SCIENCEONTHEWEB.NETtanzania.scienceontheweb.net/privatecandidates/files/MATHEMATIC… · FORM SIX, AND FOUR 2015 You downloaded free ….from education blog tanzania TABLE OF CONTENTS

FORM SIX, AND FOUR 2015

You downloaded free ….from education blog tanzania

TABLE OF CONTENTS

1. CHAPTER ONE: STATISTICS ................................................................................. 1 1.1 What is Statistics? ..................................................................................................... 1 1.2 Population and Sample ............................................................................................. 1

1.3 Measure of Central Tendency ................................................................................... 1 1.4 Measure of Dispersion .............................................................................................. 2

1.5 Frequency Distribution ............................................................................................. 3 1.5.1 The Mean for Frequency Distribution ............................................................... 4

1.5.2 The Population Standard Deviation For Frequency Table ................................ 4 1.6 Cumulative Frequency .......................................................................................... 4

1.7 Calculation Of Mean And Standard Deviation From Grouped Data ...................... 5 1.8 The Coding Method-Assumed Mean ........................................................................ 6

Exercise: ...................................................................................................................... 7 2. CHAPTER TWO: TRIGONOMETRY ...................................................................... 8

2.1 What Is Trigonometry? ............................................................................................. 8 2.1.1 Measuring Angles .............................................................................................. 8

2.1.2 Conversion between Revolutions, Degrees, and Radians .................................. 8 2.1.3 The Coordinate Plane ......................................................................................... 9

2.1.4 Quadrants ........................................................................................................... 9 2.2 Trigonometric Ratios/Functions ............................................................................. 10

2.2.1 Periodic Functions ........................................................................................... 15 2.2.2 Reference Angles ............................................................................................. 15

2.2.3 Graphs of Trigonometric Functions ................................................................. 17 2.3 Trigonometric Identities.......................................................................................... 20

2.3.1 Eight Fundamental Trigonometric Identities ................................................... 20 2.3.2 Negative Angle Identities ................................................................................ 21

2.4 Trigonometric Equations ........................................................................................ 22 2.4.1 Solving Trigonometric Equations .................................................................... 22

2.6 Compound Angle Formulae .................................................................................... 25 2.6.1 Addition and Subtraction Formulas ................................................................. 25

2.6.2 Double and Half-Angle Formulas .................................................................... 26 2.6.3 Products of Functions ...................................................................................... 27

2.6.4 Factor Theorem ................................................................................................ 27 2.7 Solving Oblique Triangles ...................................................................................... 28

2.7.1 The Law of Sines ............................................................................................. 28 2.7.2 The Law of Cosines ......................................................................................... 29

3. CHAPTER THREE: VECTOR ALGEBRA ............................................................ 30 3.1 Basic Vector Definitions ......................................................................................... 30

3.2 Representations of Vectors ..................................................................................... 31 3.2.1 Vectors as directed line segments .................................................................... 31

3.2.2 Vector notation................................................................................................. 31 3.2.3 Position vectors ................................................................................................ 31

3.2.4 Equality of vectors ........................................................................................... 32 3.3 Composition of vectors ........................................................................................... 32

3.3.1 Introduction ...................................................................................................... 32

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3.3.2 The Parallelogram Rule For Addition .............................................................. 33

3.3.3 The Triangle Rule For Addition ...................................................................... 33 3.3.4 Adding a vector to its negative ........................................................................ 35

3.3.5 Subtraction of vectors ...................................................................................... 35 3.3.6 The Head-Minus-Tail Rule .............................................................................. 36

3.3.7 Multiplication By A Scalar .............................................................................. 37 3.3.8 When Are Vectors Parallel?............................................................................. 37

3.3.9 Unit Vectors ..................................................................................................... 38 3.4 Magnitude and Direction of Vectors....................................................................... 38

3.5 Basic Vector Operations ......................................................................................... 39 3.5.1 The Scalar Product ........................................................................................... 39

3.5.2 The Vector Product .......................................................................................... 40 3.6 Vectors Applications ............................................................................................... 42

3.7 The Ratio Theorem ................................................................................................. 43 3.7.1 Internal division of a line segment ................................................................... 43

3.7.2 External division of a line segment .................................................................. 44 3.8 Lines and Planes ..................................................................................................... 45

3.8.1 Finding the vector equation of a line ............................................................... 45 3.8.2 Vector Equations of Planes .............................................................................. 47

3.8.3 Finding equations of planes using normal vectors ........................................... 48 3.8.4 Finding a normal vector ................................................................................... 50

3.4.5 Equations of planes in the form Ax + By + Cz = D......................................... 50 Exercise ..................................................................................................................... 53

3.9 Distance From a Point To a Line ............................................................................ 54 3.9.1 Shortest Distance From a Point To a Line ....................................................... 55

3.10 Distance Between Two Lines ............................................................................... 55 3.11 The angle between 2 lines ..................................................................................... 56

3.12 The angle between 2 planes .................................................................................. 56 3.13 The angle between a line and a plane ................................................................... 57

4. CHAPTER FOUR: DIFFERENTIATION .............................................................. 58 4.1 The Definition of the Derivative ............................................................................. 58

4.2 Alternate Notation ................................................................................................... 61 4.2 Differentiation Formulas ......................................................................................... 61

4.3 Product and Quotient Rule ...................................................................................... 65 4.3.1 Product Rule..................................................................................................... 65

5.3.2 Quotient Rule ................................................................................................... 66 4.4 Derivatives of Trig Functions ................................................................................. 67

4.5 Derivatives of Exponential and Logarithm Functions ............................................ 71 4.5.1 Exponential Functions ..................................................................................... 71

4.5.2 Logarithm Functions ........................................................................................ 72 4.6 Derivatives of Inverse Trig Functions .................................................................... 75

4.6.1 Alternate Notation ............................................................................................ 79 4.7 Chain Rule .............................................................................................................. 79

4.8 Implicit Differentiation ........................................................................................... 81 4.9 Related Rates .......................................................................................................... 83

4.10 Logarithmic Differentiation .................................................................................. 85

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1. CHAPTER ONE: STATISTICS 1.1 What is Statistics? Statistics can be defined as the science of collecting, organizing, presenting and analyzing numerical data for the purposes of description and decision marking. 1.2 Population and Sample Population The total body of information that we ca be concerned about in any given situation is called population. Sample A sample is a selection of items taken from a population, that is, a subset of a population. The reason for sampling is clear: a population, typically, is very large, so that the statistician cannot possibly hope to handle all the information about it, but must be content with looking only at some of the information (a sample).

1.3 Measure of Central Tendency Mean This is what people usually intend when they say "average" which is found by adding the numbers and dividing by the size of the population.

Where x is an entry and N the size of the population.

Mode

The mode is the most frequent data value. There may be no mode if no one value appears more than any other. There may also be two modes (bimodal), three modes (trimodal), or more than three modes (multi-modal). For grouped frequency distributions, the modal class is the class with the largest frequency.

Median

The data must be ranked (sorted in ascending order) first. The median is the number in the middle. To find the depth of the median, there are several formulas that could be used, the one that we will use is:

Depth of median = 0.5 * (n + 1)

Where n is the number of entries.

Example:

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A test marked out of 40 marks is given to a class of 15 students; who scored the following marks:32,27,30,20,19,20,34,26,23,30,34,26,31,24,29. Find the mean, mode and median of the population.

Solution:

Mean = (19+20+23+24+26+26+26+27+29+30+30+31+32+34+34)/15

= 27.4

Mode = 26 (occurs three times)

Median: First arrange the data in ascending order that is

19,20,23,24,26,26,26,27,29,30,30,31,32,34,34 here 27 is in the middle f the data.

1.4 Measure of Dispersion Range The largest number in the sample/population minus the smallest number. The range only involves the smallest and largest numbers, and it would be desirable to have a statistic which involved all of the data values.

Average Deviation

The first attempt one might make at this is something they might call the average deviation from the mean and define it as: The problem is that this summation is always zero. So, the average deviation will always be zero. That is why the average deviation is never used.

Population Variance

So, to keep it from being zero, the deviation from the mean is squared and called the "squared deviation from the mean". This "average squared deviation from the mean" is called the variance.

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Sample Variance = S 2 x

x

2

n 1 Example:

A test marked out of 40 marks is given to a class of 15 students; who scored the following marks:32,27,30,20,19,20,34,26,23,30,34,26,31,24,29. Find population variance.

From the above example mean=27.4

Variance 2 = ((19-27.4)

2+ (20-27.4)

2+…+ (34-27.4)

2)/15

= 19.97

And the Standard Deviation = square root of Population Variance = 4.47

1.5 Frequency Distribution Organizing /Grouping Data It is not practicable to list of the individual value when analyzing a large amount of data with repetitions (frequency). So we use a table and remove all repeating data by counting their frequency.

Example: The number of brothers and sisters of 100 St Theresa Academy students are:

No of bros and sis xi Frequency f i

0 1

1 2

2 11

3 12

4 18

5 18

6 10

7 15

8 6

9 3

10 2

11 2

12 1

18 1

3

Total Number of Computers 100

So from the table:

The Mode is 4 and 5

1.5.1 The Mean for Frequency Distribution

= x

i

fi

N Mean = ((2x1) + (2x11) +…+ (18x1))/100

= 5.04

1.5.2 The Population Standard Deviation For Frequency Table

2 =

fi

xi 2

2

N

So 2 =

3056100 5.042

= 5.16

1.6 Cumulative Frequency

This is a total frequency up to the end of a particular interval. Add the frequency of the previous data to the next data to get the cumulative frequency of the next data.

No of bros &sis frequency Cumulative Frequency

0 1 1

1 2 3

2 11 14

3 12 26

4 18 44

5 18 62

6 10 72

7 15 87

8 6 93

9 3 96

10 2 98

11 2 100

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A graph showing Cumulative Frequency will look like this

no

of b

ros &

sis

Ogive Chart

70

60

50

40

30

20

10

0 0 1 2 3 4 5

cumulative frequency

1.7 Calculation Of Mean And Standard Deviation From Grouped Data Example: 120 people set out on a sponsored walk to get money for Charity Fund. The distance walked is as follows:

Distance km Less than 10 10-15 15-20 20-30 30-50

No of Walkers 12 15 25 41 27

Find the mean distance walked and standard deviation (s.d)

Solution:

Take the middle value for each interval:

Distance km Mid class interval No of walkers

Less than 10 5.00 12

10-15 12.50 15

15-20 17.50 25

20-30 25.00 41

30-50 40.00 27

The Mean=

x

i fi =

12 5 15 12.5 25 17.5 41 25 27 40

N 120

= 23.25km

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EDUCATION BLOG TANZANIA

S.D: = 79125

23.52

120

= 10.35

1.8 The Coding Method-Assumed Mean To make calculations for easier, we sometimes guess what is a mean before performing any calculations. This is called an Assumed Mean. If the assumed mean is A then the real mean is found by

fd

A C

N

Where f=frequency, d=the distance between each mid-class level, A=assumed mean and C=d/ Example: Use an assumed mean of 67 to calculate the mean mass of a group of 100 students in kg given below:

Mass kg Frequency

60-62 5

63-65 18

66-68 42

69-70 27

72-74 8

Solution: find the mid class

Mass kg Mid-class Frequency d

60-62 61 5 -30

63-65 64 18 -54

66-68 67 42 0

69-70 70 27 51

72-74 73 8 48

The normal method is mean= 5 61 18 64 42 67 27

70 8 73 100 = 67.45

Now using the assumed mean, C=3

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fd

A C

N

67+ 3 5 2 18 1 42 0 27 1 8 2/100

= 67+0.45= 67.45 Exercise:

1. Find the mode, median, mean and standard deviation of 7, 10, 6, 11, 10 2. Find the mode, median, mean and standard deviation of 7, 8, 8, 13, 15, 10, 5, 14

3. The following table gives the distribution of marks in an exam

Marks 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90

No of 1 3 9 15 44 22 5 1

Student

Using the assumed mean A=55 find the mean and standard deviation of the given distribution. 4. The table below given the distribution of people according to their weights

weight 45- 50- 55- 60- 65- 70- 75- 80- 85- 90-95

50 55 60 65 70 75 80 85 90

frequency 2 3 5 7 9 11 7 2 3 1 a). Draw the Cumulative Frequency curve for the weight b). find the mean using appropriate assumed mean A.

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2. CHAPTER TWO: TRIGONOMETRY 2.1 What Is Trigonometry? Trigonometry is an important, fundamental step in math education. From the seemingly simple shape, the right triangle, we gain tools and insight that help us in further practical as well as theoretical endeavors. The subtle mathematical relationships between the right triangle, the circle, the sine wave, and the exponential curve can only be fully understood with a firm basis in trigonometry. 2.1.1 Measuring Angles There are three units of measure for angles: revolutions, degrees, and radians. In trigonometry, radians are used most often, but it is important to be able to convert between any of the three units. Revolution A revolution is the measure of an angle formed when the initial side rotates all the way around its vertex until it reaches its initial position. Thus, the terminal side is in the same exact position as the initial side. In trigonometry, angles can have a measure of many revolutions--there is no limit to the magnitude of a given angle. A revolution can be abbreviated "rev". Degrees A more common way to measure angles is in degrees. There are 360 degrees in one revolution. Degrees can be subdivided, too. One degree is equal to 60 minutes, and one minute is equal to 60 seconds. Therefore, an angle whose measure is one second has a measure of degrees. When perpendicularity is discussed, it is most often defined as

a situation in which a 90-degree angle exists. Often degrees are used to describe certain

triangles, like 30-60-90 and 45-45-90 triangles. As previously mentioned, however, in

most cases that concern trigonometry, radians are the most useful and manageable unit of

measure. Degrees are symbolized with a small superscript circle after the number

(measure). 360 degrees is symbolized 360o.

Radian A radian is not a unit of measure that is arbitrarily defined, like a degree. Its definition is geometrical. One radian (1 rad) is the measure of the central angle (an angle whose vertex is the center of a circle) that intercepts an arc whose length is equal to the radius of the circle. The measure of such an angle is always the same, regardless of the radius of the circle. It is a naturally occurring unit of measure, just like π is the natural ratio of the circumference of a circle and the diameter. If an angle of one radian intercepts an arc of length r, then a central angle of 2π radians would intercept an arc of length 2πr, which is the circumference of the circle. Such a central angle has a measure of one revolution.

Therefore, 1 rev = 360o = 2π rad. Also, 1 rad = ( )

o = rev.

2.1.2 Conversion between Revolutions, Degrees, and Radians Below is a chart with angle measures of common angles in revolutions, degrees, and radians. Any angle can be converted from one set of units to another using the definition

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of the units, but it will save time to memorize a few simple conversions. It is particularly important to be able to convert between degrees and radians.

2.1.3 The Coordinate Plane Angles lie in a plane. To specify the point in space where an angle lies, or where any figure exists, a plane can be assigned coordinates. Since a plane is two-dimensional, only two coordinates are required to designate a specific location for every point in the plane. One coordinate determines the length, and the other determines width. In reality, length and width are the same thing--they are used because they describe distance in two directions which are perpendicular to each other. This is all the coordinate plane is: a plane with two perpendicular axes by which distance in either of two dimensions can be measured. 2.1.4 Quadrants The axes of the coordinate plane divide the plane into four regions--these regions are called quadrants. The region in which the x-coordinate and the y-coordinate are both positive is called Quadrant I. Quadrant II is the region in which x < 0 and y > 0. Quadrant

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III is the region in which x < 0 and y < 0. Quadrant IV is the region in which x > 0 and y < 0. The quadrants are labeled in the figure below. 2.2 Trigonometric Ratios/Functions

Terms Domain - The set of all possible inputs of a function. Function - An operation that assigns a correspondence from elements of one set to elements of another set. Period - The repeating interval of a periodic function; the period of a function is a real number. Periodic Function - A function that repeats itself in regular intervals; it follows this equation: f (x + c) = f (x), where c is a constant. Range - The set of all possible outputs of a function. Reference Angle - The positive acute angle formed between the terminal side of an angle and the x-axis. Unit Circle - The circle whose center is at the origin and whose radius is one.

First let’s start with the six trig functions and how they relate to each other. Recall as well that all the trig functions can be defined in terms of a right triangle. From this right triangle we get the following definitions of the six trig functions.

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EDUCATION BLOG TANZANIA Applications The ability to solve right triangles has many applications in the real world. Many of these applications have to do with two-dimensional motion, while others concern stationary objects. We'll discuss both. Two-Dimensional Motion Two-dimensional motion can be represented by a vector. Every vector can be resolved into a vertical and a horizontal component. When a vector is combined with its vertical and horizontal component, a right triangle is formed.

A vector and its component parts form a right triangle Often the motion of a vehicle of some kind is modeled using a vector. With limited information, using right triangle solving techniques, it is possible to find out a lot about the motion of an object in a two-dimensional plane. For example, if a boat goes 12 miles

in a direction 31o north of east, how far east did it travel? If the boat began at the origin,

the problem looks like this in the coordinate plane:

c = 12 and A = 31o. Then b = c cos(A) 10.29. So the boat went slightly more than 10

miles east on its journey. The motion of a projectile in the air can also be easily modeled using a right triangle. The most common example of this is an airplane's motion. For example, if an airplane takes

off at an angle of elevation of 15o and flies in a straight line for 3 miles, how high does it

get? 3 sin(15) .78. The plane climbs about .78 miles. These types of problems use the terms angle of elevation and angle of depression, which refer to the angles created by an object's line of motion and the ground. They can be mathematically represented by a vector and a horizontal line, usually the x- axis.

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Angles of elevation and depression A zero degree angle of elevation or depression means that the object is moving along the ground--it is not in the air at all. A 90 degree angle of elevation is motion directly upward, whereas a 90 degree angle of depression is motion directly downward. Stationary Objects Stationary objects that form right triangles can also be examined and understood by using right triangle solving techniques. One of the most common examples of a right triangle seen in real life is a situation in which a shadow is cast by a tall object. For example, if a 40 ft. tree casts a 20 ft. shadow, at what angle from vertical is the sun shining?

The shadow cast by a tree forms a right triangle

As the picture shows, tan(x) = = . So x = arctan( ) 26.6o.

Whenever you use a right triangle to model a real-life situation, it is immensely helpful to draw a picture or diagram of the situation. Then labeling the parts of the right triangle is easy and the problem can be simply solved.

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EDUCATION BLOG TANZANIA Problems

1. Two boats leave the same port. One goes 10 miles due west and drops anchor. The other leaves the port 20 degrees north of west. How far must it go in a straight line to get as far west as the first boat?

2 A man flies a kite with a 100 foot string. The angle of elevation of the string is 52o.

How high off the ground is the kite?

3. An airplane takes off 200 yards in front of a 60 foot building. At what angle of elevation must the plane take off in order to avoid crashing into the building? Assume that the airplane flies in a straight line and the angle of elevation remains constant until the airplane flies over the building.

4 A 14-foot ladder is used to scale a 13-foot wall. At what angle of elevation must the ladder be situated in order to reach the top of the wall?

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5 A ramp is needed to allow vehicles to climb a 2-foot wall. The angle of elevation in

order for the vehicles to safely go up must be 30o or less, and the longest ramp available

is 5 feet long. Can this ramp be used safely?

Functions in Quadrants

The sign of a trigonometric function is dependent on the signs of the coordinates of the points on the terminal side of the angle. By knowing in which quadrant the terminal side of an angle lies, you also know the signs of all the trigonometric functions. There are eight regions in which the terminal side of an angle may lie: in any of the four quadrants, or along the axes in either the positive or negative direction (the quadrantal angles). Each situation means something different for the signs of the trigonometric functions.

Signs of Angles in Quadrants

The distance from a point to the origin is always positive, but the signs of the x and y coordinates may be positive or negative. Thus, in the first quadrant, where x and y coordinates are all positive, all six trigonometric functions have positive values. In the second quadrant, only sine and cosecant (the reciprocal of sine) are positive. In the third quadrant, only tangent and cotangent are positive. Finally, in the fourth quadrant, only cosine and secant are positive. The following diagram may help clarify.

Values of Quadrantal Angles When an angle lies along an axis, the values of the trigonometric functions are either 0, 1, -1, or undefined. When the value of a trigonometric function is undefined, it means that the ratio for that given function involved division by zero. Below is a table with the values of the functions for quadrantal angles.

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EDUCATION BLOG TANZANIA The points at which the values of a function are undefined are technically not in the domain of that function. Therefore, the domain of sine and cosine is all real numbers. The

domain of tangent and secant is all real numbers except + kπ, where k is an integer. The

domain of cosecant and cotangent is all real numbers except kπ, where k is an integer. 2.2.1 Periodic Functions A periodic function is a function whose values (outputs) repeat in regular intervals. Symbolically, a periodic function looks like this: f (x + c) = f (x), for some constant c. The constant c is called the period--it is the interval at which the function has a non-repeating pattern before repeating itself again. When we graph the trigonometric functions, we'll see that the period of sine, cosine, cosecant, and secant are 2π, and the period of tangent and cotangent is π. For now, using reference angles, we'll learn how to calculate the value of a trigonometric function of any angle just by knowing the value of the trigonometric

functions from 0 to . 2.2.2 Reference Angles The use of reference angles is a way to simplify the calculation of the values of trigonometric functions at various angles. With a calculator, it is easy to calculate the value of any function at any angle. As you get more familiar with trigonometry, though, you'll memorize the values of a few simple trigonometric equations, and with reference angles, you can extend this knowledge of a few equations to many more. A reference angle for a given angle in standard position is the positive acute angle formed by the x-axis and the terminal side of the given angle. Reference angles, by definition,

always have a measure between 0 and . Due to the periodic nature of the trigonometric

functions, the value of a trigonometric function at a given angle is always the same as its

value at that angle's reference angle, except when there is a variation in sign. Because we

know the signs of the functions in different quadrants, we can simplify the calculation of

the value of a function at any angle to the value of the function at the reference angle for

that angle.

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β is the reference angle for θ.

For example, sin( ) = ±sin( ). We know this because the angle is the reference

angle for . Because we know that the sine function is negative in the third quadrant,

we know the whole answer: sin( ) = - sin( ). Shortly, we will become very familiar

with expressions like sin( ), and, without much thinking, we'll know that the answer is

. Herein lies the usefulness of reference angles: we only need to become familiar with the values of the functions from 0 to and the signs of the functions in each quadrant to

be able to calculate the value of a function at any angle. Below is a chart that will help in the easy calculation of reference angles. For angles in the first quadrant, the reference angle β is equal to the given angle θ. For angles in other quadrants, reference angles are calculated this way: For angles greater than 2π radians, simply subtract 2π from them, and then use the chart above to calculate the accompanying reference angle. When you become familiar with

the values of certain trigonometric functions at certain common angles, like and ,

you will be capable of using reference angles to figure out the values of these functions at

an infinite number of other angles.

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EDUCATION BLOG TANZANIA 2.2.3 Graphs of Trigonometric Functions

Given a point P(x, y) on the terminal side of an angle θ in standard position, distance d from the origin, cosine(θ) = cos(θ) =

.

Cosine

Given a point P(x, y) on the terminal side of an angle θ in standard position, cotangent(θ) = cot(θ) =

.

Cotangent

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Given a point P(x, y) on the terminal side of an angle θ in standard position, distance d from the origin, secant(θ) = sec(θ) =

.

Secant

Given a point P(x, y) on the terminal side of an angle θ in standard position, distance d from the origin, sine(θ) = sin(θ) =

.

Sine

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Given a point P(x, y) on the terminal side of an angle θ in standard position, tangent(θ) = tan(θ) =

.

Tangent

The Unit Circle The unit circle is the circle whose center is at the origin and whose radius is one. The circumference of the unit circle is 2π. An arc of the unit circle has the same length as the measure of the central angle that intercepts that arc. Also, because the radius of the unit circle is one, the trigonometric functions sine and cosine have special relevance for the unit circle. If a point on the circle is on the terminal side of an angle in standard position, then the sine of such an angle is simply the y-coordinate of the point, and the cosine of the angle is the x-coordinate of the point. This relationship has practical uses concerning the length of an arc on the unit circle. If an arc has one endpoint at (1,0) and extends in the counterclockwise direction, the other

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endpoint of the arc can be determined if the arc length is known. Given an arc length s, the other endpoint of the arc is provided by the coordinates (cos(s), sin(s)). This is a common alternative way to plot the unit circle. Most often, the unit circle can be drawn

according to the equation x2 + y

2 = 1. As we have seen here, though, it can also be drawn

according to the equations x = cos(s), y = sin(s), where s is the length of the arc starting at (1,0). 2.3 Trigonometric Identities

A trigonometric identity is an equation involving trigonometric functions that can be solved by any angle. Trigonometric identities have less to do with evaluating functions at specific angles than they have to do with relationships between functions. Eight specific trigonometric identities are fundamental. These can be used to form an infinite number of other identities. We'll take a look at the eight fundamental trigonometric identities, and then some additional identities concerning negative angles (angles in which the rotation between the initial and terminal side is clockwise). 2.3.1 Eight Fundamental Trigonometric Identities The following equations are eight of the most basic and important trigonometric identities. These equations are true for any angle. From them, countless additional identities can be formed. These eight should be memorized.

csc(θ) = sec(θ) = cot(θ) = tan(θ) =

cot(θ) = (sin(θ))2 + (cos(θ))

2 = 1 1 + (tan(θ))

2 = (sec(θ))

2

1 + (cot(θ))2 = (csc(θ))

2

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EDUCATION BLOG TANZANIA 2.3.2 Negative Angle Identities Identities also exist to relate the value of a trigonometric function at a given angle to the value of that function at the opposite of the given angle. Here are these identities: sin(- θ) = - sin(θ) cos(- θ) = cos(θ) tan(- θ) = - tan(θ)

Any function f for which the equation f (- x) = - f (x) holds true is called an odd function. A function for which the equation f (- x) = f (x) holds true is called an even function. Sine and cosine are examples of odd and even functions, respectively. Odd and even functions, by definition, reflect themselves across the origin and y- axis, respectively. That is, a segment whose endpoints are f (x) and f (- x) of an odd function will always contain the origin, whereas a segment with the same endpoints on an even function will always be perpendicular to the y-axis.

Additional Trigonometric Identities

Using the eight fundamental identities and the six negative angle identities, an infinite number of new identities can be created. Remember, a trigonometric identity is any equation involving trigonometric functions and which is true for any angle. Study how to prove that the following equation is an identity by simplifying it using the eight fundamental identities:

tan(θ)cos(θ) + = sin(θ) + (csc(θ))2cos(θ) - cot(θ)sin(θ)

+ = sin(θ) + -

sin(θ) + = sin(θ) + -

cos(θ) (cot(θ))2cos(θ) = cos(θ)( - 1)

(cot(θ))2cos(θ) = cos(θ)((csc(θ))

2 - 1)

(cot(θ))2cos(θ) = cos(θ)(cot(θ))

2

In general, it is helpful to follow these steps to simplify a potential identity:

1. Express everything in terms of sine and cosine. 2. Cancel all terms possible.

21


3. Rewrite the remaining expressions using as few trigonometric expressions as

possible (this will likely involve converting sines and cosines to tangents, secants, etc.)

4. Factor anything that will result in one of the addition identities.

In following these steps, as we did in the example above, simplifying most equations will become easy. 2.4 Trigonometric Equations A trigonometric equation is any equation that contains a trigonometric function. As mentioned in Trigonometric Identities, a trigonometric equation that holds true for any angle is called a trigonometric identity. Only a few simple trigonometric equations can be easily solved without a calculator. Often one might encounter an equation like tan(x) = 3.2. Such an equation has no simple answer that can be memorized. It would be tedious to use a calculator and try numerous values for x until you found one that gave a solution close to 3.2. For problems like these, the inverse trigonometric functions are helpful. The inverse trigonometric functions are the same as the trigonometric functions, except x and y are reversed. For example, another way to say sin(y) = x is y = arcsin(x). The arcsine relation is not a function, though, because it assigns more than one element of the range to each element of the

domain. For example, sin(y) = has solutions of y = 30 degrees, 150 degrees, 390

degrees, and so on. When the range is restricted, however, then arcsine is a function, and

is written with a capital letter, Arcsine. Using the inverse trigonometric functions, it

becomes possible (with a calculator) to solve nearly any trigonometric equation without

difficulty. 2.4.1 Solving Trigonometric Equations Solutions for trigonometric equations follow no standard procedure, but there are a number of techniques that may help in finding a solution. These techniques are essentially the same as those used in solving algebraic equations, only now we are manipulating trigonometric functions: we can factor an expression to get different, more understandable expressions, we can multiply or divide through by a scalar, we can square or take the square root of both sides of an equation, etc. Also, using the eight fundamental identities, we can substitute certain functions for others, or break a functions down into two different ones, like expressing tangent using sine and cosine. In the problems below, we'll see just how helpful some of these techniques can be.

Example 1

2 cos(x) - 1 = 0 2 cos(x) = 1

22

EDUCATION BLOG TANZANIA cos(x) =

x = ,

In this problem, we came up with two solutions in the range [0, 2π) : x = , and x = .

By adding 2nπ to either of these solutions, where n is an integer, we could have an

infinite number of solutions.

Example 2

sin(x) = 2 cos2(x) - 1

sin(x) = 2(1 - sin2(x)) - 1

sin(x) = 1 - 2 sin2(x)

2 sin2(x) + sin(x) - 1 = 0

(sin(x) + 1)(2 sin(x) - 1) = 0 At this point, after factoring, we have two equations we need to deal with separately. First, we'll solve (sin(x) + 1) = 0, and then we'll solve (2 sin(x) - 1) = 0

Case1

sin(x) + 1 = 0 sin(x) = - 1

23

EDUCATION BLOG TANZANIA x =

Case2

2 sin(x) - 1 = 0 sin(x) =

x = ,

For the problem, then, we have three solutions: x = , , . All of them check. Here

is one more problem.

Example 3

sec2(x) + cos

2(x) = 2

1 + tan2(x) + 1 - sin

2(x) = 2

tan2(x) = sin

2(x)

= sin2(x)

cos2(x) = 1

cos(x) = ±1

24

EDUCATION BLOG TANZANIA x = 0, π

Formulae

arccosecant

arccosine

arccotangent

arcsecant

arcsine

arctangent

y = arccosecant of x = arccsc(x) = csc-1

(x). Another way to write x = csc(y).

y = arccosine of x = arccos(x) = cos-1

(x). Another way to write x = cos(y).

y = arccotangent of x = arccot(x) = cot-1

(x). Another way to write x = cot(y).

y = arcsecant of x = arcsec(x) = sec-1

(x). Another way to write x = sec(y).

y = arcsine of x = arcsin(x) = sin-1

(x). Another way to write x = sin(y).

y = arctangent of x = arctan(x) = tan-1

(x). Another way to write x = tan(y).

2.6 Compound Angle Formulae 2.6.1 Addition and Subtraction Formulas These formulas are especially helpful when an angle whose function values you haven't memorized can be expressed as the sum or difference of two angles whose function values you have memorized. The following formulas express the values of trigonometric functions of the sum of angles in terms of sums of the products of functions of single angles.

sin(α + β) = sin(α)cos(β) + cos(α)sin(β) cos(α + β) = cos(α)cos(β) - sin(α)sin(β)

25

EDUCATION BLOG TANZANIA tan(α + β) = The following formulas express the values of trigonometric functions of the difference of angles in terms of sums of the products of functions of single angles. sin(α - β) = sin(α)cos(β) - cos(α)sin(β) cos(α - β) = cos(α)cos(β) + sin(α)sin(β) tan(α - β) = 2.6.2 Double and Half-Angle Formulas The following identities equate trigonometric functions of double angles to expressions that involve only trigonometric functions of single angles. They are very useful in differentiation and other general simplification. sin(2x) = 2 sin(x)cos(x)

cos(2x) = cos2(x) - sin

2(x) = 1 - 2 sin

2(x) = 2 cos

2(x) - 1

tan(2x) = The following identities equate trigonometric functions of half-angles to expressions that involve only trigonometric functions of single angles.

sin = ±

cos = ±

26


tan = ± =

= If an angle in question is a variable, these formulas are sometimes the only means by which the trigonometric expression can be simplified. Even when an angle is known, these identities can be useful in simplifying expressions. They should be memorized. 2.6.3 Products of Functions The following formulas relate the products of sines and cosines to sines and cosines of multiple angles. These formulas are derived from the addition formulas. They are useful identities for simplification.

sin(α)sin(β) = - (cos(α + β) - cos(α - β))

cos(α)cos(β) = (cos(α + β) + cos(α - β))

sin(α)cos(β) = (sin(α + β) + sin(α - β))

cos(α)sin(β) = (sin(α + β) - sin(α - β)) 2.6.4 Factor Theorem The following formulas relate the sum of trigonometric functions of different angles to products of multiple angle functions. These identities are also useful for simplifying expressions. sin(α) + sin(β) = 2 sin cos

27

EDUCATION BLOG TANZANIA cos(α) + cos(β) = 2 cos cos sin(α) - sin(β) = 2 cos sin cos(α) - cos(β) = - 2 sin sin

2.7 Solving Oblique Triangles An oblique triangle is a triangle that contains no right angles. Oblique triangles are not as easy to solve as right triangles because three parts of the triangle must be known in order to solve the triangle. Also, with oblique triangles, we can no longer use two of the simplest techniques for solving right triangles--the Pythagorean Theorem and the fact that two acute angles are complementary. Instead, two new techniques are needed: the Law of Sines and the Law of Cosines. These laws are formulas that relate the parts of triangles to each other. They work for all triangles, including right triangles, but since right triangles have special properties we use simpler methods to solve them. 2.7.1 The Law of Sines The Law of Sines states that each side of a triangle is proportional to the sine of the opposite angle. It looks like this:

= = The law of sines can be used when two angles and a side of a triangle are known. Consider the following problem, in which we have two angles and the side opposite one

of them: A = 35o, B = 49

o, and a = 7. The first part we calculate is the third angle, C. C =

180o -35

o -49

o = 96

o. Then, using the Law of Sines, b and c can be calculated. =

= = = = . b 9.21, and c 12.13. Now we'll consider two angles and the side included: A = 27

o, B = 105

o, and c = 13.

First we'll calculate the measure of the third angle, C. C = 180o -27

o -105

o = 48

o. Then

using the Law of Sines, a and b can be calculated.

28


2.7.2 The Law of Cosines The law of cosines asserts that in an arbitrary triangle a

2 = b

2 + c

2 - 2bc cos , where a,

b, c and are as shown below. The formula follows by applying the Theorem of Pythagoras to the triangles ADC and DBC. Note that the red line AD has length |b cos | and the green line DB has length c - b cos . Hence by the Theorem of Pythagoras

Multiplying the right hand side out we get

Adding b2 cos

2 and rearranging we get a

2 = b

2 + c

2 - 2bc cos as required.

This law is used primarily in two situations: when two sides and their included angle are given, and when three sides are given. If two sides and their included angle are given, the next thing to calculate is the third side. The Law of Cosines, as shown above, is perfect for the situation. After the third side is calculated, the Law of Sines can be used to calculate either of the other two angles. If three sides are given, the Law of Cosines must be manipulated a bit: For this situation,

the Law of Cosines is most useful in this form: cos(A) = . Once one of the

angles is known, the next can be calculated using the Law of Sines, and the third using

subtraction, knowing that the angles of a triangle sum to 180 degrees There are three basic situations in which the area of a triangle can be calculated using trigonometric techniques.

29


3. CHAPTER THREE: VECTOR ALGEBRA 3.1 Basic Vector Definitions A vector is a quantity that has both magnitude and direction. The magnitude is a scalar quantity, a scalar being defined as a quantity which may be completely specified by a number and perhaps a unit. Common textbook representations of vectors include boldfaced letters and boldface with an arrow above them. For example a displacement vector of 30 meters east could be represented in a variety of ways:

The magnitude of the vector might vector symbol, or just

be represented by absolute value signs around the

the letter without the boldface.

Examples of vectors: Position, velocity, acceleration...

Notation: , , , ...

Coordinate representation:

a column

Length:

A vector might also be expressed in terms of unit vectors.

Unit Vectors

It is convenient to use vectors of unit length to specify the directions of vector quantities in the various coordinate systems. In Cartesian coordinates it is typical to use i, j and k to represent unit vectors in the x, y and z directions respectively. A vector which specifies a position in space with respect to the origin of the coordinate system could then be written.

30


3.2 Representations of Vectors 3.2.1 Vectors as directed line segments Vectors are represented in a natural way by directed line segments, the length of the segment measuring the magnitude of the vector and the arrowhead indicating the direction of the vector. The arrowhead is usually placed at one end (or near the end) of the line segment; this end is called the head of the vector and the other end is called the tail of the vector. Such vectors are called free vectors because they are distinguished only by their length and direction, and not by their position in space. The vector with the same magnitude as u but opposite direction, the negative of u, is labeled -u.

The zero vector (a vector with zero length and no direction) is written 0. 3.2.2 Vector notation In printed works, such as text books, it is common practice to use bold face to indicate that a symbol, such as v, represents a vector. In written work, where bold face is not available, the symbols representing vectors are underlined.

The convention for position vectors in both printed and written work is to place an arrow

over the pair of points that defines the vector, such as .

3.2.3 Position vectors

The line segment , where A and B are points in space, is called the position vector of

B relative to A. It is also known as the vector from A to B as well as the displacement of

B from A.

The vector from B to A has the same magnitude as the vector from A to B but

opposite direction. That is, is the negative of

31


The zero vector is the position vector of a point A with respect to itself; that is, 0 = .

The length of the zero vector is 0 but it has no direction. 3.2.4 Equality of vectors

Given points A and B, we may think of the position vector v = as the vector that acts

on the point A to get B. If the line segments AB and DC have the same direction and the same length, then ABCD

is a parallelogram and the position vectors and are equal; we write this as =

.

The vector v is a free vector because although it has a definite direction and length it does not have any particular position in space. When using vectors to model physical phenomena it is not always appropriate to use free vectors. For example, in order to completely describe the effect of a force we need to give a vector representing the magnitude and direction of the force as well as a point of application of the force. Once we have chosen a suitable point as origin we can describe the point of application by the position vector of the point relative to the origin. It is usual to think of the vector representing the force as being confined to a line and for this reason it is often called a line vector. Two line vectors are equal if they have the same direction and length and lie along the same line. 3.3 Composition of vectors 3.3.1 Introduction Because non-zero vectors have direction as well as magnitude, adding vectors involves more than simply adding numbers. The sum of two vectors is another vector, and so the definition of addition must give a process for determining both the magnitude and the direction of the sum vector. There are two equivalent procedures for addition of vectors, called the parallelogram rule and the triangle rule.

32

EDUCATION BLOG TANZANIA 3.3.2 The Parallelogram Rule For Addition Suppose u and v are two vectors (in the plane or in space). Translate them so that they are tail-to-tail at point O. From the head of each vector, draw a copy of the other vector to complete a

parallelogram OAPB. In this parallelogram, u = = and v = = .

3.3.3 The Triangle Rule For Addition Another way to define addition of two vectors is by a head-to-tail construction that creates two sides of a triangle. The third side of the triangle determines the sum of the two vectors, as shown below.

Place the tail of the vector v at the head of the vector u. That is, u = and v = .

33


Now construct the vector to complete the third side of the triangle OAP. This method is equivalent to the parallelogram law of addition, as can be easily seen by drawing a copy of v tail-to-tail with u, to obtain the same parallelogram as before.

Using position vector notation, the triangle rule of addition is written as follows: for any three points X, Y , Z, Both the triangle and the parallelogram rules of addition are procedures that are independent of the order of the vectors; that is, using either rule, it is always true that u + v = v + u for all vectors u and v. This is known as the commutative law of addition. There are other rules like this one, and they are discussed in the component Vector Algebra.

34


3.3.4 Adding a vector to its negative

To add u to -u, place the tail of -u = at the head of u = .

Then u + (-u) = = 0, the zero vector. By the commutative law,

3.3.5 Subtraction of vectors Subtraction of two vectors is just a special case of addition. The vector u - v is defined to be

Using the parallelogram law of addition, u - v = u + (-v) = + = . The vector v - u can be found in the same way.

35


The vectors u + v and u - v are both diagonals of the parallelogram formed by u and v.

3.3.6 The Head-Minus-Tail Rule

Given a vector and any point O, the triangle rule for addition shows us that =

+ . We can rewrite this equation as

Since points to the head of the vector and points to its tail, we call this the

“head-minus-tail” rule.

The advantage of this rule is that it allows us to rewrite all the vectors occurring in a problem as combinations of vectors emanating from a given point O (often called the origin).

36


3.3.7 Multiplication By A Scalar Using a head-to-tail procedure, a vector v can be added to itself to give the vector v + v.

We have v = and v = . It’s natural to write + = as 2v. The vector 2v

has length 2|v| and is in the same direction as v. The notion of multiplication of a vector

by a positive integer is then generalized to define the vector sv for all scalars s, as

follows. For example, -3v has three times the magnitude of v but points in the opposite direction;

v (also written ) has magnitude and has the same direction as v. Note that -1v has the same magnitude as v but has the opposite direction, and so is the same vector as the negative of v, that is, -1v = -v. 3.3.8 When Are Vectors Parallel? Two vectors u and v are said to be parallel if they have either the same direction or opposite direction. This means that each is a scalar multiple of the other: for some non-

zero scalar s, v = su and so u = v.

37


Since the zero vector can be written 0 = 0v, the zero vector is considered to be parallel to every other vector v.

3.3.9 Unit Vectors It is often useful to be able to find a unit vector (a vector of length 1) in the direction of a given non-zero vector v. A common notation used for this unit vector is .

For example, if v has length 3, then = v has length 1 and is in the same direction as v. In general, the unit vector in the same direction as v is

Since = , we also have v = |v| .

3.4 Magnitude and Direction of Vectors

If the components of a vector are known, then its magnitude and direction can be calculated with the use of the Pythagorean relationship and triangle trig. This is called the polar form of the vector.

38


3.5 Basic Vector Operations Both a magnitude and a direction must be specified for a vector quantity, in contrast to a scalar quantity which can be quantified with just a number. Any number of vector quantities of the same type (i.e., same units) can be combined by basic vector operations. 3.5.1 The Scalar Product The scalar product in terms of magnitude and angles

Suppose u and v are two non-zero vectors (in the plane or in space) with magnitudes |u| and |v|.

Translate the two vectors such that they are tail-to-tail and denote the angle between them by .

The scalar quantity is called the scalar product (or dot product) of u and v. If one (or both) vectors are zero vectors then we define u · v = 0. Note that it does not matter which way round the angle is measured as cos(360° - ) = cos

.

The scalar product in terms of components

If u = u1i + u2j + u3k and v = v1i + v2j + v3k then

For vectors in the plane you need to drop the third terms. We often call the above formula the Cartesian representation of the scalar product.

39

EDUCATION BLOG TANZANIA Proof of the formula If one (or both) vectors are zero vectors then the above formula is obvious. Hence assume that u and v are non-zero.

The vectors u, v and u - v then form a triangle.

The edges of the above triangles have length |u|, |v| and |u - v|. By the law of cosines (see below)

By definition u · v = |u||v| cos , so

(1) By the formula for the magnitude of a vector

Multiplying the right hand side out we get

If we substitute the above into (1) we get

So u · v = u1v1 + u2v2 + u3v3 as claimed.

3.5.2 The Vector Product The vector product in terms of magnitudes, angles and the “right-hand rule” The vector product is a product of two vectors in space. The result of the multiplication is a vector, hence the name vector product. Other name for the vector product is cross product.

If u and v are vectors, their vector product is denoted by u × v. We must assign that product a direction and a magnitude.

We start by specifying the direction. Suppose that u and v are two non-zero and non-parallel vectors in space. When placed tail-to-tail they define a plane. There are two possibilities to choose a unit vector perpendicular to that plane:

40

EDUCATION BLOG TANZANIA We choose n such that the triple u,v,n is right-handed. By this we mean the following. Align a screw-driver along the line perpendicular to the plane defined by u and v. Then turn the screw-driver such that u moves to v through the smaller angle. The direction of n is then given by the direction the screw moves. Note that it is irrelevant which way we align the screw-driver, the direction in which the screw moves will be the same either way.

Having established the direction it remains to define the magnitude of the vector product. We let the magnitude be |u||v| sin , where is the smaller angle between u and v.

Then we define

41


If u and v are parallel or one (or both) of them are zero vectors then we set u×v = 0. Geometrically, the magnitude equals the area of the parallelogram spanned by the vectors u and v when placed tail-to-tail. Important Note: Unlike in the definition of the scalar product it is essential that the angle between u to v is the smaller one, that is, the one between 0° and 180°.

The vector product in terms of components

If u = u1i + u2j + u3k and u = v1i + v2j + v3k are two vectors then their vector product is If you are familiar with determinants of a 3 × 3 matrix, there is an easy way to remember the above formula. Formally,

The vector product obeys the following algebraic rules 3.6 Vectors Applications Vector products and the area of a triangle

The magnitude of the product u × v is by definition the area of the parallelogram spanned by u and v when placed tail-to-tail. Hence we can use the vector product to compute the area of a triangle formed by three points A, B and C in space.

42

EDUCATION BLOG TANZANIA It follows that the area of the triangle is

3.7 The Ratio Theorem 3.7.1 Internal division of a line segment

With an assigned point O as origin, the position of any point P is given uniquely by the

vector , which is called the position vector of P relative to O.

Let P1 and P2 be any points, and let R be a point on the line P1P2 such that R divides the

line segment P1P2 in the ratio m : n. That is, R is the point such that = . Our

task is to find the position vector of R (relative to O) in terms of the position vectors of P1

and P2.

As = , we have n = m and therefore (1) Which

rearranges to give

43


When m and n are both positive, the vectors and have the same direction, since

= . This corresponds to the situation where R lies between P1 and P2, as

shown in the diagram above. R is then said to divide the line segment P1P2 internally in

the ratio m : n.

As a special case of the general formula (1) we can obtain a formula for the position

vector of the midpoint M between two points P1 and

P2. In this case, m : n = 1 : 1 and so 3.7.2 External division of a line segment We look at a situation similar to the one on the previous page, but allow m and n to have

opposite sign. Again, R is the point such that = , and we want to find the

position vector of R (relative to O) in terms of the position vectors of P1 and P2.

Since is a negative multiple of , the vectors and have opposite direction.

The point R then lies outside the line segment P1P2 (but still on the line joining P1 and

P2). In these cases, R is said to divide the line segment P1P2 externally in the ratio m : n. The formula is, as in case of an internal division,

44

EDUCATION BLOG TANZANIA 3.8 Lines and Planes 3.8.1 Finding the vector equation of a line Using vectors gives us a very neat way of writing down an equation which gives the

position vector of any point on a given straight line. This method works equally well in

two or three dimensions. Suppose we 've got a straight line like the one shown in the

drawing below. (You have to imagine that it extends infinitely far in either direction.) In order to write down the vector equation of this line, we need to know two things.

We have to know the position vector of some point which lies on the line, like a on the

diagram.

We have to know a vector that gives the direction of the line, like b in the diagram. This is called a direction vector.

Then the position vector r of any general point P on the line is given by the equation r = a + tb where t tells us how much of b we need to take in order to get from A to P. Notice that writing r = ta + b would give you a completely different line! It's important to realise that there are many possible ways of writing the vector equation of any given line. So, in the example above, any point A on the line would work equally well provided we knew its position vector, and any vector lying parallel to b would work equally well as a direction vector. (For example, We could have used 3b or - b.) Here are some examples of equations of particular lines so we can look in more detail at how they actually work. Example 1

Suppose that Line A has the equation r = i + 3k + t(2i + j + k)

so that the a in the drawing is i + 3k and b is 2i + j + k.

Different values of t give the position vectors of different points on this line.

Putting t = 0 gives r = i + 3k.

45


Putting t = 1 gives r = 3i + j + 4k.

Putting t = -1 gives r = -i - j + 2k.

Suppose there are two more lines B and C so that now we have

line A with equation r = i + 3k + t(2i + j + k)

line B with equation r = i + 3k + s(i + 4j - k)

and line C with equation r = i + j + k + u(4i + 2j + 2k). The letters s and u work in exactly the same way for their lines as t does for line A. Both

the lines B and C have special relationships with line A. Can you spot what they are?

Using the a and b of the drawing to refer to, we see that lines A and B both have a = i +

3k but they have different direction vectors. Therefore they cut each other at point A with

r = i + 3k. Line C has the same direction as line A, (its direction vector is just scaled up

by a factor of 2), so either lines A and C are parallel or they are really the same line.

Putting t = 0 for line A gives r = i + 3k but there is no value we can give to u in line C which would make r = i + 3k so therefore A and C are distinct parallel lines.

Example 2

Line D has equation r = i - j + 4k + s(i - j + k).

Line E has equation r = 2i + 4j + 7k + t(2i + j + 3k). They are not parallel since their direction vectors aren't parallel but do they cut each

other? (If not, they are what are called skew lines.) If they cut each other then the point P

where they cut must lie on both lines. We'll call its position vector p. The point P can

only exist if there are values of s and t so that p = i - j + 4k + s(i - j + k) = 2i + 4j + 7k + t(2i + j + 3k). For this equation to have a solution, the components in the i, j and k directions must each

seperately be equal. This would mean that

i + si = 2i + 2ti so giving 1 + s = 2 + 2t

-j - sj = 4j + tj so giving -1 - s = 4 + t

4k + sk = 7k + 3tk so giving 4 + s = 7 + 3t.

Is this possible?

Adding (1) and (2) gives 0 = 6 + 3t so t = -2 and 1 + s = 2 - 4 so s = -3.

THE LINES ONLY MEET if these values of s and t also fit equation (3). Putting s = -3 and t = -2 in 4 + s = 7 + 3t gives LHS = 4 - 3 =1 and RHS = 7 - 6 =1 so the 3 equations are consistent (that is, there is a solution which fits all 3 of them) and the

46


lines do cut each other. Putting s = -3 in line D's equation gives the position vector of this point of intersection as p = i - j + 4k - 3(i - j + k) = - 2i + 2j + k. You will see that putting t = -2 in line E's equation gives exactly the same result. 3.8.2 Vector Equations of Planes One way of doing this is to use a very similar method to the one for finding the vector

equation of a line. The difference is that now we want an equation that gives the position

vector of any point in a flat surface or plane. Case 1 It's easiest to explain how this works by starting with the case where the origin lies in the plane. The drawing below shows part of a plane like this.

To find the position vector of any point P, we have to know 2 non-parallel vectors which lie in the plane. I have called these s and t. It is then possible to get to P by adding together suitable multiples of s and t.

This gives us the equation of the plane as r = as + bt.

Case 2 Now suppose we have a plane which doesn't pass through the origin.The drawing below shows part of a plane like this. Again, s and t are known vectors which lie in the plane. We also now need a way of getting to the plane from the origin, so we have to know the position vector of some particular point in the plane. In my drawing, this point is M with position vector m.

47


Once we have reached the plane, we can find the position of any general point P relative to M in the same way that we did above by saying that p = as + bt. Now we can get the equation of the plane in terms of the known vectors m, s and t. We have r = m + p but p = as + bt so r = m + as + bt. In the special case where the plane passes through the origin, we can leave out the m

because it is the zero vector. If two planes are parallel, then the same s and t can be used

for both of them, since we can move these free vectors so that they lie in either plane. The

equations of the planes are different because each one must also include a position vector

from the origin to a known point in that particular plane.

Writing the equation of a plane in this way has one big disadvantage. There are infinitely

many directions that vectors lying in the plane can have. Therefore there are infinitely

many pairs like s and t to choose from. It would be much nicer if we could use a direction

that is unique to the plane. There is a direction that has exactly this property.

3.8.3 Finding equations of planes using normal vectors The direction perpendicular to a plane is unique to that plane (and any plane parallel to it). To see how we can use this to give us another form of the vector equation of a plane,

we'll start with the case where the plane passes through the origin. (It's particularly easy

to see how to do it in this case!) I've shown part of such a plane in my drawing below.

48


Suppose we know the vector N which is perpendicular to the plane. This means that it

must be perpendicular to the position vector r of any point in the plane from the origin, so

the dot product of the perpendicular vectors N and r gives us the equation N.r = 0. The vector N is called a normal vector to the plane.

Now we extend this method to find the equation of a plane which doesn't pass through

the origin. This time, we have to be able to get to the plane first from the origin, so we

must know the position vector of some particular point in the plane from the origin. In

my drawing, this point is M with position vector m. If P is any general point in the plane, so that the vector MP = p, then N and p are perpendicular to each other. Therefore N.p = 0 but p = r - m.

N.(r - m) = 0 or N.r = N.m = C where C is the number we get from working out the dot product of the two known vectors N and m.

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Now suppose we are starting with the plane below, with the 2 known vectors s and t lying in this plane.

How could we use s and t to find a vector N which would be perpendicular to this plane? 3.8.4 Finding a normal vector Here's the plane again with s and t being known vectors which lie in it. Working out the cross product of s and t will automatically give us a vector perpendicular to the plane in which s and t lie. This is the speediest method for finding a normal vector to this plane.

As an example, suppose s = 2i + j + 3k and t = 3i + 2j + 4k. Then s x t = - 2i + j + k = N, a normal vector to the plane. (The working out of s x t is described at the end of the cross product section where I used the same vectors in an example there.) It's also possible to find a normal vector by using the dot product. If we call the normal vector N, then N must be perpendicular to both s and t. Therefore N.s = 0 and N.t = 0.

Now, using algebra it is possible to find an N which fits these two equations. 3.4.5 Equations of planes in the form Ax + By + Cz = D This is another useful way to describe planes. It is known as the Cartesian form of the equation of a plane because it is in terms of the Cartesian coordinates x, y and z. .

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The form Ax + By + Cz = D is particularly useful because we can arrange things so that D gives the perpendicular distance from the origin to the plane. To get this nice result, we need to work with the unit normal vector. This is the vector of unit length which is normal to the surface of the plane. (There are two choices here, depending on which direction you choose, but one is just minus the other).

I'll call this unit normal vector n. Next we see how using n will give us D, the perpendicular distance from the origin to the plane. In the picture below, P is any point in the plane. It has position vector r from the origin O. Now we work out the dot product of r and n.

This gives us r.n = |r||n|cos A.

But |n| = 1 so we have r.n = |r|cos A = D.

This will be true wherever P lies in the plane.

Next, we split both r and n into their components.

We write r = xi + yj + zk and n = n1i + n2j + n3k.

Therefore r.n = (xi + yj + zk) . (n1i + n2j + n3k) = D so r.n = xn1 + yn2 + zn3 = D.

We see that n1, n2 and n3 (the components of the unit surface normal vector) give us the A, B and C in the equation Ax + By + Cz = D.

A numerical example I've put this in here so that you can see everything actually happening and see how it ties back to the earlier pages in this section. We start with the plane I show below. We'll let s = i - 6j + 2k and t = 2i - 2j - k

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EDUCATION BLOG TANZANIA We'll take m, the position vector of the known point M in the plane, to be

m = 2i + 3j + 5k.

P is any point in the plane, with OP = r = xi + yj + zk. First, we find N, a normal vector to the plane, by working out the cross product of s and t. This gives s x t = 10i + 5j + 10k = N.

The length of this vector is given by the square root of (102 + 5

2 + 10

2) = 15.

So the unit normal vector, n, is given by

n = 1/15(10i + 5j + 10k) = 2/3i +1/3j + 2/3k.

Now we use n.r = n.m = D to get the equation of the plane.

This gives us (2/3i +1/3j + 2/3k).(xi + yj + zk) = (2/3i +1/3j + 2/3k).(2i + 3j + 5k)

or 2/3x + 1/3y + 2/3z = 4/3 +3/3 + 10/3 = 17/3. The perpendicular distance of this plane from the origin is 17/3 units.

So what would have happened if we had found the equation of the plane using the first normal vector we found?

Using N.r = N.m gives (10i + 5j + 10k).(xi + yj + zk) = (10i + 5j + 10k).(2i + 3j + 5k)

or 10x + 5y + 10z = 20 + 15 + 50 = 85. It is exactly the same equation as the one we found above except that it is multiplied through by a factor of 15, and 85 gives us 15 times the perpendicular distance of the origin from the plane. Also, are you confident that you will get the same equation for the plane if you start out with the position vector of a different known point in it? The point L also lies in this plane. Its position vector l is given by l = 7i - 7j + 5k. Check that working with l instead of m does give you the same equation for the plane. Geometrically, you can see that this will be so.

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L and M are both just possible positions of P, so that both n.l and n.m give the distance D.

The general case This is how the working goes with letters taking the place of the numbers we have used

in the numerical example.

m is the position vector of the known point in the plane.

n is the unit surface normal to the plane.

We'll let m = x0i + y0j + z0k and n = Ai + Bj + Ck. The position vector of the general point P in the plane is given by r = xi + yj + zk where the values of x, y and z vary according to the particular P chosen. Now we use n.r = n.m = D to write down the equation of the plane. This gives us (Ai + Bj + Ck) . (xi + yj + zk)= (Ai + Bj + Ck) . (x0i + y0j + z0k). = D so Ax + By + Cz

= Ax0 + By0 + Cz0 = D or, if you prefer, you can write

A(x-x0) + B(y-y0) + A(z-z0) = 0. If you have found a normal vector which is not of unit length, you will first need to scale

it down.

Suppose you have found N = N1i + N2j + N3k. Then the length of N is given by

and n, the unit normal vector, is given by

Now, putting n = Ai + Bj + Ck, we have

Exercise 1. The two vectors

The point M also m = 2i + 4j + 7k.

s = 4i + 3k and t = 8i - lies in Q

and its position vector

j + 3k lie in plane Q. from

the origin is given by

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Show that the perpendicular distance of the origin to this plane is 2 units and find its equation.

2. If u = (-3, 2, 1) and v = (5, 0, -3) and w = (6, 1, -4) is it possible to find a scalar t

such that u + tv is parallel to w? 3. Find out if the line r = (1, 3, 8) + t (-2, 5, 7) is parallel to the plane

3x + 4y - 2z = 1.

Where does the line r = (1, 2, -5) + t (2, -3, 1) meet the plane 2x + 5y - 3z = 6?

4. Find the vector equation of the line in which the 2 planes 2x - 5y + 3z = 12 and 3x + 4y - 3z = 6 meet.

5. Finding the shortest distance between a point and aline.

6 How can we find the shortest distance between two skew lines?

3.9 Distance From a Point To a Line

Let the point be C = (Cx,Cy) and the line be AB, determined by (Ax,Ay) and (Bx,By). Let

P be the point of perpendicular projection of C on AB. The parameter r, which indicates P’s position along AB, is computed by the dot product of AC and AB divided by the square of the length of AB:

AC ⋅ AB r = -------- ||AB||²

r has the following meaning: r = 0 P = A r = 1 P = B r < 0 P is on the backward extension of AB r > 1 P is on the forward extension of AB 0 < r < 1 P is interior to AB

The length of a line segment AB in d dimensions is computed by:

L = √( (B1-A1)² + (B2-A2)² + ... + (Bd-

Ad)²) so in 2D:

L = √( (Bx-Ax)² + (By-Ay)² ) and the dot product of two vectors in d dimensions, U ⋅ V is computed:

D = (U1 * V1) + (U2 * V2) + ... + (Ud *

Vd) so in 2D: D = (Ux * Vx) + (Uy *

Vy) So r expands to: (Cx-Ax)(Bx-Ax) + (Cy-Ay)(By-

Ay) r = ------------------------------ L²

The point P can then be found:

Px = Ax + r(Bx-Ax) Py = Ay + r(By-Ay) And the distance from A to P equals r*L. Use another parameter s to indicate the location along PC, with the following

meaning: s < 0 C is left of AB s > 0 C is right of AB

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s = 0 C is on AB Compute s as follows:

(Ay-Cy)(Bx-Ax)-(Ax-Cx)(By-

Ay) s = ---------------------------- L²

Then the distance from C to P = |s|*L. 3.9.1 Shortest Distance From a Point To a Line Problem Find the shortest distance from a point P to a line L. Solution We shall take a numerical example because then I can explain both the principles and

how the working out goes. Suppose the position vector of P from the origin O is OP = p = (2, -1, 2).

Suppose that the vector equation of line L is r = (-1, 0, 7) + t (4, 1, -2). The shortest distance of P from L is given by the length of the perpendicular from P to L. Suppose this perpendicular meets L at H. Then we want to find the length of PH. There are two ways we can set about this.

Method (1) Using the dot product

Since H lies on L we can say that OH = h = (-1, 0, 7) + t (4, 1, -2) = (-1 + 4t, t, 7-2t)

for some value of t which we need to find. Also, vector PH = -p + h = - (2, -1, 2) + (-1 + 4t, t, 7 - 2t) = (-3 + 4t, 1 + t, 5 - 2t).

But PH is perpendicular to the direction vector (4, 1, -2) of line L. So the dot product of vector PH and (4, 1, -2) is zero.

So (-3 + 4t, 1 + t, 5 - 2t).(4, 1, -2) = -12 + 16t + 1 + t - 10 + 4t = - 21 + 21t = 0 so t = 1. Therefore OH = h = (3, 1, 5) and vector PH = -p + h = (1, 2, 3).

Its length |OH| is given by the square root of (1 + 4 + 9) = 3.74 to 2 sf.

Method (2) Using the equation of a plane.

Just as we found in method (1), we have h = (-1 + 4t, t, 7-2t). Now PH lies in a plane which is perpendicular to line L. So the direction vector (4, 1, -2) of L is perpendicular to this plane. Therefore it is a normal vector to the plane. Also, P lies in the plane.

Using the equation of a plane of r.n = p.n, with r = (x, y, z), we get

(x, y, z).(4, 1, -2) = (2, -1, 2).(4, 1, -2) so 4x + y - 2z = 3. But H also lies in this plane so we can say

4 (-1 + 4t) + t - 2 (7 - 2t) = 3 so 21t = 21 so t = 1 as before.

From here, the working is exactly the same as in method (1). 3.10 Distance Between Two Lines Let be points on the respective lines and unit direction vectors along the lines. Then the distance is

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EDUCATION BLOG TANZANIA 3.11 The angle between 2 lines We define the angle between 2 lines to be the angle between their direction vectors placed tail to tail. Notice that this definition works equally well if the lines don't actually cut each other since we then just slide the 2 direction vectors together until their tails meet. I show an example of this in the drawing below. The two lines have the equations r = a + tb and r = c + sd.

The angle between the lines is found by working out the dot product of b and d.

We have b.d = |b||d| cos A.

From this, knowing b and d, we can find the angle A.

suppose that b = 2i - 3j + k and d = 4i + j - 3k.

Then the working out there shows that the angle A is 84 degrees.

3.12 The angle between 2 planes It is important to choose the correct angle here. It is defined as the angle between 2 lines, one in each plane, so that they are at right angles to the line of intersection of the 2 planes (like the angle between the tops of the pages of an open book). The picture below shows part of 2 planes and the angle between them.

To find this angle, will we first have to find the equation of the line of intersection of the 2 planes, and then find 2 vectors which are in the planes and perpendicular to this?

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Fortunately no! We just need to know a normal vector to each of the planes. Then we can find the angle we want very neatly as I show in the drawing below. The angle between the planes is the same as the angle between their 2 normal vectors (sliding their tails together if necessary). Now we just use n.m = |n||m| cos A and find the angle in the same way as we did for the 2 lines.

3.13 The angle between a line and a plane Again, the neatest method is to use a normal vector to the plane. I show how this works in the drawing below.

We slide the normal vector n until its tail is at the point of intersection with the line L with the plane P. Then n and L together define a plane which is perpendicular to plane P. The angle which line L makes with plane P is defined to be the red angle A in this plane. Since A and B together make a right angle, we can find A by using the dot product of n and the direction vector b of line Lto first findcos B.

Or we can find A even more directly by using the trig identity

cos B = cos (90 - A) = sin A

so n.b = |n||b| sin A so giving A.

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4. CHAPTER FOUR: DIFFERENTIATION 4.1 The Definition of the Derivative This is such an important limit and it arises in so many places that we give it a name. We call it a derivative. Here is the official definition of the derivative.

Definition

The derivative of with respect to x is the function and is defined as,

The derivative is really a function as well. We often “read” as “f prime of x”.

Let’s compute a couple of derivatives using the definition.

Example 1 Find the derivative of the following function using the definition of the derivative. Solution So, all we really need to do is to plug this function into the definition of the derivative and do some algebra. Admittedly, the algebra will get somewhat unpleasant at times, but it’s just algebra.

So, first plug the function into the definition of the derivative. Be careful and make sure that you properly deal with parenthesis when doing the

subtracting. Now, we know from the previous chapter that we can’t just plug in h=0 since this will give us a division by zero error. So we are going to have to do some work. In this case that means multiplying everything out and distributing the minus sign through on the second term. Doing this gives,

Notice that every term in the numerator that didn’t have an h in it canceled out and we can now factor an h out of the numerator which will cancel against the h in the denominator. After that we can compute the limit.

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EDUCATION BLOG TANZANIA So, the derivative is,

Example 2 Find the derivative of the following function using the definition of the derivative.

Solution This one is going to be a little messier as far as the algebra goes. However, outside of that it will work in exactly the same manner as the previous examples. First note that we changed all the letters in the definition to match up with the given function. Next, also note that we wrote the fraction a much more compact manner to help us with the work.

As with the first problem we can’t just plug in h=0. So we will need to simplify things a little. In this case we will need to combine the two terms in the numerator into a single rational expression. Before finishing this let’s note a couple of things. First, we didn’t multiply out the denominator. Multiplying out the denominator will just overly complicate things so let’s keep it simple. Next, as with the first example, after the simplification we only have terms with h’s in them left in the numerator. So, upon canceling the h we can evaluate the limit and get the derivative.

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Example 3 Find the derivative of the following function using the derivative.

Solution First plug into the definition of the derivative as we’ve done with the previous two examples

In this problem we’re going to have to rationalize the numerator. Again, after the simplification we have only h’s left in the numerator. So, cancel the h

and evaluate the limit.

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EDUCATION BLOG TANZANIA 4.2 Alternate Notation The typical derivative notation is the “prime” notation. However, there is another

notation that is used on occasion so let’s cover that.

Given a function y=f(x) all of the following are equivalent and represent the derivative of f(x) with respect to x. Because we also need to evaluate derivatives on occasion we also need a notation for evaluating derivatives when using the fractional notation. So if we want to evaluate the derivative at x=a all of the following are equivalent.

Note as well that on occasion we will drop the (x) part on the function to simplify the notation somewhat. In these cases the following are equivalent.

4.2 Differentiation Formulas For more complex functions using the definition of the derivative would be an almost impossible task. Luckily for us we won’t have to use the definition terribly often. We will have to use it on occasion, however we have a large collection of formulas and properties that we can use to simplify our life considerably and will allow us to avoid using the definition whenever possible.

We will introduce most of these formulas over the course of the next several sections. We

will start in this section with some of the basic properties and formulas. We will give the

properties and formulas in this section in both “prime” notation and “fraction” notation.

Properties

1) OR

In other words, to differentiate a sum or difference all we need to do is differentiate the individual terms and then put them back together with the appropriate signs. Note as well that this property is not limited to two functions.

This property is easy enough to show using the definition of the derivative. We’ll do this for the sum of two functions and we’ll leave it to you to check the difference of two functions.

We first plug the sum into the definition of the derivative and rewrite the numerator a little.

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Now, break up the fraction into two pieces and recall that the limit of a sum is the sum of the limits. Using this fact we see that we end up with the definition of the derivative for each of the two functions.

2)

OR

, c is any number

So, we can factor a multiplicative constant out of a derivative if we need to.

This is also very easy to show using the definition provided you recall that we can factor a constant out of a limit.

Note that we have not included formulas for the derivative of products or quotients of two functions in these properties. The derivative of a product or quotient of two functions is not the product or quotient of the derivatives of the individual pieces. We will take a look at these in the next section.

Facts

1) If then OR

The derivative of a constant is zero.

This can be easily checked using the definition of the derivative.

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2) If then OR , n is any number. This formula is sometimes called the power rule. All we are doing here is bringing the original exponent down in front and multiplying and then subtracting one from the original exponent.

Note as well that in order to use this formula n must be a number, it can’t be a variable and the base must be a variable.

Showing this using the definition of the derivative is not terribly difficult, but is a little tedious and so we won’t show it here.

These are the only properties and formulas that we’ll give in this section. Let’s do compute some derivatives using these properties.

Example 1 Differentiate each of the following functions.

(a)

(b)

(c)

(d)

(e) Solution

(a) In this case we have the sum and difference of four terms and so we will differentiate each of the terms using the two facts and then put them back together with the proper sign. So, here is the derivative for this function.

Notice that in the third term the exponent was a one and so upon subtracting 1 from the

original exponent we get a new exponent of zero. Now recall that x0=1. Also notice that

in each term where there was a coefficient we just multiplied the coefficient times the derivative of the x term.

(b) The point of this problem is to make sure that you deal with negative exponents

correctly. Here is the derivative.

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Make sure that you correctly deal with the exponents in these cases, especially the negative exponents.

(c) Now in this function the second term is not correctly set up for us to use the power rule. The power rule requires that the term be a variable to a power only and the term must be in the numerator.

So, prior to differentiating we first need to rewrite the second term into a form that we can deal with.

Note that we left the 3 in the denominator and only moved the variable up to the numerator. This is usually what you will want to do in these cases. We can now differentiate the function. (d) All of the terms in this function have roots in them. In order to use the power rule we need to first convert all the roots to fractional exponents.

In the last two terms we combined the exponents. You should always do this. Also we moved the term in the denominator of the third term up to the numerator. We can now differentiate the function.

Make sure that you can deal with fractions. You will see a lot of them in this class.

(e) In all of the previous examples the exponents have been nice integers or fractions. That is usually what we’ll see in this class. However, the exponent only needs to be a number so don’t get excited about problems like this one. They work exactly the same.

The answer is a little messy and we won’t reduce the exponents down to decimals.

However, this problem is not terribly difficult it just looks that way initially.

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(a)

(b) Solution (a) In this function we can’t just differentiate the first term, differentiate the second term and then multiply the two back together. That just won’t work. We will discuss this in detail in the next section so if you’re not sure you believe that hold on for a bit and we’ll be looking at that soon.

It is still possible to do this derivative however. All that we need to do is convert the radical to fractional exponents (as we should anyway) and then multiply this through the parenthesis.

Now we can differentiate the function. (b) As with the first part we can’t just differentiate the numerator and the denominator and the put it back together as a fraction. Again, if you’re not sure you believe this hold on until the next section and we’ll take a more detailed look at this.

We can simplify this rational expression however as follows.

This is a function that we can differentiate. So, as we saw in this example there are a few products and quotients that we can differentiate. If we can first do some simplification the functions will sometimes simplify into a form that can be differentiated using the properties and formulas in this section.

4.3 Product and Quotient Rule To differentiate products and quotients we have the Product Rule and the Quotient Rule.

4.3.1 Product Rule

If the two functions f(x) and g(x) are differentiable (the derivative exist) then the product is differentiable and,

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(a)

(b)

Solution

a) There’s not really a lot to do here other than use the product rule. However,

before doing that we should convert the radical to a fractional exponent as always.

Now let’s take the derivative. So we take the derivative of the first function times the second then add on to that the first function times the derivative of the second function.

This is NOT what we got in the previous section for this derivative. However, with some simplification we can arrive at the same answer. (b) This one is actually easier than the previous one. Let’s just run it through the product rule.

5.3.2 Quotient Rule

If the two functions f(x) and g(x) are differentiable (the derivative exist) then the quotient is differentiable and,

Example Differentiate each of the following functions.

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EDUCATION BLOG TANZANIA Solution There’s not much to do with these other than use the quotient rule on them. 4.4 Derivatives of Trig Functions In this section we are going to take a look at the derivatives of the six trig functions. Two of the derivatives will be derived. The remaining four are left to the reader and will follow similar proofs for the two given here.

We’ll start with the derivative of the sine function. To do this we will need to use the definition of the derivative.

Since we can’t just plug in h=0 to evaluate the limit we will need to use a trig formula on the first sine in the numerator. Doing this gives us, As you can see upon using the trig formula we can combine the first and third term. We can then break up the fraction into two pieces, both of which can be dealt with separately.

Now, both of the limits here are limits as h approaches zero. In the first limit we have a sin(x) and in the second limit we have a cos(x). Both of these are only functions of x and as h moves in towards zero this has no affect on the value of x. Therefore, as far as the limits are concerned, these two functions are constants and can be factored out of their respective limits. Doing this gives,

To finish off this proof we will need the following facts

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Fact We will eventually be able to prove these limits. Ironically, the proof of these limits will be an application of the derivative that we’ll be taking a look at in the next chapter.

Using these limits gives us the final answer for the derivative of the sine function. Differentiating cosine is done in a similar fashion. but other than that is an almost identical proof. done with the proof you should get,

It will require a different trig formula,

The details will be left to you. When

With these two out of the way the remaining four are fairly simple to get. All the remaining four trig functions can be defined in terms of sine and cosine and these definitions can be used to get their derivatives.

Let’s take a look at tangent. Tangent is defined as,

Now that we have the derivatives of sine and cosine all that we need to do is use the quotient rule on this. Let’s do that.

Don’t forget that ! Now, while this is a formula for the derivative

we can go one step farther by recalling the definition of secant.

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The remaining three trig functions are also quotients involving sine and cosine and so can be differentiated in a similar manner. We’ll leave the details to you. Here are the derivatives of all six of the trig functions.

Derivatives of the six trig functions


(a)

(b)

Solution (a) There really isn’t a whole lot to this problem. We’ll just differentiate each term

using the formulas from above.

(b) In this part we’ll need to use the quotient rule. Be careful with the signs when differentiating the denominator. The negative sign we get from differentiating the cosine will cancel against the negative sign that is already there.

This appears to be done, but there is actually a fair amount of simplification that can yet be done. To do this we need to factor out a “-2” from the last two terms in the numerator

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and the make use of the fact that

. Example 2 Suppose that the amount of money in a bank account is given by

where t is in years. During the first 10 years in which the account is open when is the

amount of money in the account increasing?

Solution To determine when the amount of money is increasing we need to determine when the rate of change is positive. Since we know that the rate of change is given by the derivative that is the first thing that we need to find. Now, we need to determine where in the first 10 years this will be positive. This is equivalent to asking where in the interval [0, 10] is the derivative positive. Recall that both sine and cosine are continuous functions and so the derivative is also a continuous function. The Intermediate Value Theorem then tells us that the derivative can only change sign if it first goes through zero.

So, we need to solve the following equation.

The solution to this equation is,

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We are only interested in those solutions that fall in the range [0, 10]. Those solutions

are, So, much like solving polynomial inequalities all that we need to do is sketch in a number line and add in these points. These points will divide the number line into regions in which the derivative must always be the same sign. All that we need to do then is choose a test point from each region to determine the sign of the derivative in that region.

Here is the number line with all the information on it.

So, it looks like the amount of money in the bank account will be increasing during the following intervals.

Note that we can’t say anything about what is happening after t=10 since we haven’t done any work for t’s after that point.

4.5 Derivatives of Exponential and Logarithm Functions The next set of functions that we want to take a look at are exponential and logarithm functions. The most common exponential and logarithm functions in a calculus course

are the natural exponential function, , and the natural logarithm function, .

We will take a more general approach however and look at the general exponential and

logarithm function.

4.5.1 Exponential Functions We’ll start off by looking at the exponential function,

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We want to differentiate this. The power rule that we looked at a couple of sections ago won’t work as that required the exponent to be a fixed number and the base to be a variable. That is exactly the opposite from what we’ve got with this function. So, we’re going to have to start with the definition of the derivative.

Now, the ax is not affected by the limit since it doesn’t have any h’s in it and so is a

constant as far as the limit is concerned. We can therefore factor this out of the limit. This gives, Now let’s notice that the limit is exactly the definition of the derivative at x=0. Therefore, the derivative becomes, So, we are kind of stuck we need to know the derivative in order to get the derivative! The following fact will help with one possible value of a.

Fact

For the natural exponential function, we have .

So, provided we are using the natural exponential function we get the following.

At this point we’re missing some knowledge that will allow us to easily get the derivative for a general function. Eventually we will be able to show that for a general exponential function we have,

4.5.2 Logarithm Functions In this case we will need to start with the following fact about functions that are inverses of each other.

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EDUCATION BLOG TANZANIA Fact If f(x) and g(x) are inverses of each other then,

So, how is this fact useful to us? Well recall that the natural exponential function and the natural logarithm function are inverses of each other and we know what the derivative of the natural exponential function is!

So, if we have and then, The last step just uses the fact that the two functions are inverses of each other.

Putting this all together gives, Note that we need to require that x>0 since this is required for the logarithm and so must also be required for its derivative. In can be shown that,

Using this all we need to avoid is x=0.

In this case, unlike the exponential function case, we can actually find the derivative of the general logarithm function. All that we need is the change of base formula. Using the change of base formula we can write a general logarithm as,

Differentiation is then fairly simple.

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We took advantage of the fact that a was a constant and so ln a is also a constant and can be factored out of the derivative. Putting all this together gives,

Here is a summary of the derivatives in this section.

Okay, now that we have the derivations of the formulas out of the way let’s compute a couple of derivatives.


(a)

(b)

Solution (a) This will be the only example that doesn’t involve the natural exponential and

natural logarithm functions.

(b) Not much to this one. Just remember to use the product rule on the second term.

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4.6 Derivatives of Inverse Trig Functions In this section we are going to look at the derivatives of the inverse trig functions. In order to derive the derivatives of inverse trig functions we’ll need the formula from the last section relating the derivatives of inverse functions.

Since it’s going to be so important to our work here we’ll go ahead and give it again. If f(x) and g(x) are inverse functions then,

Recall as well that two functions are inverses if f(g(x))=x and g(f(x))=x.

Inverse Sine Let’s start with inverse sine. Here is the definition of the inverse sine.

So, evaluating an inverse trig function is the same as asking what angle (i.e. y) did we plug into the sine function to get x. Let’s work a quick example.

Example 1

Evaluate Solution So we are really asking what angle y solves the following

equation. and we are restricted to the values of y above.

From a unit circle we can quickly see that .

Note as well that since we also have . We also have

the following relationship between the inverse sine function and the sine function. In other words they are inverses of each other. This means that we can use the fact above to find the derivative of inverse sine. Let’s start with,

Then,

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This is not a very useful formula. Let’s see if we can get a better formula. Let’s start by recalling the definition of the inverse sine function.

Using the first part of this definition the denominator in the derivative becomes, Now, recall that Using this, the denominator is now, Now, use the second part of the definition of the inverse sine function. The denominator is then,

Putting all of this together gives the following derivative.

Inverse Cosine Now let’s take a look at the inverse cosine. Here is the definition for the inverse cosine.

We will also have here as we did the inverse sine and we also have the

following facts.

Once again they are inverses of each other.

So to find the derivative we’ll do the same kind of work that we did with the inverse sine above. If we start with

then, Simplifying the denominator here is almost identical to the work we did for the inverse sine and so isn’t shown here. Upon simplifying we get the following derivative.

So, the derivative of the inverse cosine is nearly identical to the derivative of the inverse sine.

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EDUCATION BLOG TANZANIA Inverse Tangent Here is the definition of the inverse tangent.

Notice that we can’t let y be either of the two endpoints in the restriction above since tangent isn’t even defined at those two points.

In this case, unlike the previous two we have no restriction on x. So, we can plug any x into the inverse trig function. This means that we can ask for the limits of the inverse tangent function as x goes to plus or minus infinity. To do this we’ll need the graph of the inverse tangent function. This is shown below.

From this graph we can see that

The tangent and inverse tangent functions are inverse functions so,

Therefore to find the derivative of the inverse tangent function we can start with

We then have,

Simplifying the denominator is similar to the inverse sine, but different enough to warrant showing the details. We’ll start with the definition of the inverse tangent.

The denominator is then,

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EDUCATION BLOG TANZANIA Now, if we start with the fact that

and divide every term by cos2 y we will get,

The denominator is then, Finally using the second portion of the definition of the inverse tangent function gives us, The derivative of the inverse tangent is then,

Summary There are three more inverse trig functions but the three shown here the most common ones. Formulas for the remaining three could be derived by a similar process as we did those above. Here are the derivatives of all six inverse trig functions.

Example Differentiate the following functions.

(a)

(b)

Solution (a) Not much to do with this one other than differentiate each term.

(b) Don’t forget to convert the radical to fractional exponents before using the product rule

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EDUCATION BLOG TANZANIA 4.6.1 Alternate Notation There is some alternate notation that is used on occasion to denote the inverse trig functions. This notation is,

4.7 Chain Rule We’ve taken a lot of derivatives over the course of the last six sections. However, if you look back they have all been functions similar to the following kinds of functions.

These are all fairly simple functions. What about functions like the following, None of our rules will work on these functions and yet some of these functions are closer to the derivatives that we’re liable to run into than the functions in the first set.

Let’s take the first one for example. Back in the section on the definition of the derivative

we actually used the definition to compute this derivative. In that section we found that, If we were to just use the power rule on this we would get, which is not the derivative that we computed using the definition. It is close, but it’s not the same. So, the power rule alone simply won’t work to get the derivative of this function. Let’s keep looking at this function and note that if we define, then we can write the function as a composition. It turns out that it’s actually fairly simple to differentiate a function composition using the Chain Rule. There are two forms of the chain rule. Here they are.

Chain Rule Suppose that we have two functions f(x) and g(x) and they are both differentiable.

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1. If we define then the derivative of F(x) is,

2. If we have y=f(u) and u=g(x) then the derivative of y is, Each of these forms have their uses, however we will work mostly with the first form in this class. Example 1

Use the Chain Rule to differentiate .

Solution We’ve already identified the two functions that we needed for the composition, but let’s write them back down anyway and take their derivatives.

So, using the chain rule we get, And this is what we got using the definition of the derivative.

Example 2 Differentiate each of the following.

(a)

(b)

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EDUCATION BLOG TANZANIA 4.8 Implicit Differentiation To this point we’ve done quite a few derivatives, but they have all been derivatives of functions of the form y=f(x). Unfortunately not all the functions that we’re going to look at will fall into this form.

Let’s take a look at an example of this.

Example 1

Find for xy=1.

Solution There are actually two solution methods for this problem.

Solution 1 : This is the simple way of doing the problem. Just solve for y to get the function in the form that we’re used to dealing with. So, that’s easy enough to do. However, there are some functions for which this can’t be done. That’s where the second solution technique comes into play.

Solution 2 : In this case we’re going to leave the function in the form that we were given. We do want to remember however that we are thinking of y as a function of x. In other words, y=y(x). Let’s rewrite the equation to note this.

Now, we will differentiate both sides with respect to x. The right side is easy. It’s just the derivative of a constant. The left side is also easy, but we’ve got to recognize that we’ve actually got a product here. So to do the derivative of the left side we’ll need to do the product rule.

Now, recall that

so, we get, Note that we dropped the (x) on the y as it’s no longer really needed. We just wanted it in the equation to recognize the product rule.

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This is not what we got from the first solution. Or at least it doesn’t look like the same derivative. Recall however, that we do know what y is in terms of x and if we plug that in we will get, which is what we got from the first solution.

The process that we used in the second solution to the previous example is called implicit differentiation. In the previous example we were able to just solve for y and avoid implicit differentiation. However, that won’t always be the case.

Let’s see an example of this.

Example 2

Find for the following function.

Solution This is just a circle and while can solve for y this would give, In other words, we would get two functions. This would cause us problems in getting the derivative and so will do us no good. We want a single function for the derivative and using this would give us two.

In this example we really are going to need to do implicit differentiation. We’ll do the same thing we did in the first example and remind ourselves that y is really a function of x and write y as y(x).

Notice that when we differentiated the y term we used the chain rule. Dropping the (x) part of the y and solving for the derivative give,

Unlike the first example we can’t just plug in for y since we wouldn’t know which of the two roots to use. Most answers from implicit differentiation will involve both x and y so don’t get excited about that when it happens.

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EDUCATION BLOG TANZANIA 4.9 Related Rates This is an application of implicit differentiation. For these related rates problems it’s usually best to just jump right into some problems and see how they work.

Example 1 Air is being pumped into a spherical balloon at a rate of 5 cm3/min.

Determine the rate at which the radius of the balloon is increasing when the diameter of the balloon is 20 cm.

Solution The first thing that we’ll need to do here is to identify what information that we’ve been given and what we want to find. Before we do that let’s notice that both the volume of the balloon and the radius of the balloon will vary with time and so are really functions of time.

Now, we know that air is being pumped into the balloon at a rate of 5 cm3/min. This is

the rate at which the volume is increasing. Recall that rates of change are nothing more than derivatives and so we know that, We want to determine the rate at which the radius is changing. Again, rates are derivatives and so it looks like we want to determine,

Note that we needed to convert the diameter to a radius.

Now that we’ve identified what we have been given and what we want to find we need to relate these two quantities to each other. In this case we can relate the volume and the radius with the formula for the volume of a sphere. We will typically not use the (t) part of things, but since this is the first time through one of these we will do that to remind ourselves that they are really functions of t.

Now we don’t really want a relationship between the volume and the radius. We really want a relationship between their derivatives. We can do this by differentiating both sides with respect to t. In other words, we will need to do implicit differentiation.

Doing this gives, At this point all that we need to do is plug in what we know and solve for what we want to find.

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EDUCATION BLOG TANZANIA We can get the units of the derivative be recalling that, The units of the derivative will be the units of the numerator (cm in the previous example) divided by the units of the denominator (min in the previous example).

Let’s work another example

Example 2 A 15 foot ladder is resting against the wall. The bottom is initially 10 feet away from the

wall and is being pushed towards the wall at a rate of ft/sec. How fast is the top of the ladder moving up the wall 12 seconds after we start pushing?

Solution The first thing to do in this case is to sketch picture that shows us what is going on.

So, we’ve defined the distance of the bottom of the latter from the wall to be x and the distance of the top of the ladder from the floor to be y. We know that the rate at which the bottom of the ladder is moving towards the wall. This is,

Note that the rate is negative since the distance from the wall, x, is decreasing. We always need to be careful with signs with these problems.

We want to find the rate at which the top of the ladder is moving away from the floor.

This is . Note as well that this quantity should be positive since y will be increasing.

As with the first example we first need a relationship between x and y. We can get this using Pythagorean theorem.

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All that we need to do at this point is to differentiate both sides with respect to t and we’ll get the relationship between the derivatives. This equation has two quantities that we don’t know and will need to know in order to finish the problem. We need to determine x and y. Both are fairly simple to get. We

know that initially x=10 and the end is being pushed in towards the wall at a rate of

ft/sec and that we are interested in what has happened after 12 seconds. This means that, To find y (after 12 seconds) all that we need to do is reuse the Pythagorean Theorem with the values of x that we just found above.

Now all that we need to do is plug into the equation and solve for . Notice that we got the correct sign. If we’d gotten a negative then we’d have known that we had made a mistake and we could go back and look for it.

4.10 Logarithmic Differentiation There is one last topic to discuss in this section. Taking the derivatives of some complicated functions can be simplified by using logarithms. This is called logarithmic differentiation.

It’s easiest to see how this works in an example. Example 1 Differentiate the function.

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Solution Differentiating this function could be done with a product rule and a quotient rule. However, that would be a fairly messy process. We can simplify things somewhat by taking logarithms of both sides. Of course, this isn’t really simpler. What we need to do is use the properties of logarithms to expand the right side as follows.

This doesn’t look all the simple. However, the differentiation process will be simpler. What we need to do at this point is differentiate both sides with respect to x. Note that this is really implicit differentiation.

To finish the problem all that we need to do is multiply both sides by y and the plug in for y since we do know what that is.

We can also use logarithmic differentiation to differentiate functions in the form. Example 2

Differentiate

Solution We’ve seen two functions similar to this at this point.

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EDUCATION BLOG TANZANIA Neither of these two will work here since both require either the base or the exponent to be a constant. In this case both the base and the exponent are variables and so we have no way to differentiate this function using only known rules from previous sections.

With logarithmic differentiation we can do this however. First take the logarithm of both sides as we did in the first example and use the logarithm properties to simplify things a little.

Differentiate both sides using implicit differentiation. As with the first example multiply by y and substitute back in for y.

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