate 1x is

rotated about the line 1y . Find the volume of the solid formed.

Explanation: V volume of the bigger cylinder – volume of the smaller one

= 22 rRheight , and yrR

Volume, then, is

44

sin1120

lim 25.0

0

y

yyyy

V units3.

Question #8: Find the volume of solid of rotation when the region bounded by 2xy and

2 xy is rotated about the line 3x . ANS: 245 units3.


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Question #9: Find the volume generated when the area between 24 xxy and the x-axis is

rotated about the line 6y . [Hint: the parabola is symmetrical about a vertical axis, hence the

height of cylinder = x24 .] ANS: 151408 units3.

Other Questions: try to do questions #1 to #6 by taking cylindrical shells.

When the Axis of Rotation is Oblique

Remember that you are taking strips perpendicular or parallel to the axis of rotation. You will

have two new oblique axes and you won‟t be integrating in terms of x or y. Questions should

have parts to guide you.

A Time-Saving Tip

You must show both the limiting sum and the definite integral. However I think it is safe to

write

5.0

00

lim

y

yAy

instead of

5.0

0

sin1120

lim

y

yyyy

in question #7, and to then

write

5.0

0dyA . Do it at your own risk.

Volumes of Solids with Cross-Sections of Similar Shapes

The process is similar to the slices method (i.e. summing up slices), except that the slice is

usually not circular. Some rather obvious tips:

Read the question slowly and visualise the situation as you read

Think of the best way to slice the object so all cross-sections are similar in shape (if it‟s not

told, like in questions #11 and #12).

Draw a picture (if it‟s not given).

Introduce some new variables and axes as necessary.

When the question gives a curve as the base of solid (such as in questions #9 and #10

below), you may need to have a third axis.

Find an expression for V , in terms of the variable required for integration, often by using

geometry.

Question #9: The base of a solid is the segment of the parabola yx 42 cut off by the line

2y . Cross-sections taken perpendicular to the axis of the parabola are right-angled isosceles

triangles with hypotenuse in the base of the solid. Find the volume of the solid. ANS: 8 cubic

units.

Question #10: A solid has an elliptical base with semi-axes

2 and 1. Cross-sections perpendicular to the major axis of

the ellipse are parabolic segments with axis passing

through the major axis of the ellipse. The height of each

such segment is determined by a bounding parabola of

height 4, as shown on the right. Find the volume of the

solid.

The slice is a parabolic segment with thickness x . Its

height is given by point z on the bounding parabola (the

parabola facing you above, not the shaded one). By noting

values, the equation of the bounding parabola is 24 xz . Now we

need to find the expression for A . It can be found by Simpson‟s rule

with two strips (note: Simpson‟s rule gives exact result for a parabola,

just like trapezium rule gives exact value for linear functions).

23

22 43

20440

6

2xx

yA . Now it‟s a matter of


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integrating. ANS: 4 units3.

Question #11: The diagram is of a cake tin with a rectangular

base with sides of 16 cm and 10 cm. Its top is also rectangular,

with sides 20 cm and 12 cm. The tin has a depth of 4 cm and

each of its 4 side faces is a trapezium. Find its volume.

We take slices parallel to the base of the tin, with thickness y ,

distant y units from the base, sides s and t. Express s and t in

terms of y. The process for s is done below.

Using similarity of triangles: 22

4 yx

x

y

But: yxs 16216

ANS: 33

2794 cm .

Question #12: Two cuts are made on a circular log of radius 8

cm, the first perpendicular to the axis of the log and the

second inclined at 60o with the first. If the two cuts meet on a

line through the centre, find the volume of the wood cut out.

The red stuff isn‟t shown in the question.

Name some axes. Slicing the shaded solid vertically, we have

rectangular slices of width x , height h, length 2y. ANS: 512π√3 – 1024/√3 cm3.

Integration Process

Keep in mind the option of integration by parts, substitutions, and partial fractions. Also look

for (i) odd integral like 011

1

2 dxxx and (ii) circular area like 2

0

21

24 dxx = 4

1 area

of a circle of radius 2.

6. Mechanics Simple Harmonic Motion

Sample typical question:

Today, low water for a harbour occurs at 3.30 am and high water at 9.45 am, the corresponding

depths being 5 m and 15 m. Find: (a) between what times during the morning a ship drawing

12.5 m of water can safely enter the harbour [hint: x = 0 is the centre of oscillation, at depth =

10] (b) the rate at which the level of water is rising or falling when the depth of water is 13 m.

Sample more creative question:

At ground level, where g = 9.81, a pendulum beats exact seconds (each half-oscillation takes

one second). If it is taken up a mountain where g = 9.80, find by how many seconds per day it

will be wrong. Assume that a pendulum of length L has period g

L2 . ANSWER: On the

ground, the period is 2 seconds 281.9

81.922

LL . It makes G =

3600245.0 oscillations per day. On the mountain, G oscillations are made in

80.92 LG seconds, but it should be in 360024 seconds if it was correct.

Resisted Motion in One Dimension (and Some Other Motion in One Dimension)

Questions on this topic will tell something about forces. Our process is essentially:

Find an expression for the net force (the sum of all forces)

Use amF (where if F is in Newton, m is in kg and a is in ms-2

) to find the expression

for a

Work from the acceleration expression

Another unit for force: 1 kg wt = 9.8 N [that‟s the gravitational force on a 1 kg object]


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Working from the Acceleration Expression

Three important expressions for acceleration ( x ), and their proofs:

dt

dv ; very obvious but anyway I‟ll write

dt

dv

dt

dx

dt

d

dt

xdx

2

2

for proof

dx

dvv ;

dt

dx

dx

dv

dt

dv [chain rule, adding extra variables that get cancelled]

2

2

1v

dx

d ;

2

2

1v

dv

d

dx

dvv

dx

dv

dt

dv [integrate v and differentiate again] then cross

out the dv‟s

Which expression for acceleration to use?

Find out what relation or function you want by interpreting the question. For example, if

the question asks “find the velocity at the end of 4 s”, then you need v as a function of t. If

it asks “find the maximum height when the ball is thrown upwards”, then you need x as a

function of v (maximum height is when v reaches zero). When you want a relation between

x and t, either (i) find two relations, between v and x and between v and t, then solve them

simultaneously OR (ii) find v as a function of t and integrate it using dtdxv .

If you want a relation involving v and x (i.e. v as a function of x OR the other way around),

the expression for acceleration needs to contain the letters v and x (note: dv contains the

letter v) – that‟s all. If you want a relation between v and t then it needs to contain v and t.

We can integrate the appropriate expression to get the relation. You may well need to

remember the technique of flipping the derivative. For example, we have 2kvgx and

we need a relation between x and v. v

kvg

dx

dvkvg

dx

dvv

22

(notice I don‟t try to

write it as kvv

g )

dvkvg

vx

kvg

v

dv

dx22

.

So, does it matter whether we choose

2

2

1v

dx

dx or

dx

dvvx ? It does, in a way. When

you have x in terms of v, using dx

dvvx is generally quicker (think of “

dx

dvvx likes

v”). When you have x in terms of x, either one is fine. You just have to use different

process as illustrated in the following example with 2xx .

dxxvxv

dx

d 2222

2

1

2

1

dxxdvvdxxdvvxdx

dvv 222

Things to revise on:

Integration techniques. Also remember that if after integration you have the relation in the

wrong order (for example, you have Cvx ln , but want v as a function of x), make sure

you find the constant first (don‟t think it‟s Dev x ).

Properties of ln and e function, such as the fact that they can cancel each other, domain of

ln, naan loglog and baba logloglog .

Finding the Net Force

Some forces acting on a body or object (with their common symbols):

Gravity (mg): always directed downwards

Friction or resistance (R or F): acts in opposite direction to motion


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Tension (T): directed along a string, moving away from the body

Normal or reaction (N): always exists when the body touches a surface or another body,

acts at right angles to the surface, is also called reaction force

The thing that governs acceleration is the sum of all forces – the net or resultant force.

Three types of common resisted motion in one dimension: horizontal, upwards and downwards.

A fourth type is sliding down a rough surface. In all of them:

The body has a non-zero initial speed but there‟s no engine in the body that pushes it

forwards after 0t

Make 0x at 0t

Take the direction of motion as positive (even when it‟s downwards)

Draw a diagram showing all forces, the positive direction, and initial conditions

( 0t , 0x , uv )

The forces we are concerned with:

Horizontal motion: friction (there are gravity and normal but they cancel out)

(Vertically) upwards or downwards motion: gravity and air friction

Sliding down on a surface: normal, friction and gravity (don‟t worry about this type until

you‟ve read the General Information on Vectors section)

A fifth type is when an object is fired vertically upwards to very high altitudes and so is subject

to decreasing gravity. The object will escape the earth when 0v as x .

Doing Questions

Terminal velocity is the maximum velocity experienced by a falling object. To find it, let

t or 0x (both conditions are basically equivalent).

Impact speed is the speed when a projectile hits the ground.

If a question involves both upwards and downwards motion, the upwards and downwards

sections need to be investigated separately (choose a new 0t , 0x and +ve direction).

Note: when an object is thrown upwards from 0x with speed V, with air resistance, it won‟t

have V speed at 0x when it comes down. The time taken won‟t be equal either.

Sample question: A particle of unit mass moves in a straight line against a resistance

proportional to 3vv . Initially, the particle is travelling at Qv . Show that

Qv

vQcx

1tan 1 for some constant c. HINT: 3vvmkF but mass is 1 unit. You will

then get vQkx 11 tantan . Notice some similarity in Qv

vQ

1 with the compound angle

result. So, take tan of both sides of vQkx 11 tantan , treating Q1tan and v1tan as

angles and using compound angle formula to simplify RHS.

Projectile Motion

Many questions will rely on some idea of the path of the projectile. At this stage I think it‟s

beneficial to commit to memory that cosVx and gtVy sin (I mean, don‟t waste

time drawing triangles).

Sample question: A particle is projected upwards over horizontal ground with velocity U,

inclined at an angle θ to the horizontal. A second particle is projected simultaneously from the

same point in the same direction with velocity UV .

(a) Show that throughout the motion, the line joining the positions of the particles makes a

constant angle with the horizontal. HINT: naming the particles 1 and 2, find 2211 ,,, yxyx and

the gradient of the line. It will be tanm so the angle is θ.


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(b) given that the range and time of flight of the first particle are respectively 2sin1 2Ug

and

sin2

Ug

, find the positions of the faster particle and the direction of its velocity vector when

the slower particle hits the ground. HINT: put sin2

Ug

t into the expression for 2x and 2y .

Find 2x and 2y . Then you‟ll get (if the velocity makes an angle with the horizontal)

tan2

tan

V

UV. Note that

V

UV 2 can be positive or negative.

tan21tan 1

V

U.

(c) If this velocity vector is directed upwards at an angle 2

1 , find V

U in terms of θ and

deduce that 2

1

4

1

V

U. HINT: tan21

2

1tan

V

U and use the t-formula. You will get

2

1sec

4

11

4

1 22TV

U but 900 .

General Information on Vectors

Force, acceleration, velocity and displacement are all vector quantities.

Vector quantities can be numerically added when their directions are on the same line (e.g.

upwards and downwards are on the vertical line). Let‟s invent the term „main line‟ and define it

as the line on which the object is moving (or more correctly, to fit with the circular motion

topic, it‟s the line associated with the net force). To be useful, force and acceleration are to be

expressed in terms of the main line and another line perpendicular to it. Any force that is on a

different line needs to be separated into two perpendicular components on those two lines. The

components are often horizontal and vertical but can be something else. For example, in the

case of sliding down on a surface, the main line is the sloping surface.

The force mg is out of line. You want a component

which is in line with the surface and another

component which is in line with the N. Using simple

geometry to determine the angles, you‟ll find the two

components on the diagram. sinmgFnet because

0cos mgN (otherwise P floats or sinks).

Note on deciding whether a component has sin or

cos :

The side (of the triangle) opposite the angle always has sin ; the other has cos .

Circular Motion, Conical Pendulum and Movement on a Banked Track

The there are two ways to express velocity and acceleration: angular expression and linear

expression. Angular expressions use the symbol and relate to radian in order for other

formulas to work.

Angular velocity:

dt

d

Linear velocity: dt

dlv (l = arc length = r ) rv , directed tangential to circle


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Angular acceleration: w , which can also be written as

2

2

1

d

d

d

d

A body moving in circle is affected by two forces: one directed towards the centre (radial

force) and another directed at tangential to a particular point on the circle (tangential force).

The radial force maintains its orbit by changing the direction of velocity, while the tangential

force changes its speed. Linear acceleration consists of: 2raR

r

v2

(putting rv to the above)

raT

Proving guideline:

Drawing a circle on a number plane,

cosrx and sinry

Use the chain rule to derive yx , :

sinr

dt

d

d

dxx

To derive yx , , use the product rule then the chain rule (remembering is a variable):

sincos2 rrx and cossin2 rry

Add two like terms to form Ra and the other two to form Ta using triangles; for example:

224222222cossinsincos rrraR

Why is it the two like terms? Consider a case when there is no tangential force (so the

angular speed is constant and 0 ): Ra contains the two like terms above.

Uniform Circular Motion

There‟s no tangential force and speed is constant. fT

22

.

Doing Questions on Uniform Circular Motion (Including Pendulum and Motion around a

Banked Track)

Radial force is provided by the tension in the string attached to the body, by the friction force,

by the normal force, or a combination of them.

As the speed increases, normal and friction decrease, and become zero when the body leaves

the surface.

Draw a dimension diagram, showing relevant data such as direction of motion, two +ve

directions (with inwards made +ve), masses, angles and relations between lengths.

Identify all forces acting on each moving body (if there are more than one), draw them in a

separate force diagram if necessary.

Resolve the forces into two perpendicular directions if they‟re not so already. One

direction will be along the radius. Draw components in dotted lines to distinguish them

from the real forces. Draw separate force resolution diagrams if necessary.

Always start with two equations: one summing the forces radially and the other summing

the forces on the other direction. Recognise that the radial net force is Ram .

Label each new equation as you go. Keep in mind the relations between lengths (such as in

question #2). The key to some problems is to be able to cleverly select some equations to

be solved simultaneously.

Question #1: A car of mass 1 tonne passes over a bridge formed by the arc of a circle of radius

10 metres.

(i) Find the force exerted by the car on the road at the top of the bridge if the car is travelling at

8 m/s.


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ANS: We need to consider the conditions at topmost point on the bridge only. Let N be the

reaction or normal force of the bridge on the car, P be the tractive force of the engine, and F be

the friction. N is the force exerted by the road on the car, but it‟s also the force exerted by the

car on the road (Newton‟s third law).

Horizontally: 10 FP Radially: 222

r

mvmgN

r

mvNmg

(ii) What speed would cause the car to be on the point of leaving the bridge?

ANS: This happens when 0N

Question #2: A particle P of mass m travels with constant speed in a horizontal circle around

the inside of a smooth hemispherical bowl of radius R, centre O. C is the centre of the circle of

motion and OP makes an angle θ with OC [note: O is directly on top of C]. Show that

cos

2

R

g and find the reaction force in terms of . Describe what would happen if

were increased.

ANS: Let N be the normal at P. Its

direction is perpendicular to (the

tangent of) the surface, so it‟s towards

O (centre of the bowl).

Vertically: 10cos mgN

[because P is not moving up or down]

Radially (towards C):

2sin 2 mrN

There are common variables in (1) and

(2). One of them is N. Let‟s make N the subject so we can equate the two equations.

From (1): cos

mgN

From (2):

sin

2mrN , but we need something with R because we need

cos

2

R

g .

3

sin

sin 22

mR

RmN . Then, equate 31 .

Now, if were increased, from

cos

2

R

g we know that cos would decrease and

would increase, hence the particle moves up.

Conical Pendulum

It is formed by a bob and a string tied to a fixed point, with the bob moving on horizontal

circle. Question #2 above is essentially a conical pendulum with an invisible string. Often, you

need to eliminate either T or N from two equations.

Question #3: a string passing through a smooth hole in a smooth table connects a particle P of

mass m on the table to a particle Q of mass M suspended below the table. P and Q are both

performing uniform circular motion with angular velocity , where Q moves in horizontal

circle at a depth h below the table. The lengths of the string above and below the table are l and

L respectively. Show that h depends only on , and that m

M

L

l .

ANS: The tension in L has equal magnitude with that in l.


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We don‟t want r in (4) but we want L, because cosLh . Use sinLr and we get:

52 MLT . Eliminate T by division, 53 : 22cos

gh

Lg

.

Equate (2) with (5) and we get m

M

L

l .

Motion on a Banked Circular Track

When a car or train turns a corner circularly, there‟s a lateral force exerted on the car towards

the centre of the circle. On a banked track, this force is provided by the horizontal components

of friction and normal forces. With a particular car travelling with a particular radius, the

friction force may act either up or down the plane depending on the speed of the car. At a

particular speed, there would be no friction. This speed is called the optimum speed. By

considering a case where there‟s no friction, it can be shown that tanRgvo (not to

memorise).

The conditions of the road (e.g. wet or not) and of the tyres influence how much friction can be

provided. If insufficient friction is provided, the car will slide. At the optimum speed, the car

won‟t slide whatsoever because no friction is needed. It can also be shown, by finding the

expression for F and combining it with Rg

vo

2

tan (also not to memorise), that when

ovv , the car will slide down and the friction acts up the slope, and when ovv the friction

acts down the slope.

In our examples, the car and train have their centre of gravity on the surface of the track. In a

train, upwards friction is provided by the inner rail and downwards friction is given by the

outer rail.

dhsin .

For small angles, less than about 15 ,

tansin .

Multiplication by sin or cos is often

required to solve simultaneously.

Question #4: A railway line around a circular arc of radius 800 m is banked by raising the outer

rail h m above the inner rail, where the distance between the rails is 1.5 m. when the train

travels around the curve at 10 ms-1

, the lateral thrust on the inner rail is equal to the lateral

thrust on the outer rail when the speed is 20 ms-1

. Calculate (i) the value of h and (ii) the speed

of the train when no lateral thrust is exerted on the rails. (Take g = 9.8 ms-2

.)

ANSWER:

When 110 msv (upwards friction):

Vertically: 1sincos 11 mgFN


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Radially (in this case horizontally): 2cossin2

11 r

mvFN

We want to equate F values of the10 ms-1

and 20 ms-1

, so we want to eliminate 1N .

sin1 : sinsincossin 2

11 mgFN

cos2 :

coscoscossin2

2

11r

mvFN

Agmr

vgmmagnitudeF

cos

8

1sincossin

2

1

When 120 msv (downwards friction):

Without going through the derivation process but noting that 2F is in the opposite direction of

1F : Bgmr

vgmmagnitudeF

sincos

2

1cossin

2

2 .

(i) Equate BA and we get g16

5tan . This means d

hg

165sin ( 5.1d ).

(ii) Optimum speed is when 0F . Use either of the two F expressions above. You‟ll get

r

vg

r

vg

22

tancossin (dividing by cos ). We know 800r , g16

5tan .

7. Polynomials Factors and Zeroes

They depend on the field we are concerned with. Example: 32 24 xxxP

Over Q (rational): 13 22 xxxP , zeroes are none

Over R (real): 133 2 xxxxP , zeroes are 3,3

Over C (complex): ixixxxxP 33 , zeroes are ii ,,3,3

Long Division Involving Complex Numbers

Example:

ix

i

ixi

xi

ixx

xxix

1

2

11

31

3

2

2

Factoring a Quadratic by Completing the Squares

Example: 232323296762222 xxxxxxx

Multiplicity

If α is a zero of multiplicity (or order) r of xP , 1r , then α is a zero of multiplicity

1r of xP' . Proof: differentiate xP , then show that xAxxPr 1

'

where

0A .

As an implication, α is a zero of multiplicity r of xP if α is a zero of 1rP , that is,

0''' 1 rPPPP .

The Division Transformation


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This refers to the writing of xP as xRxQxD , where DR degdeg .

The Fundamental Theorem of Algebra

It states that every polynomial of degree greater than or equal to one has at least one zero over

C. Implication: by mathematical induction, every polynomial of degree n has n zeroes over C

(not necessarily distinct); it can be factorised into n linear factors.

Polynomial with Integer Coefficients

Any integer zero of the polynomial is a divisor of the constant term.

Proof: let 01

2

1

1

01

1

1 aaxaxaxaxaxaxaxP n

n

n

n

n

n

n

n

Since 0P , 01

2

1

10 aaaa n

n

n

n

Rearranging, 1

2

1

1

0 aaaa n

n

n

n

But: saalland ' are integral, so 1

2

1

1 aaa n

n

n

n

is integral.

Hence, is a divisor of the constant term 0a .

The numerator of any rational zero is a divisor of the constant term, and its denominator is

a divisor of the leading coefficient.

Any non-rational zeroes occur in pairs of conjugate surds.

Polynomial with Real Coefficients (note: real includes integers)

If xP has odd degree, then it has at least one real zero.

For any xP , the non-real zeroes occur in complex conjugate pairs.

Hence, any xP can be factorised into real linear and real quadratic factors. Note:

222 Re2 xxxxxx , which is a real quadratic.

Polynomial with Non-Real Coefficients

No pattern; zeroes need not be conjugate pairs. Example: ixixixx 222 .

Sum of the Products of Roots

For 01

1

1 axaxaxaxP n

n

n

n

,

Sum of the products of roots taken r at a time = n

rnr

a

a 1

Notations for the sums: let‟s take xP of degree 4: ,,,

Some Questions

1. Suppose 322 23 xQxaxx where xQ is a polynomial. Find the value of a.

SOLN: Since LHS has 3x , Q(x) must have 2x . 322 223 cbxxxaxx .

Equating coefficient of 2x , 422 bb . Equating coefficient of x,

802 cbc . 19a .

2. a and b are real numbers such that the sum of the squares of the roots of the equation

032 ixibax is 8. Find all possible pairs of a and b. SOLN: Let the roots be

, . 862322 222222 iabbaiiba .

baab 3062 and 83 2222 bbba . Pairs (a,b): (3,1) and (-3,-1).

3. i1 is a root of the equation 0522 ibxiax , where a and b are real. Find the

values of a and b. SOLN: Put ix 1 into the equation and expand.


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Changing the Variable to Find a New Polynomial

Let 023 dcxbxax has three roots represented by , that is, 023 dcba .

To find an equation with roots 3 , we can replace x with 3x so that when 3 is substituted

as x, we still get 023 dcba . By writing the new equation in y instead of x, we

can find the substitution systematically. Examples:

1. 03223 xxx has roots ,, . Find the equation with roots 222 ,, .

SOLN: 3 . New roots are 3,3,3 . Let 33 yxxy . New equation in y:

033

233

23

yyy or 081183 23 yyy .

2. 0124 23 xxx has roots ,, . Find the equation with roots 222 ,, . SOLN:

Let yxxy 2 . The equation in y with roots 222 ,, is given by:

142012423

yyyyyy (separate the terms with y ).

Squaring, 011212142 2322 yyyyyy .

Not everything can be done by directly letting y = …x and x = …y. For example, to find the

equation with roots 444 ,, in the above two examples, we need to first find those with

roots 222 ,, , and after making it a cubic, repeat the process for 222222 ,, .

Evaluating 333

We don‟t have to find the new cubic by replacing x with its cube root. We use the fact that

023 dcba , 023 dcba and 023 dcba , and then sum

them to get: 03222333 dcba [ 222

22

if you haven‟t found the equation with roots 222 ,, ].

A quartic with symmetric coefficients abxcxbxaxxP 234 can be converted to

quadratic in x

x 1 . Method: write it as 2122 axbxcbxaxx

cxxbxxax 1222 , then use the fact that 22221 xxxx . Also note

that 0x won‟t be a zero because then 1x is not defined.

Solving Equations over C Using Complex Number Technique

Solving 01nx : find the complex nth roots of 1 . Use De Moivre‟s theorem to find the

roots in modulus-argument form. Alternatively: evennxn ,01 can be solved by treating it

as a difference of two squares; 2244 21210 xxxx then complete the square and

treat the result as a difference of two squares; etc.

Solving 0121 xxx nn : multiply by 1x to give 01nx , 1x

Solving 01121 nnn xx : multiply by 1x to give 01nx , 1x

Solving cxn : use De Moivre‟s theorem

Solving 0 cxxn: write it as 01 cxx n ; etc.

New Expressions for nandn cossin

Let sincos iz . We can equate the imaginary and real parts of the expressions of nz

using De Moivre‟s theorem and Binomial expansion. From the imaginary part, we get nsin


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equals some summation of products in the form of ba cossin ; similarly we get an

alternative expression for ncos from the real part.

Sample questions with answers:

(i) Express 5tan in terms of powers of tan , hence show that 0510 24 xx has roots

52tan,

5tan . SOLN:

5tan

cossin5cossin10cos

sincossin10cossin5

5cos

5sin4325

5234

42

53

tan5tan101

tantan10tan5

, after

dividing top and bottom by 5cos .

Then, let tanx . Pay attention to the top part of the fraction above and notice that it

resembles 510 24 xxx , hence 05tan if x is a root of 0510 24 xx .

(ii) Deduce that 55

4tan5

3tan5

2tan5

tan . SOLN:

Notice that 5

3tan5

2tan,5

4tan5

tan . From 0510 24 xx , the product of

the four roots is 5.

(iii) By solving 0510 24 xx another way, find the value of 5

tan as a surd. SOLN:

Consider 0510 24 xx as a quadratic in 2x and use the quadratic formula.

Partial Fractions

Method: Set up an identity, n

n

x

c

x

c

x

c

xQ

xP

2

2

1

1 , where

nxxxkxQ 21 , all zeroes of Q(x) distinct, and QP degdeg .

We can then rearrange this identity so that the fractions disappear by multiplying both sides by

Q(x). Now, substitute selected values of x by which we can equate coefficients of the x-terms

to determine the constants, c‟s. Useful values to substitute are ,, 21 , and if the factors are

irreducible quadratics, zero and ik (for factors like 222 ,1 kxx ) [ 1i ]

An alternative approach to equating coefficients in the identity:

[Note: This approach is mentioned in the syllabus but is very rarely asked.]

i

ii

Q

Pc

' , where ni ,,3,2,1

Proof: n

n

x

c

x

c

x

c

xQ

xP

2

2

1

1 ; let‟s investigate the case when 1i

ix , i.e. 1x in this case: n

nx

xc

x

xcc

xQ

xxP

1

2

121

1

If we take the limit of the RHS as 1x : RHS = 1c

Taking the same limit of the LHS will give 1

1

'

Q

P; how does it work?

Notice, in the LHS, that 1

11

QxQ

x

xQ

x

, since 01 Q

Also, 1

limx

1

1

1 '

Q

x

QxQ

by the first principle of differentiation

A Case Where QP degdeg

Perform division transformation with xQxD before setting up the identity. For example,

express 32

1222

23

xx

xxx as a sum of partial fractions.


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Using division transformation, 132122 223 xxxxxxx . Hence,

32

1

32

12222

23

xx

xx

xx

xxx, then

3132

1 21

2

x

c

x

c

xx

x and so on.

A Case Where Q(x) Has Irreducible Quadratic Factor(s) Over Real

Example: 112112

12

21

2

2

xx

bax

x

c

x

c

xxxx

x

[Notice: either a or b may turn out to equal zero.]

[Remember that zeroes, factors and partial fractions depend on the number field asked. For

example, there‟s no such thing as irreducible cubic factor over R, but there is over Q or Z. In

that case, let the denominator be that cubic.]

[In general, let the degree of the numerator (of each fraction on the RHS of the identity) be one

less than that of the corresponding denominator.]

A Case Where Q(x) Has Multiple Zero(es)

Example: 22

2

1111

47

x

cbx

x

a

xx

xx (same zeroes must be grouped together) which then

gives

222

1

1

1

2

1

3

1

122

1

3

1

12

1

3

xxxx

x

xx

x

x.

Generally,

21

2

1121

x

a

x

b

x

d

x

c

xx

xPnn

[The syllabus says not to discuss this case but there have been related questions on it.]

8. Harder 3U Topics There is no summary for the harder 3U topics, but we have compiled a set of typical questions

from previous HSC examinations, along with the solutions.

based on 1990 syllabus -...

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