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2020-ME1-HSC-Study-Notes-Vincent-Liu

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Mathematics Extension 1
HSC study notes
Table of Contents
Trigonometric Identities ..........................................................................................................2
Mathematical Induction ...........................................................................................................3
Co-ordinate Geometry (3 Unit) ...............................................................................................4
Polynomials ...............................................................................................................................5
Parametrics ................................................................................................................................8
Permutations and Combinations ..........................................................................................10
Circle Geometry .....................................................................................................................11
Inverse Functions and Inverse Trigonometric Functions ...................................................13
Integration (3 Unit) .................................................................................................................15
Applications of Calculus to the Physical World (3 Unit) ....................................................16
The Binomial Theorem ..........................................................................................................21
Written by Vincent Liu (2018)
1
Trigonometric Identities
Pythagorean
sin2 x + cos2 x = 1
tan2 x + 1 = sec2 x
cot 2 x + 1 = cosec2 x
Angle Sum
sin(x + y) = sin x cos y + cos x sin y
sin(x − y) = sin x cos y − cos x sin y
cos(x + y) = cos x cos y − sin x sin y
cos(x − y) = cos x cos y + sin x sin y
tan(x + y) =
tan x + tan y
1 − tan x tan y
tan(x − y) =
tan x − tan y
1 + tan x tan y
Double Angle
sin 2x = 2 sin x cos x
cos 2x = cos2 x − sin2 x
Sums to Products
OR
cos 2x = 1 − 2 sin2 x
OR
cos 2x = 2 cos2 x − 1
(usually must be proved before using)
sin x + sin y = 2 sin
x +y
x −y
cos
( 2 )
( 2 )
cos x + cos y = 2 cos
sin x − sin y = 2 cos
x +y
x −y
cos
( 2 )
( 2 )
x +y
x −y
sin
( 2 )
( 2 )
cos x − cos y = − 2 sin
x +y
x −y
sin
( 2 )
( 2 )
t-formulae
If t = tan
sin x =
x
and x ≠ π, 3π, 5π . . . then:
(2)
2t
1 + t2
cos x =
1 − t2
1 + t2
tan x =
2t
1 − t2
Transformations (Auxiliary method)
a sin x + b cos x = r sin(x + α)
a cos x + b sin x = r cos(x − α)
a sin x − b cos x = r sin(x − α)
a cos x − b sin x = r cos(x + α)
Where r =
a 2 + b 2 and α = tan−1
b
(a)
General Solutions
sin θ = sin a
θ = n π + (−1)n a
cos θ = cos a
θ = 2n π ± a
tan θ = tan a
θ = nπ + a
2
Mathematical Induction
To prove a statement or proposition S(n) true by mathematical induction:
1. We first prove it true for a base case, usually the minimum value of n.
2. We then assume that S(k) is true (that is, S(n) is true for some n = k).
3. Then, we prove that S(k + 1) is true, usually by using the assumption that S(k) is true.
If S(k) is true, then S(k + 1) is true, so therefore S((k + 1) + 1) = S(k + 2) must be true…
i.e. if S(k) is true for some integer k, then it is also true for all subsequent integers.
Proving the statement for a base value completes the proof.
Divisibility notation
The convenient notation m | n means that m divides n (that is, n is divisible by m).
3
Co-ordinate Geometry (3 Unit)
Angle between two lines
For two lines with gradients m1 and m 2, the angle θ between them is given by:
90∘
y
if m1m 2 = − 1, or
θ1
m1 − m 2
tan θ =
1 + m1m 2
if m1m 2 ≠ − 1
θ2
x
tan θ1 = m1
tan θ2 = m2
Division of Intervals
The order in which the endpoints of the interval A and B are specified is important. There are two cases:
Internal Division
n
m
B(x2, y2)
P(x, y)
A(x1, y1)
External Division
If
m
> 1,
n
i.e. | m | > | n |
m
P(x, y)
B(x2, y2)
If
m
< 1,
n
i.e. | m | < | n |
n
B(x2, y2)
A(x1, y1)
n
A(x1, y1)
P(x, y)
m
Note: these three statements are equivalent:
• external division in the ratio m : n
• division in the ratio −m : n
• division in the ratio m : − n
In either case, the point P that divides the line A B in the ratio m : n is given by the formula
x=
m x2 + n x1
m+n
y=
my2 + ny1
m+n
When using the formula with external division, one of m and n must be negative (it does not matter which).
4
Polynomials
A polynomial is an algebraic expression of the form an x n + an−1 x n−1 + an−2 x n−2 + . . . + a 0 for n ∈ ℤ+.
Using division, any polynomial can be expressed in the form P(x) ≡ D(x)Q(x) + R(x), where the polynomial
P(x) is the dividend, D(x) is the divisor, Q(x) is the quotient and R(x) is the remainder.
Polynomial Long Division
Polynomial long division occurs like normal long division.
e.g.
x + 1 | 2x 3 + 3x 2 + 2x + 1
1. Choose a term that can be multiplied with the first term of the divisor to give the first term of the
dividend. This is the first term of the quotient.
2x 2
x + 1 | 2x 3 + 3x 2 + 2x + 1
2x 3
2. Multiply the rest of the terms of the divisor by the term in step 1. and write down the result directly
underneath.
2x 2
x + 1 | 2x 3 + 3x 2 + 2x + 1
2x 3 + 2x 2
3. Perform a subtraction.
2x 2
x + 1 | 2x 3 + 3x 2 + 2x + 1
2x 3 + 2x 2
x2
4. Bring down the next term(s) (the number of terms you end up with must be the same number of terms
in the divisor) and repeat.
2x 2
x + 1 | 2x 3 + 3x 2 + 2x + 1
2x 3 + 2x 2
x 2 + 2x
Note: When dividing polynomials, the degree of the remainder is always less than that of the divisor.
5
Remainder Theorem
If a polynomial P(x) is divided by (x − α), the remainder is P(α).
More generally, if a polynomial P(x) is divided by (bx − α), the remainder is P
α
.
(b)
Proof: We use the form P(x) ≡ D(x)Q(x) + R(x) with D(x) = bx − α.
Since D(x) has degree one, R(x) must be of degree zero (that is, a constant).
P(x) = (bx − α)Q(x) + r
α
∴P
= 0 ⋅ Q(x) + r
(b)
P
α
=r
(b)
Factor Theorem
It follows from the remainder theorem that (x − α) is a factor to P(x) if and only if P(α) = 0.
This technique can be used to quickly factor polynomials by testing values of α.
More generally, (bx − α) is a factor to P(x) if and only if P
α
= 0.
(b)
Graphing Factorised Polynomials
The graph of a factorised polynomial can be sketched easily by inspecting its factors.
If a polynomial has a factor (x − α)r, then its graph has a curve of degree r at x = α.
y
y
y
x
x
x
degree 1
degree 3, 5, 7…
degree 2, 4, 6…
The exact nature of a curve can be determined by testing if the polynomial is positive or negative at a point
on the curve.
e.g. P(x) = (x − 1)(x − 2)2(x − 3)3
x = 0,
Testing points:
y
y = P(x)
y>0
x = 32 , y < 0
x = 52 , y < 0
x = 4,
y
1
2
y = P(x)
3
1
x
2
3
x
y>0
6
Roots and Coefficients of a Polynomial
Quadratic
Cubic
Quartic
a x 2 + bx + c = 0
a x 3 + bx 2 + cx + d = 0
a x 4 + bx 3 + cx 2 + d x + e = 0
−b
a
c
αβ =
a
α +β =
−b
a
c
αβ + α γ + β γ =
a
−d
αβ γ =
a
α +β +γ =
α +β +γ +δ =
−b
a
αβ + α γ + αδ + β γ + βδ + γδ =
αβ γ + αβδ + βδγ + αδγ =
αβ γδ =
e
a
−d
a
c
a
Approximation of roots
Suppose a polynomial function f (x) has unknown roots that we wish to approximate. We have two methods.
An important observation: if f (x 0 ) and f (x1) have different signs, then a root must lie between x 0 and x1.
Half-interval method
From the above observation, the first approximation using this method is given by f
x 0 + x1
as the first point.
2
x 0 + x1
Then, the second point is whichever of x 0 and x1 has a different sign to
.
2
x 0 + x1
.
( 2 )
To apply the method again, choose
Newton’s method
y
If x1 is an approximation, a better approximation x 2 is given by
x 2 = x1 −
f (x1)
f′(x1)
tan θ = f ʹ(x1)
tan θ = ______
f(x1)
(x1 - x2)
Equating these
gives Newton’s
method.
Newton’s method does not work when:
1. x1 is a stationary point, i.e. f′(x1) = 0
2. x1 is closer to another root, in which case x 2 approximates
that root instead
θ
α
x2
x1
x
In some cases, Newton’s method will oscillate or end up moving towards another root instead.
y
Here, when Newton’s method is applied on x1 to approximate α,
the new approximation x2 is too far away. Applying Newton’s
method again causes an approximation of β instead.
x3
α
β
x2
x1
x
7
Parametrics
A parametric equation is a set of two equations each describing x and y in terms of a parameter.
Parametric equations can be converted to Cartesian equations by solving and eliminating the parameter.
Parametric representation of a parabola
The Cartesian form of a parabola with the vertex at the origin is x 2 = 4a y.
The corresponding parametric equation is
x = 2at
.
{y = at 2
Note: The gradient of the tangent at a point is equal to the parameter:
dy
d y /dt
2at
=
=
=t
dx
d x /dt
2a
Equation of the chord
The gradient of the chord is
y
2
2
aq − ap
p +q
=
(since a ≠ 0).
2a q − 2a p
2
Using point-gradient form gives
y=
Q(2aq, aq2)
P(2ap, ap2)
p +q
x − a pq
( 2 )
x
If the chord is focal, it passes through the focus (0, a).
Substituting this point into the equation gives a condition for a focal chord: pq = − 1.
Equation of the tangent
y
By obtaining the gradient and then using the point-gradient
formula, we get:
Parametric: y = px − a p 2
P(2ap, ap2)
Cartesian: x x1 = 2a(y − y1)
Equation of the normal
By obtaining the gradient of the tangent, taking the negative
reciprocal and using the point-gradient formula, we get:
x
y
P(2ap, ap2)
Parametric: x + p y = a p 3 + 2a p
Cartesian: y − y1 = −
2a
(x − x1)
x1
x
8
Intersection of tangents
By simultaneously solving two parametric equations of the tangent with parameters p and q,
y
y = px − a p 2
{y = qx − a q 2
x = a( p + q) and y = a pq
Q(2aq, aq2)
P(2ap, ap2)
The intersection of two tangents is always outside the parabola.
x
Intersection of normals
By simultaneously solving two parametric equations of the normal with parameters p and q,
y
x + p y = a p 3 + 2a p
{x + q y = a q 3 + 2a q
2
2
x = − pq( p + q) and y = a( p + pq + q + 2)
Q(2aq, aq2)
P(2ap, ap2)
The intersection of two normals is always inside the parabola.
x
Chord of contact
The chord of contact is the line drawn between two points of contact on the parabola, i.e. points whose
tangents intersect.
y
Observe that the equations for both tangents are:
x1 x 0 = 2a(y1 + y0 )
{x 2 x 0 = 2a(y2 + y0 )
Both these equations are of the form
x x 0 = 2a(y + y0 )
(x2, y2)
(x1, y1)
x
(x0, y0)
A line can be unambiguously defined by two points — in this case, the two points of contact.
Since this equation represents a line and is satisfied by both points of contact, we deduce that it must
therefore be the equation for the chord of contact.
9
Permutations and Combinations
Permutations are arrangements where order is important i.e. A BC is different from CB A.
Combinations are arrangements where order is not important i.e. A BC is the same as CB A.
We introduce the factorial notation: n! = (n)(n − 1)(n − 2) . . . (1).
Multiplication Principle
If one operation can be performed in m different ways and afterwards, a second operation can be
performed in n different ways and so on, the number of ways of performing the operations are m × n × . . . .
Addition Principle
If two operations are mutually exclusive, the number of arrangements for each can be added together to
achieve the number of arrangements for either.
Circular Permutations
For every arrangement in a circle, there are n arrangements in a line which would be equivalent (e.g. for
n = 4 we have A BCD = BCDA = . . . ). Thus, the number of ways to arrange n people in a circle is
n!
.
n
Permutations with Repetition
When arranging n things with p amount of one thing, q amount of another thing and so on, the number of
permutations is
n!
.
p! q!
The notation nPr
n
Pr represents the number of permutations where r places are filled with n objects.
n
n!
(n − r)!
n
n!
(n − r)!r!
Pr =
The notation nCr
n
Cr represents the number of combinations where r places are filled with n objects.
or alternatively, the binomial notation nCr =
Cr =
n
may be used.
(r)
The Stars and Bars method
When distributing n items across m places, permutations with repetition may be used to simulate the
scenario by using dividers | to create places and stars (or similar objects) to represent items.
e.g. ⋆ | ⋆ ⋆ ⋆ | ⋆ ⋆ | ⋆ simulates distributing 7 items across 4 places, the dividers creating places.
10
Circle Geometry
Theorems
Equal angles at the centre
stand on equal chords.
The perpendicular from the centre of a circle
to a chord bisects the chord.
Equal chords in equal circles
are equidistant from centres.
(and converse)
The line from the centre to the midpoint of a
chord is perpendicular to the chord
(converse)
(and converse)
O
O
O
θ
θ
Opposite angles of a cyclic
quad. are supplementary.
The angle in a semicircle is a
right angle.
(converse to prove concyclic points)
(converse to prove diameter)
The angle at the centre is twice
the angle at the circumference
subtended by the same arc.
θ
θ
O
O
180° – θ
Angles in the same segment
are equal.
2θ
If an interval subtends equal
angles at two points on the same
side of it, the interval’s end points
and the two points are concyclic.
The angle between a tangent and
the chord is equal to the angle in
the alternate segment.
(converse to prove tangent)
θ
θ
θ
θ
θ
θ
θ
11
Tangents to a circle from an
external point are equal.
The tangent to a circle is
perpendicular to the radius
drawn to the point of contact.
When two circles intersect,
the line through their centres
bisects the common chord at
right angles.
(and converse)
O
Oʹ
O
The products of the
intercepts of two intersecting
chords are equal.
The products of the
intercepts of two secants to a
circle are equal.
The square of the tangent is
equal to the product of the
intercepts of the secant.
AX · XB = CX · XD
AX · BX = CX · DX
CX2 = AX · BX
X
A
X
B
D
B
D
X
C
A
B
C
A
C
General Tips for Solving Circle Geometry Problems
• draw the most general form of a shape
(e.g. if drawing a triangle, draw a scalene one, not isoceles)
When two circles touch, the
line joining their centres
passes through the point of
contact.
• rotate the diagram to get a new perspective
Oʹ
• highlight cyclic quadrilaterals
• construct any common chords or tangents
T
O
12
Inverse Functions and Inverse Trigonometric Functions
y
An inverse function (denoted f −1(x)) is a function that “reverses” another.
y=x
y = f -1(x)
If f (x) = y, then f −1(y) = x, and f ( f −1(x)) = x = f −1( f (x)).
To find the inverse, change f (x) to x and x to f −1(x) in the equation.
e.g. f (x) = log x + 1 ⟹ x = log f −1(x) + 1 ⟹ f −1 x = e x−1.
x
y = f (x)
When taking the inverse of a function, its domain and range are switched.
Graphing Inverse Functions
The graph of y = f −1(x) is the graph of y = f (x) mirrored across the line y = x.
This means that the graph of y = f −1(x) intersects the graph of y = f (x) at y = x.
Restricting the Domain
If f (x) is NOT a one-to-one function (that is, it does not pass the “horizontal line test”), then its inverse will
not actually be a function (it will not pass the “vertical line test”).
In this case, we can restrict the domain of f (x) such that it will have an inverse function. The domain is
usually chosen to preserve as much of the original function as possible, preferably including x = 0.
The Inverse Trigonometric Functions
By restricting the domain of the trigonometric functions, we may take their inverse.
The inverse sine function sin−1 x is obtained by restricting the domain of sin x to −
π
π
≤x≤ .
2
2
The inverse cosine function cos−1 x is obtained by restricting the domain of cos x to 0 ≤ x ≤ π.
The inverse tangent function tan−1 x is obtained by restricting the domain of tan x to −
y = sin−1 x
y = cos−1 x
y
y = tan−1 x
y
π
—
2
π
π
<x< .
2
2
y
π
π
—
2
π
—
2
-1
0
π
–—
2
1
x
0
-1
1
x
π
–—
2
Note: sin−1 x and tan−1 x are both odd functions.
13
Calculus with Inverse Trigonometric Functions
Proof: By using the property
−1
y = sin
dy
1
= dx ,
dx
dy
x
(
x = sin y
for −
π
π
<y<
2
2)
dx
= cos y
dy
=
∴
dy
=
dx
(since cos y ≥ 0 for −
1 − x2
π
π
<y< )
2
2
1
1 − x2
A similar procedure may be followed for cos−1 x and tan−1 x.
Differentiation
d
sin−1 x =
dx
(General form)
1
1 − x2
d
cos−1 x = −
dx
1
1 − x2
d
sin−1 f (x) =
dx
f′(x)
1 − f (x)2
d
cos−1 f (x) = −
dx
f′(x)
1 − f (x)2
d
1
tan−1 x =
dx
1 + x2
d
f′(x)
tan−1 f (x) =
dx
1 + f (x)2
Integration
(General form)
∫
1
1−
−1
x2
= sin
x +C
∫
1
a2 − b2 x2
1
bx
sin−1
+C
b
a
=−
= − cos−1 x + C
1
= tan−1 x + C
∫ 1 + x2
=
∫
a2
1
bx
cos−1
+C
b
a
1
1
bx
=
tan−1
+C
2
2
+b x
ab
a
14
Integration (3 Unit)
Integration by Substitution
Indefinite Integrals
Suppose we have an integral in the form I =
∫
f (g(x)) ⋅ g′(x) d x.
We may perform a substitution u = g(x), which gives
du
= g′(x) (or du = g′(x) d x).
dx
Substituting this result gives:
du
dx
dx
I=
∫
f (u)
=
∫
f (u) du
After integrating, we can undo the substitution u = g(x) to obtain the result in terms of x.
Definite Integrals
If the integral has bounds, they must also be dealt with during the substitution process.
Because definite integrals are to be evaluated numerically, substituting back for x is not necessary.
I=
a
∫b
f (g(x)) ⋅ g′(x) d x
u = g(x), so du = g′(x) d x
I=
=
=
x=a
∫x=b
f (g(x)) ⋅ g′(x) d x
u=g(a)
∫u=g(b)
g(a)
∫g(b)
f (u) du
(given that u = g(x): if x = a, then u = g(a), etc.)
f (u) du
There are two types of substitutions that can be performed — either u = g(x) or x = g(u).
When performing a substitution from one variable to another, all instances of the original variable must be
replaced with the new variable. The differential e.g. d x must also be accounted for.
15
Applications of Calculus to the Physical World (3 Unit)
Related Rates
Suppose we have two quantities x and y each defined as a function of time t. If given the rate of change of
one of the quantities and a relationship between the two quantities (e.g. having y in terms of x), we can
calculate the rate of change of the other quantity.
If y = g(t), x = h(t) and we relate the two variables y = f (x), then we can use the chain rule to obtain:
dy
dy dx
=
⋅
dt
d x dt
and
dx
dx dy
=
⋅
dt
d y dt
(note that
dx
1
= dy .)
dy
dx
Exponential Growth and Decay
Basic Exponential Growth and Decay
If the rate of change of a quantity P over time is proportional to the quantity, then
This implies that:
P = P0 e kt
dP
= k P for some k.
dt
(for constants k and P0 where P0 is the value of P at t = 0.)
dP
= kP
dt
Proof:
d
1
t=
dP
kP
1
t = log P + C
k
P = e kt−kC = (e −kC) e kt
If k is positive, then P is increasing. If k is negative, then P is decreasing.
To find the values of the constants, substitute any given conditions into the equation.
Modified Exponential Growth and Decay
If the rate of change of a quantity P over time is proportional to P − N for some N, then
This implies that:
P = N + Ae kt
dP
= k (P − N ).
dt
(for constants k, N and A.)
Proof: As before, replacing P with P − N.
For negative k,
lim P = N.
t→∞
(i.e. N is the value that P approaches as time passes.)
16
Motion
Displacement, velocity and acceleration can be expressed as functions of time.
If displacement x is a function of time t:
x = f (t)
Then velocity is its first derivative:
dx
v = x· =
dt
And acceleration is its second derivative:
d2x
dv
dv
d
1 2
a = x·· = 2 =
=v
=
v
dt
dt
dx
dx (2 )
Acceleration can be expressed in four useful forms.
dv
dv d x
dv
=
⋅
=v
Proof: x·· =
dt
d x dt
dx
1 2
dv
dv d (2 v )
d
1 2
=
⋅
=
v
and v
dx
dx
dv
dx (2 )
When integrating acceleration or velocity, we can use the initial conditions given to us (e.g. when t = 0,
v = 0) to find the value of the constant of integration.
Taking the Square Root of Velocity
If we know that the velocity of an object doesn’t change signs, then it is either positive or negative.
This means that if v 2 = f (x), then we may take either v =
f (x) or v = −
f (x) (not both at once).
Usually, if the given initial conditions for velocity are either positive or negative, then only one of these
equations can hold, so we may discard the other.
Choosing Which Form of Acceleration to Use
If acceleration is given as:
A function of time:
dv
d2x
··
use x =
or
dt
dt 2
A function of displacement:
use x·· =
A function of velocity:
check the initial conditions:
d
1 2
v
dx (2 )
dv
dt
If they are (t, v):
use x·· =
If they are (x, v):
use x·· = v
dv
dx
17
Simple Harmonic Motion (SHM)
In simple harmonic motion, acceleration is proportional to displacement, but acts in the opposite direction.
x
a
T
O
t
—a
—a
a
O
Simple harmonic motion can be modelled by a sine wave.
If the object starts from the origin:
Acceleration:
Velocity:
Displacement:
x· = − a n 2 sin(nt + α) OR x·· = − n 2 x
x· = a n cos(nt + α)
OR v 2 = n 2 (a 2 − x 2)
x = a sin(nt + α)
OR x = A sin(nt) + B cos(nt)
(by the angle sum formula)
where a is the amplitude, α is the phase shift angle, and the period T is given by T =
(if the centre of motion is not the origin, replace x with x − x 0).
2π
.
n
Proof: Given that x·· = − n 2 x,
d
1 2
v = − n2 x
dx (2 )
v 2 = n 2 (a 2 − x 2)
1 2
n2 x2
v =−
+ C1
2
2
n2a2
Let C1 =
for some constant a,
2
∴ v 2 = n 2 (a 2 − x 2)
dx
= n a2 − x2
dt
d
1
(t) =
dx
n(
a2 − x2 )
1
nt = sin−1 (x /a) + C2
∴ x = a sin (nt − C2)
⟹ x = a sin(nt + α)
Properties of SHM
Acceleration
x·· = − n 2 x
| a | max,
v=0
x
• reaches maximum magnitude | a | when x = ± a.
• is zero when x = 0.
Velocity
v = n (a − x )
2
2
2
• reaches maximum magnitude | v | when x = 0.
• is zero when x = ± a
O
t
2
| v | max,
a=0
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Choosing Which Form of SHM to Use
Which form of SHM to use depends on where the moving object is initially, at t = 0.
Centre of motion
Positive extreme
Negative extreme
x = a sin(nt) + b
x = a cos(nt) + b
x = − a cos(nt) + b
b—a
b
b+a
b—a
b
b+a
b—a
b
b+a
Note: These formulae assume positive initial velocity. If initial velocity is negative, reverse the sign.
Otherwise:
• If the conditions are given are initial,
(i.e. for t = 0)
• If the conditions given are NOT initial,
use x = A sin(nt) + B cos(nt) + b,
v = A n cos(nt) − Bn sin(nt).
use x = a sin(nt + α) + b,
v = a n cos(nt + α).
After picking an equation, substitute in the given conditions to find unknowns.
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Projectile Motion
In the 3 unit course, several assumptions are made about projectile motion:
There is no air resistance, gravity g is constant, and projectiles have negligible mass.
This leads to the two fundamental equations of projectile motion:
Fundamental Equations of Projectile Motion
x·· = 0 and
··y = − g
where g is the acceleration due to gravity.
Resolution of Velocity
.
y
v
Velocity can be resolved into horizontal and vertical components, and vice versa.
If the velocity is v and the angle of inclination is α at a point in time, then
x· = v cos α
{y· = v sin α
y·
tan α = ·
x
v 2 = x· 2 + y· 2 and
OR
are the equations that give the velocity of an object, at that point in time.
α
.
x
The velocity is always tangent
to the path of the projectile.
Deriving Projectile Motion Equations by Integrating
Suppose that the projectile starts from the origin with velocity V and angle of inclination α.
Then, by integrating and using initial conditions to find the constants of integration, we get:
Acceleration:
Velocity:
Displacement:
(Cartesian):
x·· = 0
··y = − g
x· = V cos α
y· = − gt + V sin α
x = (V cos α) t
y =−
1 2
gt + (V sin α) t
2
g sec2 α
y = x tan α −
x2
( 2V 2 )
These equations lead to some important expressions:
Time of Flight:
T =
2V sin α
g
(solve for t at y = 0)
Horizontal Range:
R=
V 2 sin 2α
g
(substitute the time of flight T into the equation for x)
Maximum Height:
H=
V 2 sin2 α
2g
(solve for t at y· = 0, substitute into equation for y)
20
The Binomial Theorem
Using the
n
n!
notation for binomial coefficients,
=
( r ) (n − r)! r!
The expansion of the binomial (a + b)n is given by
(a + b)n =
n
n
n
n
n
an +
a n−1b +
a n−2 b 2 + . . . +
a n−k b k + . . . +
bn
(0)
(1)
(2)
(k)
(n)
OR
(a + b)n =
n
a n−k b k
∑ (k)
k=0
n
In this way, the (k + 1)th term of a binomial expansion is given by Tk+1 =
n
a n−k b k.
(k)
Binomial Identities
n
n
=
( k ) (n − k)
n
n−1
n−1
=
+
(k) ( k ) (k − 1)
(generally, this is not important to know)
Pascal’s Triangle
Each row of Pascal’s triangle is made by summing adjacent numbers on the row before.
(a + b)0
(a + b)1
(a + b)2
(a + b)3
(a + b)4
…
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
…
Pascal’s triangle can be used to quickly find the coefficients of a binomial expansion.
For example, to determine the coefficients of (a + b)2 we look at the corresponding row in Pascal’s triangle
to find that the coefficients are 1, 2, 1 — i.e. (a + b)2 = 1a 2 + 2a b + 1b 2.
21
Finding the Greatest Coefficients/Terms
It’s easy to see that if
Tk+1
Tk
> 1 then Tk+1 > Tk (given that both terms are positive, otherwise see below).
Therefore, to find out which coefficients/terms are greatest, we can solve
Tk+1
Tk
(k) a
n
=
(k − 1) a
n
Tk+1
> 1 for k.
Tk
n−k k
b
n−k+1b k−1
=
n!
(n − k + 1)!(k − 1)! b
⋅
⋅
(n − k)!k!
n!
a
=
n−k +1 b
⋅
k
a
To get the greatest coefficient, discard the variable (e.g. x) and then solve
Tk+1
Tk
> 1 for k.
To get the greatest term, substitute the given value into the variable and then solve
If negative coefficients/terms are involved and only the magnitude is needed, use
Remember that k must be an integer, e.g. if k < 2.5 then k = 2, 1.
Tk+1
Tk
Tk+1
Tk
> 1 for k.
> 1 instead.
Equal Coefficients/Terms
If the final inequality involves an integer e.g. k < 5, then there are equal coefficients/terms present.
i.e. for some constant r, if k < r when
Tk+1
Tk
> 1, then k = r when
Tk+1
Tk
= 1, that is, Tr+1 = Tr .
Binomial Probability
Suppose an event has two outcomes: success (probability p) and failure (probability q).
Incidentally, means that p + q = 1 ⟹ q = 1 − p.
Then, in n independent trials, the probability of r successes is given by P(r) =
n
q n−r p r.
(r)
This can be thought of as choosing which trials to fail/succeed and multiplying by the probability of the
desired number of successes/failures happening.
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