1
HOW TO THINK
ABOUT HSC
MATHEMATICS
JW Hick, 2013
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“If you can not explain something simply, then you do not
understand it well enough.” - Albert Einstein
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Contents
HSC Mathematics
1.
2.
3.
4.
5.
6.
7.
8.
Applications of Coordinate Geometry
Geometrical Applications of Calculus
Integral Calculus
Exponential and Logarithmic Functions
Trigonometric Functions
Applications of Sequences and Series
Applications of Calculus to the Physical World
Probability
6
12
21
31
38
45
54
60
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Applications of Coordinate Geometry
Coordinate Methods in Geometry
This chapter is basically using coordinate geometry techniques to prove properties of
shapes, usually via congruent or similar triangles. It combines a few areas of the
preliminary course together as an application of knowledge, and does not introduce any
new concepts.
Know your basic shape properties, angle sum of polygons and all the preliminary
geometry facts. From this point you will use relevant coordinate geometry formulae to
prove things involving preliminary geometry content.
Formulae
Midpoint
M = x1 + x2
 2
y1 + y2 
2 


Distance between two points
2
d = (x1 – x2) + (y1 – y2)
2
Gradient
m = y2 – y1
x2 – x1
tanθ = m
Point Gradient formula (“Find the equation of the line”)
m(x1 – x) = y1 – y
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Perpendicular distance between a line and a point
d=
|ax1 + by1 + c|
2
a +b
2
Parallel lines
m1 = m2
Perpendicular lines
m1 × m2 = -1
The other concepts covered in the preliminary course you should be aware of include the
general form of a line, the point gradient form of a line, finding the intersection of two
lines and how to find the x and y intercepts of a line.
Example: The points A(3,5), B(6,1) andC(x1, y1) form an equilateral triangle.
Find, in exact form one possible value for the point C.
We need to use whatever formula we can to form equations to find the unknown
coordinate. We could use dab = dac = dbc as we know equilateral triangles have all side
lengths equal.
2
dab = (3 – 6) + (5 – 1)
2
dab = 25 = 5
We can find the equation of the locus which is equidistant from points A and B.
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2
2
2
(3 – x1) + (5 – y1) = (6 – x1) + (1 – y1)
2
2
2
(3 – x1) + (5 – y1) = (6 – x1) + (1 – y1)
2
2
2
2
2
2
9 – 6x1 + x1 + 25 – 10y1 + y1 = 36 – 12x1 + x1 + 1 – 2y1 + y1
34 – 6x1 – 10y1 = 37 – 12x1 – 2y1
6x1 – 8y1 – 3 = 0
So this is the locus of points which are equidistance from A and B.
We can construct a right angled triangle as this locus is a straight line which bisects AB
and passes through C. The locus is the perpendicular bisector of AB. We can call the
midpoint of AB, D
D =  3 + 6 , 5 + 1 
2 
 2
D =  9 , 3 
2

2
2
25 = d +  5 
 2
d=
75 = 5 3
4
2
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Hence, the distance DC is 5 3 .
2
So if we solve the locus of points which are 5 3 units from D, simultaneously with the
2
locus of points which are equidistant from A and B we will solve for the two possible
points C.
x –
 1

9  2 + (y – 3) 2 = 75 1
2 
4
Hence solve this simultaneously with,
6x1 – 8y1 – 3 = 0
y1 = 6x1 – 3 8
Substituting into gives us
2
2
 x – 9  +   6x1 – 3  – 3  = 75


 1 2

4
 8 



2
2
 x – 9  +  6x1 – 27  = 75
 1 2


8
4




2
2
x1 – 9x1 + 81 + 36x1 – 324x1 + 729 = 75
4
64
4
2
2
64x1 – 576x1 + 1296 + 36x1 – 324x1 + 729 = 1200
2
100x1 – 900x1 + 825 = 0
2
4x1 – 36x1 + 33 = 0
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Putting this into the quadratic formula gives solutions,
2
x1 = – b± b – 4ac
2a
x1 = 36± 1296 – 528
8
x1 = 36± 768
8
x1 = 36± 64 × 4 × 3
8
x1 = 36±16 3
8
x1 = 9±4 3
2
We only require on possible point for C, so let’s substitute the positive answer into the
straight line locus to find the corresponding y value.
6x1 – 8y1 – 3 = 0


6 ×  9 + 4 3  – 8y1 – 3 = 0
2


y1 = 3 + 3 3
2


Hence the positive value gives the point C  9 + 4 3 , 3 + 3 3 
2
2


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This example was particularly tough, however the process is what you should be focusing
on, rather than all the intricate algebraic manipulations. We had a geometrical problem,
and then combined our coordinate geometry formulae with our knowledge of geometry to
form equations and solve for the unknown values. There are often multiple ways to solve
or prove a question.
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Geometrical Applications of Calculus
Introduction (Gradients and Tangents)
If a function is continuous (no gaps) and smooth (no sharp bits) then you can differentiate
it.
Gradient is calculated as,
m = rise
run
For a curve, the gradient usually changes with every small change in x. Hence, the closer
the x-values are together, when applying the gradient formula, the more accurate the
approximation of the gradient will be. If you take the limit of gradient as the x values get
infinitely close together then the secant which represents the gradient becomes the
tangent. This is why the gradient of the tangent represents the gradient of the curve at a
specific value of x.
The First Derivative (The Gradient Function)
The first derivative tells us the gradient at certain x-values that we substitute into it. The
first derivative is also called the gradient function, because for each value of x we
substitute in, we will find the corresponding gradient value.
A positive gradient (increasing) will slope upward to the right, a negative gradient
(decreasing) upward to the left and a zero gradient (stationary) is a flat bit.
The gradient at a certain point on the curve is equal to the gradient of the tangent at that
point.
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3
Example: Find the points on the curve y = 3x + x + 3 where the gradient equals 13
2
y = 3x + x + 3
y ′ = 6x + 1
when the gradient function equals 13
13 = 6x + 1
x=2
y = 29
Therefore at (2,29) the gradient is 13
The gradient of 13 occurs when x = 2. This is equivalent to the gradient of the tangent at
the point x = 2, which is the red line in the above diagram.
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Stationary Points (Four types)
Stationary points occur when the first derivative equals zero, meaning the gradient value
is zero and there is a flat bit on the curve. There are 4 types of stationary points.
Turning Points
The point A is a maximum turning point. The point B is a minimum turning point. Y’ = 0
at both these points.
For a maximum turning point the gradient flows left to right as “positive, zero, negative”.
For a minimum turning point the gradient is “negative, zero, positive”.
Horizontal inflections
Both points C and D are described as a horizontal points of inflection. Y’ = 0 at both
these points.
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In the case of point C, the gradient flows left to right as “positive, zero, positive”. In the
case of point D, the gradient flows as “negative, zero, negative”.
To “determine the nature” of a stationary point, it means to figure out which one of the
four above possible stationary points it is.
You can use the first derivative to find the nature of the stationary points. This is the “test
to the left, and test to the right” method. For example if x = 3 is the only stationary point
on a curve, you could find the gradient at x = 2 and x = 4 to determine the nature of the
stationary point, as per the 4 unique descriptions of how the gradient flows in the points
A to D.
3
2
Example: Find the stationary points and determine their nature for y = 2x + 6x + 3
3
2
y = 2x + 6x + 3
2
y ′ = 6x + 12x
y ′ = 6x(x + 2)
stationary points occur when y ' = 0, hence
x = 0 or x = -2
y = 3 or y = 11
Hence stationary points are located at (0,3) and (-2,11)
Let's test the gradients on either side of the staionary point
when x= - 3, y ' = positive, when x= - 1, y ' = negative, when x=1, y ' = positive
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The Second Derivative (The Concavity Function)
The second derivative tells you information about the concavity at a certain x-value. If
you differentiate the function twice and then substitute an x-value into the second
derivative, you will find the concavity at that x-value.
We are mainly interested in whether the concavity is a positive or negative number, or
zero. We are not particularly interested in the magnitude of the number.
If the function has a positive concavity at an x-value, then it lives on an “upward
parabola”. If it has negative concavity then it lives on a “downward parabola”. Zero
concavity means there is “no curve” at that point, so there is a little “straight bit” which
may or may not be horizontal. A concavity of zero does not help us determine the nature
of a stationary point.
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Here we can see that any point “that lives on an upward parabola part” of the graph has a
positive concavity. Any point “that lives on a downward parabola part” has a negative
concavity. If there is “no curve” at that point then the concavity is zero, as indicated on
the diagram.
Curve Sketching (Finding Stationary Points and Determining their Nature)
Curve sketching brings together all of what we have learned so far. Basically to sketch a
curve “find all stationary points and determine their nature, and test for non horizontal
points of inflection”.
The first derivative will find where the stationary points occur. The second derivative will
find the nature of the stationary points, unless y’’ = 0, in which case revert back to the
“test on the left and test on the right” method using the first derivative.
3
2
Example: sketch the curve y = x – 3x + 2 showing all important features.
3
2
y = x – 3x + 2
stationary points occur when y ' = 0
2
y ′ = 3x – 6x
0 = 3x(x – 2)
Hence x = 0 , x = 2 are the values of x which produce stationary points
y ′′ = 6x – 6
When x = 0, y = 2 and y ''= 12, hence is concave up
When x = 2, y = - 2 and y ''= 0 hence there is no curve, need to use first derivative test
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When x = 1, y ' = - 3, when x = 3, y '' = 9, ie negative, zero, positive.
We have found all the stationary points, and determined their nature. Now we need to
check for any non horizontal points of inflection.
y′′ = 6x – 6
points of inflection when y '' = 0, ie x = 1
When x = 1, y = 0 and y' = - 3
We can see there is a maximum turning point at (0,2), a non horizontal point of inflection
at (1,0) and a minimum turning point at (2,-2). These constitute the “important features”
when we sketch the curve.
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Maxima and Minima Problems
These problems basically involve creating simultaneous equations, eliminating a variable
and differentiating to find the turning points and determine their nature. Instead of having
x and y, you will probably have say Area and Length for example. Same idea, solve A’ =
0 instead of y’ = 0, just using different letters.
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Example: If the surface area of a cylinder is 600π cm , find the height that will give the
maximum volume for the closed cylinder.
SA = 600π
2
2πr + 2πrh = 600π
2
r + rh = 300
r + h = 300
r
h = 300 – r r
We will regard this as our first equation. We created this from the SA information, and
simplified it down to this final line.
Now, as we are interested in “maximum turning point in the volume equation” we will
need to create a volume equation.
2
v = πr h
Now this is good, but we can’t differentiate V as it is a function of two variables.
However if we use our substitution for h in equation (1), we can eliminate the h variable.
2
v = πr ×  300 – r 
 r

v = 300πr – πr
3
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Now it is simple, we can just differentiate V with respect to r and do our normal “find the
stationary point and determine the nature” process. It is exactly the same, it is just with V
and r, as opposed to x and y.
dv = 300π – 3πr2
dr
stationary points when dv = 0 hence
dr
2
300π – 3πr = 0
r = 10, ignoring the r = - 10 result
substituting back into gives h=20
Hence we know the only turning point exists when r = 10, for r > 0 as you can not have a
negative radius.
Now we need to determine if it is a maximum turning point.
2
dv
= – 6πr
2
dr
Hence there is a constant negative concavity for all r > 0, ie maximum turning point.
Therefore the maximum possible value for the volume is generated by a height of 20cm.
Solving this problem was basically creating simultaneous equations, eliminating a
variable then using the normal process to find stationary points and determine their nature.
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Integral Calculus
Introduction - The Area under a Curve
When we differentiate a function, it is so we can find the rate of change (gradient) at a
certain value of x. When we integrate it is used to find the area under a curve.
Differentiation and Integration are inverse operations, like multiplication and division.
They will cancel out and leave you with the original function you started with.
If the area is “under” the x-axis then you will get a negative answer. As we can not have a
“negative area” you simply take the absolute value of it and make it a positive number.
This means you may have to split up your integral if some of it is above the x-axis and
some is below the x-axis.
Indefinite Integrals
An indefinite integration is simply the “reverse differentiation” process. Add one to the
index and divide by the new index. You will have a + C after you integrate as any
constant would disappear thorough differentiation.
An integral needs a “differential” to tell you what variable you are integrating with
respect to. Basically this is a dx or dy in HSC math, are you integrating “x stuff” or are
you integrating “y stuff”.
Example: Solve the following integral
⌠ 3
 2 + x – 4 dx
⌡ x
Here we need to do a little bit of index manipulation before we can commence on our
process of “adding one to the index and dividing by the new index”.
1
-2
2
⌠
 3x + x – 4 dx
⌡
Now this is in a format where we can integrate easily.
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1
-2
2
⌠
 3x + x – 4 dx =
⌡
-1
3
2
1
3x + x – 4x + c =
3
-1
1
2
3
2
-1
-3x + 2x – 4x + c
3
Note: If we differentiate this final expression we revert back to the original integral.
Definite Integrals
Definite integrals are used to find the area under a curve or between curves. Unlike
indefinite integrals, definite integrals have “bounds”, ie x-values you will substitute in
after you integrate. Integrate, then substitute both the bound numbers in and subtract the
answers.
A definite integral between two bounds will find an area. Specifically the area it finds
will be between the curve and either the x-axis or the y-axis depending on how you have
set up your integral.
If you are asked to “solve the integral” then you simply compute the answer, disregarding
whether the curve crosses into the negative or positive regions. If you are asked to “find
the area” then you have to draw the graph and ensure you break the integral up so you
can absolute value any negative regions.
Areas bounded by the x-axis
An area bounded between a curve and the x-axis will have the following features:
1. The integral bounds will be x-values.
2. The function you are integrating is “make y the subject and integrate the function
of x”.
3. There will be a dx after the integral.
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2
2
⌠
Example: Find  x – x + 1 dx =
⌡-1
We have been asked to solve the integral, hence we simply calculate this value.
2
 x3 x2

 – + x =
3 2
 -1
 8 – 4 + 2  –  -1 – 1 – 1  =
3 2
 3 2


 

4.5
Therefore the value of the integral is 4.5. Unless we determine that the entire region is
above the x-axis by drawing the graph, then we are not sure if this figure is also the area
under the curve between x = -1 and x = 2.
As we can see, the entire region of the integral is above the x-axis, hence the integral
value we calculated is in fact also the area under the curve between x = -1 and x = 2.
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Areas bounded by the y-axis
An area bounded between a curve and the y-axis will have the following features:
4. The integral bounds will be y-values.
5. The function you are integrating is “make x the subject and integrate the function
of y”.
6. There will be a dy after the integral.
Note: An area left of the y-axis will produce a negative area if you are integrating with
respect to y.
3
Example: Find the area enclosed between the y-axis, y = -1 and y = 2 for the function y = x
We note that the area is bounded by the y-axis, and also the question has specifically
asked for the area. Hence we need to draw the graph to examine any negative regions,
and make x the subject to integrate the “y stuff”.
Note: Some of the region is in the negative region, hence we will need to split up the
integral and absolute value the negative part.
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x=y
⌠ 0

⌡-1





1
3
y
1
3

dy 

0
  4 
  3 
  3y  


  4   -1
2
1
3
⌠
+  y dy =
⌡0
2

  4 

  3 

  3y  
 + 
 =

  4 0
   4  4  
  4  4 
3 ×    03  –  -13    + 3 ×   23  –  03   =
     
    
4
4
  4

3 × |(0 – 1)| + 3 ×   23  – (0)  =
 

4
4
4
3 + 3 × 23 =
4 4
2.64 (2dp)
This was not a particularly easy example, but the process remains the same. We made x
the subject then integrated the “y stuff” and we did this between the “y bounds”. We
noted some of the area was in the negative region, so we had to split up the integral and
absolute value this part. After this was done, we could integrate as per normal and
compute our area.
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Areas between Curves
To find the area between curves, we basically just subtract the curves in the integral
between the two x values, or y-values depending on what you are integrating with respect
to. Again you can do this with respect to either the x-axis or the y-axis.
2
Example: Find the area bounded by y = x + 1 and y = 2x + 4
The first thing we should do is draw a diagram, so we can get a feel for what we are
doing.
Solving simultaneously we get the points of intersection to be (-1,2) and (3,10). As we
will integrate with respect to the x-axis, this means we want our integral bounds in terms
of x.
Hence the integral will simply be the difference between the two curves, between x = -1
and x = 3.
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3
3
2
⌠
⌠
 2x + 4 dx –  x + 1 dx =
⌡-1
⌡-1
3
2
⌠
 ( 2x + 4) – ( x + 1) dx =
⌡-1
3
2
⌠
 2x + 3 – x dx =
⌡-1
3
3
 2

 x + 3x – x  =
3

 -1
 9 + 9 – 27  –  1 – 3 + 1  =

3  
3 

= 32
3
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Volumes of Revolution
Volumes of revolution is based on “adding up a whole lot of circle slices” to find a
2
volume. The integral is adding all the πr slices between the two bounds.
Volumes of revolution around the x-axis
b
2
⌠
V = π  y dx
⌡a
If the area is rotated around the x-axis there will be a dx in the integral, meaning we will
need to change everything to be “x stuff” in the integral. No y’s.
Volumes of revolution around the y-axis
b
2
⌠
V = π  x dy
⌡a
If the area is rotated around the y-axis there will be a dy in the integral, meaning we will
need to change everything to be “y stuff” in the integral. No x’s.
Basically to do these volumes of revolution problems, we identify the relevant formula
depending on if it is around the x-axis or the y-axis. From here, we will substitute in for
2
2
either the x or y . If it is around the x-axis we will have the integral with “dx” at the end
2
of it. This means we want everything in terms of x. Hence we have to replace the y with
the equivalent x expression.
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Example: Find the volume if y = x + 1 is rotated around the x-axis between x = 3 and x = 5
5
⌠ 2
V = π  y dx
⌡3
5
2
⌠
V = π  (x + 1) dx
⌡3
5
⌠ 2
V = π  x + 2x + 1 dx
⌡3

3

5
V = π x + x + x
3
3
2
V = π   125 + 25 + 5  –  27 + 9 + 3  
 3
  3

V = 152π
3
We identified the relevant formula for volumes around the x-axis and the substituted in
2
for the y , as we want everything in terms of dx, to match the dx in our integral. Once
everything was in terms of x, it was just a normal integration process.
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Trapezoidal and Simpson’s Rule
Basically these are slower and less accurate methods of integration. Know the formula,
and simply substitute into it, there are no real tricks here. Both methods are very similar,
with the only real difference being slight variations in the formulae. Other than that, the
process is pretty much the same.
Note: Function values represent the lines (height of y at a certain x value), whilst columns
represent the entire bar, formed by two lines.
Trapezoidal Rule
b
⌠
h
 f(x) dx ≈ (y0 + yL + 2(y1 + y2 + . . . + yL – 1)
2
⌡a
h is the constant “jump” between the x values. The y values are the heights of the y
values at that corresponding x value.
“The area under the curve approximately equals the distance between the x points divided
by 2, multiplied by the first plus last y values (line height), plus two times the sum of all
the other line heights.”
Simpson’s Rule
b
⌠
h
 f(x) dx ≈ { (y0 + yL + 4(y1 + y3 + . . . ) + 2(y2 + y4 + . . . )}
3
⌡a
“The area under the curve approximately equals the distance between the x points divided
by 3, multiplied by the first plus last y values (line height), plus four times the odd
heights, plus two times the even heights.”
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Exponential and Logarithmic Functions
Index laws
a
b
a+b
a
b
a–b
y ×y =y
y ÷y =y
a b



y



x y
y


a
–a
a
b
y =
=y
b c
a×c
= x


b×c
y
1
=
b
a×b
a
y
a
y
Basically if we multiply the same “base” then we add the indices, if we divide then we
subtract the indices. If we do not have the same base, we can not add or subtract the
indices.
If there is a “power to a power” then we multiply into the bracket powers.
A negative power means “invert it and make it positive”
For a fractional power follow the format above, with the number on the bottom coming
out the front indicating the magnitude of the root.
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Logarithms and logarithm laws
A logarithm is a different way to express an index equation.
c
loga b = c ⇔ a = b
c
This means an index equation, say a = b can be rewritten using logarithms. We haven’t
changed anything, we have just written the same thing, in a new format. This new
logarithm format allows us to manipulate the index equation in new ways.
To remember the order the numbers appear in the logarithm, use the sentence “to turn A
into B you need a power of C.
Example: Write the following index equation in logarithmic form,
5
2 = 32
To turn a 2 into 32 you need a power of 5. Hence
log2 32 = 5
Logarithmic Laws
loga x + loga y ⇔ loga (x × y)
loga x – loga y ⇔ loga (x ÷ y)
loga x
y
⇔ y × (loga x)
These are the main logarithmic laws from which everything is derived. Note that you
need a common base in order to use the first two rules. The third rule states that if the x is
to the power of something, you can bring that power out the front as a multiple. This is
NOT the same as the entire logarithm being raised to a power. Any power that only
applies to the x can be taken out the front. All the rules work in reverse, so for example,
any multiple out the front can be taken up to be a power of x.
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2
Example: Solve log2 x + log2 9x = -9
2
log2 x + log2{ 3x} = -9
log2 x + 2log2 x = -9
3log2 x = -9
log2 x = -3
-3
Hence 2 = x
x=1
8
Here we noted that we could write the second log as “something to a power” hence we
could take the index out the front. We then added the logs together to get three of them
and used the basic definition of a logarithm to solve the equation. You could have done it
another way, using the “adding logs” law which results in multiplication as outlined in
the first log law above.
Change of base
When using logarithms we can use any base we like. The most useful base is the
irrational number “e”, which is referred to as the “natural base” of a logarithm. In order to
change the base of a logarithm, use the following formula,
loga b = logc b
logc a
All this means is, you choose a new base, which we called c. The ‘a’ and ‘b’ go in those
respective positions and you have changed the base of the logarithm.
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Example: Change the following to base 'e' log7 19
log7 19 = loge 19
loge 7
Graphs of Exponential and Logarithmic Functions
Logarithmic functions and exponential (e) functions are inverses of each other. Inverse
functions are reflected in the line y = x.
x
Red: y = e
Blue: y = ln(x)
Black: y = x
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Differentiation of the logarithmic function
When we differentiate a logarithm, our answer will be a fraction which does not contain a
logarithm.
y = ln{ f (x)}
y ′ = f ′(x)
f (x)
So the function becomes the denominator, and the numerator is the derivative of the
function which has the logarithm applied to it.
2


Example: Find dy if y = ln  3x + x 
dx
dy = f ′(x)
dx
f (x)
dy = 6x + 1
2
dx
3x + x
Integration involving logarithms
We are not interested in “integrating logarithms”, we “integrate fractions which turn into
logarithms”.
⌠ f ′(x)
dx = ln{ f (x)} + c

⌡ f (x)
You will often have to “manipulate a fraction” to get it into this basic format, where the
numerator is the derivative of the denominator. You are allowed to manipulate the
fraction, but you can not change the value of it.
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⌠
3x + 1
Example: Solve the following integral, 
dx
2
⌡ 3x + 2x + 7
Notice this is “almost” in the form where the derivative is on the top. What we can do is
double the top so it is the derivative of the numerator, and “balance it out” by putting a
“times 0.5” out the front. The overall result of doubling and multiplying by a half will
cancel out, meaning we didn’t change the fraction, we just manipulated it.
6x + 2
1 × ⌠
dx =

2
⌡ 3x2 + 2x + 7
1 ln  3x2 + 2x + 7  + c

2 
“Whatever is on the denominator goes into the logarithm function.”
Differentiation of the exponential function
When we differentiate an exponential function, everything stays the same except you will
multiply by f ’(x), which means the “derivative of the index”.
y=e
f( x)
y ′ = f ′(x) × e
f( x)
2
Example: Find y ' if y = 3e

x + 4

x +4
– 4x
2
y ′ = 3 ×  2x × e

–4
2
y ′ = 6xe
x +4
–4
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Here we do our normal differentiation processes, noting that when we differentiate the
exponential function, that we use the above rule. The derivative of the exponential is in
the brackets to better delineate the process.
Integration of the exponential function
When we integrate an exponential function, we set it up to reverse the differentiation
process. We need the integral in the general format, which is,
f(x)
f(x)
⌠
+c
 f ′(x) × e dx = e
⌡
1
x
⌠
Example: Find  xe + 4 dx
⌡0
1
2
x
⌠
 xe + 4 dx =
⌡0
2
1
1 ⌠ 2xex + 8 dx =

2 ⌡0
2
1
1  ex + 8x  =
0
2
2
1 × { (e + 8) – (1 + 0)} =
2
e+7
2
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Trigonometric Functions
Introduction
This chapter closely follows from the preliminary trigonometry chapter. The main
concepts in the preliminary chapter are:
•
•
•
•
•
Knowledge of the shape of trigonometric graphs
The exact value triangles
Angles of any magnitude
Solving trigonometric equations
Manipulating trigonometric identities
Radians
We can measure angles using “radians” instead of “degrees”. After being introduced to
radians, expect to do almost all of your work in radians instead of degrees.
π radians = 180°
Hence this basically means you can replace π with 180 degrees when converting.
Arc length and sector area
L = θr
A = 1 θr
2
2
If you have any doubts, you can substitute in 2π for θ and confirm that it all makes
sense.
In harder questions you may have to find the area of a segment by subtracting the area of
a triangle from the area of a sector.
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Example: Find the area of the minor segment.
Area of the sector is
As = 1 × 2π × 3
2 3
2
As = 3π
Area of the triangle is
At = 1 absinC
2
At = 1 × 3 × 3 × sin  2π 
2
 3
At = 9 3
4
2
Hence area of the segment is As – At = 3π – 9 3 cm
4
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Manipulation of Trigonometric Graphs
In the preliminary course we learned the basic trigonometric graphs. Here we will look at
manipulating them.
Manipulation of Sine and Cosine Graphs
We will use the example of Sine, however the principles are identical in their application
to the Cosine graph.
(y + b) = a × sin(dθ + c)
a → represents the amplitude
b → causes vertical shift of b units
c → causes horizontal shift of c units
Period = 2π
d
So basically the ‘a’ tells you how high and low the graph “bounces”, ‘b’ and ‘c’ will
cause shifts in the graph but will not alter the shape of it. The period, “how compressed”
the graph is will be determined by ‘d’, the number in front of our angle.
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Example: Graph y = 3sin(2θ – π) + 1
Putting this in general format we get, (y – 1) = 3sin(2θ – π) hence,
amplitude (bounce) is 3 units up and down
the graph is shifted up 1 unit and to the right
π units
the graph is twice as compressed, as period = 2π = π
2
This was a fairly challenging example, but it is simply an application of all the relevant
manipulations at once. It is uncommon to be asked to manipulate a tan function, but you
may be asked to draw a ‘vertically or horizontally shifted tan function’.
To graph the reciprocal functions, graph the original function and graph the reciprocal
function going in the opposite direction with asymptotes when y = 0 on the original
function.
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Example: Graph y = cosec θ
Here we see the blue graph of the original function y = sinθ , and we draw the black
asymptotes where y = 0. From here it is easy to draw the inverse function, which is done
in red.
Differentiation of Trigonometric Functions
The process for differentiation of trigonometric functions can be explained as
“What function does it turn into and what do I multiply by?”
y = sin f(x) → y ′ = f ′(x) cos f(x)
y = cos f(x) → y ′ = – f ′(x) sin f(x)
2
y = tan f(x) → y ′ = f ′(x) sec f(x)
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So basically you multiply by the derivative of the function which has the trigonometric
function applied to it. Then you will change the trigonometric function into a new one.
Note that cosine is the odd one out, it is the only one which will differentiate to a
negative function.
4
 2

Example: Find dy if y = 3tan  x + x  – x
dx


2 2
3
y ′ = 3 ×  (2x + 1) × sec  x + x   – 4x
2 2

3
y ′ = (6x + 3)sec  x + x  – 4x
Here the expression contained a trigonometric function. We did our normal
differentiation to the non trigonometric part, and the trigonometric part in represented
inside the large brackets in the second line. We multiplied by the derivative of f (x) ,
2
which is 2x + 1 and the tan turned into sec .
Integration of Trigonometric Functions
This is the reverse process of the differentiation. Hence it is basically a case of “put into
general derivative format and reverse the process through integrating”.
⌠
 f ′(x) cos f(x) dx = sin f(x) + c
⌡
⌠
 f ′(x) sin f(x) dx = – cos f(x) + c
⌡
2
⌠
 f ′(x) sec f(x) dx = tan f(x) + c
⌡
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⌠
Example: Find  sin(4x + 2) dx
⌡
⌠
 sin(4x + 2) dx =
⌡
1 ⌠ 4sin(4x + 2) dx =

4 ⌡
– 1 cos (4x + 2) + c =
4
What differentiates to give sin? – cos does, so we will turn our function into – cos when
we integrate. We just need to manipulate it into the general format to integrate.
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Applications of Sequences and Series
Sequences and Series
This chapter is about number patterns, a set of numbers (terms) which follow a certain
rule. A collection of terms is called a sequence, if we are adding this collection of terms
up, it is called a series.
For example we could have 1,4,7,11,… which is a sequence, and the corresponding series
is 1+4+7+11+…
A convenient way of writing a sum is with “sigma notation”.
∞
Σ
(1 + (n – 1) × 3)
1
This means we start with n = 1 and get that number. Then we let n = 2 and get that
number. We continue forever as the top number tells us what we stop at. We then add all
these answers together. That’s what sigma notation does, it adds all the stuff together.
Arithmetic Sequences and Series
If the pattern of terms is a “plus or minus” pattern then it is called an “arithmetic series”.
For example,
5,7,9,11…or -3,-4,-5,-6…
In both these cases we have to plus or minus by a “common difference” to get to the next
term.
Term Formula
To find the value of a specific term we use the following formula
Tn = a + (n – 1)d
Where a represents the ‘first term’ and d represents the ‘common difference’.
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Summation Formula
Sn = n (2a + (n – 1)d)
2
Or
Sn = n (a + l)
2
Where l is the “last term”. Both formulae will work, it is just a matter of choosing the
most convenient one for the question.
Example, for the following sequence find the 7th term and the sum of the first 5 terms
21, 18, 15, 12,...
First of all we can note that this is an arithmetic sequence as there is a “common
difference” of negative three. This means that our arithmetic formula apply, so it is
simply a matter of using them.
So, to find the 7th term we use the term formula with a = 21 (our first term), d = -3
(common difference) and n = 7 as there are 7 terms.
Tn = a + (n – 1)d
T7 = 21 + (7 – 1) × -3
T7 = 3
So the 7th term is 3
Now to find the sum of the first 5 terms,
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Sn = n (2a + (n – 1) × d)
2
Sn = 5 × (2 × 21 + 4 × -3)
2
Sn = 75
So if we added the first 5 terms together we would get 75.
Geometric Sequences and Series
If the pattern of terms is a “multiply or divide” then it is called a geometric sequence.
For example, 2,6,18,54, … or 12,6,3,1.5, …
Term Formula
All the pronumerals mean the same as with the arithmetic formula, but now we have r
(common ratio) instead of d (common difference).
Tn = ar
n–1
Summation Formula
 n



Sn = a  r – 1 
r–1
Some books will have two formulae for the geometric summation. They are equivalent. If
you use this formula then you will get the correct answer every time.
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Example: If the first term of a geometric progression is 2 and the 6th term is 486, what
is the common ratio?
This example requires us to use the formula to create an equation. Sometimes you will be
required to create two equations and solve simultaneously.
a=2
now we need to substitute into the formula
Tn = ar
n–1
486 = 2 × r
243 = r
5
6–1
5
243 = r
r=3
So basically we just substituted the information into the equation and solved for r.
Limiting Sum
If the common ratio of a geometric series has a magnitude of less than 1, so it is between
positive and negative one, excluding zero, then a ‘limiting sum’ exists.
The limiting sum, also called infinite sum is the number you will get very close to, but
never hit, if you add up all the terms. It is an asymptote.
To calculate the limiting sum, we use the following formula
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S∞ =
a
1–r
Example: Does the following series have a limiting sum? If so calculate it.
2, 1, 0.5, 0.25, . . .
First of all it does have a limiting sum as the common ratio has a magnitude less than 1, it
is 0.5.
To find the limiting sum, we substitute into the formula,
S∞ =
a
1–r
S∞ =
2
1 – 0.5
S∞ = 4
Therefore, if we added all these terms, they would get very close to 4, but never quite
touch it.
Problems involving series
This chapter follows on from the preliminary chapter “sequences and series”. You need
to know all the relevant formulae from there, and apply it to “wordy questions”. The vast
majority of these questions tend to be financial questions. Basically all you have to do is
turn the “wordy question” into a series and use the appropriate AP or GP summation
formula.
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Financial applications
Most questions in this chapter tend to be either superannuation or loan repayment based
questions.
Superannuation
The idea with these questions is that at the start of every time period (usually year) you
invest a certain amount of money. You will usually be asked how much it will be worth
at the end of a certain amount of time.
The compound interest formula is often used in these questions.
n
A = P  1 + r 
100 

The process can be described as follows,
1. Form a GP using the amount of money invested per time period as the terms
2. Create a summation formula to find what it is all worth added together over a
certain amount of time.
Example: Sonny invests $10000 at the start of each year. He receives 8% compound
interest p.a. How much will this investment be worth after 10 years?
We need to form a GP by listing out all the terms. From there we can identify what our
“a” and “r” are going to be, hence we can use the summation formula to find out what the
ten terms all add up to.
A1 = 10000(1 + 0.08)
10
A2 = 10000(1 + 0.08)
9
A3 = 10000(1 + 0.08)
8
So we can see the pattern here. Our first term is 10000(1 + 0.08)
10
and our common ratio,
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-1
(1 + 0.08) .
Now all we have to do is take the sum of ten terms.
 n



S = a r – 1
r–1
S10 =



10000(1.08)
10  


1.08
-10

– 1 
-1
1.08 – 1
S10 = 156454.87 2dp
Therefore the total investment would be worth $156454.87 over the ten year period.
Loan Repayments
These questions tend to be a bit harder than the superannuation questions. Usually these
type of questions involve borrowing lots of money and making regular periodic
repayments.
1. For the general expression for An
2. Solve An = 0
3. Use GP summation formula to simplify the equation
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Example: Adrian borrows $10000 at 8% p.a. reducible. He pays the loan off in
equal yearly instalments over 6 years. Calculate the size of the yearly instalment.
So let’s create an expression for the amount owing after every year, then solve the
amount owing after 6 years as equal to zero. The GP summation will only play a small
role in this process, but it can greatly simplify the working out.
Let A1 be the amount owing after the first instalment is paid and M be the amount of the
instalment, hence,
A1 = 10000(1 + 0.08) – M
A2 = A1 × (1.08) – M
2
A2 = 10000(1.08) – (1.08)M – M
3
2
A3 = 10000(1.08) – (1.08) M – (1.08)M – M
We can generalise this to generate An.
In this question we could keep going until we get to A6 however many questions
will require a number much higher than 6, so we need to get used to finding An
n
An = 10000(1.08) – (1.08)
n

n–1
An = 10000(1.08) – M  1.08
M – (1.08)
n–1
+ 1.08
n–2
n–2
M – ... – M

+ . . . + 1 
This is the general expression for the amount owing after the ‘nth’ repayment. Now we
need to set this expression equal to zero when n = 6 and solve. The GP summation
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formula will be used to simplify the expression in the brackets where M was factorised
out from.
A6 = 0
6

5

0 = 10000(1.08) – M  1.08 + . . . + 1 
Use the GP summation formula for the bracket sequence and solve the equation for M.
 1.085  1.08-6 – 1  

 1.08-1 – 1 
M
= 10000(1.08)
6
M = 2163.15 2dp
Therefore the amount of the yearly repayment is $2163.15.
So basically this chapter is about using GP summations to solve “wordy” questions,
which are usually financially based and include the compound interest formula.
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Applications of Calculus to the Physical
World
Rates of Change
Whenever you differentiate a function, you obtain the derivative or gradient function.
This is also known as the “rate of change” function. The rate of change function tells you
“how fast” something is happening. On a Cartesian plane (xy axis) we refer to the rate of
change as the gradient.
Instead of getting a function of x and y, we now get functions linking time, temperature,
etc and other real world variables. The process is exactly the same as when you have a
function of x and y. Differentiate, and you have the rate of change function. If you
integrate the rate of change function you obtain the original function.
Example: If the amount of corn produced on a farm can be calculated using,
2
c = 3a – a, where c is kgs of corn and a is area in hectares being farmed.
Find the rate at which the amount of corn production is increasing when a = 10 ha
So basically you need to ask yourself is this a “how much” question (use the original
function) or a “how fast” question (use the derivative function). This is a “how fast”
question, so we need to find the first derivative of our function and then substitute in the
relevant value of ‘a’.
dc = 6a – 1
da
when a equals 10,
dc = 59
da
Hence the rate of increase is 59kgs per hectare.
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In some questions they may give you the rate of change function and you will have to
integrate it to get the original, function. Also, you may be required to use the second
derivative to describe the rate of change. This is the same idea as “determining the
concavity” from the Applications of Calculus chapter.
Exponential Growth and Decay
Whenever the rate of change of a function is “in proportion to itself” then you are dealing
with exponential growth and decay. It is the same concept as normal rates of change, but
there is a specific format these functions follow.
dQ = kQ
dt
kt
Q = Q0 e
Basically these questions require you to recognise that it is an exponential growth and
decay question (“proportional” is the key word usually in the question), and then use the
basic format of the formula. To find the value of the unknown constants you will usually
have to substitute in a point and solve the resulting equation.
Q and t are the variables, quantity and time
Q0 is the constant, initial quantity
k is also an unknown constant
Example: In a town the population was 10000 in 1986. In 1996 the population
was 14000. Assuming the rate of growth is proportional to the population, what
will the population be in 2002?
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This question told us that the rate of change is proportional to the population, ie the rate
of change is proportional to itself, so we can use the exponential growth and decay
formula.
Q = Q0 e
kt
Let 1986 be when time is zero, hence 10000 is the initial quantity
Q = 10000e
kt
Now we just need to find the constant k, and we do this by substituting in a point. When
time = 10, Quantity = 14000.
14000 = 10000e
k × 10
ln  14  = 10k × lne
 10 
k = 0.034 3dp
Now we have found all the constant, we have our equation and we can use this to find the
value of Q when t = 16,
Q = 10000 e
0.034 × 16
Q = 17229 to the nearest whole number
So the basic process can be seen above. Identify that it is an exponential growth and
decay question, find all the unknown constants and use the formula.
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Particle Motion
The motion of a particle can be described in terms of rates of change. We start off with
the equation linking displacement (distance) with time. If we differentiate this with
respect to time we get the velocity equation. If we differentiate the velocity equation we
obtain the acceleration equation.
So basically you differentiate to move down and you integrate to move upwards in the
diagram. You will do all your differentiation and integration with respect to time.
1. Identify what function the question requires you to use, displacement, velocity or
acceleration.
2. Obtain that equation, usually through differentiation or integration.
3. Solve for the required value through substitution.
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Example: The acceleration of a particle moving along the x-axis is given by dv = 10t – 2
dt
Initially the particle is 3 meters to the right of the origin with a velocity of 1m/s. Find the
displacement after 3 seconds.
Here we can see that we are given the acceleration equation, and are required to use the
displacement equation. If we integrate the acceleration equation twice, with respect to
time, we will obtain the displacement equation.
a = 10t – 2
2
v = 5t – 2t + c
When we integrate we will get a + c that we need to deal with. We have to substitute in a
point to find this value. We know that when t = 0, v = 1
2
1 = 5 × (0) – 2 × (0) + c
c=1
2
Hence v = 5t – 2t + 1
3
x = 5t – t + t + c
3
2
When t = 0, x = 3. Hence
3 = 5 × (0) – (0) + 0 + c
3
3
2
c=3
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So we have found our displacement equation after integrating our acceleration equation
twice with respect to time and substituting to find the value of the unknown constants.
3
x = 5t – t + t + 3
3
2
When t = 3
3
x = 5 × (3) – (3) + 3 + 3
3
2
x = 42
Therefore the displacement is 42 meters to the right of the origin.
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Probability
Introduction
Probability of a Single Event
The probability that something will occur is given by the number of ways it could occur,
divided by the number of possible outcomes.
P(E) = number of favorible outcomes
total number of outcomes
Complementary Events
If events are mutually exclusive, then the probability of something not happening is 1
minus the probability it does happen. Events are mutually exclusive if “either one
happens or the other happens”.
Example: If there is a 2 chance of rolling a number greater than four on a dice
6
what is the chance of not rolling a number greater than four?
As these are mutually exclusive events, the probability is one minus 2 = 2
6
3
Multi-Staged Events
Multi-staged events are when ‘more than one thing’ happens. You multiply the
consecutive probabilities to find the final probability.
Important terms include:
Or: One thing happens or the other happens. You add the probabilities
And: Both things need to happen
At least: The chance of something and everything greater than it happening
At most: The chance of something and everything less than it happening
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Another thing to be aware of is “replacement” and “non replacement” questions.
Basically this means that if something happens, does this reduce the chance of it
happening the next time? If it is a replacement question then it will not reduce the
probability of it happening again, if it is a non replacement question it will reduce the
probability of it happening again.
Tree Diagrams
Tree diagrams provide a graphical representation of multi-staged events. It can be
thought of as “multiplying down braches and adding between branches” to help simplify
the process.
Example: The names of 7 boys and 4 girls are placed in a hat. Two names are drawn out to
form a group. What is the probability there will be at least one girl in the group?
So this means we can count GG, GB and BG. We can add all these probabilities together.
We can also find 1 – BB and get the same answer.
P(1 – BB) = 1 –  7 × 6 
 11 10 
P(1 – BB) = 68
110
Basically the chance of getting boy, boy is the multiplication, noting that once you pick
one boy there are only 6 left out of a possible 10 people now. This is a non replacement
question, once we picked a boy, there was one left to pick from hence it lowered our
chance of getting another boy.
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Summary
This provides you with an overview of all the important concepts. You should read over
each of these and then see if you can recall all of the formulae or principles. After you do
so you can look back in the book to check if you were correct. This tests your ability to
recall information, in much the same way as test does.
Applications of Coordinate Geometry
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Formulae
Coordinate Methods in Geometry
Geometrical Applications of Calculus
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Introduction (Gradients and Tangents)
The First Derivative (The Gradient Function)
Stationary Points (Four types)
The Second Derivative (The Concavity Function)
Curve Sketching (Finding Stationary Points and Determining their Nature)
Maxima and Minima Problems
Integral Calculus
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Introduction - The Area under a Curve
Indefinite Integrals
Definite Integrals
Areas between Curves
Volumes of Revolution
Trapezoidal and Simpson’s Rule
Exponential and Logarithmic Functions
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Index laws
Logarithms and logarithm laws
Graphs of Exponential and Logarithmic Functions
Differentiation of the logarithmic function
Integration involving logarithms
Differentiation of the exponential function
Integration of the exponential function
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Trigonometric Functions
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Introduction
Radians
Arc length and sector area
Manipulation of Trigonometric Graphs
Differentiation of Trigonometric Functions
Integration of Trigonometric Functions
Applications of Sequences and Series
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Arithmetic term and summations
Geometric term and summations
Limiting sums
Problems involving series
Financial applications (super and loan repayments)
Applications of Calculus to the Physical World
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Rates of Change
Exponential Growth and Decay
Particle Motion
Probability
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Introduction
Multi-Staged Events
JW Hick, 2013