2U Maths
Short Description
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Description
Epping Boys High School
Mathematics HSC
Vinodh Thanabalasingham, Lawrence Feng, Pasan Waidyasekara, Amit Dharamdasani 2014
Contents CONTENTS Basic Arithemetic and Algebra
2
Real Functions
7
Trigonometric Ratios
9
Linear Functions
25
The Quadratic Polynomial and the Parabola
40
Plane Geometry
42
Tangent to a Curve and Derivative of a Function
46
Geometrical Applications of Differentiation
48
Integration
50
Trignometric Functions
81
Exponential and Logarithmic Functions
84
Application of Calculus to the Physical World
97
Probability
100
Series
107
MATHEMATICS | PAGE 1 OF 121
Basic Arithemetic and Algebra BASIC ARITHEMETIC AND ALGEBRA OVERVIEW This is essentially the fundamentals of mathematics, and is basically learnt through high school. This topic is basically a review of previous years.
BASIC OPERATIONS There are 4 basic operations that are applied: Operations
Symbol
Result
Addition
+
Sum
Subtraction
Difference
Multiplication
×
Division
÷
Quotient
Product
There is an order in certain questions in which the operations must be done.
BIDMAS Brackets Indices Division Multiplication Addition Subtraction
This rule is follow from left to right, or top to bottom.
Example 1
Simplify 3 + 15 ÷ 3 + 4 × (3 − 1)2 3 + 15 ÷ 3 + 4 × (3 − 1)2 = 3 + 15 ÷ 3 + 4 × 4 = 3 + 15 ÷ 3 + 16 = 3 + 5 + 16 = 24
INTEGERS Integers are values that can be expressed as a whole number. Such as: 1, 1 000 000, etc.
MATHEMATICS | PAGE 2 OF 121
Basic Arithemetic and Algebra FRACTIONS Fractions are numbers written in the form
𝑛 𝑑
where n is the numerator, and d is the
denominator. Both the numerator and denominator are integers. NOTE: 0 =0 𝑑 ∴ 𝑖𝑡 𝑖𝑠 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑛 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑 0 To perform addition or subtractions on fractions, we must modify them so that they have a common denominator. Then we can continue as if they were integers, and express them in the simplest form. When attempting to multiply fractions, we may simply multiply the numerators and denominators. i.e.
𝑎 𝑐 𝑎𝑐 × = 𝑏 𝑑 𝑏𝑑
When dividing fractions, we may simply multiply by the reciprocal of the divisor. i.e.
𝑎 𝑐 𝑎 𝑑 𝑎𝑑 ÷ = × = 𝑏 𝑑 𝑏 𝑐 𝑏𝑐
Example
Simplify a)
1 4
1
+5 1 1 5 4 9 + = + = 4 5 20 20 20 1
1
3
4
b) 3 −
1 1 10 1 40 3 37 3 − = − = − = 3 4 3 4 12 12 12 c)
1 4
3
×4 1 3 3 × = 4 4 16
d)
8 9
3
÷4 8 3 8 4 32 ÷ = × = 9 4 9 3 37
MATHEMATICS | PAGE 3 OF 121
Basic Arithemetic and Algebra FRACTIONS, DECIMALS AND PERCENTAGES Sometimes it is needs to convert fractions into decimals and percentages, and vice versa. such as: 0.3 =
3 = 30% 10
To convert recurring decimals, we use algebra. Step 1: Identify the number of repeating digits, and refer to this as ’n’ Step 2: Multiply the decimal by 10n Step 3: Subtract the original decimal Step 4: Simplify the expression Example
Express 0. 3̇4̇ as a fraction 𝑥 = 0.34343434343434 … There are 2 recurring digits. 100𝑥 = 34.34343434343434 … 100𝑥 − 𝑥 = 34.34343434343434 … − 0.34343434343434 … 99𝑥 = 34 34 𝑥= 99 Therefore, it can denote as
34 99
INDICES: POWERS AND ROOTS Some rules to remember: (-ve)even =+ve (-ve)odd =-ve 𝑥 𝑎 × 𝑥 𝑏 = 𝑥 𝑎+𝑏 𝑥𝑎 = 𝑥 𝑎−𝑏 𝑥𝑏 (𝑥 𝑎 )𝑏 = 𝑥 𝑎𝑏 (𝑥𝑦)𝑎 = 𝑥 𝑎 𝑦 𝑎
𝑥 𝑎 𝑥𝑎 ( ) = 𝑎 𝑦 𝑦 𝑥1 = 𝑥 𝑥0 = 1 𝑥 −𝑎 = 1
1 𝑥𝑎
𝑏
𝑥 𝑏 = √𝑥 𝑎
𝑏
𝑥 𝑏 = √𝑥 𝑎
MATHEMATICS | PAGE 4 OF 121
Basic Arithemetic and Algebra SCIENTIFIC NOTATION Scientific notation is a shorter form for writing down large or small numbers, it is in the form: 𝑎 × 10𝑏 Where 1 ≤ 𝑎 < 10 and b is an integer. Often ask to a certain decimal or significant figure.
RATIONAL NUMBERS A rational number is a number than can be expressed as a fraction
𝑝 𝑞
where p and q are
integers and 𝑞 ≠ 0
IRRATIONAL NUMBERS/S URDS A irrational number is a number than cannot be expressed as a fraction
𝑝 𝑞
where p and q are
integers and 𝑞 ≠ 0 Surd rules include: √𝑎 + √𝑎 = 2 √𝑎 √𝑎 × √𝑏 = √𝑎𝑏 √𝑎 √𝑏
= √
𝑎 𝑏
𝑛
( 𝑛√𝑎 ) = 𝑎 Conjugate surds are two surds which multiply together to give a rational number. 𝑎
Rationalising the denominator of a fraction is done to simplify a fraction of the form 𝑏+
√𝑐
To rationalise it, just multiply by
𝑏−√𝑐 𝑏−√𝑐
.
SIMPLIFICATION AND S UBSTITUTION OF ALGEBRA Done in the same matter as arithmetic, but involves using like terms.
MATHEMATICS | PAGE 5 OF 121
Basic Arithemetic and Algebra FACTORISATION OF SIMPLE EXPRESSIONS This is standard factorisations or expansions that should be remembered. 𝑎2 + 2𝑎𝑏 + 𝑏2 = (𝑎 + 𝑏)2 𝑎2 − 2𝑎𝑏 + 𝑏2 = (𝑎 − 𝑏)2 𝑎2 − 𝑏2 = (𝑎 + 𝑏)(𝑎 − 𝑏) 𝑎3 + 3𝑎2 𝑏 + 3𝑎𝑏2 + 𝑏3 = (𝑎 + 𝑏)3 𝑎3 − 3𝑎2 𝑏 + 3𝑎𝑏2 + 𝑏3 = (𝑎 − 𝑏)3 𝑎3 + 𝑏3 = (𝑎 + 𝑏)(𝑏2 − 𝑎𝑏 − 𝑏2 ) 𝑎3 − 𝑏3 = (𝑎 − 𝑏)(𝑏2 + 𝑎𝑏 − 𝑏2 )
MATHEMATICS | PAGE 6 OF 121
Real Functions REAL FUNCTIONS RELATIONS AND FUNCTIONS A relation (such as an equation in x and y) is a function if it passes the vertical line test when graphed. That is, no vertical lines cut the curve at more than one point.
Even Function
:
f x f x
Odd Function
:
f x f x
TRANSFORMATION OF KNOWN GRAPHS To stretch the graph in a vertical direction by a factor of a, replace y by y/a. To stretch the graph in a horizontal direction by a factor of a, replace x by x/a. For example,
y 4 sin x is
y sin x stretched vertically by a factor of 4. Replacing x by x-a shifts the graph to the right by a units (think of x-a being zero when x=a). Similarly replacing y by y-a shifts it a units up. Replacing x by –x reflects it in y-axis (just visualise it). Replacing y by –y reflects in x-axis.
LOCUS AND ALGEBRAIC REPRESENTATION Example 1:
Point P(x,y) moves so that its distance from A(1,0) is two times its distance from B(7,0). What is the Cartesian equation of the locus of P?
x 12 y 02
2 x 7 y 0 x 1 y 0 4x 7 4 y 0 2
2
2
2
2
2
After expansion and simplification, locus is 3 x 54 x 3 y 195 0 . 2
2
Example 2: 2 2 Find the centre and radius of the following circle: x 4 x y 6 y 12 .
Completing the square,
x 2 4x 4 y 2 6 y 9 12 4 9 x 2 y 3 25 . 2
2
Centre is (2, -3) and radius is 5. Example 3: 2 2 What is the locus whose equation is x y 0 ?
x 2 y 2 x y so it’s either the line y x or y x .
MATHEMATICS | PAGE 7 OF 121
Real Functions REGION AND INEQUALIT Y To draw the region y x 1 , draw the line y x 1 and the region will be above the line. Similarly y x is above the parabolic curve, so is y x . Now, 2
2
x 12 y 12 25 is
outside the circle (and excludes the circle).
MATHEMATICS | PAGE 8 OF 121
Trigonometric Ratios TRIGONOMETRIC RATIOS OVERVIEW The rules of trigonometry arises from trigonometric ratios.
RIGHT-ANGLE TRIANGLES There are three main rules for right-angle triangles:
Sine: sin 𝜃 =
opposite hypotenuse
Cosine: cos 𝜃 =
adjacent hypotenuse
Tangent: tan 𝜃 =
opposite adjacent
An easy way to remember these ratios is the acronym SOH CAH TOA: S ine (is)
C osine (is)
T angent (is)
O pposite (over)
A djacent (over)
O pposite (over)
H ypotenuse
H ypotenuse
A djacent
The inverse of these ratios are:
Cosecant: cosec 𝜃 =
1 sin 𝜃
Secant: sec 𝜃 =
1 cos 𝜃
Cotangent: cot 𝜃 =
1 tan 𝜃
Complementary angles arise from this. They are derived from looking at the relationship between the ratios above, and the same ratios when measuring from the other acute angle of the triangle (i.e. sin(90° − 𝜃), cos(90° − 𝜃)… etc.). From this, the following ratios can be found: sin 𝜃 = cos(90° − 𝜃) cos 𝜃 = sin(90° − 𝜃)
sec 𝜃 = cosec(90° − 𝜃) cosec 𝜃 = sec(90° − 𝜃)
tan 𝜃 = cot(90° − 𝜃) cot 𝜃 = tan(90° − 𝜃)
MATHEMATICS | PAGE 9 OF 121
Trigonometric Ratios EXAMPLES Example 1
If 𝑠𝑖𝑛 𝜃 =
2
, find the exact ratios of 𝑐𝑜𝑠 𝜃 , 𝑡𝑎𝑛 𝜃, and 𝑐𝑜𝑡 𝜃 .
7
Solution By Pythagoras’ theorem: 𝑐 2 = 𝑎2 + 𝑏2 72 = 𝑎 2 + 22 49 = 𝑎2 + 4 𝑎2 = 45 ∴ 𝑎 = √45 cos 𝜃 =
√45 7
tan 𝜃 =
2
√45 √45 cot 𝜃 = 2
Example 2
Simplify 𝑡𝑎𝑛 50° − 𝑐𝑜𝑡 40°. Solution tan 50° = cot(90° − 50°) = cot 40° ∴ tan 50° − cot 40° = tan 50° − tan 50° =0 Example 3
Find the value of 𝑥, correct to 1 decimal place.
𝑥 11.8 𝑥 = 11.8 cos 23° 29′
cos 23° 49′ =
≈ 10.8 cm (to 1 decimal point)
MATHEMATICS | PAGE 10 OF 121
Trigonometric Ratios ANGLES IN A CALCULATOR On a calculator, the value of an angle will often be given as a decimal. Sometimes, the angle may be required to be represented in degrees, minutes and seconds. 60 minutes = 1 degree (60′ = 1°) 60 seconds = 1 minute (60′′ = 1′ ) Angles are rounded off to the nearest degree by rounding up if there are 30 minutes or more. Rounding from seconds to minutes is done in the same way. Most calculators will represent angles in decimal form by default. To change to the degrees, minutes and seconds form, press the DMS or
button on the calculator.
EXAMPLES Example 1
Round off 23° 12′ 22′′ to the nearest minute. Solution 23° 12′ 22′′ = 23° 12′ Example 2
Round off 59° 34′ 41′′ to the nearest minute. Solution 59° 34′ 41′′ = 59° 35′ Example 3
Round off 16° 54′ 30′′ to the nearest degree. Solution 16° 54′ 30′′ = 17° Example 4
Find the value of 𝛼 in degrees and minutes. 2.1
Solution tan 𝛼 =
4.9
4.9 2.1
𝛼 = tan−1 (
4.9 ) 2.1
≈ 66° 48′
MATHEMATICS | PAGE 11 OF 121
Trigonometric Ratios APPLICATIONS OF RIGHT-ANGLE TRIGONOMETRY Angle of Elevation The angle of elevation, 𝜃, is the angle measured when looking from the ground up to the top of an object, assuming the ground is horizontal. General right-angle trigonometry ratios apply. See Example 1.
Angle of Depression The angle of depression, 𝜃, is the angle measured when looking down from the horizontal to an object below. General right-angle trigonometry ratios apply. See Example 2.
Bearings Bearings can be described in two different ways: N70°W means starting at North, measure 70° towards West. 070° is known as a true bearing. The angle given is always measured clockwise from north, and is always 3 digits. See Examples 3 and 4.
EXAMPLES Example 1
The angle of elevation of a tree from a point 50 m out from its base is 38° 14′. Find the height of the tree to the nearest metre. Solution The tree is assumed vertical.
ℎ 50 ℎ = 50 tan 38° 14′
tan 38° 14′ =
≈ 39 m
MATHEMATICS | PAGE 12 OF 121
Trigonometric Ratios Example 2
A bird sitting on top of an 8 m tall tree looks down at a possum 3.5 m out from the base of the tree. Find the angle of elevation to the nearest minute. Solution
B
Since 𝐴𝐵 ∥ 𝐷𝐶
(horizontal lines)
∠𝐵𝐷𝐶 = 𝜃
(alternate angles)
tan 𝜃 =
8 3.5
𝜃 = tan−1 (
A
8m 8 ) 3.5
≈ 66° 22′
C
3.5 m
D
Example 3
Sketch a diagram of bearing N70°W. Solution
Example 4 𝑋 is on a bearing of 030° from 𝑌. Sketch this diagram. Solution
MATHEMATICS | PAGE 13 OF 121
Trigonometric Ratios EXACT RATIOS By utilising a unit circle (i.e. a circle with radius 1 unit), various exact ratios can be derived. These are useful when simplifying trigonometric functions. 45° TRIANGLE
sin 45° =
1
cos 45° =
√2
1 √2
tan 45° = 1
Proof: 𝑐 2 = 𝑎2 + 𝑏2 A
𝐴𝐶 2 = 12 + 12 =2 𝐴𝐶 = √2
1
45° B
C
1
60° TRIANGLE
sin 60° =
√3 2
cos 60° =
1 2
tan 60° = √3
sin 30° =
1 2
cos 30° =
√3 2
tan 30° =
1 √3
Proof: 𝑎2 = 𝑐 2 + 𝑏2
A
𝐴𝐵2 = 22 − 12 30°
=3
2
𝐴𝐵 = √3
60° B
1
C
MATHEMATICS | PAGE 14 OF 121
Trigonometric Ratios EXAMPLES Example 1
Find the exact value of sec 45°. Solution sec 45° =
1 cos 45°
1 1 √2 = √2 =
Example 2
A boat ramp is to be made with an angle of 30° and base length 5m. What is the exact length of the surface of the ramp? Solution Let 𝑥 be the length of the surface of the ramp. cos 30° =
5 𝑥
𝑥 cos 30° = 5 𝑥= =
5 cos 30° 5 √3 2
=5×
2 √3
= 10√3 =
10√3 m 3
MATHEMATICS | PAGE 15 OF 121
Trigonometric Ratios ANGLES OF ANY MAGNITUDE A circle can be used to find trigonometric ratios of any angles. Angles are measured in an anti-clockwise direction. A circle is divided up into four quadrants, where the 1 st quadrant is from 0° to 90°, 2nd quadrant is from 90° to 180°, 3rd quadrant is from 180° to 270°, and 4th quadrant is from 270° to 360°. 90°
2nd
1st 0° 360°
180° 3rd
4th
270° Angles in the first quadrant are given as 𝜃. Angles in the second quadrant are given as 180° − 𝜃. Angles in the third quadrant are given as 180° + 𝜃. Angles in the fourth quadrant are given as 360° − 𝜃. For example: 90° 2nd
90° 1st
2nd
180° − 𝜃
360° − 𝜃 0° 360°
𝜃
180°
3rd
1st
180°
4th 270°
𝜃
3rd
0° 360°
4th 270°
The sign of trigonometric ratios change in each quadrant: 1st:
All ratios are positive
2nd:
Sin is positive (i.e. cos and tan are negative)
MATHEMATICS | PAGE 16 OF 121
Trigonometric Ratios 3rd:
Tan is positive
4th:
Cos is positive
From this, the rule “All Stations to Central” can be used to remember which ratios are positive and negative. 90° 2nd
1st S
A 0° 360°
180°
T
C
3rd
4th 270°
The following general formulas apply as a result: sin 𝜃 = sin(180° − 𝜃)
cos 𝜃 = − cos(180° − 𝜃)
tan 𝜃 = − tan(180° − 𝜃)
sin 𝜃 = − sin(180° + 𝜃)
cos 𝜃 = − cos(180° + 𝜃)
tan 𝜃 = tan(180° + 𝜃)
sin 𝜃 = − sin(360° − 𝜃)
cos 𝜃 = cos(360° − 𝜃)
tan 𝜃 = − tan(360° − 𝜃)
When a negative angle is given (e.g. sin(−𝜃), cos(−𝜃), tan(−𝜃)), it is measured in a clockwise direction beginning from the same position. Other rules still apply. The domain given will change the value of an angle.
EXAMPLES Example 1
Find the exact ratio of tan 330°. Solution 330° is in the 4th quadrant ∴ tan will be negative. tan 330° = − tan(360° − 330°) = − tan 30° 1 =− √3
MATHEMATICS | PAGE 17 OF 121
Trigonometric Ratios Example 2
Find the exact value of cos 510°. Solution As 510° is greater than 360°, it is measured around the circle more than once. 𝑦
510° 30°
150°
𝑥
510° − 360° = 150° 510° = 360° + 150° The angle is in the 2nd quadrant, i.e. cos is negative. The triangle has 30° in it. cos 510° = − cos 30° =−
√3 2
Example 3
Simplify 𝑐𝑜𝑠(180° + 𝑥 ). Solution 180° + 𝑥 is an angle in the 3rd quadrant where cos is negative. As cos 𝜃 = − cos(180° + 𝜃), cos(180° + 𝑥 ) = − cos 𝑥 Example 4
Find the exact value of 𝑐𝑜𝑡(−120°). Solution The angle is in the 3rd quadrant, with 60° in the triangle. Tan is positive in the 3rd quadrant. cot(−120°) = =
1 tan 60° 1 √3
MATHEMATICS | PAGE 18 OF 121
Trigonometric Ratios Example 5
Solve 𝑐𝑜𝑠 𝑥 =
3 in the domain 0° ≤ 𝑥 ≤ 360°. 2
Solution √3 is positive and cos is positive in the 1st and 4th quadrants. 2 √3 2 𝑥 = 30°, 360° − 30°
cos 𝑥 =
= 30°, 330° Example 6
Solve 𝑡𝑎𝑛 𝑥 = √3 for −180° ≤ 𝑥 ≤ 180°. Solution Tan is positive in 1st and 3rd quadrants. In the domain −180° ≤ 𝑥 ≤ 180°, positive angles are used for 0° ≤ 𝑥 ≤ 180° while negative angles are used for −180° ≤ 𝑥 ≤ 0°. tan 𝑥 = √3 𝑥 = 60°, −(180° − 60°) = 60°, −120° Example 7
Solve 2 𝑠𝑖𝑛 2𝑥 − 1 = 0 for 0° ≤ 𝑥 ≤ 360°. Solution The angle is 2𝑥 but the domain is for 𝑥. The domain must be changed to 2𝑥, i.e. 0° ≤ 2𝑥 ≤ 720°. 2 sin 2𝑥 − 1 = 0 2 sin 2𝑥 = 1 1 sin 2𝑥 = 2 1 sin 30° = 2 Sin is positive in the 1st and 2nd quadrants. 2𝑥 = 30°, 180° − 30°, 360° + 30°, 360° + (180° − 30°) = 30°, 150°, 390°, 510° ∴ 𝑥 = 15°, 75°, 195°, 255°
MATHEMATICS | PAGE 19 OF 121
Trigonometric Ratios TRIGONOMETRIC IDENTITIES Again using a unit circle, various trigonometric identities can be proven. tan 𝜃 =
sin 𝜃 cos 𝜃
cot 𝜃 =
cos 𝜃 sin 𝜃
Proof:
tan 𝜃 = =
𝑦 𝑥 sin 𝜃 cos 𝜃
cot 𝜃 =
1 tan 𝜃
1 sin 𝜃 cos 𝜃 cos 𝜃 = sin 𝜃 =
PYTHAGOREAN IDENTITIES cos2 𝜃 + sin2 𝜃 = 1 Proof: A circle has equation 𝑥 2 + 𝑦 2 = 1. Substitute 𝑥 = cos 𝜃 and 𝑦 = sin 𝜃: ∴ (cos 𝜃)2 + (sin 𝜃 )2 = 1 ∴ cos 2 𝜃 + sin2 𝜃 = 1 1 + tan2 𝜃 = sec 2 𝜃 Proof: cos 2 𝜃 + sin2 𝜃 = 1 cos 2 𝜃 sin2 𝜃 1 + = cos 2 𝜃 cos 2 𝜃 cos 2 𝜃 1 + tan2 𝜃 = sec 2 𝜃 cot 2 𝜃 + 1 = cosec 2 𝜃 Proof: cos 2 𝜃 + sin2 𝜃 = 1 cos 2 𝜃 sin2 𝜃 1 + 2 = 2 sin 𝜃 sin 𝜃 sin2 𝜃 cot 2 𝜃 + 1 = cosec 2 𝜃
MATHEMATICS | PAGE 20 OF 121
Trigonometric Ratios EXAMPLES Example 1
Simplify 𝑠𝑖𝑛 𝜃 𝑐𝑜𝑡 𝜃 . Solution sin 𝜃 cot 𝜃 = sin 𝜃 ×
cos 𝜃 sin 𝜃
= cos 𝜃 Example 2
Simplify √𝑠𝑖𝑛4 𝜃 + 𝑠𝑖𝑛2 𝜃 𝑐𝑜𝑠 2 𝜃 . Solution √sin4 𝜃 + sin2 𝜃 cos 2 𝜃 = √sin2 𝜃 (sin2 𝜃 + cos 2 𝜃 ) = √sin2 𝜃 (1) = √sin2 𝜃 = sin 𝜃
NON-RIGHT-ANGLED TRIANGLES Sine Rule: 𝑎 𝑏 𝑐 = = sin 𝐴 sin 𝐵 sin 𝐶 Proof: In ∆𝐴𝐵𝐶, draw perpendicular 𝐴𝐷 and call it ℎ. From ∆𝐴𝐵𝐷, sin 𝐵 =
ℎ 𝑐
∴ ℎ = 𝑐 sin 𝐵 From ∆𝐴𝐶𝐷, sin 𝐶 =
ℎ 𝑏
∴ ℎ = 𝑏 sin 𝐶 ∴ 𝑐 sin 𝐵 = 𝑏 sin 𝐶 sin 𝐵 sin 𝐶 = 𝑏 𝑐 Similarly, drawing a perpendicular from 𝐶, it can be proven that sin 𝐴 sin 𝐵 = . 𝑎 𝑏
MATHEMATICS | PAGE 21 OF 121
Trigonometric Ratios Cosine Rule: 𝑐 2 = 𝑎2 + 𝑏2 − 2𝑎𝑏 cos 𝐶 Proof: A
b
C
c
p
x
D
a−x
B
In ∆𝐴𝐵𝐶, draw perpendicular 𝐶𝐷 with length 𝑝 and let 𝐶𝐷 = 𝑥. Since 𝐵𝐶 = 𝑎, 𝐵𝐷 = 𝑎 − 𝑥 𝑏2 = 𝑥 2 + 𝑝2 cos 𝐶 =
𝑥 𝑏
∴ 𝑥 = 𝑏 cos 𝐶 𝑐 = 𝑝 2 + ( 𝑎 − 𝑥 )2 2
= 𝑝2 + 𝑎2 − 2𝑎𝑥 + 𝑥 2 = 𝑝2 + 𝑥 2 + 𝑎2 − 2𝑎𝑥 = 𝑏2 + 𝑎2 − 2𝑎𝑥 = 𝑏2 + 𝑎2 − 2𝑎(𝑏 cos 𝐶 ) = 𝑏2 + 𝑎2 − 2𝑎𝑏 cos 𝐶
EXAMPLES Example 1
Find the value of 𝜃 , in degrees and minutes. Solution sin 𝜃 sin 86° 11′ = 6.7 8.3 sin 𝜃 =
6.7 sin 86° 11′ 8.3
𝜃 = sin−1 (
6.7 sin 86° 11′ ) 8.3
≈ 53° 39′
MATHEMATICS | PAGE 22 OF 121
Trigonometric Ratios Example 2
Find the value of 𝑥, correct to the nearest whole number. Solution 𝑐 2 = 𝑎2 + 𝑏2 − 2𝑎𝑏 cos 𝐶 𝑥 2 = 5.62 + 6.42 − 2(5.6)(6.4) cos 112° 32′ ≈ 99.79 𝑥 ≈ √99.79 ≈ 10 Example 3
Evaluate ∠𝐵𝐴𝐶 in degrees and minutes.
A
Solution
4.5 cm
6.1 cm
𝑐 2 = 𝑎2 + 𝑏2 − 2𝑎𝑏 cos 𝐶 B
𝑐 2 + 2𝑎𝑏 cos 𝐶 = 𝑎2 + 𝑏2
8.4 cm
C
2𝑎𝑏 cos 𝐶 = 𝑎2 + 𝑏2 − 𝑐 2 𝑎2 + 𝑏2 − 𝑐 2 cos 𝐶 = 2𝑎𝑏 4.52 + 6.12 − 8.42 cos ∠𝐵𝐴𝐶 = 2(4.5)(6.1) 4.52 + 6.12 − 8.42 ∠𝐵𝐴𝐶 = cos −1 ( ) 2(4.5)(6.1) ≈ 103° 48′ Example 4
The angle of elevation of a tower from point 𝐴 is 72°. From point 𝐵, 50 m further away from the tower than 𝐴, the angle of elevation is 47°. a) Find the exact length of 𝐴𝑇. Solution ∠𝐵𝐴𝑇 = 180° − 72°
(straight angle)
= 108° ∠𝐵𝑇𝐴 = 180° − (47° + 108°) (angle sum of ∆) = 25° 𝐴𝑇 50 = sin 47° sin 25° 50 sin 47° ∴ 𝐴𝑇 = sin 25°
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Trigonometric Ratios b) Hence, or otherwise, find the height ℎ of the tower to 1 decimal place. Solution sin 72° =
ℎ 𝐴𝑇
∴ ℎ = 𝐴𝑇 sin 72° =
50 sin 47° × sin 72° sin 25°
≈ 82.3 m
AREA OF A TRIANGLE The area of a triangle is often calculated using 𝐴 =
1 𝑏ℎ, however it can also be calculated 2
using trigonometry. 1 𝐴 = 𝑎𝑏 𝑠𝑖𝑛 𝐶 2 Proof: From ∆𝐵𝐶𝐷, sin 𝐶 =
ℎ 𝑎
∴ ℎ = 𝑎 sin 𝐶 1 𝐴 = 𝑏ℎ 2 1 = 𝑏𝑎 sin 𝐶 2
EXAMPLE Example
Find the area of ∆𝐴𝐵𝐶 correct to 2 decimal places. Solution 1 𝐴 = 𝑎𝑏 sin 𝐶 2 1 = (4.3)(5.8) sin 112° 34′ 2 ≈ 11.52 units 2
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Linear Functions LINEAR FUNCTIONS OVERVIEW Linear functions are straight-line graphs.
DISTANCE The distance between two points can be found using the following formula: 𝑑 = √(𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2 Proof:
Let 𝐴 = (𝑥1 , 𝑦1 ) and 𝐵 = (𝑥2 , 𝑦2 ) Length 𝐴𝐶 = 𝑥2 − 𝑥1 and length 𝐵𝐶 = 𝑦2 − 𝑦1 By Pythagoras’ theorem 𝐴𝐵2 = 𝐴𝐶 2 + 𝐵𝐶 2 𝑑 2 = (𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2 ∴ 𝑑 = √(𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2
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Linear Functions EXAMPLES Example 1
Find the distance between the points (1, 3) and (−3, 0). Solution Let (1, 3) be (𝑥1 , 𝑦1 ) and (−3, 0) be (𝑥2 , 𝑦2 ) 𝑑 = √(−3 − 1)2 + (0 − 3)2 = √(−4)2 + (−3)2 = √16 + 9 = √25 =5 So the distance is 5 units. Example 2
Show that the points 𝑋(3, −3), 𝑌(7, 4) and 𝑍(−4, 1) form the vertices of an isosceles rightangled triangle. Solution 𝑋𝑌 = √(7 − 3)2 + [4 − (−3)]2 = √42 + 72 = √16 + 49 = √65 𝑌𝑍 = √(−4 − 7)2 + (1 − 4)2 = √(−11)2 + (−3)2 = √121 + 9 = √130 𝑋𝑍 = √(−4 − 3)2 + [1 − (−3)]2 = √(−7)2 + 42 = √49 + 16 = √65 Since 𝑋𝑌 = 𝑋𝑍, triangle 𝑋𝑌𝑍 is isosceles. 𝑋𝑌 2 + 𝑋𝑍 2 = 65 + 65 = 130 = 𝑌𝑍 2 So triangle 𝑋𝑌𝑍 is right angled. (Pythagoras’ theorem)
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Linear Functions MIDPOINT The midpoint is the point halfway between two other points. It can be found using this formula: 𝑀=(
𝑥1 + 𝑥2 𝑦1 + 𝑦2 , ) 2 2
Proof:
Find the midpoint of points 𝐴(𝑥1 , 𝑦1 ) and 𝐵(𝑥2 , 𝑦2 ). Let 𝑀 = (𝑥, 𝑦) Then ∆𝐴𝑃𝑄 ||| ∆𝐴𝐵𝑅 𝐴𝑄 𝐴𝑃 = 𝐴𝑅 𝐴𝐵 𝑥 − 𝑥1 1 ∴ = 𝑥2 − 𝑥1 2 ∴
2(𝑥 − 𝑥1 ) = 𝑥2 − 𝑥1 2𝑥 − 2𝑥1 = 𝑥2 − 𝑥1 2𝑥 = 𝑥1 + 𝑥2 𝑥1 + 𝑥2 2 𝑦1 + 𝑦2 Similarly, 𝑦 = 2 𝑥=
EXAMPLES Example 1
Find the midpoint of (−1, 4) and (5, 2). Solution −1 + 5 4 + 2 , ) 2 2 4 6 =( , ) 2 2
Midpoint = (
= (2, 3)
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Linear Functions Example 2
Find the values of 𝑎 and 𝑏 if (2, −3) is the midpoint between (−7, −8) and (𝑎, 𝑏). Solution Midpoint = (2, −3) ∴ 𝑥 = 2, 𝑦 = −3 ∴
−7 + 𝑎 =2 2 −7 + 𝑎 = 4 𝑎 =4+7
= 11 −8 + 𝑏 ∴ = −3 2 −8 + 𝑏 = −6 𝑏 = −6 + 8 =2 So 𝑎 = 11 and 𝑏 = 2. Example 3
A circle with centre (−2, 5) has one end of a diameter at (−4,3). Find the coordinates of the other end of the diameter. Solution Centre (−2, 5) ∴ Midpoint at (−2, 5) Let coordinates at other end of diameter be (𝑥1 , 𝑦1 ) ∴
4 + 𝑥1 = −2 2 4 + 𝑥1 = −4 𝑥1 = −4 − 4
= −8 −3 + 𝑦1 ∴ =5 2 −3 + 𝑦1 = 10 𝑦1 = 10 + 3 = 13 ∴ coordinates of other end of diameter are (−8, 13).
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Linear Functions GRADIENT The gradient of a straight line is its slope, i.e. it compares its vertical rise with the horizontal run. This is a measure of change of 𝑦 with respect to 𝑥.
When the line rises as 𝑥 increases, it has a positive gradient.
When the line lowers as 𝑥 increases, it has a negative gradient.
When the horizontal run and vertical rise is known, the gradient can be found using: 𝑚=
rise run
The gradient of a line between two points (𝑥1 , 𝑦1 ) and (𝑥2 , 𝑦1 ) can be found using: 𝑚=
𝑦2 − 𝑦1 𝑥2 − 𝑥1
Proof: 𝐵𝐶 = 𝑦2 − 𝑦1 and 𝐴𝐶 = 𝑥2 − 𝑥1 Gradient = =
rise run 𝑦2 − 𝑦1 𝑥2 − 𝑥1
The gradient of a line when given the angle the line makes with the 𝑥-axis in the positive direction can be found using: 𝑚 = 𝑡𝑎𝑛 𝜃 Where: 𝜃 = angle the line makes with the 𝑥-axis in the positive direction
EXAMPLES Example 1
Find the gradient of the line between points (2, 3) and (−3, 4).
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Linear Functions Solution 𝑚=
𝑦2 − 𝑦1 𝑥2 − 𝑥1
=
4−3 −3 − 2
=
1 −5
=−
1 5
Example 2
Prove that points (2, 3), (−2, −5) and (0, −1) are collinear. Solution Collinear points have the same gradient. Gradient of the interval between (−2, −5) and
Gradient of the interval between (0, −1) and
(0, −1):
(2, 3):
−1 − (−5) 0 − (−2) −1 + 5 = 2 4 = 2
𝑚=
=2
3 − (−1) 2−0 3+1 = 2 4 = 2
𝑚=
=2
As the gradient of both intervals is the same, the points are collinear. Example 3
Find the gradient of the straight line that makes an angle of 42° 51′ with the 𝑥-axis to 2 significant figures. Solution 𝑚 = tan 42° 51′ ≈ 0.93 (2 sig. figs.) Example 4
Find the gradient of the line that makes an angle of 87° 14′ with the 𝑥-axis to 2 significant figures. Solution 𝑚 = tan 87° 14′ = − tan(90° − 87° 14′ ) ≈ 21
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Linear Functions EQUATION OF A STRAIGHT LINE Intercept Form:
General Form:
𝑥 𝑦 + =1 𝑎 𝑏
𝑎𝑥 + 𝑏𝑦 + 𝑐 = 0
Gradient Form:
Where: 𝑎 = 𝑥-intercept 𝑏 = 𝑦-intercept
𝑦 = 𝑚𝑥 + 𝑏 Where: 𝑚 = gradient 𝑏 = 𝑦-intercept
Point-gradient Formula: 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ) Where: (𝑥1 , 𝑦1 ) lies on the line with gradient 𝑚
EXAMPLES Example 1
Find the equation of the straight line with gradient –4 and passing through the point (−2, 3). Solution 𝑚 = −4 At (−2, 3), 𝑦 − 3 = −4[𝑥 − (−2)] = −4(𝑥 + 2) = −4𝑥 − 8 ∴ 𝑦 = −4𝑥 − 5 or 4𝑥 + 𝑦 + 5 = 0
(gradient form) (general form)
Example 2
Find the equation of the line with 𝑥-intercept 3 and 𝑦-intercept 2. Solution 𝑎 = 3, 𝑏 = 2 𝑥 𝑦 ∴ + =1 3 2 2𝑥 + 3𝑦 = 6 ∴ 2𝑥 + 3𝑦 − 6 = 0
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Linear Functions PARALLEL AND PERPENDICULAR LINES If two lines are parallel, they will have the same gradient. i.e.: 𝑚1 = 𝑚2 For two lines that are parallel, the following general equations can be found: 𝑎𝑥 + 𝑏𝑦 + 𝑐1 = 0 𝑎𝑥 + 𝑏𝑦 + 𝑐2 = 0 Proof: 𝑎 𝑏 𝑎 𝑎𝑥 + 𝑏𝑦 + 𝑐2 = 0 has the gradient 𝑚2 = − 𝑏 𝑎𝑥 + 𝑏𝑦 + 𝑐1 = 0 has the gradient 𝑚1 = −
Since 𝑚1 = 𝑚2 , the two lines are parallel. For perpendicular lines, the gradients of each are negative reciprocals of each other, i.e.: 𝑚1 𝑚2 = −1 1 i.e. 𝑚2 = − 𝑚1 Proof: Let line 𝐴𝐵 have gradient 𝑚1 = tan 𝛼. Let line 𝐶𝐷 have gradient 𝑚2 = tan 𝛽. tan 𝛽 =
𝐸𝐵 𝐸𝐶
∠𝐶𝐵𝐸 = 180° − 𝛼
(straight angle)
𝐸𝐶 𝐸𝐵 𝐸𝐵 ∴ cot(180° − 𝛼 ) = 𝐸𝐶 tan(180° − 𝛼 ) =
∴ tan 𝛽 = cot(180° − 𝛼 ) = − cot 𝛼 1 tan 𝛼 1 𝑆𝑜 𝑚2 = − 𝑚1 =−
𝑜𝑟 𝑚1 𝑚2 = −1 For two lines that are perpendicular, the following general equations can be found: 𝑎𝑥 + 𝑏𝑦 + 𝑐1 = 0 𝑏𝑥 − 𝑎𝑦 + 𝑐2 = 0
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Linear Functions EXAMPLES Example 1
Prove that the straight lines 5𝑥 − 2𝑦 − 1 = 0 and 5𝑥 − 2𝑦 + 7 = 0 are parallel. Solution 5𝑥 − 2𝑦 − 1 = 0 5𝑥 − 1 = 2𝑦 5 1 𝑥− =𝑦 2 2 5 ∴ 𝑚1 = 2 5𝑥 − 2𝑦 + 7 = 0 5𝑥 + 7 = 2𝑦 5 7 𝑥+ =𝑦 2 2 5 ∴ 𝑚2 = 2 5 𝑚1 = 𝑚2 = 2 ∴ the lines are parallel. Example 2
Find the equation of a straight line parallel to the line 2𝑥 − 𝑦 − 3 = 0 and passing through (1, −5). Solution 2𝑥 − 𝑦 − 3 = 0 2𝑥 − 3 = 𝑦 ∴ 𝑚1 = 2 For parallel lines, 𝑚1 = 𝑚2 ∴ 𝑚2 = 2 Equation:
𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ) 𝑦 − (−5) = 2(𝑥 − 1) 𝑦 + 5 = 2𝑥 − 2 0 = 2𝑥 − 𝑦 − 7
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Linear Functions Example 3
Show that the lines 3𝑥 + 𝑦 − 11 = 0 and 𝑥 − 3𝑦 + 1 = 0 are perpendicular. Solution 3𝑥 + 𝑦 − 11 = 0 𝑦 = −3𝑥 + 11 ∴ 𝑚1 = −3 𝑥 − 3𝑦 + 1 = 0 𝑥 + 1 = 3𝑦 1 1 𝑥+ =𝑦 3 3 1 ∴ 𝑚2 = 3 𝑚1 𝑚2 = −3 ×
1 3
= −1 ∴ the lines are perpendicular. Example 4
Find the equation of the straight line through (2, 3) perpendicular to the line that passes through (−1, 7) and (3, 3). Solution Line through (−1, 7) and (3, 3) has gradient: 7−3 −1 − 3 4 = −4
𝑚1 =
= −1 For perpendicular lines, 𝑚1 𝑚2 = 1 i.e.
− 1 × 𝑚2 = −1 𝑚2 = 1
Equation through (2, 3): 𝑦 − 3 = 1 (𝑥 − 2 ) =𝑥−2 0 =𝑥−𝑦+1
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Linear Functions INTERSECTION OF LINES The intersecting point of two lines can be found using simultaneous equations. To find the equation of a line through the intersection of two other lines, the following formula can be used: (𝑎1 𝑥 + 𝑏1 𝑦 + 𝑐1 ) + 𝑘(𝑎2 𝑥 + 𝑏2 𝑦 + 𝑐1 ) = 0 Where: 𝑎1 𝑥 + 𝑏1 𝑦 + 𝑐1 = equation of the first line 𝑎2 𝑥 + 𝑏2 𝑦 + 𝑐2 = equation of the second line 𝑘 = a constant Proof: Let 𝑙1 have equation 𝑎1 𝑥 + 𝑏1 𝑦 + 𝑐1 = 0. Let 𝑙2 have equation 𝑎2 𝑥 + 𝑏2 𝑦 + 𝑐2 = 0. Let the point of intersection of 𝑙1 and 𝑙2 be 𝑃 (𝑥1 , 𝑦1 ). Then 𝑃 satisfies 𝑙1 i.e. 𝑎1 𝑥1 + 𝑏1 𝑦1 + 𝑐1 = 0 𝑃 also satisfies 𝑙2 i.e. 𝑎2 𝑥1 + 𝑏2 𝑦1 + 𝑐2 = 0 Substitute 𝑃 into
(𝑎1 𝑥 + 𝑏1 𝑦 + 𝑐1 ) + 𝑘(𝑎2 𝑥 + 𝑏2 𝑦 + 𝑐2 ) = 0 (𝑎1 𝑥1 + 𝑏1 𝑦1 + 𝑐1 ) + 𝑘(𝑎2 𝑥1 + 𝑏2 𝑦1 + 𝑐2 ) = 0 0 + 𝑘 (0) = 0 0=0
∴ if point 𝑃 satisfies both equations 𝑙1 and 𝑙2 then it satisfies 𝑙1 + 𝑘𝑙2 = 0.
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Linear Functions EXAMPLES Example 1
Find the point of intersection between lines 2𝑥 − 3𝑦 − 3 = 0 and 5𝑥 − 2𝑦 − 13 = 0. Solution Solve simultaneous equations: 2𝑥 − 3𝑦 − 3 = 0 3𝑦 = 2𝑥 − 3 2 3 𝑦= 𝑥− 3 3 2 = 𝑥−1 3 5𝑥 − 2𝑦 − 13 = 0 2𝑦 = 5𝑥 − 13 5 13 𝑦= 𝑥− 2 2
2 5 13 ∴ 𝑥−1 = 𝑥− 3 2 2 4𝑥 − 6 = 15𝑥 − 39 15𝑥 − 4𝑥 = −6 + 39 11𝑥 = 33 𝑥=
33 11
=3 2 ∴ 𝑦 = (3) − 1 3 = 2−1 =1 ∴ point of intersection at (3, 1)
Example 2
Show that the lines 3𝑥 − 𝑦 + 1 = 0, 𝑥 + 2𝑦 + 12 = 0 and 4𝑥 − 3𝑦 − 7 = 0 are concurrent (i.e. the pass through the same point). Solution 3𝑥 − 𝑦 + 1 = 0 𝑦 = 3𝑥 + 1 𝑥 + 2𝑦 + 12 = 0 2𝑦 = −𝑥 − 12 1 12 𝑦=− 𝑥− 2 2 1 =− 𝑥−6 2 1 ∴ 3𝑥 + 1 = − 𝑥 − 6 2 6𝑥 + 2 = −𝑥 − 12
∴ 𝑦 = 3(−2) + 1 = −6 + 1 = −5 ∴ point of intersection is (−2, −5). At (2, −5), LHS = 4𝑥 − 3𝑦 − 7 = 4(−2) − 3(−5) − 7 = −8 + 15 − 7 =0 = RHS ∴ the lines are concurrent.
6𝑥 + 𝑥 = −12 − 2 7𝑥 = −14 𝑥 = −2
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Linear Functions Example 3
Find the equation of the line through (−1, 2) that passes through the intersection of lines 2𝑥 + 𝑦 − 5 = 0 and 𝑥 − 3𝑦 + 1 = 0. Solution 2𝑥 + 𝑦 − 5 ≡ 𝑎1 𝑥 + 𝑏1 𝑦 + 𝑐1 𝑥 − 3𝑦 + 1 ≡ 𝑎2 𝑥 + 𝑏2 𝑦 + 𝑐2 ∴ (2𝑥 + 𝑦 − 5) + 𝑘(𝑥 − 3𝑦 + 1) = 0 The line passes through (−1, 2), therefore At (−1, 2), [2(−1) + 2 − 5] + 𝑘 [−1 − 3(2) + 1] = 0 −5 − 6𝑘 = 0 6𝑘 = −5 5 𝑘=− 6 5 ∴ (2𝑥 + 𝑦 − 5) + (− ) (𝑥 − 3𝑦 + 1) = 0 6 6(2𝑥 + 𝑦 − 5) − 5(𝑥 − 3𝑦 + 1) = 0 12𝑥 + 6𝑦 − 30 − 5𝑥 + 15𝑦 − 5 = 0 7𝑥 + 21𝑦 − 35 = 0 𝑥 + 3𝑦 − 5 = 0
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Linear Functions PERPENDICULAR DISTANCE Perpendicular distance is the shortest distance between a point and a line, given by: 𝑑=
|𝑎𝑥1 + 𝑏𝑦1 + 𝑐 | √𝑎2 + 𝑏2
which is the perpendicular distance between point (𝑥1 , 𝑦1 ) and the line 𝑎𝑥 + 𝑏𝑦 + 𝑐 = 0.
Proof: Let 𝑑 be the perpendicular distance of (𝑥1 , 𝑦1 ) from the line 𝑎𝑥 + 𝑏𝑦 + 𝑐 = 0. 𝑐 𝐴 = (− , 0) 𝑎 𝑐 𝐶 = (0, − ) 𝑏 −𝑎𝑥1 − 𝑐 𝑅 = (𝑥1 , ) 𝑏 𝑐2 𝑐2 In ∆𝐴𝐶𝑂, 𝐴𝐶 = √ 2 + 2 𝑎 𝑏 𝑐 2 𝑏2 + 𝑐 2 𝑎2 =√ 𝑎2 𝑏2 𝑐√𝑎2 + 𝑏2 𝑎𝑏 −𝑎𝑥1 − 𝑐 𝑃𝑅 = 𝑦1 − ( ) 𝑏 =
𝑎𝑥1 + 𝑏𝑦1 + 𝑐 𝑏
=
∆𝐴𝐶𝑂 is similar to ∆𝑃𝑅𝑄 ∴
𝑃𝑄 𝑃𝑅 = 𝐴𝑂 𝐴𝐶 𝑃𝑄 =
𝐴𝑂 ∙ 𝑃𝑅 𝐴𝐶
∴𝑑=
𝑐 𝑎𝑥1 + 𝑏𝑦1 + 𝑐 𝑐√𝑎2 + 𝑏2 × ÷ 𝑎 𝑏 𝑎𝑏
= =
𝑐(𝑎𝑥1 + 𝑏𝑦1 + 𝑐 ) 𝑎𝑏 × 𝑎𝑏 𝑐√𝑎2 + 𝑏2 𝑎𝑥1 + 𝑏𝑦1 + 𝑐 √𝑎2 + 𝑏2
Now distance must be positive, ∴𝑑=
|𝑎𝑥1 + 𝑏𝑦1 + 𝑐 | √𝑎2 + 𝑏2
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Linear Functions EXAMPLES Example 1
Find the perpendicular distance of (4, −3) from the line 3𝑥 − 4𝑦 − 1 = 0. Solution 𝑥1 = 4, 𝑦1 = −3, 𝑎 = 3, 𝑏 = −4, 𝑐 = −1 𝑑= = =
|𝑎𝑥1 + 𝑏𝑦1 + 𝑐 | √𝑎2 + 𝑏2 |3(4) + (−4)(−3) + (−1)| √33 + (−4)2 |12 + 12 − 1| √25
23 = 5 = 4.6 units Example 2
Prove that the line 6𝑥 + 8𝑦 + 20 = 0 is a tangent to the circle 𝑥 2 + 𝑦 2 = 4. Solution 𝑟=2 Centre = (0, 0) 𝑑= =
|6(0) + 8(0) + 20| √62 + 82 |20|
√100 20 = 10 =2 𝑟=𝑑 ∴ the line is a tangent to the circle.
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The Quadratic Polynomial and the Parabola THE QUADRATIC POLYNOMIAL AND THE PARABOLA OVERVIEW The axis of symmetry and the vertex are always midway between the x-intercepts. The quadratics whose zeroes (x-intercepts) are α and β form a family with equation
y ax x .
x
Axis of Symmetry
:
Zeroes
: x
Vertex
:
b (think of quadratic formula with zero discriminant) 2a
b , b 2 4ac 2a
b , 4a 2a
THEORY OF DISCRIMINANT If Δ > 0, there are two distinct real zeroes. If Δ = 0, there are two equal real zeroes (or one real double zero). If Δ < 0, there are two distinct unreal zeroes.
DEFINITE AND INDEFIN ITE QUADRATICS A quadratic is definite is it has no real zeroes, being positive definite if it is always positive, and negative definite if it is always negative. A quadratic is indefinite if it has at least one zero.
Sum of Roots (Zeroes) :
b a
Product of Roots (Zeroes)
:
c a
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The Quadratic Polynomial and the Parabola QUADRATIC IDENTITY F(x) is identical to G(x) when they give the same value for all values of x. If two quadratics intersect at more than two points then they are identical. Example: Express 2 x 3x 6 in the form 2
We set up an identity,
ax 1 bx 1 c . 2
2x 2 3x 6 ax 1 bx 1 c . 2
We can expand the RHS and equate coefficients. RHS is ax 2a b x a b c so a 2 2
, 2a b 3 and a b c 6 . Or, we can use substitution. When x 1 LHS = 5 and RHS = c, and so on.
Sample question #2: Given that a, b and c are three distinct constants, prove that
x a x b x bx c x b2 x a c . Instead of expanding it is suffice to show
that LHS = RHS when x = a, x = b and x = c.
PARABOLA A parabola is the locus of a point which moves so that its distance to a fixed point (the focus) is equal its perpendicular distance to a fixed line (the directrix) not passing through the focus. A locus of P is the set of points, all denoted by P, which fit the given conditions. In a sense P is movable. The axis of parabola is the line through focus and vertex. The latus rectum is the line parallel to the directrix, passing the focus. Note that all parabolas are similar (i.e. have the same shape, only enlarged or shrunk).
STANDARD NOTATIONS Parabolas with vertex at the origin: 𝑥 2 = 4𝑎𝑦 ,
x 2 4ay,
y 2 4ax,
y 2 4ay , where a is
the focal length. The first concaves up, the second concaves down, the third concaves right and the fourth concaves left. As with other functions, parabolas can be translated by doing this kind of thing:
x h2 4a y k has vertex at (h, k) and concaves up.
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Plane Geometry PLANE GEOMETRY OVERVIEW Acceptable symbols include the following: is parallel to ||
is similar to |||
is perpendicular to I therefore .’. is congruent to
≡
because ‘.’
Collinear: when three or more points lie on the same line
Concurrent: when three or more lines intersect at the same point
Line: technically, a line with no endpoints
Interval AB: a line with two endpoints, A and B
Intercept: a section of an interval, for example, a point P on an interval AB divides the interval into two intercepts, AP and PB
Transversal: a line that crosses two or more other lines
Ray AB: a line with one end, A, extending towards B onwards
Opposite ray AB: a line with one end, A, but extending not towards B
Straight angle: has a size of 180 degrees
Reflex angle: has a size of more than 180 degrees [and less than 360 degrees]
Complementary angle: the angle which adds up to 90 degrees with another angle
Supplementary angle: the angle which adds up to 180 degrees with another angle
Vertically opposite angles: a pair of angles, one behind the other, created by the same two lines; they are equal in size
The Three Lines
Name: example in
(the two lines don’t need to be parallel) and Angles on Them
diagram [property if the two lines are parallel]
Corresponding angles: the two angles facing up on the right of the “/” [equal] Alternate angles: the angle facing down on the left of the “/” and the angle facing up on the right of the “/”; technically they must be inside the “=” [equal] Co-interior angles: they are inside the “=” and to the right of the “/” [supplementary] Conversely, if one of the parallel properties above is satisfied, the two lines are parallel.
INTERIOR AND EXTERIOR ANGLES Interior angle: the angle inside a polygon at its vertex (when a question mentions only “angle” then it’s the interior angle) Exterior angle: the angle between a side and the extension of another side, located at the vertex and outside the polygon; each vertex actually has two exterior angles but they are equal
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Plane Geometry SUM OF INTERIOR ANGLES Triangle: anything with 3 vertices. Sum of interior angles: 180 degrees. Quadrilateral: anything with 4 vertices. Sum of interior angles: 360 degrees. Polygon: any closed figure with vertices [including triangle and quadrilateral]. Sum of interior angles:
n 2180 ,
n is the number of sides or of vertices [hint: think of triangle and
quadrilateral and follow the pattern]. Proof: draw up n triangles with one side of each triangle being a side of the polygon and all triangles having a common vertex inside the polygon. Sum of angles of the triangles is n 180 degrees. But we have 360 around the common vertex.
CONVEX AND NON-CONVEX POLYGON A convex polygon is the one you’re probably imagining just now. A non-convex polygon has at least one interior angle greater than 180° (try to draw one).
Sum of exterior angles of any convex polygon : is 360 degrees. Proof: use the fact that at each vertex, the interior and exterior angles form a straight line and add to 180 degrees and that the sum of interior angles is
n 2180
degrees.
Regular polygon: has equal sides, and so equal interior angles and equal exterior angles
TRIANGLES STANDARD CONGRUENCE TESTS FOR TRIANGLES (NOTATION: ) SSS; SAS [A is the included angle]; AAS; RHS [the hypotenuse and one other side of a triangle are equal to those of another triangle] Note: there is no ASS test for congruence (unless the angle is obtuse). It is a good idea to name the vertices, sides and angles in corresponding order. When
A F , B D and C E , write ABC FDE instead of ABC DEF . Two vertices in two triangles are called corresponding if the angles at those vertices are equal.
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Plane Geometry STANDARD SIMILARITY TESTS FOR TRIANGLES (NOTATION: ||| ) SSS; SAS; AA [or AAA]; RHS [hypotenuse and another side are proportional in the ’s] [A is for equal angles, S is for proportional sides] Again, you should write the vertices in corresponding order.
Test for a triangle to be isosceles : if two angles of that triangle are equal Test for a triangle to be equilateral (has equal sides): if the 3 angles are equal, each being 60° Median: the line joining a vertex to the midpoint of the opposite side. Altitude: the line perpendicular from a vertex to the opposite side. The three medians are concurrent. The three altitudes are also concurrent. The size of an exterior angle in a Δ is equal to the sum of the two interior opposite angles.
Ratios of Intercepts on Sides of Triangle: suppose we have triangle ABC and a line parallel to BC cuts the triangle at P and Q, P is on side AB. If AP : PB k : l , then AQ : QC k : l and
PQ : BC k : k l . Proof: use similarity of triangles APQ and ABC. TRANSVERSALS TO THREE PARALLEL LINES The previous result can be used to obtain the following general result. If we have 3 parallel lines and a number of transversals (lines) that cross the three of them, the ratio of the intercepts on one transversal is the same as the ratio on any other transversals. The transversals don’t need to be parallel. Conversely, if the ratios are equal, the lines are parallel.
SPECIAL QUADRILATERALS A trapezium is a quadrilateral with at least one pair of opposite sides parallel. A parallelogram is a quadrilateral with both pairs of opposite sides parallel. A rhombus is a parallelogram with a pair of adjacent sides equal. A rectangle is a parallelogram with one angle a right angle. A square is both a rectangle and a rhombus.
Tests for parallelogram: Opposite angles are equal OR Opposite sides are equal OR One pair of opposite sides are equal and parallel OR The diagonals bisect each other
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Plane Geometry Tests for rhombus: All sides are equal OR The diagonals bisect each other at right angles OR The diagonals bisect each vertex angle
Tests for rectangle: All angles are equal [so they are 90 degrees] OR The diagonals are equal and bisect each other
Tests for square: Must be a rhombus and also a rectangle Area of Plane Figures
Square: (side length)2
Rectangle: (length) x (breath)
Parallelogram: (base) x (perpendicular height)
Triangle: ½ x (base) x (perpendicular height)
Rhombus: ½ x (product of the diagonals)
Trapezium: (average of parallel sides) x (perpendicular height)
Note: some figures belong to other more general figures. For example, a rhombus belongs to the parallelogram family, and so you can use the formula for parallelogram.
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Tangent to a Curve and Derivative of a Function TANGENT TO A CURVE AND DERIVATIVE OF A FUNCTION OVERVIEW The derivative represents the instantaneous rate of change of something with respect to
dy
another thing. For example,
dx shows how y is changing as x changes.
THE DERIVATIVE AS A LIMIT OF THE GRADIENT OF A CHORD
f ' x lim
ho
f x h f x and h
dy y lim x 0 dx x
Note that it is actually a two-sided limit and needs a continuity around that point. Also, suppose that f(x) and g(x) are both continuous. Then f(x) + g(x), f(x) - g(x) and f(x)g(x) will all be continuous.
If f x x n , then f ' x n x n1 If f x K , then f ' x 0
Rule 1 Rule 2
If f x K g x , then f ' x K g ' x
Rule 3
If f x g x hx , then f ' x g ' x h' x
Product Rule
If y u v, then y' u' v u v'
Quotient Rule
If y
u ' v u v' u , then y' v v2
(if you forget which one to differentiate first, just do a quick test with something like
y x2 / x )
Chain Rule
:
dy dy du (also called function of a function rule) dx du dx
The chain rule can be used for algebraic and non-algebraic functions (learnt later).
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Tangent to a Curve and Derivative of a Function Example 1:
4 Differentiate y 3 x x
Let u 3x x . 4
3
2
with respect to x.
dy 3 12 3 u 12 x 3 1 12 x 3 1 3x 4 x . dx 2 2
Example 2:
Differentiate y ln sin x with respect to x. Let u sin x .
dy 1 cos x cos x cot x . dx u sin x
TRIVIAL NOTES ON NOTATIONS Example 1:
Find the derivative of x 2 when x 1 Correct answers:
Let y x , then y ' 2 x and at x 1, y' 2 2
Let f x x , then 2
f ' x 2 x and f ' 1 2
NOTATION OF SECOND DERIVATIVE:
f ' ' x OR y ' ' OR
d2y d dy (think of , d is squared and dx is too) 2 dx dx dx
Example 1:
Find the second derivative of x3 when x 1 Y = x3, then y’ = 3x2 then y’’ = 6x y’’ = 6(1) and at x = 1 y’’ = 6
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Geometrical Applications of Differentiation GEOMETRICAL APPLICATIONS OF DIFFERENTIATION Geometrically the derivative represents the gradient of the tangent to a curve at a point. This is often called the gradient of the curve. A normal is the line through that point perpendicular to the tangent. There are three types of stationary points: maxima, minima and horizontal points of inflexion. Maxima and minima are also called turning points. There are global (absolute) and local (relative) maximum and minimum. Any local maximum or minimum is called an extremum. A function can only change from increasing to decreasing (or vice versa) at the critical points. Critical points ate points where the derivative is zero or undefined.
APPLICATION OF THE S ECOND DERIVATIVE If
f ' ' x 0 , the graph of f(x) is concave up at that point.
If
f ' ' x 0 , the graph of f(x) is concave down at that point.
If
f ' ' x 0 and there is a change in f ' ' x around that point, it is a point of inflexion.
SKETCHING CURVES (IF ONE COLUMN APPLIES, THE OTHERS APPLY TOO) Function f(x)
Derivative f’(x)
Second Derivative f’’(x)
Stationary point*
f’(x) = 0
N/A
Increasing**
f’(x) > 0
N/A
Decreasing**
f’(x) < 0
N/A
Concave up
f’(x) is increasing
f’’(x) > 0
Concave down
f’(x) is decreasing
f’’(x) < 0
Point of inflexion
f’(x) has a turning point
f’’(x) = 0***
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Geometrical Applications of Differentiation Example: Find any stationary points on the curve y = x3 – 48x – 7 For stationary points y’ = 0 3𝑥 2 – 48 = 0 3𝑥 2 = 48 i.e. 𝑥 2 = 16 𝑥 = ±4
Therefore
When x = 4
When x = -4, 𝑦 = (4)3 − 48(4) − 7 = −135
𝑦 = (−4)3 − 48(−4) − 7 = 121
So the stationary points are (4,-135) and (-4, 121) * When it’s a horizontal point of inflexion, the graph of f’(x) will only touch the x-axis (be tangential to x-axis). When it is a turning point, it will cut the x-axis. ** At a horizontal point of inflexion, we can still call the curve increasing or decreasing. *** f’’(x) = 0 doesn’t automatically mean columns 1 and 2 apply. But if column 2 applies then column 1 applies.
VOLUME AND SURFACE AREA OF SOLIDS (FOR MAXIMUM AND MINIMUM PROBLEMS AND FOR RATES OF CHANGE LATER ON)
For a cone :
For a sphere :
For a cylinder :
V r
V r h
4 3
3
A 4 r 2
2
A 2 r 2 2 r h
V 13 r 2 h A r2 rl (l r 2 h 2 )
For a pyramid : V 13 base height A sum of faces
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Integration INTEGRATION Integration is essentially reverse of differentiation. It is the process of find a primitive function. We use integration to find the area and volume under a curve. New notation 𝑖𝑓,
𝑑 2 (𝑥 ) = 2𝑥 𝑑 𝑡ℎ𝑒𝑛, ∫ 2𝑥 = 𝑥 2 + 𝑐
METHOD To integrate a standard term, we add one to the power and divide by the new power. This is the reverse of differentiation. ∫ 𝑥 𝑛 𝑑𝑥 =
𝑥 𝑛+1 +𝑐 𝑛+1
The ‘c’ accounts for any constants that could have been differentiated to zero, such as: 𝑑 2 (𝑥 ) = 2𝑥 𝑑𝑥 𝑑 (𝑥 2 + 5) = 2𝑥 𝑑𝑥 Example
Integrate 4𝑥 3 4𝑥 4 +𝑐 4 4 =𝑥 +𝑐
∫ 4𝑥 3 𝑑𝑥 =
When integrating with respect to x, this is an operation involving x only, thus any constant terms that do not involve x can be removed. ∫ 𝑎 × 𝑓 (𝑥 )𝑑𝑥 = 𝑎 ∫ 𝑓(𝑥 ) 𝑑𝑥 Therefore the previous question can also be done as: ∫ 4𝑥 3 𝑑𝑥 = 4 ∫ 𝑥 3 𝑑𝑥 𝑥4 + 𝑐) 4 = 𝑥4 + 𝑐 = 4(
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Integration When integrating a sum or difference, integrate them term by term. That is, ∫ 𝑓(𝑥 ) ± 𝑔(𝑥 )𝑑𝑥 = ∫ 𝑓(𝑥) 𝑑𝑥 ± ∫ 𝑔(𝑥 )𝑑𝑥 Example Find a. ∫(3𝑥 2 + 𝑥 ) 𝑑𝑥 ∫(3𝑥 2 + 𝑥 ) 𝑑𝑥 = ∫(3𝑥 2 ) 𝑑𝑥 + ∫(𝑥) 𝑑𝑥 3𝑥 3 𝑥 2 + +𝑐 3 2 1 2 3 =𝑥 + 𝑥 +𝑐 2 =
b. ∫(4𝑥 5 − 6) 𝑑𝑥 ∫(4𝑥 5 − 6) 𝑑𝑥 = ∫(4𝑥 5 ) 𝑑𝑥 − ∫(6) 𝑑𝑥 4𝑥 6 + 6𝑥 + 𝑐 6 2𝑥 6 = + 6𝑥 + 𝑐 3 =
CHAIN RULE FOR INTEGRATION The chain rule of integration only works when the expression inside the brackets is linear. ∫(𝑎𝑥 + 𝑏)𝑛 =
(𝑎𝑥 + 𝑏)𝑛 +𝑐 𝑎(𝑛 + 1)
Example
Using chain rule for integration, find: a) ∫(𝑥 + 2)4 𝑑𝑥 (𝑥 + 2)5 +𝑐 5×1 (𝑥 + 2)5 = +𝑐 5
∫(𝑥 + 2)4 𝑑𝑥 =
b) ∫(3𝑥 − 8)10 𝑑𝑥 (3𝑥 − 8)11 +𝑐 3 × 11 (3𝑥 − 8)10 = +𝑐 33
∫(3𝑥 − 8)10 𝑑𝑥 =
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Integration TRIGONOMETRIC INTEGR ATION When integrating trigonometric functions, remember: ∫ cos(𝑎𝑥 + 𝑏) 𝑑𝑥 =
∫ sin(𝑎𝑥 + 𝑏) 𝑑𝑥 =
sin(𝑎𝑥 + 𝑏) +𝑐 𝑎
−cos(𝑎𝑥 + 𝑏) +𝑐 𝑎
∫ sec 2 (𝑎𝑥 + 𝑏) 𝑑𝑥 =
tan(𝑎𝑥 + 𝑏) +𝑐 𝑎
Example Find: a) ∫ sin(2𝑥 + 1) 𝑑𝑥 ∫ sin(2𝑥 + 1) 𝑑𝑥 =
b) ∫ sec 2 (3 − 4𝑥 ) 𝑑𝑥
−cos(2𝑥 + 1) +𝑐 2
∫ sec 2 (3 − 4𝑥 ) 𝑑𝑥 =
−tan(3 − 4𝑥) +𝑐 4
DEFINITE INTEGRALS Fundamental Theorem of Calculus Given that 𝐹 ′ (𝑥 ) = 𝑓(𝑥): 𝑏
∫ 𝑓(𝑥)𝑑𝑥 = 𝐹 (𝑏) − 𝐹 (𝑎) 𝑎
Proof
a) Let 𝐹(𝑥) be any primitive of 𝑓(𝑥) and note that 𝐹(𝑥) and 𝐴(𝑥) (the specific primitive) must differ by a constant (proven below). Write 𝐹(𝑥) in terms of 𝐴(𝑥) and 𝐶 . 𝐴 ( 𝑥 ) = 𝐹 (𝑥 ) + 𝐶
b) By letting 𝑥 = 𝑎, find the value of 𝐶 . Since 𝐴(𝑎) = 0 (the area from 𝑎 to 𝑎 is 0), we have: 𝐴 ( 𝑎 ) = 𝐹 (𝑎 ) + 𝐶 0 = 𝐹 (𝑎 ) + 𝐶 𝐶 = −𝐹 (𝑎)
c) Hence, find the area from 𝑎 to 𝑏 (that is, find 𝐴(𝑏)) in terms of 𝐹.
𝐴 (𝑏 ) = 𝐹 (𝑏 ) + 𝐶 = 𝐹 (𝑏 ) − 𝐹 (𝑎 )
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Integration Example 3
Evaluate ∫1 (𝑥 + 2) 𝑑𝑥 3 3 𝑥2 ∫ (𝑥 + 2) 𝑑𝑥 = [ + 2𝑥] 2 1 1 9 1 = ( + 6) − ( + 2) 2 2 21 5 = − 2 2 16 = =8 2
NOTE: we can swap the limits of integrating by negating an integral, i.e. 𝑎
𝑏
∫ 𝑓(𝑥 ) 𝑑𝑥 = − ∫ 𝑓 (𝑥 ) 𝑑𝑥 𝑏
𝑎
THE AREA BENEATH A CURVE The area of region about the x-axis and is bounded by the curve y=f(x) and the axis for 𝑎 ≤ 𝑥 ≤ 𝑏 is given by 𝑏
𝐴𝑟𝑒𝑎 = ∫ 𝑓(𝑥 ) 𝑑𝑥 𝑎 𝑏
= ∫ 𝑦 𝑑𝑥 𝑎
NOTE: Integrals give the ‘signed’ area. That is, areas above the axes are counted as positive whilst areas underneath the axes are counted as negative. Take the absolute value of all areas underneath the curve, since areas are positive. Add the areas of the region to find the total area.
STEPS TO FOLLOW TO FIND AREA STEP 1: Sketch the curve in question and identify the area to be found. STEP 2: Split the area up into regions which lie above the 𝑥 axis and regions which lie below the 𝑥 axis. STEP 3: Find the area of each region by taking the absolute value of definite integral in that region. STEP 4: Add up each of the areas found.
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Integration Example 1
Find the area enclosed by the curve 𝑦 = 2𝑥(𝑥 + 1)(𝑥 − 3) and the 𝑥 axis
STEP 1: Sketch the curve, noting in particular its roots, and identify the area to be found. Note that the curve has roots at 𝑥 = −1, 0, 3 and since it is a cubic, the graph is sketched below:
STEP 2: Split the area up into regions which lie above the 𝑥 axis and regions which lie below the 𝑥 axis.
The area to be found is the area of 𝐴 plus the area of 𝐵 since these are the only regions enclosed by the curve and the 𝑥 axis.
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Integration STEP 3: Find the area of each region by taking the absolute value of definite integral in that region. The definite integral for the area of 𝐴 is:
0
0
𝐴: ∫ 2𝑥(𝑥 + 1)(𝑥 − 3)𝑑𝑥 = ∫ (2𝑥 3 − 4𝑥 2 − 6𝑥)𝑑𝑥 −1
−1 0
=[
2𝑥 4 4𝑥 3 − − 3𝑥 2 ] 4 3 −1
1 4 = [0] − [ + − 3] 2 3 7 = 6 |𝐴 | =
7 6
And for 𝐵, the definite integral required is: 3
3
𝐵: ∫ 2𝑥(𝑥 + 1)(𝑥 − 3)𝑑𝑥 = ∫ (2𝑥 3 − 4𝑥 2 − 6𝑥)𝑑𝑥 0
0 3
=[
2𝑥 4 4𝑥 3 − − 3𝑥 2 ] 4 3 0
81 − 36 − 27] − [0] 2 45 =− 2 =[
|𝐵 | =
45 2
STEP 4: Add up each of the areas found.
7 45 + 6 2 142 = 6 71 = 3 2 = 23 3
𝐴+𝐵 =
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Integration Example 2
Find the area bounded by the 𝑥 axis, the curve 𝑦 = 𝑥 2 and the line 𝑦 = −𝑥 + 6
The area is made up of two separate areas: One under the graph of 𝑦 = 𝑥 2 from 𝑥 = 0 to 𝑥 = 2, and the other under the graph of 𝑦 = −𝑥 + 6 from 𝑥 = 2 to 𝑥 = 6 2
6
Total Area = ∫0 𝑥 2 𝑑𝑥 + ∫2 −𝑥 + 6 𝑑𝑥 𝑥3
2
= [ ] + [−
𝑥2
3 0
8
= 3 − 0 + (−
2
6
+ 6𝑥]
36 2
2 4
+ 36) − (− 2 + 12)
8
= + 18 − 10 3
=
32 3
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Integration AREAS BOUNDED BY THE Y-AXIS An area enclosed by the Y-axis is found by interchanging y and x in the formula before. The area of region about the y-axis and is bounded by the curve y=f(x) and the axis for 𝑐 ≤ 𝑦 ≤ 𝑑 is given by 𝑑
𝐴𝑟𝑒𝑎 = ∫ 𝑥 𝑑𝑦 𝑐 𝑑
= ∫ 𝑥 𝑑𝑦 𝑐
NOTE: Integrals give the ‘signed’ area. That is, areas right of the axes are counted as positive whilst areas left the axes are counted as negative. Take the absolute value of all areas underneath the curve, since areas are positive. Add the areas of the region to find the total area.
STEPS TO FOLLOW TO FIND AREA To find the area bounded by a curve and the 𝑦 axis, we must first following the following steps: STEP 1: Sketch the curve in question and identify the area to be found.
STEP 2: Express the curve in the form𝑥 = 𝑔(𝑦). (As opposed to 𝑦 = 𝑓(𝑥))
STEP 3: Identify the definite integral required which will be in terms of 𝑦 (ensure that the limits of integration are in terms of 𝑦)
STEP 4: Evaluate the definite integral and take the absolute value if necessary.
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Integration Example
Find the area bounded by the curve 𝑦 = 3√𝑥, the 𝑦 axis and the lines 𝑦 = 1, 𝑦 = 4. STEP 1: Sketch the curve and identify the area to be found.
STEP 2: Rearrange the equation given to be in the form 𝑥 = 𝑔(𝑦) Rearranging: 𝑦 = 3√𝑥 𝑥 = 𝑦3 STEP 3: Identify the definite integral required which will be in terms of 𝑦 (ensure that the limits of integration are in terms of 𝑦) Hence the integral required is: 4
∫ 𝑦 3 𝑑𝑦 1
STEP 4: Evaluate the definite integral and take the absolute value if necessary. 4
4
𝑦4 ∫ 𝑦 𝑑𝑦 = [ ] 4 1 1 3
1 = [64] − [ ] 4 3 = 63 4
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Integration Example 2
Find the area bounded by the curve 𝑦 = √𝑥 + 2, the 𝑥 axis and the 𝑦 axis.
Since the area lies to the left of the 𝑦-axis, the absolute value of the integral must be taken √2
Area = |∫0 𝑥 𝑑𝑦| 𝑦 = √𝑥 + 2 𝑦2 = 𝑥 + 2 𝑥 = 𝑦2 − 2
√2
Area = | ∫0 𝑦 2 − 2 𝑑𝑦| √2
𝑦3
= |[ 3 − 2𝑦]
0
2√2 3
=|
= |− =
|
− 2√2|
4√2 | 3
4√2 3
MATHEMATICS | PAGE 59 OF 121
Integration AREAS BETWEEN CURVES The area between two curves, upper curve f(x) and lower curve g(x) with points of intersection at x=a and x=b is given by: 𝑏
∫ (𝑓(𝑥) − 𝑔(𝑥 ))𝑑𝑥 𝑎
STEPS TO FOLLOW TO FIND AREA To find the area between two curves 𝑦 = 𝑓(𝑥) and 𝑔(𝑥) as above, follow the following steps:
STEP 1: Sketch the curves on the same graph and identify the area to be found.
STEP 2: If required, split the area up into different regions if the curves cross over each other.
STEP 3: For each region, find the area by evaluating the definite integral: 𝑏
∫ (𝑓(𝑥) − 𝑔(𝑥 ))𝑑𝑥 𝑎
Where 𝑓(𝑥) is above 𝑔(𝑥) in that region.
STEP 4: Add up the areas of each of the regions to find the total area.
MATHEMATICS | PAGE 60 OF 121
Integration Example 1
Find the area enclosed by the curve 𝑦 = 2𝑥(𝑥 + 1)(𝑥 − 3) and the 𝑥 axis STEP 1: Sketch the curve, noting in particular its roots, and identify the area to be found. Note that the curve has roots at 𝑥 = −1, 0, 3 and since it is a cubic, the graph is sketched below:
STEP 2: Split the area up into regions which lie above the 𝑥 axis and regions which lie below the 𝑥 axis. The area to be found is the area of 𝐴 plus the area of 𝐵 since these are the only regions enclosed by the curve and the 𝑥 axis. STEP 3: Find the area of each region by taking the absolute value of definite integral in that region. The definite integral for the area of 𝐴 is: 0
0
𝐴: ∫ 2𝑥(𝑥 + 1)(𝑥 − 3)𝑑𝑥 = ∫ (2𝑥 3 − 4𝑥 2 − 6𝑥)𝑑𝑥 −1
−1 0
=[
2𝑥 4 4𝑥 3 − − 3𝑥 2 ] 4 3 −1
1 4 = [0] − [ + − 3] 2 3 7 = 6 |𝐴 | =
7 6
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Integration And for 𝐵, the definite integral required is: 3
3
𝐵: ∫ 2𝑥(𝑥 + 1)(𝑥 − 3)𝑑𝑥 = ∫ (2𝑥 3 − 4𝑥 2 − 6𝑥)𝑑𝑥 0
0
3
2𝑥 4 4𝑥 3 =[ − − 3𝑥 2 ] 4 3 0 81 = [ − 36 − 27] − [0] 2 45 =− 2 |𝐵 | =
45 2
STEP 4: Add up each of the areas found. 7 45 + 6 2 142 = 6 71 = 3 2 = 23 3
𝐴+𝐵 =
Example 2
Find the area bounded by the 𝑥 axis, the curve 𝑦 = 𝑥 3 + 1 and the lines 𝑥 = −2 and 𝑥 = 0
The area is made up of two areas, one below and one above the 𝑥-axis −1
0
Total area = |∫−2 𝑥 3 + 1 𝑑𝑥 | + ∫−1 𝑥 3 + 1 𝑑𝑥 −1
𝑥4
𝑥4
0
= |[ 4 + 𝑥] | + [ 4 + 𝑥] −2
1
−1
16
1
= |(4 − 1) − ( 4 − 2)| + 0 − (4 − 1) 3
3
= |− 4 − 2| + 4 =
11 4
+
3 4
7
= 2 units2
MATHEMATICS | PAGE 62 OF 121
Integration VOLUMES FORMED BY ROTATING A CURVE Integration can be used to find the volume of solids formed by rotating a curve about a line. These solids are known as solids of revolution
SOLIDS OF REVOLUTION ABOUT THE 𝒙 AXIS Given the solid of revolution formed by rotating the curve 𝑦 = 𝑓(𝑥) between 𝑥 = 𝑎 and 𝑥 = 𝑏 about the 𝑥 axis, its volume is: 𝑏
𝑉 = 𝜋 ∫ 𝑦 2 𝑑𝑥 𝑎
Since we are integrating 𝑦 2 it doesn’t matter where the curve is negative or positive when making this calculation. So to find the volume of a solid of revolution about the 𝑥 axis, follow the following steps: STEP 1: Sketch the curve and the solid of revolution STEP 2: Identify the limits of integration and write down the definite integral using the formula above. STEP 3: Evaluate the definite integral. Proof
a) Give the general formula for the area of a circle, and the volume of a cylinder. 𝐴 = 𝜋𝑟 2 𝑉 = 𝜋𝑟 2 ℎ
b) Consider the cylinder shown below. Express the radius and height of the cylinder in terms of 𝑥 and 𝑦. Hence, find the volume, 𝛿𝑉, of the cylinder 𝑅𝑎𝑑𝑖𝑢𝑠 = 𝛿𝑥 𝐻𝑒𝑖𝑔ℎ𝑡 = 𝑦 𝛿𝑉 = 𝜋𝑟 2 ℎ 𝑦
= 𝜋𝑦 2 𝛿𝑥
𝛿𝑥
MATHEMATICS | PAGE 63 OF 121
Integration c) Now, by recognizing that the total volume, V, is the sum of volumes of all the small cylinders from 𝑥 = 𝑎 to 𝑥 = 𝑏, give a formula for V: 𝑥=𝑏
𝑉 ≈ ∑ Δ𝑉 𝑥=𝑎 𝑏
= ∑ 𝜋𝑦 2 Δ𝑥 𝑎
d) By taking the limit as 𝛥𝑥 → 0 (i.e. the cylinders become infinitely thin),give a formula for the exact volume of solid of revolution. 𝑏
𝑉 = lim ∑ 𝜋𝑦 2 δ1𝑥 δx→0
𝑎
𝑏
= ∫ 𝜋𝑦 2 𝑑𝑥 𝑎 𝑏
= 𝜋 ∫ 𝑦 2 𝑑𝑥 𝑎
Example
Find the volume of the solid of revolution formed when the curve 𝑦 = 3 is rotated about the 𝑥 axis between 𝑥 = 1 and 𝑥 = 4. STEP 1: Sketch the curve and the solid of revolution
𝑦
3
1
4 𝑥
MATHEMATICS | PAGE 64 OF 121
Integration STEP 2: Identify the limits of integration and write down the definite integral using the formula. 𝑦2 = 9 4
𝑉 = 𝜋 ∫1 9 𝑑𝑥 STEP 3: Evaluate the definite integral. = 𝜋[9𝑥]14 = 𝜋(36 − 9) = 27𝜋 Find the volume of this solid using a different method [Hint: what type of solid is it?]. Are these two answers equal? The solid of revolution is a cylinder, of height 3 and radius 3. Hence, the volume 𝑉 is 𝑉 = 32 𝜋 × 3 = 27𝜋
Example 2
Find the volume of a cone of radius 𝑟 and height ℎ. a) Let the cone have its apex (point) at the origin and let it be formed by rotating a line about the 𝑥 axis. Find the equation of this line.
𝑦 𝑟
𝑥
ℎ
The point (ℎ, 𝑟) lies on the line, and the gradient of the line is Hence, the equation of the line is 𝑦 =
ℎ 𝑟
𝑟𝑥 ℎ
MATHEMATICS | PAGE 65 OF 121
Integration b) Find the definite integral that represents the volume of the cone and evaluate the integral 1
to show that the volume of a cone is 3 𝜋𝑟 2 ℎ. 𝑦2 =
𝑟 2𝑥 2 ℎ2 ℎ 𝑟 2𝑥 2
𝑉 = 𝜋 ∫0
𝑑𝑥
ℎ2
𝑟 2𝑥3
ℎ
= 𝜋 [ 3ℎ2 ]
0
=𝜋×
𝑟 2ℎ3 3ℎ 2
1
= 3 𝜋𝑟 2 ℎ Example
Find the volume of a sphere of radius 𝑟. a) Write down the equation of a circle centred at the origin with radius 𝑟. 𝑥 2 + 𝑦 2 = 𝑟2 b) Consider a sphere as the solid of revolution formed by rotating the upper semi-circle from 4
part (a) about the 𝑥 axis. Show that the volume of the sphere is 3 𝜋𝑟 3 . 𝑦
−𝑟
𝑟
𝑥
𝑦 2 = 𝑟2 − 𝑥 2 𝑟
𝑉 = 𝜋 ∫−𝑟 𝑟 2 − 𝑥 2 𝑑𝑥 = 𝜋 [𝑟 2 𝑥 − = 𝜋 [(𝑟 3 − 2𝑟 3
=𝜋×[
3
𝑥3
𝑟
]
3 −𝑟 𝑟3 3
) − (𝑟 3 −
− (−
2𝑟 3 3
𝑟3 3
)]
)]
4
= 3 𝜋𝑟 3
MATHEMATICS | PAGE 66 OF 121
Integration SOLIDS OF REVOLUTION ABOUT THE 𝒚 AXIS Just like when finding the area about the y-axis, volume about the y-axis can be found by interchanging the y and x. Given the solid of revolution formed by rotating the curve 𝑥 = 𝑔(𝑦) between 𝑦 = 𝑐 and 𝑦 = 𝑑about the 𝑦 axis, its volume is: 𝑑
𝑉 = 𝜋 ∫ 𝑥 2 𝑑𝑦 𝑐
Example 1
The region bounded by the curve 𝑦 = 𝑥 2 , the 𝑦 axis and the lines 𝑦 = 1 and 𝑦 = 4 is rotated about the 𝑦 axis. Find the volume of the solid formed. STEP 1: Sketch the curve and the solid of revolution 𝑦 4
2
𝑥
STEP 2: Identify the limits of integration and write down the definite integral using the formula. 𝑥2 = 𝑦 4
𝑉 = 𝜋 ∫2 𝑦 𝑑𝑦 STEP 3: Evaluate the definite integral. 𝑦2
4
𝑉 = 𝜋[2]
2 16
4
= 𝜋 ( 2 − 2) = 6𝜋
MATHEMATICS | PAGE 67 OF 121
Integration Example 2
An ellipse has the equation 9𝑥 2 + 𝑦 2 = 36 and is sketched below.
𝑦
𝑥
A football-like shape is formed by rotating this ellipse about the 𝑦 axis.
a) Find the 𝑦 intercepts of the ellipse When 𝑥 = 0 𝑦 2 = 36 𝑦 = ±6
𝑦 intercepts at (0, ±6)
b) Hence, find the volume of the football. 9𝑥 2 = 36 − 𝑦 2 𝑥2 = 4 − 6
𝑦2 9
𝑉 = 𝜋 ∫−6 4 −
𝑦2
𝑑𝑦
9 𝑦3
6
= 𝜋 [4𝑦 − 27 ]
−6
= 𝜋[(24 − 8) − (−24 + 8)] = 32𝜋
MATHEMATICS | PAGE 68 OF 121
Integration VOLUMES FROM MULTIPLE CURVES Harder questions will ask for volumes formed by multiple curves. Step 1: Find the volume of the solid of revolution formed by rotating the curve 𝑦 = 𝑓(𝑥) about the 𝑥 axis. Step 2: Find the volume of the solid of revolution formed by rotating the curve 𝑦 = 𝑔(𝑥) about the 𝑥 axis. Noting that the volume we wish to find is the difference of the two above volumes, find the volume of this solid.
𝑏
𝑉 = 𝜋 ∫ (𝑓(𝑥 )2 − 𝑔(𝑥 )2 )𝑑𝑥 𝑎
NOTE: for multiple curves about the y –axis, once again interchange the x and y values as previously done.
Proof Sometimes a solid may be formed by rotating two curves about an axis and taking the inside of these two figures. Suppose the area between the following two curves is rotated about the 𝑥 axis forming a solid of revolution:
MATHEMATICS | PAGE 69 OF 121
Integration Find the volume of the solid of revolution formed by rotating the area above about the 𝑥 axis.
a) Find the volume of the solid of revolution formed by rotating the curve 𝑦 = 𝑓(𝑥) about the 𝑥 axis. The volume of the solid is: 𝑏
𝑉 = 𝜋 ∫ 𝑓 (𝑥 )2 𝑑𝑥 𝑎
b) Find the volume of the solid of revolution formed by rotating the curve 𝑦 = 𝑔(𝑥) about the 𝑥 axis. The volume of the solid is: 𝑏
𝑉 = 𝜋 ∫ 𝑔(𝑥 )2 𝑑𝑥 𝑎
c) Noting that the volume we wish to find is the difference of the two above volumes, find the volume of this solid. The volume of the resulting solid is: 𝑏
𝑉 = 𝜋 ∫ (𝑓(𝑥 )2 − 𝑔(𝑥 )2 )𝑑𝑥 𝑎
MATHEMATICS | PAGE 70 OF 121
Integration Example 1
Find the volume of the solid formed when the region enclosed by the curves 𝑦 = 𝑥 2 and 𝑦 = 3𝑥 + 4 is rotated about the 𝑥 axis.
a) Sketch a graph, indicating the area to be rotated and identifying the end points of the area by solving for the points of intersection. 𝑦 𝑦 = 𝑥2
(4, 16)
𝑦 = 3𝑥 + 4
(−1, 1) 𝑥
Solving simultaneously, 𝑥 2 = 3𝑥 + 4 𝑥 2 − 3𝑥 − 4 = 0 (𝑥 − 4)(𝑥 + 1) = 0 ∴ 𝑥 = 4 or 𝑥 = −1
b) Hence, find the volume of the solid formed when this area is rotated about the 𝑥 axis. Let 𝑓(𝑥 ) = 3𝑥 + 4, [𝑓 (𝑥 )]2 = (3𝑥 + 4)2 Let 𝑔(𝑥 ) = 𝑥 2 , [𝑔(𝑥 )]2 = 𝑥 4 4
𝑉 = 𝜋 ∫−1(3𝑥 + 4)2 − 𝑥 4 𝑑𝑥 4
= 𝜋 ∫−1 9𝑥 2 + 24𝑥 + 16 − 𝑥 4 𝑑𝑥 = 𝜋 [3𝑥 3 + 12𝑥 2 + 16𝑥 − = 𝜋 [(192 + 192 + 64 −
𝑥5
4
]
5 −1
1024 5
1
) − (−3 + 12 − 16 + 5)]
= 250𝜋
MATHEMATICS | PAGE 71 OF 121
Integration Example 2 Question 15
Consider an ice-cream cone shaped solid of revolution formed by rotating the area bounded by the 𝑦 axis, the line 𝑦 = 3𝑥 − 3 and the first quadrant of the circle 𝑥 2 + 𝑦 2 = 1 about the 𝑦 axis. a) Sketch a graph showing the area to be rotated. 𝑦 1
𝑦 = 3𝑥 − 3
𝑥
−3
b) Consider the volume in two different parts and find the definite integral for each part. Consider the volume generated by the circle, 𝑉1 Rotating the first quadrant around the 𝑦-axis is equivalent to rotating the top half of the circle about the 𝑦-axis 𝑥2 = 1 − 𝑦2 1
𝑉1 = 𝜋 ∫0 1 − 𝑦 2 𝑑𝑦 Consider the volume generated by the line, 𝑉2 3𝑥 = 𝑦 + 3 𝑦
𝑥 = 3+1 𝑦
𝑥 2 = (3 + 1) =
𝑦2 9
+
2𝑦 3
0 𝑦2
𝑉2 = 𝜋 ∫−3
2
9
+1
+
2𝑦 3
+ 1 𝑑𝑦
MATHEMATICS | PAGE 72 OF 121
Integration c) Evaluate each definite integral and add them to find the total volume of the solid. 𝑉1 = 𝜋 [𝑦 −
𝑦3
1
]
3 0 1
= 𝜋 (1 − 3) =
2𝜋 3
𝑦3
𝑉2 = 𝜋 [27 +
0
𝑦2
+ 𝑦]
3
−3
= 𝜋[0 − (−1 + 3 − 3)] =𝜋 Total Volume =
2𝜋 3
=
+𝜋 5𝜋 3
Example 2
A spinning top is formed from a solid of revolution. The area bounded by the curves 𝑦 = 𝑥 2 , 𝑦 = −𝑥 + 2 and the 𝑥 axis is rotated about the 𝑥 axis. Find the volume of the spinning top.
𝑦
(−2, 4)
(1, 1) 2
𝑥
Consider the volume generated by 𝑦 = 𝑥 2 𝑦2 = 𝑥4 1 𝑉 = 𝜋 ∫0 𝑥 4 𝑑𝑥 𝑥5
1
= 𝜋[5] =
0
𝜋 5
Consider the volume generated by 𝑦 = −𝑥 + 2 𝑦 2 = (−𝑥 + 2)2 = 𝑥 2 − 4𝑥 + 4 2
𝑉 = 𝜋 ∫1 𝑥 2 − 4𝑥 + 4 𝑑𝑥 𝑥3
2
= 𝜋 [ 3 − 2𝑥 2 + 4𝑥] 8
1 1
= 𝜋 [(3 − 8 + 8) − (3 − 2 + 4)] =
𝜋 3
MATHEMATICS | PAGE 73 OF 121
Integration APPROXIMATE METHODS OF INTEGRATION Sometimes an integral is too difficult to find directly. However it is still possible to approximate the value of the integral by finding the approximate area enclosed by the curve.
TRAPEZOIDAL RULE In this method, approximation of the area underneath a curve is done by a number of trapeziums.
𝑦
𝑦 = 𝑓(𝑥)
𝑥1
𝑎 = 𝑥0
𝑥
𝑥2 = 𝑏
The general form of the trapezoidal rule is given below: Given that 𝑥1 , 𝑥2 , … , 𝑥𝑛−1 are equally spaced with width ℎ along the interval from 𝑎 = 𝑥0 to 𝑏 = 𝑥𝑛 , then the integral of 𝑓(𝑥) is approximately: 𝑏
∫ 𝑓(𝑥)𝑑𝑥 ≈ 𝑎
ℎ (𝑓(𝑥0 ) + 2𝑓 (𝑥1 ) + 2𝑓 (𝑥2 ) + ⋯ + 2𝑓 (𝑥𝑛−1 ) + 𝑓 (𝑥𝑛 )) 2
NOTE: An easy way to remember the trapezoidal rule is: 𝑏
∫ 𝑓(𝑥)𝑑𝑥 ≈ 𝑎
ℎ (1𝑠𝑡 + 2(𝑟𝑒𝑠𝑡) + 𝑙𝑎𝑠𝑡) 2
MATHEMATICS | PAGE 74 OF 121
Integration When specifying the number of trapeziums to use in your approximation, the question might ask for:
Function values
Sub-intervals
Strips
It is important to note that a question asking for 3 strips/sub-intervals/trapeziums requires 4 function values. NOTE: h is an equal spacing so to find h, find the distance between the furthest two values and divide by the number of sub intervals.
ℎ=
𝑏−𝑎 𝑛𝑜. 𝑜𝑓 𝑠𝑢𝑏𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑠
Proof
Approximate the area underneath the curve 𝑦 = 𝑓(𝑥) from 𝑎 to 𝑏 by two trapeziums as shown above. Note that 𝑥1 is the midpoint of the interval from 𝑎 to 𝑏. a) Write down the formula for a trapezium of sides of length a and b and width h.
𝐴𝑟𝑒𝑎 =
𝑎+𝑏 ×ℎ 2
b) Find the area of the trapezium shaded below
The height of the trapezium is ℎ = 𝑥1 − 𝑥0 (letting 𝑎 = 𝑥0). The length of the two parallel sides is 𝑓(𝑥0 ) and 𝑓(𝑥0 )
𝑓(𝑥1 )
𝑓(𝑥1 ). Hence the area is: 𝐴 = (𝑥1 − 𝑥0 )
𝑓(𝑥0 ) + 𝑓 (𝑥1 ) 2
𝑥1 − 𝑥0
MATHEMATICS | PAGE 75 OF 121
Integration c) Similarly, find the area of the second trapezium. The height of the trapezium is ℎ = 𝑥2 − 𝑥1 (letting 𝑏 = 𝑥2 ). The length of the two parallel sides is 𝑓(𝑥1 ) and 𝑓(𝑥2 ). Hence the area is: 𝐴 = (𝑥2 − 𝑥1 )
𝑓 (𝑥1 ) + 𝑓 (𝑥2 ) 2
d) Find the approximate area underneath the curve by adding the areas from (a) and (b). Let ℎ be the height of the trapeziums in your answer. Since 𝑥1 is the midpoint of the interval, ℎ = 𝑥1 − 𝑥0 = 𝑥2 − 𝑥1 . Hence, the area underneath the curve is approximately:
𝐴≈ℎ =
𝑓(𝑥0 ) + 𝑓 (𝑥1 ) 𝑓 (𝑥1 ) + 𝑓 (𝑥2 ) +ℎ 2 2
ℎ (𝑓(𝑥0 ) + 2𝑓(𝑥1 ) + 𝑓 (𝑥2 )) 2
Example 1 3
Approximate the integral ∫ √𝑥 2 + 1𝑑𝑥 using the trapezoidal rule with 4 function values. 0 a) Find the width, ℎ, of each sub-interval. Four function values corresponds to 3 subintervals The total length of the interval is 3, and this must be divided into 3 equal subintervals Each subinterval is 1 unit in length ∴ℎ=1 The function values are then 0, 1, 2, and 3
MATHEMATICS | PAGE 76 OF 121
Integration b) Complete the following table of function values. 𝑥
0
1
2
3
𝑓(𝑥)
1
√2
√5
√10
c) Hence, apply the formula for the trapezoidal rule to approximate the integral. 3
1
∫0 √𝑥 2 + 1 𝑑𝑥 ≈ 2 [𝑓 (0) + 2𝑓(1) + 2𝑓(2) + 𝑓 (3)] 1
= 2 (1 + 2√2 + 2√5 + √10)
=5.73142037 (8dp) d) The correct value of this integral is 5.65263972 (8 dp). How does your approximation compare to this? The approximation differs by 0.078780649, which is a fairly good approximation
Example 2 5
Approximate the integral ∫
−1
1 𝑥 2 +1
𝑑𝑥 using the trapezoidal rule with 3 sub-intervals.
The total length of the interval is 6, and this must be divided into 3 equal subintervals Each subinterval is 2 units in length ∴ℎ=2 Three subintervals require 4 function values The function values are then −1, 1, 3, and 5 𝑥
−1
1
3
5
𝑓(𝑥)
1 2
1 2
1 10
1 26
5
∫ −1
𝑥2
1 2 𝑑𝑥 ≈ [𝑓(−1) + 2𝑓 (1) + 2𝑓(3) + 𝑓 (5)] +1 2 =
1 1 1 1 + 2( )+2( )+ 2 2 10 26
=
113 or 1.74 (to 2 dp) 65
MATHEMATICS | PAGE 77 OF 121
Integration SIMPSON’S RULE The trapezoidal rule can be fairly inaccurate though because it approximates curves by straight lines. A more accurate approximation method is called Simpson’s Rule which uses parabolas to approximate a curve.
𝑦
𝑦 = 𝑓(𝑥)
𝑎 = 𝑥0
𝑥1
𝑥2 =𝑏
𝑥
Because a parabola requires three points to be specified, Simpson’s rule requires at least 2 sub-intervals to begin with. Given that 𝑥1 is the midpoint of the interval and that ℎ is the width of each strip, i.e. ℎ = 𝑥1 − 𝑎 = 𝑏 − 𝑥1 , Simpson’s Rule states that: 𝑏
∫ 𝑓(𝑥 )𝑑𝑥 ≈ 𝑎
𝑏−𝑎 𝑎+𝑏 (𝑓(𝑎) + 4𝑓 ( ) + 𝑓(𝑏)) 6 2
Simpson’s rule can also be generalized and can be used whenever there are an even number of sub-intervals (i.e. an odd number (≠ 1) of function values). The general formula is below: Given that 𝑥1 , 𝑥2 , … 𝑥𝑛−1 are evenly spaced along the interval from 𝑎 = 𝑥0 to 𝑏 = 𝑥𝑛 and that ℎ is the width of each sub-interval: 𝑏
∫ 𝑓 (𝑥 )𝑑𝑥 ≈ 𝑎
ℎ (𝑓(𝑥0 ) + 4𝑓 (𝑥1 ) + 2𝑓 (𝑥2 ) + 4𝑓 (𝑥3 ) + ⋯ + 𝑓 (𝑥𝑛 )) 3
NOTE: Easy way to remember the Simpson’s rule is 𝑏
∫ 𝑓(𝑥 )𝑑𝑥 ≈ 𝑎
ℎ (𝑓𝑖𝑟𝑠𝑡 + 4(𝑒𝑣𝑒𝑛𝑠) + 2(𝑜𝑑𝑑𝑠) + 𝑙𝑎𝑠𝑡) 3
NOTE: The Simpson’s provides the exact answer for quadratic and cubic functions as opposed to just an approximation
MATHEMATICS | PAGE 78 OF 121
Integration Example 1 4
Approximate the value of the integral ∫
𝑥
√𝑥 2 +4 0
𝑑𝑥 using Simpson’s rule with 5 function
values. a) Find the width, ℎ of each sub-interval. Five function values corresponds to 4 subintervals The total length of the interval is 4, and this must be divided into 4 equal subintervals Each interval is 1 unit in length ∴ℎ=1 The function values are then 0, 1, 2, 3, and 4
b) Fill out the following table with the function values. 𝑥
0
𝑓(𝑥)
0
1
2
3
4
1
1
3
2
√5
√2
√13
√5
c) Apply the formula for Simpson’s rule to find an approximation for the integral. 4
∫ 0
𝑥 √𝑥 2
1 𝑑𝑥 ≈ [𝑓 (0) + 4𝑓 (1) + 2𝑓(2) + 4𝑓 (3) + 𝑓(4)] 3 +4
1 4 2 12 2 = (0 + + + + ) 3 √5 √2 √13 √5 =2.475232104 (to 8 dp)
d) The correct value of this integral is 2.47213595 (8 dp). Find the percentage error of your calculation.
%𝑒𝑟𝑟𝑜𝑟 =
|2.475232104 − 2.47213595| × 100% = 0.1% 2.47213595
MATHEMATICS | PAGE 79 OF 121
Integration Example 2 2
Approximate the value of the integral ∫
1
1 𝑥
𝑑𝑥 using Simpson’s rule with 5 function values.
Five function values corresponds to 4 subintervals The total length of the interval is 1, and this must be divided into 4 equal subintervals Each interval is 0.25 unit in length 1
∴ ℎ = 0.25 = 4 The function values are then 1, 1.25, 1.5, 1.75, and 2 𝑥
1
1.25
1.5
1.75
2
𝑓(𝑥)
1
1 1.25
1 1.5
1 1.75
1 2
2
∫ 1
=
1 1 1 𝑑𝑥 ≈ × [𝑓 (1) + 4𝑓 (1.25) + 2𝑓 (1.5) + 4𝑓(1.75) + 𝑓 (2)] 𝑥 4 3
1 4 2 4 1 (1 + + + + ) 12 1.25 1.5 1.75 2
= 0.6933 (to 4 dp)
MATHEMATICS | PAGE 80 OF 121
Trignometric Functions TRIGNOMETRIC FUNCTIONS TRIG. RATIOS
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
𝜃 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
𝒔𝒊𝒏𝜽 =
𝒐𝒑𝒑𝒐𝒔𝒊𝒕𝒆 𝒉𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆
𝒄𝒐𝒔𝜽 =
𝒂𝒅𝒋𝒂𝒄𝒆𝒏𝒕 𝒉𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆
𝒕𝒂𝒏𝜽 =
𝒐𝒑𝒑𝒐𝒔𝒊𝒕𝒆 𝒂𝒅𝒋𝒂𝒄𝒆𝒏𝒕
𝒄𝒐𝒔𝒆𝒄𝜽 =
𝟏 𝒔𝒊𝒏𝜽
𝒔𝒆𝒄𝜽 =
𝟏 𝒄𝒐𝒔𝜽
𝒄𝒐𝒕𝜽 =
𝟏 𝒕𝒂𝒏𝜽
DERIVATION OF TRIG. IDENTITIES
𝑠𝑖𝑛𝜃
1 𝜃 𝑐𝑜𝑠𝜃
𝐵𝑦 𝑃𝑦𝑡ℎ𝑎𝑔𝑜𝑟𝑎𝑠 ∴ 𝒔𝒊𝒏𝟐 𝜽 + 𝒄𝒐𝒔𝟐𝜽 = 𝟏 ∴ 𝒕𝒂𝒏𝟐 𝜽 + 𝟏 = 𝒔𝒆𝒄𝟐 𝜽 (𝑑𝑖𝑣𝑖𝑑𝑒 𝑏𝑦 𝑐𝑜𝑠 2 𝜃)
∴ 𝟏 + 𝒄𝒐𝒕𝟐 𝜽 = 𝒄𝒐𝒔𝒆𝒄𝟐 𝜽 (𝑑𝑖𝑣𝑖𝑑𝑒 𝑏𝑦 𝑠𝑖𝑛2 𝜃)
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Trignometric Functions LENGTH OF AN ARC
AREA OF AN ARC
RADIANS 𝟑𝟔𝟎° = 𝟐𝝅 To convert degrees into radians: ×
𝜋 180
To convert radians into degrees: ×
180 𝜋
AREA OF A TRIANGLE
MATHEMATICS | PAGE 82 OF 121
Trignometric Functions DERIVING COSINE RULE
DERIVING SINE RULE
AREA OF A SEGMENT
MATHEMATICS | PAGE 83 OF 121
Exponential and Logarithmic Functions EXPONENTIAL AND LOGARITHMIC FUNCTIONS DEFINITION OF A LOG A log allows the index to be written as the subject of the equation:
𝑎 𝑦 = 𝑥 ↔ 𝑦 = log 𝑎 𝑥 Examples Example 1 24 = 16 log 2 16 = 4 Example 2 log 3 9 = 2 32 = 9 Example 3 log 𝑥 49 = 2 𝑥 2 = 49 𝑥=7
EXPONENTIAL AND LOG LAWS
Index Laws
Log Laws 𝒂𝒙 × 𝒂𝒚 = 𝒂𝒙+𝒚 𝒂𝒙 ÷ 𝒂𝒚 = 𝒂𝒙−𝒚
log 𝑎 (𝑥𝑦) = log𝑎 𝑥 + log 𝑎 𝑦 log 𝑎
𝑥 = log 𝑎 𝑥 − log 𝑎 𝑦 𝑦
(𝒂𝒙 )𝒚 = 𝒂𝒙𝒚
log 𝑎 𝑥 𝑦 = 𝑦 log 𝑎 𝑥
𝒂𝟎 = 𝟏
log 𝑎 1 = 0
𝒂𝟏 = 𝒂
log 𝑎 𝑎 = 1
𝒂−𝒙 =
𝟏 𝒂𝒙
log 𝑎
1 = −𝑥 𝑎𝑥
MATHEMATICS | PAGE 84 OF 121
Exponential and Logarithmic Functions Example 1 log 3 + log 4 − log 6 = log(3 × 4 ÷ 6) = log 2 Example 2 254𝑥−3 = 5𝑥+7 52(4𝑥−3) = 5𝑥+7 8𝑥 − 6% = 𝑥 + 7 7𝑥 = 13 13 𝑥= 7
CHANGE OF BASE FORMULA Change of base formula is used to evaluate complex logs which cannot be inputted in to calculators. log 𝑐 𝑥 =
log 𝑥 log 𝑐
Example 1 log 3 5 = log 5 log 3 = 1.46 =
Example 2 log 𝜋 √2 =
log √2 log 𝜋
= 0.30 log 𝜋 √2 = log √2 log 𝜋 = 0.30 =
MATHEMATICS | PAGE 85 OF 121
Exponential and Logarithmic Functions THE DERIVATIVE OF THE LOGARITHMIC FUNCTION FORMULAE Derivative of 𝑙𝑜𝑔𝑒 𝑥
𝑑 1 log 𝑒 𝑥 = 𝑑𝑥 𝑥
Derivative of 𝑙𝑜𝑔𝑒 𝑢
𝑑 1 𝑑𝑢 log 𝑒 𝑢 = × 𝑑𝑥 𝑢 𝑑𝑥
Derivative of 𝑙𝑜𝑔𝑒 (𝑎𝑥 + 𝑏)
𝑑 𝑎 log 𝑒 (𝑎𝑥 + 𝑏 ) = 𝑑𝑥 𝑎𝑥 + 𝑏 PROOF Find
𝑑𝑦 𝑑𝑥
if 𝑦 = log 𝑒 (𝑎𝑥 + 𝑏)by using the chain rule:
Let 𝑢 = 𝑎𝑥 + 𝑏 𝑑𝑢 =𝑎 𝑑𝑥 So 𝑦 = log 𝑢 𝑑𝑦 1 = 𝑑𝑢 𝑢 Using the chain rule, 𝑑𝑦 𝑑𝑦 𝑑𝑢 = × 𝑑𝑢 𝑑𝑢 𝑑𝑥 1 = ×𝑎 𝑢 𝑎 = 𝑎𝑥 + 𝑏
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Exponential and Logarithmic Functions Derivative of 𝑙𝑜𝑔 𝑓(𝑥)
𝑑 𝑓′(𝑥) log 𝑓(𝑥) = 𝑑𝑥 𝑓(𝑥) PROOF If 𝑦 = log 𝑓 (𝑥 ) find
𝑑𝑦 𝑑𝑥
Let 𝑢 = 𝑓(𝑥) 𝑑𝑢 = 𝑓 ′ (𝑥) 𝑑𝑥 So 𝑦 = log 𝑢 𝑑𝑦 1 1 = = 𝑑𝑢 𝑢 𝑓(𝑥) Using the chain rule, 𝑑𝑦 𝑑𝑦 𝑑𝑢 = × 𝑑𝑢 𝑑𝑢 𝑑𝑥 1 = × 𝑓 ′ (𝑥) 𝑓(𝑥) 𝑓 ′ (𝑥) = 𝑓(𝑥)
Example 1
Find the derivative of 𝑙𝑜𝑔 3𝑥 =
3 3𝑥
=
1 𝑥
Example 2
Find the derivative of 𝑙𝑜𝑔(10𝑥 + 3) =
10 10𝑥 + 3
MATHEMATICS | PAGE 87 OF 121
Exponential and Logarithmic Functions Example 3
Find the derivative of 𝑦 = 𝑙𝑜𝑔 √𝑥 1 𝑦 = log 𝑥 2 𝑑𝑦 1 = 𝑑𝑥 2𝑥
Example 4
Differentiate 𝑦 = 3 𝑙𝑜𝑔7 4𝑥 Tip: If we want to differentiate logs with a base other than 𝑒, use the change of base formula to express the log in terms of log 𝑒 𝑥 , and then differentiate. 𝑦=
3 log 4𝑥 log 7
𝑑𝑦 3 4 = × 𝑑𝑥 log 7 4𝑥 =
3 𝑥 log 7
Example 5
Differentiate 𝑥 𝑙𝑛 𝑥 𝑑 1 (𝑥 ln 𝑥 ) = ln 𝑥 + 𝑥 × 𝑑𝑥 𝑥 = ln 𝑥 + 1
APPLICATIONS OF DIFFERENTIATION Tangents and Normals STEP 1: Find the y co-ordinate of the point STEP 2: Find the derivative 𝑓′(𝑥 ), and then find the gradient of the tangent or normal (Remember 𝑚 𝑇 × 𝑚𝑁 = −1) STEP 3: Find the equation of the line using point-gradient form
MATHEMATICS | PAGE 88 OF 121
Exponential and Logarithmic Functions Example 1
Find the normal to the curve 𝑦 = 𝑙𝑜𝑔 𝑥 at (1,0) STEP 1: Find the y co-ordinate of the point 𝑦=0 STEP 2: Find the derivative 𝑓′(𝑥 ), and then find the gradient of the tangent or normal 𝑑𝑦 1 = 𝑑𝑥 𝑥 At 𝑥 = 1,
𝑑𝑦 𝑑𝑥
=1
𝑚 𝑇 × 𝑚𝑁 = −1 𝑚𝑁 = −1 STEP 3: Find the equation of the line using point-gradient form Equation of normal: 𝑦 − 0 = −1(𝑥 − 1) 𝑦 = −𝑥 + 1 Example 2
Find the equation of the tangent to the curve 𝑦 = 𝑙𝑜𝑔 𝑥 at the point 𝑥 = 1. STEP 1: Find the y co-ordinate of the point 𝑦(1) = log 1 =0 So the coordinate is (1, 0) STEP 2: Find the derivative 𝑓′(𝑥 ), and then find the gradient of the tangent or normal 𝑦 = log 𝑥 1 𝑦′ = 𝑥 𝑦 ′ = 1 𝑓𝑜𝑟 𝑥 = 1 STEP 3: Find the equation of the line using point-gradient form 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ) 𝑦 − 0 = 1(𝑥 − 1) 𝑦=𝑥−1
MATHEMATICS | PAGE 89 OF 121
Exponential and Logarithmic Functions CURVE SKETCHING In sketching curve involving compound logarithmic functions, such as 𝑥 log 𝑥, you will need to understand which function ‘dominates’. For example, consider the function 𝑥 log 𝑥 a) What happens to the curve 𝑦 = 𝑥 as 𝑥 → 0 As 𝑥 → 0, 𝑦 → 0 b) What happens to the curve 𝑦 = 𝑙𝑜𝑔 𝑥 as 𝑥 → 0 As 𝑥→0, 𝑦→−∞ c) What happens to the curve 𝑦 = 𝑥 𝑙𝑜𝑔 𝑥, as 𝑥 → 0 To answer the last question, we must recognise that: 𝑥 dominates log 𝑥 𝑥 𝑘 dominates log 𝑥 This means that 𝑥 beats log 𝑥, or that the curve follows the behavior of 𝑥 in priority of log 𝑥. Hence lim 𝑥 log 𝑥 = 0
𝑥→0
Example
Consider the function 𝑦 =
𝑙𝑜𝑔 𝑥 𝑥
a) Find the domain of the function b) Find any vertical or horizontal asymptotes c) Find any stationary points and determine their nature d) Find the point of inflexion e) Sketch the curve, and hence find the range of the function. a) Domain: 𝑥 > 0 b) Vertical asymptote exists at 𝑥 = 0 Since 𝑥 dominates log 𝑥, 𝑦 =
log 𝑥 𝑥
1
behaves like 𝑦 = 𝑥 for large positive and negative values
of 𝑥 ∴Horizontal asymptote exists at 𝑦 = 0
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Exponential and Logarithmic Functions 1
c)
×𝑥−log 𝑥
𝑦′ = 𝑥
=
𝑥2
1 − log 𝑥 𝑥2
Stationary points occur when 𝑦′ = 0 1 − log 𝑥 =0 𝑥2 log 𝑥 = 1 𝑥=𝑒 1
When 𝑥 = 𝑒, 𝑦 = 𝑒
1
Stationary point occurs at (𝑒, 𝑒)
′′
𝑦 =
1 − 𝑥 × 𝑥 2 − 2𝑥(1 − log 𝑥 ) 𝑥4
=
2𝑥 log 𝑥 − 𝑥 − 2𝑥 𝑥4
=
2 log 𝑥 − 3 𝑥3
When 𝑥 = 𝑒, 𝑦 ′′ < 0 1
Relative maximum occurs at (𝑒, 𝑒) d) Point of inflexion occurs when 𝑦 ′′ = 0 2 log 𝑥 − 3 =0 𝑥3 3 2
log 𝑥 =
3
𝑥 = 𝑒2 3
3
3
When 𝑥 = 𝑒 2 , 𝑦 = 2 𝑒 −2 3
3
3
Point of inflexion occurs at (𝑒 2 , 2 𝑒 −2 ) e)
𝑦
1 (𝑒, ) 𝑒
(1,0)
𝑥
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Exponential and Logarithmic Functions INTEGRATION OF THE RECIPROCAL FUNCTION We previously defined: 𝑑 1 log 𝑥 = 𝑑𝑥 𝑥 So, integrating both sides gives: Integral of
𝟏 𝒙
1 ∫ 𝑑𝑥 = log 𝑥 𝑥 Since: 𝑑 𝑎 log(𝑎𝑥 + 𝑏) = 𝑑𝑥 𝑎𝑥 + 𝑏 Integrating both sides gives: log(𝑎𝑥 + 𝑏) = ∫
𝑎 𝑑𝑥 𝑎𝑥 + 𝑏
Divide both sides by 𝑎: 1 1 log(𝑎𝑥 + 𝑏) = ∫ 𝑑𝑥 (𝑎𝑥 + 𝑏) 𝑎 Integral of
𝟏 𝒂𝒙+𝒃
∫
1 1 𝑑𝑥 = log(𝑎𝑥 + 𝑏 ) (𝑎𝑥 + 𝑏 ) 𝑎
We know that: 𝑑 𝑓 ′ (𝑥 ) log 𝑓(𝑥) = 𝑑𝑥 𝑓 (𝑥 ) Integrating both sides, gives: log 𝑓 (𝑥 ) = ∫
𝑓 ′ (𝑥 ) 𝑑𝑥 𝑓 (𝑥 )
Standard Form of Integration
Example 1
∫
𝑓 ′ (𝑥) 𝑑𝑥 = log 𝑓 (𝑥) 𝑓 (𝑥) MATHEMATICS | PAGE 92 OF 121
Exponential and Logarithmic Functions 1 ∫ 𝑑𝑥 = 𝑥 = log 𝑥 + 𝑐 Example 2 ∫
√2 𝑑𝑥 = 3𝑥 + 1 1 √2 ∫ 𝑑𝑥 3 𝑥+1 1 = √2 × log(3𝑥 + 1) + 𝐶 3 √2 = log(3𝑥 + 1) + 𝐶 3 =
Example 3 a) Differentiate 𝑥 𝑙𝑜𝑔 𝑥 b) Hence, find ∫ 𝑙𝑜𝑔 𝑥 𝑒3
c) Evaluate ∫𝑒 2 𝑙𝑜𝑔 𝑥 𝑑𝑥 a)
𝑑 𝑑𝑥
1
(𝑥 log 𝑥 ) = log 𝑥 + 𝑥 × 𝑥 = log 𝑥 + 1
𝑑
b) ∫ 𝑑𝑥 (𝑥 log 𝑥 ) 𝑑𝑥 = ∫ log 𝑥 𝑑𝑥 + ∫ 1 𝑑𝑥 ∫ log 𝑥 𝑑𝑥 = ∫ 𝑑 (𝑥 log 𝑥 ) − ∫ 𝑑𝑥 = 𝑥 log 𝑥 − 𝑥 + 𝐶
3
c) = [𝑥 log 𝑥 − 𝑥 ]𝑒𝑒 2 = (𝑒 3 log 𝑒 3 − 𝑒 3 ) − (𝑒 2 log 𝑒 2 − 𝑒 2 ) = 3𝑒 3 log 𝑒 − 𝑒 3 − 2𝑒 2 log 𝑒 + 𝑒 2 = 3𝑒 3 − 𝑒 3 − 2𝑒 2 + 𝑒 2 = 2𝑒 3 − 𝑒 2 Example 4
Find the area bounded by the curve 𝑦 = 𝑥, the 𝑥 − 𝑎𝑥𝑖𝑠, and the lines 𝑥 = 1 and 𝑥 = 2 1
2
1 𝑑𝑥 1 𝑥 2 = [log 𝑥 ] 1 = log 2 − log 1 = log 2
𝐴𝑟𝑒𝑎 = ∫
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Exponential and Logarithmic Functions THE DERIVATIVE OF THE EXPONENTIAL FUNCTION DERIVATIVE OF 𝒆𝒙 To find the derivative of 𝑦 = 𝑒 𝑥 , following the steps below. Let 𝑦 = 𝑒 𝑥 Express 𝑦 = 𝑒 𝑥 in logarithmic form: 𝑥 = ln 𝑦 Differentiate 𝑥 with respect to 𝒚: 𝑑𝑥 1 = 𝑑𝑦 𝑦 Find
𝑑𝑦 𝑑𝑥
𝑑𝑦 =𝑦 𝑑𝑥 = 𝑒 𝑥 [𝑢𝑠𝑖𝑛𝑔 𝑦 = 𝑒 𝑥 ]
Derivative of 𝒆𝒙
𝑑 𝑥 𝑒 = 𝑒𝑥 𝑑𝑥 If 𝑦 = 𝑒 𝑎𝑥+𝑏 use the chain rule to find
𝑑𝑦 𝑑𝑥
:
Let 𝑢 = 𝑎𝑥 + 𝑏 𝑑𝑢 =𝑎 𝑑𝑥 So 𝑦 = 𝑒 𝑢 𝑑𝑦 = 𝑒𝑢 𝑑𝑢 Using the Chain rule: 𝑑𝑦 𝑑𝑦 𝑑𝑢 = × 𝑑𝑥 𝑑𝑢 𝑑𝑥 = 𝑒 𝑢 × 𝑎 = 𝑎𝑒 𝑎𝑥+𝑏
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Exponential and Logarithmic Functions
Derivative of 𝒆𝒂𝒙+𝒃
𝑑 𝑎𝑥+𝑏 𝑒 = 𝑎 ∙ 𝑒 𝑎𝑥+𝑏 𝑑𝑥
If 𝑦 = 𝑒 𝑓(𝑥) , use chain rule to find
𝑑𝑦 𝑑𝑥
Let 𝑢 = 𝑓(𝑥) 𝑑𝑢 = 𝑓 ′ (𝑥) 𝑑𝑥 So 𝑦 = 𝑒 𝑢 𝑑𝑦 = 𝑒𝑢 𝑑𝑢 Using the Chain rule: 𝑑𝑦 𝑑𝑦 𝑑𝑢 = × 𝑑𝑥 𝑑𝑢 𝑑𝑥 𝑢 = 𝑒 × 𝑓 ′ (𝑥 ) = 𝑓 ′ (𝑥 )𝑒 𝑓(𝑥)
Derivative of 𝒆𝒇(𝒙) 𝑑 𝑓(𝑥) 𝑒 = 𝑓 ′ (𝑥) ∙ 𝑒 𝑓(𝑥) 𝑑𝑥
Example 1 𝑑 3𝑥 𝑒 = 𝑑𝑥 = 3𝑒 3𝑥 Example 2 𝑑 𝑥 2−3𝑥 𝑒 = 𝑑𝑥 2 = (2𝑥 − 3)𝑒 𝑥 −3𝑥
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Exponential and Logarithmic Functions Example 3 𝑑 (1 + 𝑒 𝑥 ) 𝑑 𝑑𝑥 𝑥 (log(1 + 𝑒 )) = (1 + 𝑒 𝑥 ) 𝑑𝑥 𝑒𝑥 = 1 + 𝑒𝑥 Example 4 𝑑 𝑒 𝑥 − 𝑒 −𝑥 ( )= 𝑑𝑥 𝑒 𝑥 + 𝑒 −𝑥
(𝑒 𝑥 − (−1)𝑒 −𝑥 )(𝑒 𝑥 + 𝑒 −𝑥 ) − (𝑒 𝑥 + (−1)𝑒 −𝑥 )(𝑒 𝑥 − 𝑒 −𝑥 ) (𝑒 𝑥 + 𝑒 −𝑥 )2 𝑥 −𝑥 𝑥 −𝑥 (𝑒 + 𝑒 )(𝑒 + 𝑒 ) − (𝑒 𝑥 − 𝑒 −𝑥 )(𝑒 𝑥 − 𝑒 −𝑥 ) = (𝑒 𝑥 + 𝑒 −𝑥 )2 𝑒 2𝑥 + 2 + 𝑒 −2𝑥 − (𝑒 2𝑥 − 2 + 𝑒 −2𝑥 ) = (𝑒 𝑥 + 𝑒 −𝑥 )2 2𝑥 𝑒 + 2 + 𝑒 −2𝑥 − (𝑒 2𝑥 − 2 + 𝑒 −2𝑥 ) = (𝑒 𝑥 + 𝑒 −𝑥 )2 =
MATHEMATICS | PAGE 96 OF 121
Application of Calculus to the Physical World APPLICATION OF CALCULUS TO THE PHYSICAL WORLD OVERVIEW Once you grasp the following concepts, much of the wording of questions that come under this topic become easier to understand. It is important to understand these concepts thoroughly. Firstly, the idea of distance and displacement must be understood.
Displacement: the position of an object relative to a point (usually the origin) Distance: the space between an object and a point Distance, is different to displacement, because distance cannot be negative. However displacement can. If you imagine the origin as a point, and then a particle moving from it to the left 50 metres, we would say that the particle has been displaced by -50 metres. (Convention in mathematics: to the left normally means negative displacement). 𝑥 = Horizontal Displacement 𝑥̇ = Velocity 𝑥̈ = Acceleration
Displacement, Velocity and acceleration are all vector quantites and thus you need to state the direction of the moving particle.
MATHEMATICS | PAGE 97 OF 121
Application of Calculus to the Physical World EXAMPLES Question 1
The displacement of a particle which moves along the x axis is given by 𝑥 = 4𝑡 2 + 6𝑡 + 5 i) Find where the particle is initially ii) Find an expression for its velocity at time t iii) Find its velocity after 2 seconds iv) Find at what time the particle is stationary Answers i) To find where the particle is initially, we have to find it when it hasn’t even moved, yet, when nothing's happened, at the very beginning of everything. Thus, to find where the particle is initially, we substitute the value 𝑡 = 0 because that's the beginning, thus the particle's initial position. Thus whenever the words initial appear, you now know it means when t (time) = 0 𝑥 = 4𝑡 2 + 6𝑡 + 5 when 𝑡 = 0 𝑥 = 4 (0 ) 2 + 6 (0 ) + 5 𝑥 = 5 Thus, the particle's initial position (when 𝑡 = 0) is 𝑥 = 5
ii) Velocity if we recall, is just the derivative of x. Therefore 𝑥 = 4𝑡 2 + 6𝑡 + 5
iii) To find the velocity after 2 seconds, we just plug in 𝑡 = 2 into the equation of the velocity: 𝑣 = 8𝑡 + 6 when 𝑡 = 2 𝑣 = 8 (2 ) + 6 𝑣 = 22 Thus the velocity, after 2 seconds, is 22 units/s (When we state velocities, we use the notation units/s and for acceleration, we use units/s.)
MATHEMATICS | PAGE 98 OF 121
Application of Calculus to the Physical World iv) To find the time when the particle is stationary, we have to find the time when the velocity is 0. When the particle is stationary, it’s not moving. The original equation 𝑥 = 4𝑡 2 + 6𝑡 + 5 only tells us how far the particle moves, or its displacement. The derivative of it tells us how fast the particle is moving, which is the velocity. But if the particle has stopped moving, then obviously the speed it'll be going at is 0. That will be the velocity. Think of it like a car, if you want to know when the car's stopped, that'll be when it’s got no speed, or it's going at 0km/h. That's why we use when 𝑣 = 0: So when 𝑣 = 0 𝑣 = 8𝑡 + 6 = 0 Thus 8𝑡 = − 6 𝑡 = −
6 8
Therefore, the particle is stationary when the velocity = 0, which is at 𝑡 = −
6 8
Question 2
The acceleration of a particle is given by 𝑎 = 2𝑡 − 6. When 𝑡 = 6, the particle is stationary. Find its velocity when 𝑡 = 2. Answers We know that the derivative of X will give us V. And we know that the derivative of V gives us A (where X is displacement, V is velocity and A is acceleration). So therefore, the integral of A should give us V. And the integral of V will give us X! So if we know that A = something, then the integral of that something will give us V 𝑎 = 2𝑡 – 6 Therefore 𝑽 = ∫ 𝒂 𝒅𝒙 Which is the same as 𝑽 = ∫ 𝟐𝒕 − 𝟔 𝒅𝒙, Therefore 𝑣 =
(2𝑡 2 ) 2
− 6𝑡 + 𝐶 where 𝐶
is a constant Simplifying, we get 𝑣 = 𝑡 2 + 6𝑡 + 𝐶 where 𝐶 is a constant Thus, we now have a constant to deal with, because the integral we solved was an INDEFINITE integral. Using the information we have of course - When 𝑡 = 6, the particle is stationary. That's the same as saying when 𝑡 = 6, the velocity = 0 𝑣 = 𝑡 2 − 6𝑡 + 𝐶 When 𝑡 = 6, 𝑣 = 0 0 = ((6)2) − 6(6) + 𝐶 Therefore 0 = 0 + 𝐶 Therefore 𝐶, the constant = 0! So now we know that the Velocity equation is actually: 𝑣 = 𝑡 2 − 6𝑡 + 0 So we can now use the equation to figure out what the velocity is when 𝑡 = 2. Therefore, plugging it in, we get the velocity being = –56 units per second.
MATHEMATICS | PAGE 99 OF 121
Probability PROBABILITY OVERVIEW Probability is the study of how likely an event will occur.
MUTUALLY EXCLUSIVE EVENTS Mutually exclusive events are where if one event is to occur, another cannot. For example, when rolling a die, a 6 cannot occur at the same time as a 2. If 𝑃(𝐸 ) is the probability of an event occurring, then: 𝑃 (𝐸 ) =
no. of favourable outcomes total no. of outcomes
If 𝑃(𝐸 ) = 0, then the event is impossible. If 𝑃(𝐸 ) = 1, then the event is certain to happen 0 ≤ 𝑃 (𝐸 ) ≤ 1 The sum of all mutually exclusive probabilities is 1.
EXAMPLES Example 1
A container holds 5 blue, 3 white and 7 yellow marbles. If one marble is selected at random, find the probability of getting a white marble. Solution 𝑃 (𝑊 ) =
3 1 = 15 5
Example 2
The probability that a traffic light will turn green as a car approaches it is estimated to be 5 . A taxi goes through 192 intersections where there are traffic lights. How many of these 12
would be expected to turn green as the taxi approached? Solution It is expected that
5 of the traffic lights would turn green. 12
5 × 192 = 80 12 ∴ it would be expected that 80 traffic lights turn green as the taxi approaches them.
MATHEMATICS | PAGE 100 OF 121
Probability COMPLEMENTARY EVENTS The complement of an event happening is the event not happening, i.e. the complement of 𝑃(𝐸 ) is 𝑃 (not 𝐸 ). This is written as 𝑃(𝐸́ ). 𝑃 (𝐸 ) + 𝑃(𝐸́ ) = 1 Proof: Let 𝑒 be the number of ways 𝐸 can happen out of a total of 𝑛 events. Then the number of ways 𝐸 will not happen is 𝑛 − 𝑒. 𝑒 𝑃 (𝐸 ) = 𝑛 𝑃(𝐸́ ) =
𝑛−𝑒 𝑛
=
𝑛 𝑒 − 𝑛 𝑛
=1−
𝑒 𝑛
= 1 − 𝑃 (𝐸 )
EXAMPLES Example 1
The probability of a win in a raffle is
1 . What is the probability of losing? 350
Solution 𝑃(lose) = 1 − 𝑃(win) =1− =
1 350
349 350
Example 2
The probability of a tree surviving a fire is 72%. Find the probability failing to survive a fire. Solution 𝑃(fail) = 1 − 𝑃(survive) = 100% − 72% = 28%
MATHEMATICS | PAGE 101 OF 121
Probability NON-MUTUALLY EXCLUSIVE EVENTS Non-mutually exclusive events occur where more than one event occurs at the same time. Venn diagrams prove useful when this occurs. In addition, a formula can be used for these events: 𝑃(𝐴 or 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 and B)
EXAMPLES Example 1
In a class of 32 students, 18 do music and 16 do sport. Find the probability that if a student is selected at random, that: a) They do sport Solution 18 + 16 = 34 There are only 32 students in the class. ∴ 2 students do both music and sport. M
S
16
𝑃 (S ) = =
2
32
14
16 32 1 2
b) They do music only Solution
c) They do both Solution
𝑃(M only) = =
16 32 1 2
𝑃(both) =
2 32
=
1 16
MATHEMATICS | PAGE 102 OF 121
Probability Example 2
One card is drawn from a set of cards numbered 1 to 10. Find the probability of drawing out an odd number or a multiple of 3. Solution The odd cards are 1, 3, 5, 7 and 9. The multiples of 3 are 3, 6 and 9. The numbers of 3 and 9 are both odd and multiples of 3. So there are 6 numbers that are odd or multiples of 3: 1, 3, 5, 6, 7 and 9 8
odd
multiples of 3
3
5 1 7
2
6
9
4
10
∴ the probability of drawing out an odd number or a multiple of 3 will be all the numbers in the circles. 𝑃(odd or multiple of 3) = =
6 10 3 5
Example 3
From 100 cards, numbered from 1 to 100, one is selected at random. Find the probability that the card selected is even or less than 20. Solution Some cards are both even and less than 20 (i.e. 2, 4, 6, 8, 10, 12, 14, 16, 18). 𝑃(even and < 20) =
9 100
𝑃(even) =
50 100
19 100 𝑃(even or < 20) = 𝑃(even) + 𝑃(< 20) − 𝑃(even and < 20) 𝑃(< 20) =
=
50 19 9 + − 100 100 100
=
60 100
=
3 5
MATHEMATICS | PAGE 103 OF 121
Probability MULTI-STAGE EVENTS Each possible outcome (sample space) is represented in tables and tree diagrams. Two methods of writing the possible outcomes:
Table –
List all the outcomes
Probability Trees –
Shows the probability of each event
Two general rules:
“and” means × (chances are decreased when more than one condition is satisfied)
“or” means + (chances are increased when one or the other condition can be satisfied)
This means that: 𝑃(𝐴 and 𝐵) = 𝑃 (𝐴) ∙ 𝑃(𝐵) 𝑃(𝐴 or 𝐵) = 𝑃(𝐴) + 𝑃(𝐵)
EXAMPLES Example 1
Find the sample space and the probability of each outcome for each question by using a table or tree diagram. Solution Using a table gives:
Using a tree diagram gives:
H
T
H
HH
HT
T
TH
TT
H H T
H T T Since there are four possible outcomes (HH, HT, TH, TT), each outcome has a probability of Remember that each outcome when tossing 1 coin is Notice that
1 . 4
1 . 2
1 1 1 × = . 2 2 4
MATHEMATICS | PAGE 104 OF 121
Probability Example 2
The probability of a missile hitting a target is
7 . Find the probability that two missiles fired 8
will: a) Hit two targets
b) Miss both targets
Solution
Solution
𝑃(H & H) = =
7 7 × 8 8
𝑃 (M & M) =
49 64
=
1 1 × 8 8 1 64
Example 3
Find the probability of the upper most face being a 3 or 4 if a die is rolled. Solution 𝑃(3 or 4) = 𝑃 (3) + 𝑃 (4) =
1 1 + 6 6
=
1 3
Example 4
Bob buys 5 tickets in a raffle in which 95 tickets are sold. There are 2 prizes to be won. Find the probability that he wins: a) Both prizes
b) At least one prize
Solution
Solution
𝑃(W & W) =
5 4 × 95 94
𝑃(at least one W) = 1 − 𝑃(L & L) As one raffle has
20 = 8930
been taken out, the
2 = 893
probability has decreased
=1−( =
90 89 × ) 95 94
92 893
MATHEMATICS | PAGE 105 OF 121
Probability Example 5
If 2 coins are tossed, find the probability of tossing a head and a tail. Solution 𝑃(H and T) = 𝑃 (H & T) or 𝑃(T & H) 1
1
1
1
= (2 × 2) + (2 × 2) 1
1
=4+4 1
=2 Example 6
A person has probability of 0.2 of winning a prize in a competition. If he enters 3 competitions, find the probability of his winning 2 competitions. Solution
Probability of losing is 1 − 0.2 = 0.8 𝑃(2W) = 𝑃(W & W & L) + 𝑃(W & L & W) + 𝑃(L & W & W) = (0.2 × 0.2 × 0.8) + (0.2 × 0.8 × 0.2) + (0.8 × 0.2 × 0.2) = 0.096
MATHEMATICS | PAGE 106 OF 121
Series SERIES OVERVIEW A series (also known as a progression) is a sequence of numbers which follow a certain pattern. There are two types of series: arithmetic and geometric.
ARITHMETIC SERIES In an arithmetic series, each term is a constant amount greater or less than the previous term. This constant is called the common difference. If 𝑇1 , 𝑇2 and 𝑇3 are consecutive terms of an arithmetic series (AP), then: 𝑑 = 𝑇2 − 𝑇1 = 𝑇3 − 𝑇2 To calculate the value of a particular term within an AP, the following formula can be used: 𝑇𝑛 = 𝑎 + (𝑛 − 1)𝑑 Where: 𝑎 = first term in the series 𝑛 = position of the term in the series, i.e. the 𝑛th term 𝑑 = common difference To calculate the sum of an arithmetic series, two formulas can be used. If the last term (i.e. 𝑇𝑛 ) is known, 𝑆𝑛 =
𝑛 (𝑎 + 𝑙 ) 2
Where: 𝑙 = last or 𝑛th term If the last term is unknown, 𝑆𝑛 =
𝑛 [2𝑎 + (𝑛 − 1)𝑑 ] 2
MATHEMATICS | PAGE 107 OF 121
Series EXAMPLES Example 1
If 5 + 𝑥 + 31 + ⋯ is an arithmetic series, find 𝑥. Solution For an AP: 𝑇2 − 𝑇1 = 𝑇3 − 𝑇2 i.e. 𝑥 − 5 = 31 − 𝑥 2𝑥 − 5 = 31 2𝑥 = 36 ∴ 𝑥 = 18 Example 2 a) Evaluate 𝑘 if (𝑘 + 2) + (3𝑘 + 2) + (6𝑘 − 1) + ⋯ is an arithmetic series. Solution For an arithmetic series, 𝑇2 − 𝑇1 = 𝑇3 − 𝑇2 So 3𝑘 + 2 − (𝑘 + 2) = (6𝑘 − 1) − (3𝑘 + 2) 3𝑘 + 2 − 𝑘 − 2 = 6𝑘 − 1 − 3𝑘 − 2 2𝑘 = 3𝑘 − 3 0=𝑘−3 3=𝑘 b) Write down the first 3 terms of the AP. Solution Substituting 𝑘 = 3: 𝑇1 = 𝑘 + 2 =3+2 =5 𝑇2 = 3𝑘 + 2 = 3×3+2 = 11
𝑇3 = 6𝑘 − 1 = 6×3−1 = 17
c) Find the common difference 𝑑 . Solution 𝑑 = 𝑇2 − 𝑇1 = 𝑇3 − 𝑇2 = 11 − 5 = 17 − 11 =6
MATHEMATICS | PAGE 108 OF 121
Series Example 3
The 5th term of an arithmetic progression is 37 and the 8th term is 55. Find the common difference and the first term of the series. Solution ∴ 37 − 4𝑑 = 55 − 7𝑑
𝑇𝑛 = 𝑎 + (𝑛 − 1)𝑑 𝑇5 = 37
3𝑑 = 18
i.e. 𝑎 + (5 − 1)𝑑 = 37
𝑑=6 ∴ 𝑎 = 37 − 4(6) = 37 − 24 = 13
𝑎 + 4𝑑 = 37 𝑎 = 37 − 4𝑑 𝑇9 = 55
∴ 𝑑 = 6, 𝑎 = 13
i.e. 𝑎 + (8 − 1)𝑑 = 55 𝑎 + 7𝑑 = 55 𝑎 = 55 − 87 Example 4
Evaluate 9 + 14 + 19 + ⋯ + 224. Solution 𝑎 = 9, 𝑑 = 5
𝑛 (𝑎 + 𝑙 ) 2 44 (9 + 224) = 2 = 22 × 233 = 5126
∴ 𝑆𝑛 = 𝑆44
Example 5
The 6th term of an AP is 23 and the sum of the first 10 terms is 210. Find the sum of 20 terms. Solution 𝑇6 = 23
∴ 2(23 − 5𝑑 ) + 9𝑑 = 42
∴ 𝑎 + (6 − 1)𝑑 = 23
46 − 10𝑑 + 9𝑑 = 42 𝑑 = 46 − 42
𝑎 + 5𝑑 = 23 𝑎 = 23 − 5𝑑 𝑆10 = 210 ∴
10 [2𝑎 + (10 − 1)𝑑 ] = 210 2 5(2𝑎 + 9𝑑 ) = 210
=4 ∴ 𝑎 = 23 − 5(4) =3
2𝑎 + 9𝑑 = 42
MATHEMATICS | PAGE 109 OF 121
Series 20 [2(3) + (20 − 1)4] 2 = 10[6 + 19(4)]
𝑆20 =
= 10 × 82 = 820 Example 6
A stack of cans on a display at a supermarket has 5 cans on the top row. The next row down has 2 more cans and the next one has 2 more cans and so on. a) Calculate the number of cans in the 11 th row down. Solution 𝑎 = 5, 𝑑 = 2 𝑇11 = 5 + (11 − 1)2 = 5 + 2(10) = 25 b) If there are 320 cans altogether, how many rows are there? Solution 𝑆𝑛 = 320 𝑛 = [ 2(5) + (𝑛 − 1 )2] 2 𝑛(10 + 2𝑛 − 2) = 640 2𝑛2 + 8𝑛 = 640 𝑛2 + 4𝑛 − 320 = 0 (𝑛 − 16)(𝑛 + 2) = 0 𝑛 − 16 = 0, 𝑛 + 20 = 0 𝑛 = 16, − 20 Because 𝑛 must be a positive integer, 𝑛 = 16. ∴ 16 rows.
MATHEMATICS | PAGE 110 OF 121
Series GEOMETRIC SERIES In a geometric series, each term is formed by multiplying the previous term by a constant. This constant is known as the common ratio. If 𝑇1 , 𝑇2 and 𝑇3 are consecutive terms of a geometric series, then: 𝑟=
𝑇2 𝑇3 = 𝑇1 𝑇2
To calculate the value of a particular term in a GP, the following formula can be used: 𝑇𝑛 = 𝑎𝑟 𝑛−1 Where: 𝑎 = first term in the series 𝑛 = position of the term in the series, i.e. the 𝑛th term 𝑟 = common ratio To calculate the sum of a geometric series, two formulas can be used depending on the value of 𝑟: 𝑆𝑛 =
𝑎(𝑟 𝑛 − 1) for |𝑟| > 1 𝑟−1
𝑆𝑛 =
𝑎(1 − 𝑟 𝑛 ) for |𝑟| < 1 1−𝑟
When |𝑟| < 1, there is a limiting sum, i.e. the sum of the series will eventually reach a limit. This formula can be used to find the limit of a series: 𝑆∞ =
𝑎 where |𝑟| < 1 1−𝑟
EXAMPLES Example 1
Find 𝑥 if 5 + 𝑥 + 45 + ⋯ is a geometric series. Solution 𝑟=
𝑇2 𝑇3 = 𝑇1 𝑇2
𝑥 45 ∴ = 5 𝑥
𝑥 = ±√225 = ±15 If 𝑥 = 15, the series is 5 + 15 + 45 + ⋯ where 𝑟 = 3. If 𝑥 = −15, the series is 5 − 15 + 45 − ⋯ where 𝑟 = −3.
MATHEMATICS | PAGE 111 OF 121
Series Example 2
Find the 10th term of the series −5 + 10 − 20 + ⋯. Solution 𝑎 = −5, 𝑟 = −2, 𝑛 = 10 𝑇𝑛 = 𝑎𝑟 𝑛−1
𝑇10 = −5(−2)10−1 = −5(−2)9 = −5(−512) = 2160
Example 3
Find the first value for 𝑛 for which the terms of the series
1 +1 + 5 + ⋯ exceed 3000. 5
Solution 1 𝑎 = ,𝑟 = 5 5 When 𝑇𝑛 > 3000 𝑎𝑟 𝑛−1 > 3000 1 𝑛−1 (5 ) > 3000 5 5𝑛−1 > 15000
log 15000 log 5 log 15000 𝑛> +1 log 5 𝑛 > 6.97
𝑛−1 >
∴𝑛=7 ∴ the 7th term will be the first to exceed 3000.
log 5𝑛−1 > log 15000 (𝑛 − 1) log 5 > log 15000 Example 4
Find the sum of the first 10 terms of the series 3 + 12 + 48 + ⋯. Solution 𝑎=3 𝑟=
12 48 = 3 12
=4 𝑛 = 10
|𝑟 | > 1 3(410 − 1) 4−1 = 410 − 1
∴ 𝑆10 =
= 1 048 575
MATHEMATICS | PAGE 112 OF 121
Series Example 5
The sum of 𝑛 terms of 1 + 4 + 16 + ⋯ is 21,845. Find the value of 𝑛. Solution 𝑎=1 𝑟=
∴
4 16 = 1 4
=4 𝑆𝑛 = 21 845
1 (4𝑛 − 1) = 21 845 4−1 𝑛 4 −1 = 21 845 3 𝑛 4 − 1 = 65 535 4𝑛 = 65 536
|𝑟 | > 1
𝑛 = log 4 65 536 ∴𝑛=8
Example 6
Find the sum of the series 6 + 2 +
2 + ⋯. 3
Solution 𝑎=6 𝑟= =
2 2 3 = 6 2 1 3
|𝑟| < 1, ∴ limiting sum exists 𝑆∞ = =
6 1 1−3 6 2 3
=6×
3 2
=9
MATHEMATICS | PAGE 113 OF 121
Series Example 7
A layer of tinting for a car window lets in 95% of light. How many layers will let in 40% of light? Solution 𝑎 = 0.95, 𝑟 = 0.95 When 𝑇𝑛 = 0.4,
0.95(0.95)𝑛−1 = 0.4 0.95𝑛 = 0.4 𝑛 = log 0.95 0.4 ≈ 17.9
∴ approximately 18 layers of tinting will let in 40% of light. Example 8
Write 0. 5̇ as a fraction. Solution 0. 5̇ = 0.555555 … 5 5 5 = + + +⋯ 10 100 1000 This is a GP with: 5 1 1 = ,𝑟 = 10 2 10 1 2 𝑆∞ = 1 1 − 10 1 = 2 9 10 5 = 9 𝑎=
MATHEMATICS | PAGE 114 OF 121
Series Example 9
A ball is dropped from a height of 1 metre and bounces up to
1 of its height on each 3
bounce until it eventually reaches the ground. What is the total vertical distance it travels? Solution The first distance is 1m because it is dropped from that height. From there onwards, when the ball bounces up, it also comes back down, meaning it travels the same distance twice. 1 1 m, therefore 𝑎 = . 3 3 1 1 The difference between the distance of each bounce is of the previous, therefore 𝑟 = . 3 3 The first bounce up is a distance of
As a result: 1 1 1 1 1 1 Total Distance = 1 + 2 [ + ( × ) + ( × × ) + ⋯ ] 3 3 3 3 3 3 1 1 1 = 1 + 2( + + + ⋯) 3 9 27 |𝑟| < 1, ∴ limiting sum exists. 𝑆∞ =
1 3
1 1−3
1 =3 2 3 =
1 2
1 ∴ Total Distance = 1 + 2 ( ) 2 =1+1 =2
MATHEMATICS | PAGE 115 OF 121
Series SIGMA NOTATION Sigma notation is a way of representing the sum of a series. Sigma notation is generally expressed as: 𝑞
∑ 𝑓 (𝑛 ) 𝑛=𝑝
Where:
𝑝 = first term in the series 𝑞 = last term in the series 𝑓 (𝑛) = the formula for a term in the series
The series 𝑓 (𝑛) will be in the following format: 𝑞
∑ 𝑓(𝑛) = 𝑓 (𝑝) + 𝑓 (𝑝 + 1) + 𝑓 (𝑝 + 2) + ⋯ + 𝑓 (𝑞 ) 𝑛=𝑝
The number of terms in the series is: 𝑝−𝑞+1
EXAMPLES Example 1 5 2 Evaluate ∑ 𝑟 𝑟=1
Solution 5
∑ 𝑟 2 = 12 + 22 + 32 + 42 + 5 2 𝑟=1
= 1 + 4 + 9 + 16 + 25 = 55 Example 2 7
Evaluate ∑(2𝑛 + 5) 3
Solution 7
∑(2𝑛 + 5) = (2 × 3 + 5) + (2 × 4 + 5) + (2 × 5 + 5) + (2 × 6 + 5) + (2 × 7 + 5) 3
= 11 + 13 + 15 + 17 + 19 = 75
MATHEMATICS | PAGE 116 OF 121
Series Example 3
Write 7 + 11 + 15 + ⋯ + (4𝑘 + 3) in sigma notation. Solution Formula is 4𝑛 + 3 where the last term is at 𝑛 = 𝑘. 𝑎=7 ∴ 4𝑛 + 3 = 7 4𝑛 = 4 𝑛=1 𝑘
∴ ∑(4𝑛 + 3) 𝑛=1
Example 4 100
Find the number of terms in the series ∑(3𝑛 − 7) 1
Solution number of terms = 𝑝 − 𝑞 + 1 = 100 − 1 + 1 = 100
MATHEMATICS | PAGE 117 OF 121
Series Example 5
Write the series 1 +
1 1 1 + +⋯+ in sigma notation. 2 4 512
Solution 1 1 1 1 + + + ⋯+ 2 4 512 Series is a GP 1 1 ∴𝑟= 2= 4 1 1 2 1 = 2 𝑇𝑙 =
1 512
1 𝑛−1 1 ∴ 1×( ) = 2 512 1 =𝑛−1 512 2
log 1
𝑛 = log0.5
1 +1 512
= 10 10
1 ∴ ∑( ) 2
9
𝑛−1
=∑
𝑛=1
𝑛=0
1 2𝑛 9
∴ 1+
1 1 1 1 + +⋯+ =∑ 𝑛 2 4 512 2 𝑛=0
Example 6
Write 1 + 8 + 27 + 64 + 125 in sigma notation. Solution 1 + 8 + 27 + 64 + 125 Now
1 = 13 8 = 23 27 = 33 64 = 43 125 = 53 5
∴ 1 + 8 + 27 + 64 + 125 = ∑ 𝑛3 𝑛=1
MATHEMATICS | PAGE 118 OF 121
Series COMPOUND INTEREST Compound interest is where the interest on an investment is calculated from the original interest as well as the interest that may have already been accumulated. It is a form of geometric series. In general, the compound interest formula is: 𝐴 = 𝑃 (1 + 𝑟 )𝑛 Where: 𝐴 = final amount 𝑃 = principal (initial investment) 𝑟 = interest rate as decimal 𝑛 = number of time periods
EXAMPLE Find the amount that will be in the bank after 6 years if $2000 is invested at 12% p.a. with interest paid yearly and monthly. Solution 𝑃 = 2000 For yearly: 𝑟 = 12% = 0.12 𝑛=6 𝐴 = 𝑃 (1 + 𝑟 )𝑛 = 2000(1 + 0.12)6 = 2000(1.12)6 = 3947.65 For monthly: 𝑟 must be divided in to the rate for each month, i.e. divide by 12. 𝑟 = 0.12 ÷ 12 = 0.01 In addition, the number of time periods must change because interest is paid 12 times each year. ∴ 𝑛 = 6 × 12 = 72 months 𝐴 = 2000(1 + 0.01)72 = 2000(1.01)72 = 4094.20
MATHEMATICS | PAGE 119 OF 121
Series SUPERANNUATION Superannuation attracts interest when an initial investment is put in, however it is different to compound interest. Each time money is invested, it attracts its own interest separate from the other payments that are already there. It is important to take note of when the money is invested, usually the beginning or end of a time period. The compound interest formula is used.
EXAMPLES Example 1
A sum of $1500 is invested at the beginning of each year in a superannuation fund. If interest is paid at 6% p.a., how much money will be available at the end of 25 years? Solution 𝑃 = 1500, 𝑟 = 6% = 0.06 𝐴 = 𝑃 (1 + 𝑟 )𝑛 = 1500(1 + 0.06)𝑛 = 1500(1.06)𝑛 At the end of the first year: 𝐴1 = 1500(1.06) At the end of the second year: 𝐴2 = [𝐴1 ](1.06) + 1500(1.06) = 1500(1.06)2 + 1500(1.06) At the end of the third year: 𝐴3 = [𝐴2 ](1.06) + 1500(1.06) = 1500(1.06)3 + 1500(1.06)2 + 1500(1.06) = 1500(1.06 + 1.062 + 1.063 ) And so this continues on to create this formula: 𝐴𝑛 = 1500(1.06 + 1.062 + 1.063 + ⋯ + 1.06𝑛 ) ∴ 𝐴25 = 1500(1.06 + 1.062 + 1.063 + ⋯ + 1.0625 ) Now (1.06 + 1.062 + 1.063 + ⋯ + 1.0625 ) is a geometric series with 𝑎 = 1.06, 𝑟 = 1.06 and 𝑛 = 25. 1.06(1.0625 − 1) 1.06 − 1 1.06(1.0625 − 1) ] = 1500 [ 0.06
∴ 𝑆24 = ∴ 𝐴25
≈ 87 234.57 ∴ at the end of 25 years, there will be $87 234.57
MATHEMATICS | PAGE 120 OF 121
Series Example 2
An amount of $50 is put into an investment account at the end of each month. If interest is paid at 12% p.a. paid monthly, how much is in the account at the end of 10 years? Solution 𝑃 = 50 𝑟 = 12% ÷ 12 = 0.01 𝑛 = 12 × 10 = 120 𝐴1 = 50 𝐴2 = [𝐴1 ](1 + 0.01) + 50 = 50(1.01) + 50 𝐴3 = [𝐴2 ](1.01) + 50 = 50(1.01)2 + 50(1.01) + 50 = 50(1 + 1.01 + 1.012 ) ∙ ∙ ∙ ∴ 𝐴𝑛 = 50(1 + 1.01 + 1.012 + ⋯ + 1.01𝑛−1 ) ∴ 𝐴120 = 50(1 + 1.01 + 1.012 + ⋯ + 1.01119 ) 𝑆120 = =
1(1.01120 − 1) 1.01 − 1 1.01120 − 1 0.01
∴ 𝐴120 = 50 (
1.01120 − 1 ) 0.01
= 11501.93 ∴ after 10 years, there is $11 501.93
MATHEMATICS | PAGE 121 OF 121
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