FREE FORM FOUR MATHEMATICS NOTES
Read all the form 4 notes here. You can also download a copy of the pdf notes on this link; MATH FORM FOUR NOTES
Specific Objectives
By the end of the topic the learner should be able to:
(a) Relate image and object under a given transformation on the Cartesian
Plane;
(b) Determine the matrix of a transformation;
(c) Perform successive transformations;
(d) Determine and identify a single matrix for successive transformation;
(e) Relate identity matrix and transformation;
(f) Determine the inverse of a transformation;
(g) Establish and use the relationship between area scale factor and determinant of a matrix;
(h) Determine shear and stretch transformations;
(i) Define and distinguish isometric and nonisometric transformation;
(j) Apply transformation to real life situations.
Content
(a) Transformation on the Cartesian plane
(b) Identification of transformation matrix
(c) Successive transformations
(d) Single matrix of transformation for successive transformations
(e) Identity matrix and transformation
(f) Inverse of a transformations
(g) Area scale factor and determinant of a matrix
(h) Shear and stretch (include their matrices)
(i) Isometric and nonisometric transformations
(j) Application of transformation to real life situations.
Matrices of transformation
A transformation change the shape, position or size of an object as discussed in book two.
Pre –multiplication of any 2 x 1 column vector by a 2 x 2 matrix results in a 2 x 1 column vector
Example
If the vector is thought of as apposition vector that is to mean that it is representing the points with coordinates (7, 1) to the point (17, 9).
Note;
The transformation matrix has an effect on each point of the plan. Let’s make T a transformation matrix T Then T maps points (x, y) onto image points
T
Finding the Matrix of transformation
The objective is to find the matrix of given transformation.
Examples
Find the matrix of transformation of triangle PQR with vertices P (1, 3) Q (3, 3) and R (2, 5).The vertices of the image of the triangle sis.
Solution
Let the matrix of the transformation be
=
Equating the corresponding elements and solving simultaneously
2a= 2
2c= 0
Therefore the transformation matrix is
Example
A trapezium with vertices A (1 ,4) B(3,1) C (5,1) and D(7,4) is mapped onto a trapezium whose vertices are .Describe the transformation and find its matrix
Solution
Let the matrix of the transformation be
Equating the corresponding elements we get;
Solve the equations simulteneously
11b = 11 hence b =1 or a = 0
3c + d =3
The matrix of the transformation is therefore
The transformation is positive quarter turn about the origin
Note;
Under any transformation represented by a 2 x 2 matrix, the origin is invariant, meaning it does not change its position.Therefore if the transformtion is a rotation it must be about the origin or if the transformation is reflection it must be on a mirror line which passses through the origin.
The unit square
The unit square ABCD with vertices A helps us to get the transformation of a given matrix and also to identify what trasformation a given matrix represent.
Example
Find the images of I and J under the trasformation whose matrix is;
Solution
NOTE;
The images of I and J under transformation represented by any 2 x 2 matrix i.e., are
Example
Find the matrix of reflection in the line y = 0 or x axis.
Solution
Using a unit square the image of B is ( 1, 0) and D is (0 , 1 ) .Therefore , the matrix of the transformation is
Example
Show on a diagram the unit square and it image under the transformation represented by the matrix
Solution
Using a unit square, the image of I is ( 1 ,0 ), the image of J is ( 4 , 1),the image of O is ( 0,0) and that of K is
Successive transformations
The process of performing two or more transformations in order is called successive transformation eg performing transformation H followed by transformation Y is written as follows YH or if A , b and C are transformations ; then ABC means perform C first ,then B and finally A , in that order.
The matrices listed below all perform different rotations/reflections:
This transformation matrix is the identity matrix. When multiplying by this matrix, the point matrix is unaffected and the new matrix is exactly the same as the point matrix.
This transformation matrix creates a reflection in the xaxis. When multiplying by this matrix, the x coordinate remains unchanged, but the y coordinate changes sign.
This transformation matrix creates a reflection in the yaxis. When multiplying by this matrix, the y coordinate remains unchanged, but the x coordinate changes sign.
This transformation matrix creates a rotation of 180 degrees. When multiplying by this matrix, the point matrix is rotated 180 degrees around (0, 0). This changes the sign of both the x and y coordinates.
This transformation matrix creates a reflection in the line y=x. When multiplying by this matrix, the x coordinate becomes the y coordinate and the yordinate becomes the x coordinate.
This transformation matrix rotates the point matrix 90 degrees clockwise. When multiplying by this matrix, the point matrix is rotated 90 degrees clockwise around (0, 0).
This transformation matrix rotates the point matrix 90 degrees anticlockwise. When multiplying by this matrix, the point matrix is rotated 90 degrees anticlockwise around (0, 0).
This transformation matrix creates a reflection in the line y=x. When multiplying by this matrix, the point matrix is reflected in the line y=x changing the signs of both coordinates and swapping their values.
Inverse matrix transformation
A transformation matrix that maps an image back to the object is called an inverse of matrix.
Note;
If A is a transformation which maps an object T onto an image ,then a transformation that can map back to T is called the inverse of the transformation A , written as image .
If R is a positive quarter turn about the origin the matrix for R is and the matrix for is hence
Example
T is a triangle with vertices A (2, 4), B (1, 2) and C (4, 2).S is a transformation represented by the matrix
 Draw T and its image under the transformation S
 Find the matrix of the inverse of the transformation S
Solution
 Using transformation matrix S =
 Let the inverse of the transformation matrix be. This can be done in the following ways
Therefore
Equating corresponding elements and solving simultaneously;
Therefore
Area Scale Factor and Determinant of Matrix
The ratio of area of image to area object is the area scale factor (A.S.F)
Are scale factor =
Area scale factor is numerically equal to the determinant. If the determinant is negative you simply ignore the negative sign.
Example
Area of the object is 4 cm and that of image is 36 cm find the area scale factor.
Solution
If it has a matrix of
Shear and stretch
Shear
The transformation that maps an object (in orange) to its image (in blue) is called a shear
The object has same base and equal heights. Therefore, their areas are equal. Under any shear, area is always invariant ( fixed)
A shear is fully described by giving;
 The invariant line
 A point not on the invariant line, and its image.
Example
A shear X axis invariant
Example
A shear Y axis invariant
Note;
Shear with x axis invariant is represented by a matrix of the form under this trasnsformation,J (0, 1) is mapped onto .
Likewise a shear with y – axis invariant is represented by a matrix of the form ( ). Under this transformation, I (0,1) is mapped onto .
Stretch
A stretch is a transformation which enlarges all distance in a particular direction by a constant factor. A stretch is described fully by giving;
 The scale factor
 The invariant line
Note;
i.)If K is greater than 1, then this really is a stretch.
ii.) If k is less than one 1, it is a squish but we still call it a stretch
iii.)If k = 1, then this transformation is really the identity i.e. it has no effect.
Example
Using a unit square, find the matrix of the stretch with y axis invariant ad scale factor 3
Solution
The image of I is therefore the matrix of the stretch is
Note;
The matrix of the stretch with the yaxis invariant and scale factor k is and the matrix of a stretch with x – axis invariant and scale factor k is
Isometric and Non Isometric Transformation
Isometric transformations are those in which the object and the image have the same shape and size (congruent) e.g. rotation, reflection and translation
Non isometric transformations are those in which the object and the image are not congruent e.g., shear stretch and enlargement
End of topic
Did you understand everything?
If not ask a teacher, friends or anybody and make sure you understand before going to sleep! 
Past KCSE Questions on the topic.
 Matrix p is given by 1 2
4 3
(a) Find P^{1}
(b) Two institutions, Elimu and Somo, purchase beans at Kshs. B per bag and
maize at Kshs m per bag. Elimu purchased 8 bags of beans and 14 bags of maize for Kshs 47,600. Somo purchased 10 bags of beans and 16 of maize for Kshs. 57,400
(c) The price of beans later went up by 5% and that of maize remained constant. Elimu bought the same quantity of beans but spent the same total amount of money as before on the two items. State the new ratio of beans to maize.
 A triangle is formed by the coordinates A (2, 1) B (4, 1) and C (1, 6). It is rotated
clockwise through 90^{0} about the origin. Find the coordinates of this image.
 On the grid provided on the opposite page A (1, 2) B (7, 2) C (4, 4) D (3, 4) is a trapezium
(a) ABCD is mapped onto A’B’C’D’ by a positive quarter turn. Draw the image A’B’C’D on the grid
(b) A transformation 2 1 maps A’B’C’D onto A”B” C”D” Find the coordinates
0 1 of A”B”C”D”
 A triangle T whose vertices are A (2, 3) B (5, 3) and C (4, 1) is mapped onto triangle T^{1} whose vertices are A^{1} (4, 3) B^{1} (1, 3) and C^{1} (x, y) by a
Transformation M = a b
c d
 a) Find the: (i) Matrix M of the transformation
(ii) Coordinates of C_{1}
 b) Triangle T^{2 }is the image of triangle T^{1} under a reflection in the line y = x.
Find a single matrix that maps T and T_{2}
 Triangles ABC is such that A is (2, 0), B (2, 4), C (4, 4) and A”B”C” is such that A” is (0, 2), B” (4 – 10) and C “is (4, 12) are drawn on the Cartesian plane
Triangle ABC is mapped onto A”B”C” by two successive transformations
R = a b
c d Followed by P = 0 1
1 0
(a) Find R
(b) Using the same scale and axes, draw triangles A’B’C’, the image of triangle ABC under transformation R
Describe fully, the transformation represented by matrix R
 Triangle ABC is shown on the coordinate’s plane below
(a) Given that A (6, 5) is mapped onto A (6,4) by a shear with y axis invariant
 Draw triangle A’B’C’, the image of triangle ABC under the shear
 Determine the matrix representing this shear
(b) Triangle A B C is mapped on to A” B” C” by a transformation defined by the matrix 1 0
1½ 1
(i) Draw triangle A” B” C”
(ii) Describe fully a single transformation that maps ABC onto A”B” C”
 Determine the inverse T^{‑1} of the matrix 1 2
1 1
Hence find the coordinates to the point at which the two lines
x + 2y = 7 and x – y =1
 Given that A = 0 1 and B = 1 0
3 2 2 4
Find the value of x if
(i) A 2x = 2B
(ii) 3x – 2A = 3B
(iii) 2A – 3B = 2x
 The transformation R given by the matrix
A = a b maps 17 to 15 and 0 to 8
c d 0 8 17 15
(a) Determine the matrix A giving a, b, c and d as fractions
(b) Given that A represents a rotation through the origin determine the angle of rotation.
(c) S is a rotation through 180 about the point (2, 3). Determine the image of (1, 0) under S followed by R.
CHAPTER FIFTY SEVEN
Specific Objectives
By the end of the topic the learner should be able to:
(a) State the measures of central t e n d e n c y;
(b) Calculate the mean using the assumed mean method;
(c) Make cumulative frequency table,
(d) Estimate the median and the quartiles b y
– Calculation and
– Using ogive;
(e) Define and calculate the measures of dispersion: range, quartiles,interquartile range, quartile deviation, variance and standard deviation
(f) Interpret measures of dispersion
Content
(a) Mean from assumed mean:
(b) Cumulative frequency table
(c) Ogive
(d) Meadian
(e) Quartiles
(f) Range
(g) Interquartile range
(h) Quartile deviation
(i) Variance
(j) Standard deviation
These statistical measures are called measures of central tendency and they are mean, mode and median.
Mean using working (Assumed) Mean
Assumed mean is a method of calculating the arithmetic men and standard deviation of a data set. It simplifies calculation.
Example
The masses to the nearest kilogram of 40 students in the form 3 class were measured and recorded in the table below. Calculate the mean mass
Mass kg  47  48  49  50  51  52  53  
Number of employees  2  0  1  2  3  2  5 
54  55  56  57  58  59  60 
6  7  5  3  2  1  1 
Solution
We are using assumed mean of 53
Mass x kg  t= x – 53  f  ft 
47
48 49 50 51 52 53 54 
6
5 4 3 2 1 0 1

2
0 1 2 3 2 5 6

12
0 4 6 6 2 0 6 
55  2  7  14 


56  3  5  15 
57  4  3  12 
58  5  2  10 
60  7  1
1 
6
7 
Σf = 40  Σft = 40 
Mean of t
Mean of x = 53 + mean of t
= 53 + 1
= 54
Mean of grouped data
The masses to the nearest gram of 100 eggs were as follows
Marks  100 103  104 107  108 111  112115  116119  120123 
Frequency  1  15  42  31  8  3 
Find the mean mass
Solution
Let use a working mean of 109.5.
class  Midpoint x  t= x – 109.5  f  f t 
100103  101.5  8  1  – 8 
104107  105.5  4  15  – 60 
108111  109.5  0  42  0 
112115  113.5  4  31  124 
116 119
120 123 
117.5
121.5 
8
12 
8
3 
64
36 
Σf= 100  Σft = 156 
Mean of t =
Therefore,mean of x = 109.5 + mean of t
= 109.5 + 1.56
= 111.06 g
To get the mean of a grouped data easily,we divide each figure by the class width after substracting the assumed mean.Inorder to obtain the mean of the original data from the mean of the new set of data, we will have to reverse the steps in the following order;
 Multiply the mean by the class width and then add the working mean.
Example
The example above to be used to demonstrate the steps
class  Midpoint x  t=  f  f t 
100103  101.5  2  1  – 2 
104107  105.5  1  15  – 15 
108111  109.5  0  42  0 
112115  113.5  1  31  31 
116 119
120 123 
117.5
121.5 
2
3 
8
3 
16
9 
Σf= 100  Σft = 39 
= 0.39
Therefore = 0.39 x 4 + 109.5
= 1.56 + 109.5
= 111.06 g
Quartiles, Deciles and Percentiles
A median divides a set of data into two equal part with equal number of items.
Quartiles divides a set of data into four equal parts.The lower quartile is the median of the bottom half.The upper quartile is the median of the top half and the middle coincides with the median of the whole set od data
Deciles divides a set of data into ten equal parts.Percentiles divides a set of data into hundred equal parts.
Note;
For percentiles deciles and quartiles the data is arranged in order of size.
Example
Height in cm  145 149  150154  155159  160
164 
165169  170174  175179 
frquency  2  5  16  9  5  2  1 
Calculate the ;
 Median height
 )Lower quartile
 ii) Upper quartile
 80^{th} percentile
Solution
 There are 40 students. Therefore, the median height is the average of the heights of the 20^{th} and 21^{st}
class  frequency  Cumulative frequency 
145149  2  2 
150 – 154  5  7 
155 – 159  16  23 
160 – 164
165 – 169 
9
5 
32
37 
170 – 174
175 – 179 
2
1 
39
40 
Both the 20^{th}and 21^{st}students falls in the 155 159 class. This class is called the median class. Using the formula m = L +
Where L is the lower class limit of the median class
N is the total frequency
C is the cumulative frequency above the median class
I is the class interval
F is the frequency of the median class
Therefor;
Height of the 20^{th} student = 154.5 +
= 154.5 + 4.0625
=158.5625
Height of the 21^{st} = 154.5 +
= 154.5 + 4.375
=158.875
Therefore median height =
= 158.7 cm
 (I ) lower quartile = L +
The 10^{th} student fall in the in 155 – 159 class
= 154.5 +
5 + 0.9375
4375
(ii) Upper quartile= L +
The 10^{th} student fall in the in 155 – 159 class
= 159.5 +
5 + 3.888
3889
Note;
The median corresponds to the middle quartile or the 50^{th} percentile
 the 32^{nd} student falls in the 160 164 class
= L +
= 159.5 +
5 + 5
Example
Determine the upper quartile and the lower quartile for the following set of numbers
5, 10 ,6 ,5 ,8, 7 ,3 ,2 ,7 , 8 ,9
Solution
Arranging in ascending order
2, 3, 5,5,6, 7,7,8,8,9,10
The median is 7
The lower quartile is the median of the first half, which is 5.
The upper quartile is the median of the second half, which is 8.
Median from cumulative frequency curve
Graph for cumulative frequency is called an ogive. We plot a graph of cumulative frequency against the upper class limit.
Example
Given the class interval of the measurement and the frequency,we first find the cumulative frequency as shown below.
Then draw the graph of cumulative frequency against upper class limit
Arm Span (cm)  Frequency (f)  Cumulative Frequency 
140 ≤ x ‹ 145  3  3 
145 ≤ x ‹ 150  1  4 
150 ≤ x ‹ 155  4  8 
155 ≤ x ‹ 160  8  16 
160 ≤ x ‹ 165  7  23 
165 ≤ x ‹ 170  5  28 
170 ≤ x ‹ 175  2  30 
Total:  30 
Solution
Example
The table below shows marks of 100 candidates in an examination
110  1120  2130  3140  4150  5160  6170  7180  8190  91100 
4  9  16  24  18  12  8  5  3  1 
Marks
FRCY
 Determine the median and the quartiles
 If 55 marks is the pass mark, estimate how many students passed
 Find the pass mark if 70% of the students are to pass
 Determine the range of marks obtained by
(I) The middle 50 % of the students
(ii) The middle 80% of the students
Solution
110  1120  2130  3140  4150  5160  6170  7180  8190  91100 
4  9  16  24  18  12  8  5  3  1 
Marks
Frqcy
Cumulative 4 13 29 53 71 83 91 96 99 100
Frequency
Solution
 Reading from the graph
The median = 39.5
The Lower quartile
The upper quartile
 23 candidates scored 55 and over
 Pass mark is 31 if 70% of pupils are to pass
 (I) The middle 50% include the marks between the lower and the upper quartiles i.e. between 28.5 and 53.5 marks.
(II) The middle 80% include the marks between the first decile and the 9^{th} decile i.e between 18 and 69 marks
Measure of Dispersion
Range
The difference between the highest value and the lowest value
Disadvantage
It depends only on the two extreme values
Interquartile range
The difference between the lower and upper quartiles. It includes the middle 50% of the values
Semi quartile range
The difference between the lower quartile and upper quartile divided by 2.It is also called the quartile deviation.
Mean Absolute Deviation
If we find the difference of each number from the mean and find their mean , we get the mean Absolute deviation
Variance
The mean of the square of the square of the deviations from the mean is called is called variance or mean deviation.
Example
Deviation from mean(d)  +1  1  +6  4  2  11  +1  10 
f_{i}  1  1  36  16  4  121  1  100 
Sum
Variance =
The square root of the variance is called the standard deviation.It is also called root mean square deviation. For the above example its standard deviation =
Example
The following table shows the number of children per family in a housing estate
Number of childred  0  1  2  3  4  5  6 
Number of families  1  5  11  27  10  4  2 
Calculate
 The mean number of children per family
 The standard deviation
Solution
Number of children  Number of  fx  Deviations  f  
(x)  Families (f)  d= x m  
o  1  0  – 3  9  9 
1  5  5  – 2  4  20 
2  11  22  1  1  11 
3  27  81  0  0  0 
4
5 6 
10
4 2 
40
20 12 
1
2 3 
1
4 9 
10
16 18 
Σf = 60  Σf= – 40 
 Mean =
 Variance =
Example
The table below shows the distribution of marks of 40 candidates in a test
Marks  110  1120  2130  3140  4150  5160  6170  7180  8190  91100 
frequency  2  2  3  9  12  5  2  3  1  1 
Calculate the mean and standard deviation.
Marks  Midpoint ( x)  Frequency (f)  fx  d= x – m  f  
110  5.5  2  11.0  – 39.5  1560.25  3120.5 
1120  15.5  2  31.0  29.5  870.25  1740.5 
2130  25.5  3  76.5  19.5  380.25  1140.75 
31 40  35.5  9  319.5  9.5  90.25  812.25 
41 50  45.5  12  546.0  0.5  0.25  3.00 
5160  55.5  5  277.5  10.5  110.25  551.25 
61 70
7180 81 90 91 100 
65.5
75.5 85.5 95.5 
2
3 1 1 
131.0
226.5 85.5 95.5 
20.5
30.5 40.5 50.5 
420.25
930.25 1640.25 2550.25 
840.5
2790.75 1640.25 2550.25 
Σf= 40  Σf x=1800  Σf= 15190 
Mean
Variance =
= 379.8
Standard deviation =
= 19.49
Note;
Adding or subtracting a constant to or from each number in a set of data does not alter the value of the variance or standard deviation.
More formulas
The formula for getting the variance
=
Example
The table below shows the length in centimeter of 80 plants of a particular species of tomato
length  152156  157161  162166  167171  172 176  177181 
frequency  12  14  24  15  8  7 
Calculate the mean and the standard deviation
Solution
Let A = 169
Length  Midpoint x  x169  t=  f  ft  
152 156  154  15  3  12  36  108 
157 161  159  10  2  14  28  56 
162 166  164  5  1  24  24  24 
167 171  169  0  0  15  0  0 
172176  174  5  1  8  8  8 
177181  179  10  2  7  14  28 
=
Therefore
= 4.125 + 169
= 164.875 ( to 4 s.f)
Variance of t =
=
= 2.8 – 0.6806
= 2.119
Therefore , variance of x = 2.119 x
= 52.975
= 52.98 ( 4 s.f)
Standard deviation of x =
= 7.279
= 7.28 (to 2 d.p)
End of topic
Did you understand everything?
If not ask a teacher, friends or anybody and make sure you understand before going to sleep! 
Past KCSE Questions on the topic
 Every week the number of absentees in a school was recorded. This was done for 39 weeks these observations were tabulated as shown below
Number of absentees  0.3  4 7  8 11  12 – 15  16 – 19  20 – 23 
(Number of weeks)  6  9  8  11  3  2 
Estimate the median absentee rate per week in the school
 The table below shows high altitude wind speeds recorded at a weather station in a period of 100 days.
Wind speed ( knots)  0 – 19  20 – 39  40 – 59  6079  80 99  100 119  120139  140159  160179 
Frequency (days)  9  19  22  18  13  11  5  2  1 
(a) On the grid provided draw a cumulative frequency graph for the data
(b) Use the graph to estimate
(i) The interquartile range
(ii) The number of days when the wind speed exceeded 125 knots
 Five pupils A, B, C, D and E obtained the marks 53, 41, 60, 80 and 56 respectively. The table below shows part of the work to find the standard deviation.
Pupil  Mark x  x – a  ( xa)^{2} 
A
B C D E 
53
41 60 80 56 
5
17 2 22 2 
(a) Complete the table
(b) Find the standard deviation
 In an agricultural research centre, the length of a sample of 50 maize cobs were measured and recorded as shown in the frequency distribution table below.
Length in cm  Number of cobs 
8 – 10
11 – 13 14 – 16 17 – 19 20 – 22 23 – 25 
4
7 11 15 8 5 
Calculate
 The mean
 (i) The variance
(ii) The standard deviation
 The table below shows the frequency distribution of masses of 50 new born calves in a ranch
Mass (kg)Frequency
15 – 18 2
19 22 3
23 – 26 10
27 – 30 14
31 – 34 13
35 – 38 6
39 – 42 2
(a) On the grid provided draw a cumulative frequency graph for the data
(b) Use the graph to estimate
(i) The median mass
(ii) The probability that a calf picked at random has a mass lying between 25 kg and 28 kg.
 The table below shows the weight and price of three commodities in a given period
Commodity Weight Price Relatives
X 3 125
Y 4 164
Z 2 140
Calculate the retail index for the group of commodities.
 The number of people who attended an agricultural show in one day was 510 men, 1080 women and some children. When the information was represented on a pie chart, the combined angle for the men and women was 216^{0}. Find the angle representing the children.
 The mass of 40 babies in a certain clinic were recorded as follows:
Mass in Kg No. of babies.
1.0 – 1.9 6
2.0 – 2.9 14
3.0 3.9 10
4.0 – 4.9 7
5.0 – 5.9 2
6.0 – 6.9 1
Calculate
(a) The inter – quartile range of the data.
(b) The standard deviation of the data using 3.45 as the assumed mean.
 The data below shows the masses in grams of 50 potatoes
Mass (g)  25 34  3544  45 – 54  55 64  65 – 74  7584  8594 
No of potatoes  3  6  16  12  8  4  1 
(a) On the grid provide, draw a cumulative frequency curve for the data
(b) Use the graph in (a) above to determine
(i) The 60^{th} percentile mass
(ii) The percentage of potatoes whose masses lie in the range 53g to 68g
 The histogram below represents the distribution of marks obtained in a test.
The bar marked A has a height of 3.2 units and a width of 5 units. The bar marked B has a height of 1.2 units and a width of 10 units
If the frequency of the class represented by bar B is 6, determine the frequency of the class represented by bar A.
 A frequency distribution of marks obtained by 120 candidates is to be represented in a histogram. The table below shows the grouped marks. Frequencies for all the groups and also the area and height of the rectangle for the group 30 – 60 marks.
Marks  010  1030  3060  6070  70100 
Frequency  12  40  36  8  24 
Area of rectangle  180  
Height of rectangle  6 
(a) (i) Complete the table
(ii) On the grid provided below, draw the histogram
(b) (i) State the group in which the median mark lies
(ii) A vertical line drawn through the median mark divides the total area of the histogram into two equal parts
Using this information or otherwise, estimate the median mark
 In an agriculture research centre, the lengths of a sample of 50 maize cobs were measured and recorded as shown in the frequency distribution table below
Length in cm  Number of cobs 
8 – 10
11 13 14 – 16 17 19 20 – 22 23 25 
4
7 11 15 8 5 
Calculate
(a) The mean
(b) (i) The variance
(ii) The standard deviation
 The table below shows the frequency distribution of masses of 50 newborn calves in a ranch.
Mass (kg)  Frequency 
15 – 18
19 22 23 – 26 27 – 30 31 34 35 – 38 39 – 42 
2
3 10 14 13 6 2 
(a) On the grid provided draw a cumulative frequency graph for the data
(b) Use the graph to estimate
(i) The median mass
(ii) The probability that a calf picked at random has a mass lying
between 25 kg and 28 kg
 The table shows the number of bags of sugar per week and their moving averages
Number of bags per week  340  330  x  343  350  345 
Moving averages  331  332  y  346 
(a) Find the order of the moving average
(b) Find the value of X and Y axis
CHAPTER FIFTY EIGHT
Specific Objectives
By the end of the topic the learner should be able to:
(a) State the geometric properties of common solids;
(b) Identify projection of a line onto a plane;
(c) Identify skew lines;
(d) Calculate the length between two points in three dimensional geometry;
(e) Identify and calculate the angle between
(i) Two lines;
(ii) A line and a plane;
(ii) Two planes.
Content
(a) Geometrical properties of common solids
(b) Skew lines and projection of a line onto a plane
(c) Length of a line in 3dimensional geometry
(d) The angle between
 i) A line and a line
 ii) A line a plane
iii) A plane and a plane
 iv) Angles between skewlines.
Introduction
Geometrical properties of common solids
 A geometrical figure having length only is in one dimension
 A figure having area but not volume is in two dimension
 A figure having vertices ( points),edges(lines) and faces (plans) is in three dimension
Examples of three dimensional figures
Rectangular Prism
A threedimensional figure having 6 faces, 8 vertices, and 12 edges
Triangular Prism
A threedimensional figure having 5 faces, 6 vertices, and 9 edges.
Cone
A three dimensional figure having one face.
Sphere
A three dimensional figure with no straight lines or line segments
Cube
A three dimensional figure that is measured by its length, height, and width.
It has 6 faces, 8 vertices, and 12 edges
Cylinder
A three dimensional figure having 2 circular faces
Rectangular Pyramid
A threedimensional figure having 5 faces, 5 vertices, and 8 edges
Angle between a line and a plane
The angle between a line and a plane is the angle between the line and its projection on the plane
The angle between the line L and its projection or shadow makes angle A with the plan. Hence the angle between a line and a plane is A.
Example
The angle between a line, r, and a plane, π, is the angle between r and its projection onto π, r’.
height is 4 m
Example
Suppose r’ is 10 cm find the angle
Solution
To find the angle we use tan
Angle Between two planes
Any two planes are either parallel or intersect in a straight line. The angle between two planes is the angle between two lines, one on each plane, which is perpendicular to the line of intersection at the point
Example
The figure below PQRS is a regular tetrahedron of side 4 cm and M is the mid point of RS;
 Show that PM is cm long, and that triangle PMQ is isosceles
 Calculate the angle between planes PSR and QRS
 Calculate the angle between line PQ and plane QRS
Solution
 Triangle PRS is equilateral. Since M,is the midpoint of RS , PM is perpendicular bisector
cm
= cm
Similar triangle MQR is right angled at M
cm
= cm
 The required angle is triangle PMQ .Using cosine rule
 The required angle is triangle PQM
Since triangle PMQ is isosceles with triangle PMQ =
<PQM
(109.46)
End of topic
Did you understand everything?
If not ask a teacher, friends or anybody and make sure you understand before going to sleep! 
Past KCSE Questions on the topic.
 The diagram below shows a right pyramid VABCD with V as the vertex. The base of the pyramid is rectangle ABCD, WITH ab = 4 cm and BC= 3 cm. The height of the pyramid is 6 cm.
(a) Calculate the
 Length of the projection of VA on the base
 Angle between the face VAB and the base
(b) P is the mid point of VC and Q is the mid – point of VD.
Find the angle between the planes VAB and the plane ABPQ
 The figure below represents a square based solid with a path marked on it.
Sketch and label the net of the solid.
 The diagram below represents a cuboid ABCDEFGH in which FG= 4.5 cm, GH = 8 cm and HC = 6 cm
Calculate:
(a) The length of FC
(b) (i) The size of the angle between the lines FC and FH
(ii) The size of the angle between the lines AB and FH
(c) The size of the angle between the planes ABHE and the plane FGHE
 The base of a right pyramid is a square ABCD of side 2a cm. The slant edges VA, VB, VC and VD are each of length 3a cm.
(a) Sketch and label the pyramid
(b) Find the angle between a slanting edge and the base
 The triangular prism shown below has the sides AB = DC = EF = 12 cm. the ends are equilateral triangles of sides 10cm. The point N is the mid point of FC.
Find the length of:
(a) (i) BN
(ii) EN
(b) Find the angle between the line EB and the plane CDEF
CHAPTER FIFTY NINE
Specific Objectives
By the end of the topic the learner should be able to:
(a) Recall and define trigonometric ratios;
(b) Derive trigonometric identity sin2x+cos2x = 1;
(c) Draw graphs of trigonometric functions;
(d) Solve simple trigonometric equations analytically and graphically;
(e) Deduce from the graph amplitude, period, wavelength and phase angles.
Content
(a) Trigonometric ratios
(b) Deriving the relation sin2x+cos2x =1
(c) Graphs of trigonometric functions of the form
y = sin x y = cos x, y = tan x
y = a sin x, y = a cos x,
y = a tan x y = a sin bx,
y = a cos bx y = a tan bx
y = a sin(bx ± 9)
y = a cos(bx ± 9)
y = a tan(bx ± 9)
(d) Simple trigonometric equation
(e) Amplitude, period, wavelength and phase angle of trigonometric functions.
Introduction
Consider the right – angled triangle OAB
AB = r
OA = r
Since triangle OAB is right angled
Divide both sides by gives
Example
If tanshow that;
Solution
Factorize the numerator gives and since
But
Therefore, =
Example
Show that
Removing the brackets from the expression gives
Using
Also
Therefore
Example
Given that
Solution using the right angle triangle below.
 cos
therefore=
 =
 =1
Waves
Amplitude
This is the maximum displacement of the wave above or below the x axis.
Period
The interval after which the wave repeats itself
Transformations of waves
The graphs of y = sin x and y = 3 sin x can be drawn on the same axis. The table below gives the corresponding values of sin x and 3 sin x for
0  30  60  90  120  150  180  210  240  270  300  330  360  
Sin x  0  0.50  0.87  1.00  0.87  0.50  0  0.50  0.87  0.50  0.87  0.50  0 
3 sin x  0  1.50  2.61  3.00  2.61  1.50  0  1.50  2.61  1.50  2.61  1.50  0 
390  420  450  480  510  540  570  600  630  660  690  720 
0.5  0.87  1.00  0.87  0.50  0  0.50  0.87  1.00  0.87  0.50  0 
1.50  2.61  3.00  2.61  1.50  0  1.50  2.61  3.00  2.61  2.61  0 
The wave of y = 3 sin x can be obtained directly from the graph of y = sin x by applying a stretch scale factor 3 , x axis invariant .
Note;
 The amplitude of y= 3sinx is y =3 which is three times that of y = sin x which is y =1.
 The period of the both the graphs is the same that is or 2
Example
Draw the waves y = cos x and y = cos . We obtain y = cos from the graph y = cos x by applying a stretch of factor 2 with y axis invariant.
Note;
 The amplitude of the two waves are the same.
 The period of y = cos is that is, twice the period of y = cos x
Trigonometric Equations
In trigonometric equations, there are an infinite number of roots. We therefore specify the range of values for which the roots of a trigonometric equation are required.
Example
Solve the following trigonometric equations:
 Sin 2x = cos x, for
 Tan 3x = 2, for
Solution
 Sin 2 x = cos x
Sin 2x = sin (90 – x)
Therefore 2 x = 90 – x
X =
For the given range, x =.
 Tan 3x = 2
From calculator
3x =.
In the given range;
Sin sin
End of topic
Did you understand everything?
If not ask a teacher, friends or anybody and make sure you understand before going to sleep! 
Past KCSE Questions on the topic
 (a) Complete the table for the function y = 2 sin x
x  0^{0}  10^{0}  20^{0}  30^{0}  40^{0}  50^{0}  60^{0}  70^{0}  80^{0}  90^{0}  100^{0}  110^{0}  120^{0} 
Sin 3x  0  0.5000  08660  
y  0  1.00  1.73 
(b) (i) Using the values in the completed table, draw the graph of
y = 2 sin 3x for 0^{0} ≤ x ≤ 120^{0} on the grid provided
(ii) Hence solve the equation 2 sin 3x = 1.5
 Complete the table below by filling in the blank spaces
X^{0}  0^{0}  30^{0}  60^{0}  90^{0}  120^{0}  150^{0}  180^{0}  210^{0}  240^{0}  270^{0}  300^{0}  330^{0}  360^{0} 
Cos x^{0}  1.00  0.50  0.87  0.87  
2 cos ½ x^{0}  2.00  1.93  0.52  1.00  2.00 
Using the scale 1 cm to represent 30^{0} on the horizontal axis and 4 cm to represent 1 unit on the vertical axis draw, on the grid provided, the graphs of y = cosx^{0} and y = 2 cos ½ x^{0} on the same axis.
(a) Find the period and the amplitude of y = 2 cos ½ x^{0}
(b) Describe the transformation that maps the graph of y = cos x^{0} on the graph of y = 2 cos ^{1}/_{2} x^{0}
^{ }
 (a) Complete the table below for the value of y = 2 sin x + cos x.
X  0^{0}  30^{0}  45^{0}  60^{0}  90^{0}  120^{0}  135^{0}  150^{0}  180^{0}  225^{0}  270^{0}  315^{0}  360^{0} 
2 sin x  0  1.4  1.7  2  1.7  1.4  1  0  2  1.4  0  
Cos x  1  0.7  0.5  0  0.5  0.7  0.9  1  0  0.7  1  
Y  1  2.1  2.2  2  1.2  0.7  0.1  1  2  0.7  1 
(b) Using the grid provided draw the graph of y=2sin x + cos x for 0^{0}. Take 1cm represent 30^{0} on the x axis and 2 cm to represent 1 unit on the axis.
(c) Use the graph to find the range of x that satisfy the inequalities
2 sin x cos x > 0.5
 (a) Complete the table below, giving your values correct to 2 decimal places.
x  0  10  20  30  40  50  60  70 
Tan x  0  
2 x + 300  30  50  70  90  110  130  150  170 
Sin ( 2x + 30^{0})  0.50  1 
 b) On the grid provided, draw the graphs of y = tan x and y = sin ( 2x + 30^{0}) for 0^{0} ≤ x 70^{0}
Take scale: 2 cm for 100 on the x axis
4 cm for unit on the y axis
Use your graph to solve the equation tan x sin ( 2x + 30^{0} ) = 0.
 (a) Complete the table below, giving your values correct to 2 decimal places
X^{0}  0  30  60  90  120  150  180 
2 sin x^{0}  0  1  2  1  
1 – cos x^{0}  0.5  1 
(b) On the grid provided, using the same scale and axes, draw the graphs of
y = sin x^{0} and y = 1 – cos x^{0} ≤ x ≤ 180^{0}
Take the scale: 2 cm for 30^{0} on the x axis
2 cm for I unit on the y axis
(c) Use the graph in (b) above to
(i) Solve equation
2 sin x^{o} + cos x^{0} = 1
 Determine the range of values x for which 2 sin x^{o}> 1 – cos x^{0}
 (a) Given that y = 8 sin 2x – 6 cos x, complete the table below for the missing
values of y, correct to 1 decimal place.
X  0^{0}  15^{0}  30^{0}  45^{0}  60^{0}  75^{0}  90^{0}  105^{0}  120^{0} 
Y = 8 sin 2x – 6 cos x  6  1.8  3.8  3.9  2.4  0  3.9 
(b) On the grid provided, below, draw the graph of y = 8 sin 2x – 6 cos for
0^{0} ≤ x ≤ 120^{0}
Take the scale 2 cm for 15^{0} on the x axis
2 cm for 2 units on the y – axis
(c) Use the graph to estimate
(i) The maximum value of y
(ii) The value of x for which 4 sin 2x – 3 cos x =1
 Solve the equation 4 sin (x + 30^{0}) = 2 for 0 ≤ x ≤ 360^{0}
 Find all the positive angles not greater than 180^{0} which satisfy the equation
Sin^{2} x – 2 tan x = 0
Cos x
 Solve for values of x in the range 0^{0} ≤ x ≤ 360^{0} if 3 cos^{2} x – 7 cos x = 6
 Simplify 9 – y^{2} where y = 3 cos θ
y
 Find all the values of Ø between 0^{0} and 360^{0} satisfying the equation 5 sin Ө = 4
 Given that sin (90 – x) = 0.8. Where x is an acute angle, find without using mathematical tables the value of tan x^{0}
 Complete the table given below for the functions
y= 3 cos 2x^{0} and y = 2 sin (^{3x}/_{2}^{0} + 30) for 0 ≤ x ≤ 180^{0}
X^{0}  0^{0}  20^{0}  40^{0}  60^{0}  80^{0}  100^{0}  120^{0}  140^{0}  160^{0}  180^{0} 
3cos 2x^{0}  3.00  2.30  0.52  1.50  2.82  2.82  1.50  0.52  2.30  3.00 
2 sin (3 x^{0} + 30^{0})  1.00  1.73  2.00  1.73  1.00  0.00  1.00  1.73  2.00  1.73 
Using the graph paper draw the graphs of y = 3 cos 2x^{0} and y = 2 sin (^{3x}/_{2}^{0} + 30^{0})
(a) On the same axis. Take 2 cm to represent 20^{0} on the x axis and 2 cm to represent one unit on the y – axis
(b) From your graphs. Find the roots of 3 cos 2 x^{0} + 2 sin (^{3x}/_{2}^{0} + 30^{0}) = 0
 Solve the values of x in the range 0^{0} ≤ x ≤ 360^{0} if 3 cos^{2}x – 7cos x = 6
 Complete the table below by filling in the blank spaces
x^{0}  0^{0}  30^{0}  60^{0}  90  1^{0}  150^{0}  180  210  240  270  300  330  360 
Cosx^{0}  1.00  0.50  0.87  0.87  
2cos ½ x^{0}  2.00  1.93  0.5 
Using the scale 1 cm to represent 30^{0} on the horizontal axis and 4 cm to represent 1 unit on the vertical axis draw on the grid provided, the graphs of y – cos x^{0} and y = 2 cos ½ x^{0} on the same axis
(a) Find the period and the amplitude of y =2 cos ½ x^{0}
Ans. Period = 720^{0}. Amplitude = 2
 Describe the transformation that maps the graph of y = cos x^{0} on the graph of y = 2 cos ½ x^{0}
CHAPTER SIXTY
Specific Objectives
By the end of the topic the learner should be able to:
(a) Define the great and small circles in relation to a sphere (including the
Earth);
(b) Establish the relationship between the radii of small and great circles;
(c) Locate a place on the earth’s surface in terms of latitude and longitude;
(d) Calculate the distance between two points along the great circles and small circles (longitude and latitude) in nautical miles (nm) an kilometers (km);
(e) Calculate time in relation to longitudes;
(f) Calculate speed in knots and kilometers per hour.
Content
(a) Latitude and longitude (great and small circles)
(b) The Equator and Greenwich Meridian
(c) Radii of small and great circles
(d) Position of a place on the surface of the earth
(e) Distance between two points along the small and great circles in nautical miles and kilometers
(f) Distance in nautical miles and kilometres along a circle of latitude
(g) Time and longitude
(h) Speed in knots and Kilometres per hour.
Introduction
Just as we use a coordinate system to locate points on a number plane so we use latitude and longitude to locate points on the earth’s surface.
Because the Earth is a sphere, we use a special grid of lines that run across and down a sphere. The diagrams below show this grid on a world globe and a flat world map.
Great and Small Circles
If you cut a ‘slice’ through a sphere, its shape is a circle. A slice through the centre of a sphere is called a great circle, and its radius is the same as that of the sphere. Any other slice is called a small circle, because its radius is smaller than that of a great circle.Hence great circles divides the sphere into two equal parts
Latitude
Latitudes are imaginary lines that run around the earth and their planes are perpendicular to the axis of the earth .The equator is the latitude tha divides the earth into two equal parts.Its the only great circles amoung the latitudes. The equator is , 0°.
The angle of latitude is the angle the latitude makes with the Equator at the centre, O, of the Earth. The diagram shows the 50°N parallel of latitude. Parallels of latitude range from 90°N (North Pole) to 90°S (South Pole).
The angle 5 subtended at the centre of the earth is the is the is the latitude of the circle passing through 5 north of equator.The maximum angle of latitude is 9 north or south of equator.
Longitudes /meridians
They are circles passing through the north and south poles
They can also be said that they are imaginary semicircles that run down the Earth. They are ‘half’ great circles that meet at the North and South Poles. The main meridian of longitude is the prime meridian, 0°. It is also called the Greenwich meridian since it runs through the Royal Observatory at Greenwich in London, England. The other meridians are measured in degrees east or west of the prime meridian.
The angle of longitude is the angle the meridian makes with the prime meridian at the centre, O, of the Earth. The diagram shows the 35°E meridian of longitude.
Meridians of longitude range from 180°E to 180°W. 180°E and
180°W are actually the same meridian, on the opposite side of the Earth to the prime meridian. It runs through the Pacific Ocean, east of Fiji.
Note
 If P is north of the equator and Q is south of the quator , then the difference in latitude between them is given by
 If P and Q are on the same side of the equator , then the difference in latitude is
Position Coordinates
Locations on the Earth are described using latitude (°N or °S) and longitude (°E or °W) in that order. For example, Nairobi has coordinates (1°S, 37°E), meaning it is position is 1° south of the Equator and 37° east of the prime meridian.
EG
Great Circle Distances
Remember the arc length of a circle is where θ is the degrees of the central angle, and the radius of the earth is 6370 km approx.
On a flat surface, the shortest distance between two points is a straight line. Since the Earth’s surface is curved, the shortest distance between A and B is the arc length AB of the great circle that passes through A and B. This is called the great circle distance and the size of angle ∠AOB where O is the centre of the Earth is called the angular distance.
Note
 The length of an arc of a great circle subtending an angle of (one minute) at the centre of the earth is 1 nautical mile nm.
 A nautical mile is the standard international unit from measuring distances travelled by ships and aeroplanes 1 nautical mile (nm) = 1.853 km
If an arc of a great circle subtends an angle at the centre of the earth,the arcs length is nautical miles.
Example
Find the distance between points P() and Q and express it in;
 Nm
 Km
Solution
 Angle subtended at the centre is
Is subtended by 60 nm
Is subtended by; 60 x 60.5 = 3630 nm
 The radius of the earth is 6370 km
Therefore, the circumference of the earth along a great circle is;
Angle between the points is .Therefore, we find the length of an arch of a circle which subtends an angle of at the centre is is subtended by arc whose length is
Therefore, 60. Is subtended by ;
Example
Find the distance between points A ( and express it in ;
 Km
Solution
 The two points lie on the equator, which is great circle. Therefore ,we are calculating distance along a great circle.
Angle between points A and B is (
 Distance in km =
Distance along a small Circle (circle of latitude)
The figure below ABC is a small circle, centre X and radius r cm.PQST is a great circle ,centre O,radius R cm.The angle is between the two radii.
From the figure, XC is parallel to OT. Therefore, angle COT = angle XCO=.Angle CXO =9 (Radius XC is perpendicular to the axis of sphere).
Thus, from the right angled triangle OXC
Therefore, r = R cos
This expression can be used to calculate the distance between any two points along the small circle ABC, centre X and radius r.
Example
Find the distance in kilometers and nautical miles between two points (.
Solution
Figure a shows the position of P and Q on the surface of the earth while figure b shows their relative positions on the small circle is the centre of the circle of latitude with radius r.
The angle subtended by the arc PQ centre C is .So, the length of PQ
The length of PQ in nautical miles
=
In general, if the angle at the centre of a circle of latitude then the length of its arc is 60 where the angle between the longitudes along the same latitude.
Shortest distance between the two points on the earths surface
The shortest distance between two points on the earths surface is that along a great circle.
Example
P and Q are two points on latitude They lie on longitude respectively. Find the distance from P to Q :
 Along a parallel of latitude
 Along a great circle
Solution
The positions of P and Q on earths surface are as shown below
 The length of the circle parallel of latitude is 2 km, which is 2.The difference in longitude between P and Q is
PQ
 The required great circle passes via the North Pole. Therefore, the angle subtended at the centre by the arc PNQ is;
– 2 x
Therefore the arc PNQ
=
=
Note;
Notice that the distance between two points on the earth’s surface along a great circle is shorter than the distance between them along a small circle
Longitude and Time
The earth rotates through 36 about its axis every 24 hours in west – east direction. Therefore for every change in longitude there is a corresponding change in time of 4 minutes, or there is a difference of 1 hour between two meridians apart.
All places in the same meridian have the same local time. Local time at Greenwich is called Greenwich Mean Time .GMT.
All meridians to the west of Greenwich Meridian have sunrise after the meridian and their local times are behind GMT.
All meridian to the east of Greenwich Meridian have sunrise before the meridian and their local times are ahead of GMT. Since the earth rotates from west to east, any point P is ahead in time of another point Q if P is east of Q on the earth’s surface.
Example
Find the local time in Nairobi ( ), when the local time of Mandera (Nairobi ( ) is 3.00 pm
Solution
The difference in longitude between Mandera and Nairobi is (, that is Mandera is .Therefore their local time differ by; 4 x 5 = 20 min.
Since Nairobi is in the west of Mandera, we subtract 20 minutes from 3.00 p.m. This gives local time for Nairobi as 2.40 p.m.
Example
If the local time of London ( ), IS 12.00 noon, find the local time of Nairobi ( ),
Solution
Difference in longitude is ( ) =
So the difference in time is 4 x 37 min = 148 min
= 2 hrs. 28 min
Therefore , local time of Nairobi is 2 hours 28 minutes ahead that of London that is,2.28 p.m
Example
If the local time of point A () is 12.30 a.m, on Monday,Find the local time of a point B ( ).
Solution
Difference in longitude between A and B is
In time is 4 x 340 = 1360 min
= 22 hrs. 40 min.
Therefore local time in point B is 22 hours 40 minutes behind Monday 12:30 p.m. That is, Sunday 1.50 a.m.
Speed
A speed of 1 nautical mile per hour is called a knot. This unit of speed is used by airmen and sailors.
Example
A ship leaves Mombasa (and sails due east for 98 hours to appoint K Mombasa (in the indian ocean.Calculate its average speed in;
 Km/h
 Knots
Solution
 The length x of the arc from Mombasa to the point K in the ocean
=
=
Therefore speed is
 The length x of the arc from Mombasa to the point K in the ocean in nautical miles
Therefore , speed =
= 25.04 knots
End of topic
Did you understand everything?
If not ask a teacher, friends or anybody and make sure you understand before going to sleep! 
Past KCSE Questions on the topic.
 An aeroplane flies from point A (1^{0} 15’S, 37^{0} E) to a point B directly North of A. the arc AB subtends an angle of 45^{0} at the center of the earth. From B, aeroplanes flies due west two a point C on longitude 23^{0} W.)
(Take the value of π ^{22}/ _{7} as and radius of the earth as 6370km)
(a) (i) Find the latitude of B
(ii) Find the distance traveled by the aeroplane between B and C
(b) The aeroplane left at 1.00 a.m. local time. When the aeroplane was leaving B, what was the local time at C?
 The position of two towns X and Y are given to the nearest degree as X (45^{0} N, 10^{0}W) and Y (45^{0} N, 70^{0}W)
Find
(a) The distance between the two towns in
 Kilometers (take the radius of the earth as 6371)
 Nautical miles (take 1 nautical mile to be 1.85 km)
(b) The local time at X when the local time at Y is 2.00 pm.
 A plane leaves an airport A (38.5^{0}N, 37.05^{0}W) and flies dues North to a point B on latitude 52^{0}N.
(a) Find the distance covered by the plane
(b) The plane then flies due east to a point C, 2400 km from B. Determine the position of C
Take the value π of as ^{22}/_{7} and radius of the earth as 6370 km
 A plane flying at 200 knots left an airport A (30^{0}S, 31^{0}E) and flew due North to an airport B (30^{0 }N, 31^{0}E)
(a) Calculate the distance covered by the plane, in nautical miles
(b) After a 15 minutes stop over at B, the plane flew west to an airport C (30^{0 }N, 13^{0}E) at the same speed.
Calculate the total time to complete the journey from airport C, though airport B.
 Two towns A and B lie on the same latitude in the northern hemisphere.
When its 8 am at A, the time at B is 11.00 am.
 a) Given that the longitude of A is 15^{0} E find the longitude of B.
 b) A plane leaves A for B and takes 3^{1}/_{2} hours to arrive at B traveling along a parallel of latitude at 850 km/h. Find:
(i) The radius of the circle of latitude on which towns A and B lie.
(ii) The latitude of the two towns (take radius of the earth to be 6371 km)
 Two places A and B are on the same circle of latitude north of the equator. The longitude of A is 118^{0}W and the longitude of B is 133^{0} E. The shorter distance between A and B measured along the circle of latitude is 5422 nautical miles.
Find, to the nearest degree, the latitude on which A and B lie
 (a) A plane flies by the short estimate route from P (10^{0}S, 60^{0} W) to Q (70^{0} N,
120^{0} E) Find the distance flown in km and the time taken if the aver age speed is 800 km/h.
(b) Calculate the distance in km between two towns on latitude 50^{0}S with long longitudes and 20^{0} W. (take the radius of the earth to be 6370 km)
 Calculate the distance between M (30^{0}N, 36^{0}E) and N (30^{0} N, 144^{0} W) in nautical miles.
(i) Over the North Pole
(ii) Along the parallel of latitude 30^{0} N
 (a) A ship sailed due south along a meridian from 12^{0} N to 10^{0}30’S. Taking
the earth to be a sphere with a circumference of 4 x 10^{4} km, calculate in km the distance traveled by the ship.
(b) If a ship sails due west from San Francisco (37^{0} 47’N, 122^{0} 26’W) for distance of 1320 km. Calculate the longitude of its new position (take the radius of the earth to be 6370 km and π = 22/7).
CHAPTER SIXTY ONE
Specific Objectives
By the end of the topic the learner should be able to:
(a) Form linear inequalities based on real life situations;
(b) Represent the linear inequalities on a graph;
(c) Solve and interpret the optimum solution of the linear inequalities,
(d) Apply linear programming to real life situations.
Content
(a) Formation of linear inequalities
(b) Analytical solutions of linear inequalities
(c) Solutions of linear inequalities by graphs
(d) Optimisation (include objective function)
(e) Application of quadratic equations to real life situations.
Forming linear inequalities
In linear programing we are going to form inequalities representing given conditions involving real life situation.
Example
Esha is five years younger than his sister. The sum of their age is less than 36 years. If Esha’s age is x years, form all the inequalities in x for this situation.
Solution
The age of Esha’s sister is x +5 years.
Therefore, the sum of their age is;
X + (x +5) years
Thus;
2x +5 < 36
2x < 31
X > 15.5
X > 0 ( age is always positive)
Linear programming
Linear programming is the process of taking various linear inequalities relating to some situation, and finding the “best” value obtainable under those conditions. A typical example would be taking the limitations of materials and labor, and then determining the “best” production levels for maximal profits under those conditions.
In “real life”, linear programming is part of a very important area of mathematics called “optimization techniques”. This field of study are used every day in the organization and allocation of resources. These “real life” systems can have dozens or hundreds of variables, or more. In algebra, though, you’ll only work with the simple (and graph able) twovariable linear case.
The general process for solving linearprogramming exercises is to graph the inequalities (called the “constraints”) to form a walledoff area on the x,yplane (called the “feasibility region”). Then you figure out the coordinates of the corners of this feasibility region (that is, you find the intersection points of the various pairs of lines), and test these corner points in the formula (called the “optimization equation”) for which you’re trying to find the highest or lowest value.
Example
Suppose a factory want to produce two types of hand calculators, type A and type B. The cost, the labor time and the profit for every calculator is summarized in the following table:
Type  Cost  Labor Time  Profit 
A  Sh 30  1 (hour)  Sh 10 
B  Sh 20  4 (hour)  Sh 8 
Suppose the available money and labors are ksh 18000 and 1600 hours. What should the production schedule be to ensure maximum profit?
Solution
Suppose is the number of type A hand calculators and is the number of type B hand calculators and y to be the cost. Then, we want to maximize subject to
whereis the total profit.
Solution by graphing
Solutions to inequalities formed to represent given conditions can be determined by graphing the inequalities and then reading off the appropriate values ( possible values)
Example
A student wishes to purchase not less than 10 items comprising books and pens only. A book costs sh.20 and a pen sh.10.if the student has sh.220 to spend, form all possible inequalities from the given conditions and graph them clearly, indicating the possible solutions.
Solution
Let the number of books be x and the number of pens then, the inequalities are;
This simplifies to
 .
All the points in the unshaded region represent possible solutions. A point with coordinates ( x ,y) represents x books and y pens. For example, the point (3, 10 ) means 3 books and 10 pens could be bought by the students.
Optimization
The determination of the minimum or the maximum value of the objective function ax + by is known as optimization. Objective function is an equation to be minimized or maximized .
Example
A contractor intends to transport 1000 bags of cement using a lorry and a pick up. The lorry can carry a maximum of 80 bags while a pick up can carry a maximum of 20 bags. The pick up must make more than twice the number of trips the lorry makes and the total number of trip to be less than 30.The cost per trip for the lorry is ksh 2000, per bag and ksh 900 for the pick up.Find the minimum expenditure.
Solution
If we let x and y be the number of trips made by the lorry and the pick up respectively. Then the conditions are given by the following inequalities;
The total cost of transporting the cement is given by sh 2000x + 900y.This is called the objective function.
The graph below shows the inequalities.
From the graph we can identify 7 possibilities
Note;
Coordinates stands for the number of trips. For example (7, 22) means 7 trips by the lorry and 22 trips by the pickup. Therefore the possible amount of money in shillings to be spent by the contractor can be calculated as follows.
We note that from the calculation that the least amount the contractor would spend is sh.32200.This is when the lorry makes 8 trips and the pick up 18 trips. When possibilities are many the method of determining the solution by calculation becomestedious. The alternative method involves drawing the graph of the function we wish to maximize or minimize, the objective function. This function is usually of the form ax +by , where a and b ar constants.
For this ,we use the graph above which is a convenient point (x , y) to give the value of x preferably close to the region of the possibilities. For example the point ( 5, 10) was chosen to give an initial value of thus ,2000x + 900y = 19000.we now draw the line 2000x + 900y=19000.such a line is referred to us a search line.
Using a ruler and a set square, slide the set square keeping one edge parallel to until the edge strikes the feasible point nearest ( see the dotted line ) From the graph this point is (8,18 ),which gives the minimum expenditure as we have seen earlier.The feasible point furthest from the line gives the maximum value of the objective function.
The determination of the minimum or the maximum value of the objective function ax + by is known as optimization.
Note;
The process of solving linear equations are as follows
 Forming the inequalities satisfying given conditions
 Formulating the objective function .
 Graphing the inequalities
 Optimizing the objective function
This whole process is called linear programming .
Example
A company produces gadgets which come in two colors: red and blue. The red gadgets are made of steel and sell for ksh 30 each. The blue gadgets are made of wood and sell for ksh 50 each. A unit of the red gadget requires 1 kilogram of steel, and 3 hours of labor to process. A unit of the blue gadget, on the other hand, requires 2 board meters of wood and 2 hours of labor to manufacture. There are 180 hours of labor, 120 board meters of wood, and 50 kilograms of steel available. How many units of the red and blue gadgets must the company produce (and sell) if it wants to maximize revenue?
Solution
The Graphical Approach
Step 1. Define all decision variables.
Let: x_{1} = number of red gadgets to produce (and sell)
x_{2} = number of blue gadgets to produce (and sell)
Step 2. Define the objective function.
Maximize R = 30 x_{1}+ 50 x_{2} (total revenue in ksh)
Step 3. Define all constraints.
(1) x_{1} £ 50 (steel supply constraint in kilograms)
(2) 2 x_{2} £120 (wood supply constraint in board meters)
(3) 3 x_{1} + 2 x_{2 }£ 180 (labor supply constraint in man hours)
x_{1} , x_{2}³ 0 (nonnegativity requirement)
Step 4. Graph all constraints.
Then determine area of feasible study
Note;
 The area under the line marked blue is the needed area or area of feasible solutions.
 We therefore shade the unwanted region out the trapezium marked blue
Optimization
List all corners (identify the corresponding coordinates), and pick the best in terms of the resulting value of the objective function.
(1) x_{1} = 0 x_{2} = 0 R = 30 (0) + 50 (0) = 0
(2) x_{1} = 50 x_{2} = 0 R = 30 (50) + 50 (0) = 1500
(3) x_{1} = 0 x_{2} = 60 R = 30 (0) + 50 (60) = 3000
(4) x_{1} = 20 x_{2} = 60 R = 30 (20) + 50 (60) = 3600 (the optimal solution)
(5) x_{1} = 50 x_{2} = 15 R = 30 (50) + 50 (15) = 2250
End of topic
Did you understand everything?
If not ask a teacher, friends or anybody and make sure you understand before going to sleep! 
Past KCSE Questions on the topic.
 A school has to take 384 people for a tour. There are two types of buses available, type X and type Y. Type X can carry 64 passengers and type Y can carry 48 passengers. They have to use at least 7 buses.
(a) Form all the linear inequalities which will represent the above information.
(b) On the grid [provide, draw the inequalities and shade the unwanted region.
(c) The charges for hiring the buses are
Type X: Ksh 25,000
Type Y Ksh 20,000
Use your graph to determine the number of buses of each type that should be hired to minimize the cost.
 An institute offers two types of courses technical and business courses. The institute has a capacity of 500 students. There must be more business students than technical students but at least 200 students must take technical courses. Let x represent the number of technical students and y the number of business students.
(a) Write down three inequalities that describe the given conditions
(b) On the grid provided, draw the three inequalities
(c) If the institute makes a profit of Kshs 2, 500 to train one technical students and Kshs 1,000 to train one business student, determine
 The number of students that must be enrolled in each course to maximize the profit
 The maximum profit.
 A draper is required to supply two types of shirts A and type B.
The total number of shirts must not be more than 400. He has to supply more type A than of type B however the number of types A shirts must be more than 300 and the number of type B shirts not be less than 80.
Let x be the number of type A shirts and y be the number of types B shirts.
 Write down in terms of x and y all the linear inequalities representing the information above.
 On the grid provided, draw the inequalities and shade the unwanted regions
 The profits were as follows
Type A: Kshs 600 per shirt
Type B: Kshs 400 per shirt
 Use the graph to determine the number of shirts of each type that should be made to maximize the profit.
 Calculate the maximum possible profit.
 A diet expert makes up a food production for sale by mixing two ingredients N and S. One kilogram of N contains 25 units of protein and 30 units of vitamins. One kilogram of S contains 50 units of protein and 45 units of vitamins. The food is sold in small bags each containing at least 175 units of protein and at least 180 units of vitamins. The mass of the food product in each bag must not exceed 6kg.
If one bag of the mixture contains x kg of N and y kg of S
 Write down all the inequalities, in terms of x and representing the information above ( 2 marks)
 On the grid provided draw the inequalities by shading the unwanted regions ( 2 marks)
(c) If one kilogram of N costs Kshs 20 and one kilogram of S costs Kshs 50, use the graph to determine the lowest cost of one bag of the mixture.
 Esha flying company operates a flying service. It has two types of aeroplanes. The smaller one uses 180 litres of fuel per hour while the bigger one uses 300 litres per hour.
The fuel available per week is 18,000 litres. The company is allowed 80 flying hours per week.
(a) Write down all the inequalities representing the above information
(b) On the grid provided on page 21, draw all the inequalities in (a) above by
shading the unwanted regions
(c) The profits on the smaller aeroplane is Kshs 4000 per hour while that on the
bigger one is Kshs. 6000 per hour. Use your graph to determine the maximum profit that the company made per week.
 A company is considering installing two types of machines. A and B. The information about each type of machine is given in the table below.
Machine  Number of operators  Floor space  Daily profit 
A  2  5m^{2}  Kshs 1,500 
B  5  8m^{2}  Kshs 2,500 
The company decided to install x machines of types A and y machines of type B
(a) Write down the inequalities that express the following conditions
 The number of operators available is 40
 The floor space available is 80m^{2}
 The company is to install not less than 3 type of A machine
 The number of type B machines must be more than one third the number of type A machines
(b) On the grid provided, draw the inequalities in part (a) above and shade the
unwanted region.
(c) Draw a search line and use it to determine the number of machines of each
type that should be installed to maximize the daily profit.
CHAPTER SIXTY TWO
Specific Objectives
By the end of the topic the learner should be able to:
(a) Define Locus;
(b) Describe common types of Loci;
(c) Construct;
 i) Loci involving inequalities;
 ii) Loci involving chords;
iii) Loci involving points under given conditions;
 iv) Intersecting loci.
Content
(a) Common types of Loci
(b) Perpendicular bisector loci
(c) Locus of a point at a given distance from a fixed point
(d) Angle bisector loci
(e) Other loci under given condition including intersecting loci
(f) Loci involving inequalities
(g) Loci involving chords (constant angle loci).
Introduction
Locus is defined as the path, area or volume traced out by a point, line or region as it moves according to some given laws
In construction the opening between the pencil and the point of the compass is a fixed distance, the length of the radius of a circle. The point on the compass determines a fixed point. If the length of the radius remains the same or unchanged, all of the point in the plane that can be drawn by the compass from a circle and any points that cannot be drawn by the compass do not lie on the circle. Thus the circle is the set of all points at a fixed distance from a fixed point. This set is called a locus.
Common types of Loci
Perpendicular bisector locus
The locus of a point which are equidistant from two fixed points is the perpendicular bisector of the straight line joining the two fixed points. This locus is called the perpendicular bisector locus.
So to find the point equidistant from two fixed points you simply find the perpendicular bisector of the two points as shown below.
Q is the midpoint of M and N.
In three Dimensions
In three dimensions, the perpendicular bisector locus is a plane at right angles to the line and bisecting the line into two equal parts. The point P can lie anywhere in the line provided its in the middle.
The Locus of points at a Given Distance from a given straight line.
In two Dimensions
In the figure below each of the lines from the middle line is marked a centimeters on either side of the given line MN.
The ‘a’ centimeters on either sides from the middle line implies the perpendicular distance.
The two parallel lines describe the locus of points at a fixed distance from a given straight line.
In three Dimensions
In three dimensions the locus of point ‘a’ centimeters from a line MN is a cylindrical shell of radius ‘a’ c, with MN as the axis of rotation.
Locus of points at a Given Distance from a fixed point.
In two Dimension
If O is a fixed point and P a variable point‘d’ cm from O,the locus of p is the circle O radius ‘d’ cm as shown below.
All points on a circle describe a locus of a point at constant distance from a fixed point. In three dimesion the locus of a point ‘d’ centimetres from a point is a spherical shell centre O and radius d cm.
Angle Bisector Locus
The locus of points which are equidistant from two given intersecting straight lines is the pair of perpendicular lines which bisect the angles between the given lines.
Conversely ,a point which lies on a bisector of given angle is equidistant from the lines including that angle.P C
Line PB bisect angle ABC into two equal parts.
Example
Construct triangle PQR such that PQ= 7 cm, QR = 5 cm and angle PQR = .Construct the locus L of points equidistant from RP and RQ.
Solution
L is the bisector of Angle PRQ.
P
L
Constant angle loci
A line PQ is 5 cm long, Construct the locus of points at which PQ subtends an angle of .
Solution
 Draw PQ = 5 cm
 Construct TP at P such that angle QPT =
 Draw a perpendicular to TP at P( radius is perpendicular to tangent)
 Construct the perpendicular bisector of PQ to meet the perpendicular in (iii) at O
 Using O as the centre and either OP or OQ as radius, draw the locus
 Transfer the centre on the side of PQ and complete the locus.
 Transfer the centre on the opposite sides of PQ and complete the locus as shown below.
 To are of the same radius,
 Angle subtended by the same chord on the circumference are equal ,
 This is called the constant angle locus.
Intersecting Loci
 Construct triangle PQR such that PQ =7 cm, OR = 5 cm and angle PQR = 3
 Construct the locus of points equidistant from P and Q to meet the locus of points equidistant from Q and R at M .Measure PM
Solution
In the figure below
 is the perpendicular bisector of PQ
 is the perpendicular bisector of PQ
 By measurement, PM is equal to 3.7 cm
Loci of inequalities
An inequality is represented graphically by showing all the points that satisfy it.The intersection of two or more regions of inequalities gives the intersection of their loci.
Remember we shed the unwanted region
Example
Draw the locus of point ( x, y) such that x + y < 3 , y – x and y > 2.
Solution
Draw the graphs of x + y = 3 ,y –x =4 and y = 2 as shown below.
The unwanted regions are usually shaded. The unshaded region marked R is the locus of points ( x ,y ), such that x + y < 3 , y – X 4 and y > 2.
The lines of greater or equal to ad less or equal to ( ) are always solid while the lines of greater or less (<>) are always broken.
Example
P is a point inside rectangle ABCD such that APPB and Angle DAP Angle BAP. Show the region on which P lies.
Solution
A B
Draw a perpendicular bisector of AP=PB and shade the unwanted region. Bisect <DAB (< DAP = < BAP) and shade the unwanted region lies in the unshaded region.
Example
Draw the locus of a point P which moves that AP 3 cm.
Solution
 Draw a circle, centre A and radius 3 cm
 Shade the unwanted region.
Locus involving chords
The following properties of chords of a circle are used in construction of loci
(I)Perpendicular bisector of any chord passes through the centre of the circle.
(ii) The perpendicular drawn from a centre of a circle bisects the chord.
(III) If chords of a circle are equal, they are equidistant from the centre of the circle and vice versa
(IV) In the figure below, if chord AB intersects chord CD at O, AO = x ,BO = y, CO = m and DO =n then
End of topic
Did you understand everything?
If not ask a teacher, friends or anybody and make sure you understand before going to sleep! 
Past KCSE Questions on the topic.
 Using a ruler and a pair of compasses only,
 Construct a triangle ABC such that angle ABC = 135^{o}C, AB = 8.2cm and BC = 9.6cm
 Given that D is a position equidistant from both AB and BC and also from B and C
 Locate D
 Find the area of triangle DBC.
 (a) Using a ruler, a pair of compasses only construct triangle XYZ such that XY = 6cm,
YZ = 8cm and ÐXYZ = 75^{o}
(b) Measure line XZ and ÐXZY
(c) Draw a circle that passes through X, Y and Z
(d) A point M moves such that it is always equidistant from Y and Z. construct the locus of M and define the locus
 (a) (i) Construct a triangle ABC in which AB=6cm, BC = 7cm and angle ABC = 75^{o}
Measure:
(i) Length of AC
(ii) Angle ACB
(b) Locus of P is such that BP = PC. Construct P
(c) Construct the locus of Q such that Q is on one side of BC, opposite A and angle
BQC = 30^{o }
(d) (i) Locus of P and locus of Q meet at X. Mark x
(ii) Construct locus R in which angle BRC 120^{o}
(iii) Show the locus S inside triangle ABC such that XS ³ SR
 Use a ruler and compasses only for all constructions in this question.
 a) i) Construct a triangle ABC in which AB=8cm, and BC=7.5cm and ÐABC=112½°
 ii) Measure the length of AC
 b) By shading the unwanted regions show the locus of P within the triangle ABC such that
 i) AP ≤ BP
 ii) AP >3cm
Mark the required region as P
 c) Construct a normal from C to meet AB produced at D
 d) Locate the locus of R in the same diagram such that the area of triangle ARB is ¾ the area of the triangle ABC.
 On a line AB which is 10 cm long and on the same side of the line, use a ruler and a pair of compasses only to construct the following.
 a) Triangle ABC whose area is 20 cm^{2} and angle ACB = 90^{o}
 b) (i) The locus of a point P such that angle APB = 45^{o}.
(ii) Locate the position of P such that triangle APB has a maximum area and calculate this area.
 A garden in the shape of a polygon with vertices A, B, C, D and E. AB = 2.5m, AE = 10m,
ED = 5.2M and DC=6.9m. The bearing of B from A is 030º and A is due to east of E
whileD is due north of E, angle EDC = 110º,
 a) Using a scale of 1cm to represent 1m construct an accurate plan of the garden
 b) A foundation is to be placed near to CD than CB and no more than 6m from A,
 i) Construct the locus of points equidistant from CB and CD.
 ii) Construct the locus of points 6m from A
 c) i) shade and label R ,the region within which the foundation could be placed in the garden
 ii) Construct the locus of points in the garden 3.4m from AE.
iii) Is it possible for the foundation to be 3.4m from AE and in the region?
 a) Using a ruler and compasses only construct triangle PQR in which QR= 5cm, PR = 7cm and angle PRQ = 135°
 b) Determine < PQR
 c) At P drop a perpendicular to meet QR produced at T d) Measure PT
 e) Locate a point A on TP produced such that the area of triangle AQR is equal to one and – a – half times the area of triangle PQR
 f) Complete triangle AQR and measure angle AQR
 Use ruler and a pair of compasses only in this question.
(a) Construct triangle ABC in which AB = 7 cm, BC = 8 cm and ∠ABC = 60^{0}.
(b) Measure (i) side AC (ii) ∠ ACB
(c) Construct a circle passing through the three points A, B and C. Measure the radius of the circle.
(d) Construct ∆ PBC such that P is on the same side of BC as point A and ∠ PCB = ½ ∠ ACB,∠ BPC = ∠ BAC measure ∠ PBC.
 Without using a set square or a protractor:
(a) Construct triangle ABC in which BC is 6.7cm, angle ABC is 60^{o} and ÐBAC is 90^{o}.
(b) Mark point D on line BA produced such that line AD =3.5cm
(c) Construct:
(i) A circle that touches lines AC and AD
(ii) A tangent to this circle parallel to line AD
Use a pair of compasses and ruler only in this question;
(a) Draw acute angled triangle ABC in which angle CAB = 37½ ^{o}, AB = 8cm and CB = 5.4cm. Measure the length of side AC (hint 37½ ^{o} = ½ x 75^{o})
(b) On the triangle ABC below:
(i) On the same side of AC as B, draw the locus of a point X so that angle Ax C = 52½ ^{o}
(ii) Also draw the locus of another point Y, which is 6.8cm away from AC and on the same side as X
(c) Show by shading the region P outside the triangle such that angle APC ³ 52 ½^{o} and
P is not less than 6.8cm away from AC
CHAPTER SIXTY THREE
Specific Objectives
By the end of the topic the learner should be able to:
(a) Find average rates of change and instantaneous rates of change;
(b) Find the gradient of a curve at a point using tangent;
(c) Relate the delta notation to rates of change;
(d) Find the gradient function of a function of the form y = x n (n is a positive
Integer);
(e) Define derivative of a function, derived function of a polynomial anddifferentiation;
(f) Determine the derivative of a polynomial;
(g) Find equations of tangents and normal to the curves;
(h) Sketch a curve;
(i) Apply differentiation in calculating distance, velocity and acceleration;
(j) Apply differentiation in finding maxima and minima of a function.
Content
(a) Average and instantaneous rates of change
(b) Gradient of a curve at a point
(c) Gradient of y= x n (where n is a positive integer)
(d) Delta notation ( A ) or 5
(e) Derivative of a polynomial
(f) Equations of tangents and normals to the curve
(g) Stationery points
(h) Curve sketching
(i) Application of differentiation in calculation of distance, velocity andacceleration
(j) Maxima and minima
Introduction
Differentiation is generally about rate of change
Example
If we want to get the gradient of the curve y = at a general point ( x ,y ).We note that a general point on the curve y = will have coordinates of the form ( x )The gradient of the curve y= at a general point ( x, y ) can be established as below.
If we take a small change in x , say h. This gives us a new point on the curve with coordinates
[(x +h), (x + h]. So point Q is [(x +h), (x + h] while point P is ( x ).
To find the gradient of PQ =
Change in y = (x + h
Change in x = ( x + h ) – x
Gradient =
=
=
= 2x + h
By moving Q as close to p as possible, h becomes sufficiently small to be ignored. Thus, 2x +h becomes2x.Therefore, at general point ( x,y)on the curve y =,the gradient is 2x.
2x is called the gradient function of the curve y = .We can use the gradient function to determine the gradient of the curve at any point on the curve.
In general, the gradient function of y = is given by ,where n is a positive integer. The gradient function is called the derivative or derived function and the process of obtaining it is called differentiation.
The function
Delta Notation
A small increase in x is usually denoted bysimilarly a small increase in y is denoted by .Let us consider the points P ( x ,y ) and Q [ (x + ),(y + ) on the curve y =
Note;
X is a single quantity and not a product of and x .similarly is a single quantity.
The gradient of PQ, =
=
= 2x +
As tends to zero;
 can be ignored
 gives the derivative which is denoted by
 thus
When we find , we say we are differentiating with respect to x, For example given y =; then
In general the derivatives of y = e.g. y =
Derivative of a polynomial.
A polynomial in x is an expression of the form where are constants
To differentiate a polynomial function, all you have to do is multiply the coefficients of each variable by their corresponding exponents/powers, subtract each exponent/powers by one , and remove any constants.
Steps involved in solving polynomial areas follows
Identify the variable terms and constant terms in the equation.
A variable term is any term that includes a variable and a constant term is any term that has only a number without a variable. Find the variable and constant terms in this polynomial function: y = 5x^{3} + 9x^{2} + 7x + 3
 The variable terms are 5x^{3}, 9x^{2}, and 7x
 The constant term is 3
Multiply the coefficients of each variable term by their respective powers.
Their products will form the new coefficients of the differentiated equation. Once you find their products, place the results in front of their respective variables. For example:
 5×3 = 5 x 3 = 15
 9×2 = 9 x 2 = 18
 7x = 7 x 1 = 7
Lower each exponent by one.
To do this, simply subtract 1 from each exponent in each variable term. Here’s how you do it:
 5
Replace the old coefficients and old exponents/powers with their new counterparts.
To finish differentiating the polynomial equation, simply replace the old coefficients with their new coefficients and replace the old powers with their values lowered by one. The derivative of constants is zero so you can omit 3, the constant term, from the final result.
The derivative of the polynomial y =
In general, the derivative of the sum of a number of terms is obtained by differentiating each term in turn.
Examples
Find the derived function of each of the following
 S=t ) A =
Solution
Equations of tangents and Normal to a curve.
The gradient of a curve is the same as the gradient of the tangent to the curve at that point. We use this principle to find the equation of the tangent to a curve at a given point.
Find the equation of the tangent to the curve;
at
Solution
At the point the gradient is 3 x + 2 = 5
We want the equation of straight line through (1, 4) whose gradient is 5.
Thus
A normal to a curve at appoint is the line perpendicular to the tangent to the curve at the given point.
In the example above the gradient of the tangent of the tangent to the curve at (1, 4) is 5. Thus the gradient of the normal to the curve at this point is.
Therefore, equation of the normal is:
5(y – 4) = – 1( x – 1 )
Example
Find the equation of the normal to the curve y =
Solution
At the point ( 1,2) gradient of the tangent line is 1.Therefore the gradient of the normal is 1.the required equation is
The equation of the normal is y = x 1
Stationary points
Note;
 In each of the points A ,B and C the tangent is horizontal meaning at these points the gradient is zero.so .
 Any point at which the tangent to the graph is horizontal is called a stationary point. We can locate stationary points by looking for points at which = 0.
Turning points
The point at which the gradient changes from negative through zero to positive is called minimum point while the point which the gradient changes from positive through zero to negative is called maximum point .In the figure above A is the maximum while B is the minimum.
Minimum point .
Gradient moves from negative through zero to positive.
Maximum point
Gradient moves from positive through zero to negative.
The maximum and minimum points are called turning points.
A point at which the gradient changes from positive through zero to positive or from negative zero to negative is called point of inflection.
Example
Identify the stationary points on the curve y =for each point, determine whether it is a maximum, minimum or a point of inflection.
Solution
At stationary point,
Thus
3
3
Therefore, stationary points are ( 1 , 4 ) and (1 ,0).
Consider the sign of the gradient to the left and right of x = 1
x  0  1  2 
3  0  9  
Diagrammatic representation  \  / 
Therefore ( 1 , 0 ) is a minimum point.
Similarly, sign of gradient to the left and right of x = 1 gives
x  2  1  0 
9  0  3  
Diagrammatic representation  /  ___  \ 
Therefore ( 1 , 4 ) is a maximum point.
Example
Identify the stationary points on the curve y =.Determine the nature of each stationary point.
Solution
y =
At stationary points,
Stationary points are (0, 1) and (3, 28)
Therefore (0, 1) is a point of inflection while (3, 28) is a maximum point.
Application of Differentiation in calculation of velocity and acceleration.
Velocity
If the displacement, S is expressed in terms of time t, then the velocity is v =
Example
The displacement, S metres, covered by a moving particle after time, t seconds, is given by
S =.Find:
 Velocity at :
 t= 3
 Instant at which the particle is at rest.
Solution
S =
The gradient function is given by;
V =
=
 velocity
 at t = 2 is ;
v =
= 24 + 16 – 8
=32m/s
 at t = 3 is ;
v =
= 54 + 24 – 8
=70m/s
 the particle is at rest when v is zero
It is not possible to have t = 2
The particle is therefore at rest at seconds
Acceleration
Acceleration is found by differentiating an equation related to velocity. If velocity v , is expressed in terms of time, t , then the acceleration, a, is given by a =
Example
A particle moves in a straight line such that is its velocity v m after t seconds is given by
v = 3 + 10 t – .
Find
 the acceleration at :
 t =1 sec
 t =3 sec
 the instant at which acceleration is zero
Solution
 At t = 1 sec a = 10 – 2 x 1
 At t = 3 sec a = 10 – 2 x 3
 Acceleration is zero when
Therefore, 10 – 2t = 0 hence t = 5 seconds
Example
A closed cylindrical tin is to have a capacity of 250π ml. if the area of the metal used is to be minimum, what should the radius of the tin be?
Solution
Let the total surface area of the cylinder be A ,radius r cm and height h cm.
Then, A = 2
Volume = 2
Making h the subject, h =
=
Put h = in the expression for surface area to get;
A = 2
=2
For minimum surface area,
= 5
Therefore the minimum area when r = 5 cm
Example
A farmer has 100 metres of wire mesh to fence a rectangular enclosure. What is the greatest area he can enclose with the wire mesh?
Solution
Let the length of the enclosure be x m. Then the width is
Then the area A of the rectangle is given by;
A = x (50 –x)
= 50x –
For maximum or minimum area,
Thus, 50 – 2x = 0
The area is maximum when x = 25 m
That is A = 50 X 25 – (25
= 625 .
CHAPTER SIXTY FOUR
Specific Objectives
By the end of the topic the learner should be able to:
(a) Carry out the process of differentiation;
(b) Interpret integration as a reverse process of differentiation;
(c) Relate integration notation to sum of areas of trapezia under a curve;
(d) Integrate a polynomial;
(e) Apply integration in finding the area under a curve,
(f) Apply integration in kinematics.
Content
(a) Differentiation
(b) Reverse differentiation
(c) Integration notation and sum of areas of trapezia
(d) Indefinite and definite integrals
(e) Area under a curve by integration
(f) Application in kinematics.
Introduction
The process of finding functions from their gradient (derived) function is called integration
Suppose we differentiate the function y=x^{2}. We obtain
Integration reverses this process and we say that the integral of 2x is .
From differentiation we know that the gradient is not always a constant. For example, if = 2x, then this comes from the function of the form y=, Where c is a constant.
Example
Find y if is:
Solution
Then, y =
Then, y =
Note;
To integrate we reverse the rule for differentiation. In differentiation we multiply by the power of x and reduce the power by 1.In integration we increase the power of x by one and divide by the new power.
If ,then, where c is a constant and n.since c can take any value we call it an arbitrary constant.
Example
Integrate the following expression
 2x +4
Solution
Then, y =
=
=
B.)
Then, y =
=
= –
C.) 2x +4
Then, y =
=
=
Example
Find the equation of a line whose gradient function is and passes through (0,1)
Solution
Since ,the general equation is y =.The curve passes through ( 0,1).Substituting these values in the general equation ,we get 1 = 0 + 0 + c
1 = c
Hence, the particular equation is y =
Example
Find v in terms of h if and V =9 when h=1
Solution
The general solution is
V =
=
V= 9 when h= 1.Therefore
9 = 5 + c
4 = C
Hence the particular solution is ;
V
Definite and indefinite integrals
It deals with finding exact area.
Estimate the area shaded beneath the curve shown below
The area is divided into rectangular strips as follows.
The shaded area in the figure above shows an underestimated and an overestimated area under the curve. The actual area lies between the underestimated and overestimated area. The accuracy of the area can be improved by increasing the number of rectangular strips between x = a and x = b.
The exact area beneath the curve between x = a and b is given by
The symbol
Thus means integrate the expression for y with respect to x.
The expression ,where a and b are limits , is called a definite integral. ‘a’ is called the lower limit while b is the upper limit. Without limits, the expression is called an indefinite integral.
Example
The following steps helps us to solve it
 Integrate with respect to x , giving
 Place the integral in square brackets and insert the limits, thus
 Substitute the limits ;
X = 6 gives
x = 6 gives
 Subtract the results of the lower limit from that of upper limit, that is;
(162 + c) – (
We can summarize the steps in short form as follows:
=
=
=150
Example
 Find the indefinite integral
 Evaluate
Solution
Evaluate
4 + 10 4 ) – ( )
= (27 – 18 +15) – (8 – 8 +10)
= 14
Area under the curve
Find the exact area enclosed by the curve y = ,the axis, the lines x = 2 and x = 4
Solution
2 4
The area is given by;
Example
Find the area of the region bounded by the curve , the x axis x = 1 and x = 2
Solution
The area is given by;
= (4 – 8 + 4) – (
= 0 – =
Note;
The negative sign shows that the area is below the x – axis. We disregard the negative sign and give it as positive as positive .The answer is .
Example
Find the area enclosed by the curvethe x – axis and the lines x = 4 and x =10.
Solution
The required area is shaded below.
Area =
Example
Find the area enclosed by the curve y and the line y =x.
Solution
The required area is
To find the limits of integration, we must find the x coordinates of the points of intersection when;
The required area is found by subtracting area under y = x from area under y =
The required area =
Application in kinematics
The derivative of displacement S with respect to time t gives velocity v, while the derivative of velocity with respect to time gives acceleration, a
Differentiation. Integration
Displacement. displacement
Velocity. Velocity
Acceleration. Acceleration
Note;
Integration is the reverse of differentiation. If we integrate velocity with respect to time we get displacement while if velocity with respect to time we get acceleration.
Example
A particle moves in a straight line through a fixed point O with velocity ( 4 – 1)m/s.Find an expression for its displacement S from this point, given that S = when t = 0.
Solution
Since
S =
Substituting S = 4, t = 0 to get C;
4 = C
Therefore.
Example
A ball is thrown upwards with a velocity of 40 m s
 Determine an expression in terms of t for
 Its velocity
 Its height above the point of projection
 Find the velocity and height after:
 2 seconds
 5 seconds
 8 seconds
 Find the maximum height attained by the ball. (Take acceleration due to gravity to be 10 m/.
Solution
 = 10 ( since the ball is projected upwards)
Therefore, v =10 t + c
When t = 0, v = 40 m/s
Therefore, 40 = 0 + c
40 = c
 The expression for velocity is v = 40 – 10t
 Since
When t = 0 , S = 0
C = 0
The expression for displacement is ;
 Since v = 40 – 10t
 When t = 2
v = 40 – 10 (2)
= 40 – 20
= 20 m/s
S =40t –
= 40 (2) – 5 (
= 80 – 20
= 60 m
 When t = 5
V = 40 – 10 (5)
= 10 m/s
S
= 75 m
 When t = 8
V = 40
S
= 320 – 320
= 0
 Maximum height is attained when v = 0.
Thus , 40 – 10t = 0
t= 4
Maximum height S = 160 – 80
= 80 m
Example
The velocity v of a particle is 4 m/s. Given that S = 5 when t =2 seconds:
 Find the expression of the displacement in terms of time.
 Find the :
 Distance moved by the particle during the fifth second.
 Distance moved by the particle between t =1 and t =3.
Solution
S=4t + c
Since S = 5 m when t =2;
5 = 4 (2) + C
5 – 8 = C
3 = C
Thus, S =4t – 3
 )
II.)
CHAPTER SIXTY FIVE
Specific Objectives
By the end of the topic the learner should be able to:
(a) Approximate the area of irregular shapes by counting techniques;
(b) Derive the trapezium rule;
(c) Apply trapezium rule to approximate areas of irregular shapes;
(d) Apply trapezium rule to estimate areas under curves;
(e) Derive the midordinate rule;
(f) Apply midordinate rule to approximate area under curves.
Content
(a) Area by counting techniques
(b) Trapezium rule
(c) Area using trapezium rule
(d) Midordinate
(e) Area by the midordinate rule
Introduction
Estimation of areas of irregular shapes such as lakes, oceans etc. using counting method. The following steps are followed
 Copy the outline of the region to be measured on a tracing paper
 Put the tracing on a one centimeter square grid shown below
 Count all the whole squares fully enclosed within the region
 Count all the partially enclosed squares and take them as half square centimeter each
 Divide the number of half squares by two and add it to the number of full squares.
Number of compete squares = 4
Number of half squares = 16/ 2 = 8
Therefore the total number of squares = 25 + 8
= 33
The area of the land mass on the paper is therefore 33
Note;
The smaller the subdivisions, the greater the accuracy in approximating area.
Approximating Area by Trapezium Method.
Find the area of the region shown, the region may be divided into six trapezia of uniform as shown
The area of the region is approximately equal to the sum of the areas of the six trapezia.
Note;
The width of each trapezium is 2 cm, and 4 and 3.5 are the lengths of the parallel sides of the first trapezium.
The area of the trapezium A =
Area of the trapezium B =
Area of the trapezium C =
Area of the trapezium D =
Area of the trapezium E =
Area of the trapezium F =
Therefore, the total area of the region is
If the lengths of the parallel sides of the trapezia (ordinates) are
Note;
In trapezium rule, except for the first and last lengths, each of the other lengths is counted twice. Therefore, the expression for the area can be simplified to:
In general, the approximate area of a region using trapezium method is given by:
;
Where h is the uniform width of each trapezium, are the first and last length respectively. This method of approximating areas of irregular shape is called trapezium rule.
Example
A car start from rest and its velocity is measured every second from 6 seconds.
Time (t)  0  1  2  3  4  5  6 
Velocity v ( m/s  0  12  24  35  41  45  47 
Use the trapezium rule to calculate distance travelled between t = 1 and t = 6
Note;
The area under velocity – time graph represents the distance covered between the given times.
To find the required displacement, we find the area of the region bounded by graph, t =1 and t =6
0 1 2 3 4 5 6
Solution
Divide the required area into five trapezia, each of with 1 unit. Using the trapezium rule;
;
The required displacement =
m
Example
Estimate the area bounded by the curve y = , the x – axis, the line x =1 and x = 5 using the trapezium rule.
Solution
To plot the graph y = , make a table of values of x and the corresponding values of y as follows:
x  0  1  2  3  4  5 
Y =  5  5.5  7  9.5  13  17.5 
By taking the width of each trapezium to be 1 unit, we get 4 trapezium .A, B , C and D .The area under curve is approximately;
= sq.units
The Mid ordinate Rule
The area OPQR is estimated:
The area of OPQR is estimated as follows
 Divide the base OR into a number of strips, each of their width should be the same .In the example we have 5 strips where h =
 From the midpoints of OE ,EF ,FG ,GH and HR , draw vertical lines ( mid ordinates) to meet the curve PQ as shown above
 Label the midordinates
 We take the area of each trapezium to be equal to area of a rectangle whose width is the length of interval (h) and the length is the value of mid –ordinates. Therefore, the area of the region OPQR is given by;
This the mid –ordinate rule.
Note:
The midordinate rule for approximating areas of irregular shapes is given by ;
Area = (width of interval) x (sum of mid – ordinates)
Example
Estimate the area of a semicircle of radius 4 cm using the mid – ordinate rule with four equal strips, each of width 2 cm.
Solution
The above shows a semicircle of radius 4 cm divided into 4 equal strips, each of width 2 cm. The dotted lines are the midordinates whose length are measured.
By mid ordinate rule;
= 2 (2.6 + 3.9 + 3.9 + 2.6)
= 2 x 13
= 26
The actual area is
= 25.14 to 4 s.f
Example
Estimate the area enclosed by the curve y = and the x – axis using the midordinate rule.
Solution
Take 3 strips. The dotted lines are the mid – ordinate and the width of each of the 3 strips is 1 unit.
By calculation, are obtained from the equation;
y =
When x = 0.5,
When x = 1.5,
When x = 2.5,
Using the mid ordinate rule the area required is
A = 1
= 1 (1.125 + 2.125 + 4.125)
= 7.375 square units
End of topic
Did you understand everything?
If not ask a teacher, friends or anybody and make sure you understand before going to sleep! 
Past KCSE Questions on the topic.
 The shaded region below represents a forest. The region has been drawn to scale where 1 cm represents 5 km. Use the mid – ordinate rule with six strips to estimate the area of forest in hectares. (4 marks)
 Find the area bounded by the curve y=2x^{3} – 5, the xaxis and the lines x=2 and x=4.
 Complete the table below for the function y=3x^{2} – 8x + 10 (1 mk)
x  0  2  4  6  8  10 
y  10  6  70  230 
Using the values in the table and the trapezoidal rule, estimate the area bounded by the curve y= 3x^{2} – 8x + 10 and the lines y=0, x=0 and x=104. Use the trapezoidal rule with intervals of 1 cm to estimate the area of the shaded region below
 (a) Find the value of x at which the curve y= x 2x^{2} – 3 crosses the x axis
(b) Find ò(x^{2} – 2x – 3) dx
(c) Find the area bounded by the curve y = x^{2} – 2x – 3, the axis and the lines x= 2 and x = 4.
 The graph below consists of a non quadratic part (0 ≤ x ≤ 2) and a quadrant part (2 ≤ x 8). The quadratic part is y = x^{2} – 3x + 5, 2 ≤ x ≤ 8
(a) Complete the table below
x  2  3  4  5  6  7  8 
y  3 
(1mk)
(b) Use the trapezoidal rule with six strips to estimate the area enclosed by the
curve, x = axis and the line x = 2 and x = 8 (3mks)
(c) Find the exact area of the region given in (b) (3mks)
(d) If the trapezoidal rule is used to estimate the area under the curve between
x = 0 and x = 2, state whether it would give an under estimate or an over estimate. Give a reason for your answer.
 Find the equation of the gradient to the curve Y= (x^{‑2} + 1) (x – 2) when x = 2
 The distance from a fixed point of a particular in motion at any time t seconds is given by
S = t^{3} – 5t^{2} + 2t + 5
2t^{2}
Find its:
(a) Acceleration after 1 second
(b) Velocity when acceleration is Zero
 The curve of the equation y = 2x + 3x^{2}, has x = 2/3 and x = 0 and x intercepts.
The area bounded by the axis x = 2/3 and x = 2 is shown by the sketch below.
Find:
(a) (2x + 3 x^{2}) dx
(b) The area bounded by the curve x – axis, x = – ^{2}/_{3} and x =2
 A particle is projected from the origin. Its speed was recorded as shown in the table below
Time (sec)  0  5  10  15  20  25  39  35 
Speed (m/s)  0  2.1  5.3  5.1  6.8  6.7  4.7  2.6 
Use the trapezoidal rule to estimate the distance covered by the particle within the 35 seconds.
 (a) The gradient function of a curve is given by dy = 2x^{2} – 5
dx
Find the equation of the curve, given that y = 3, when x = 2
(b) The velocity, vm/s of a moving particle after seconds is given:
v = 2t^{3} + t^{2} – 1. Find the distance covered by the particle in the interval 1 ≤ t ≤ 3
 Given the curve y = 2x^{3} + ^{1}/_{2}x^{2} – 4x + 1. Find the:
 i) Gradient of curve at {1, –^{1}/_{2}}
 ii) Equation of the tangent to the curve at {1, – ^{1}/_{2}}
 The diagram below shows a straight line intersecting the curve y = (x1)^{2} + 4
At the points P and Q. The line also cuts xaxis at (7, 0) and y axis at (0, 7)
 a) Find the equation of the straight line in the form y = mx +c.
 b) Find the coordinates of p and Q.
 c) Calculate the area of the shaded region.
 The acceleration, a ms^{2}, of a particle is given by a =25 – 9t^{2}, where t in seconds after the particle passes fixed point O.
If the particle passes O, with velocity of 4 ms^{1}, find
(a) An expression of velocity V, in terms of t
(b) The velocity of the particle when t = 2 seconds
 A curve is represented by the function y = ^{1}/_{3} x^{3}+ x^{2} – 3x + 2
(a) Find: dy
dx
(b) Determine the values of y at the turning points of the curve
y =^{ 1}/_{3}x^{3} + x^{2} – 3x + 2
(c) In the space provided below, sketch the curve of y = ^{1}/_{3} x^{3} + x^{2} – 3x + 2
 A circle centre O, ha the equation x^{2} + y^{2} = 4. The area of the circle in the first quadrant is divided into 5 vertical strips of width 0.4 cm
(a) Use the equation of the circle to complete the table below for values of y
correct to 2 decimal places
X  0  0.4  0.8  1.2  1.6  2.0 
Y  2.00  1.60  0 
(b) Use the trapezium rule to estimate the area of the circle
 A particle moves along straight line such that its displacement S metres from a given point is S = t^{3} – 5t^{2} + 4 where t is time in seconds
Find
(a) The displacement of particle at t = 5
(b) The velocity of the particle when t = 5
(c) The values of t when the particle is momentarily at rest
(d) The acceleration of the particle when t = 2
 The diagram below shows a sketch of the line y = 3x and the curve y = 4 – x^{2} intersecting at points P and Q.
(a) Find the coordinates of P and Q
(b) Given that QN is perpendicular to the x axis at N, calculate
(i) The area bounded by the curve y = 4 – x2, the x axis and the line QN (2 marks)
(ii) The area of the shaded region that lies below the x axis
(iii) The area of the region enclosed by the curve y = 4x^{2}, the line
y – 3x and the yaxis.
 The gradient of the tangent to the curve y = ax^{3} + bx at the point (1, 1) is 5
Calculate the values of a and b.
2007
 The diagram on the grid below represents as extract of a survey map showing
two adjacent plots belonging to Kazungu and Ndoe.
The two dispute the common boundary with each claiming boundary along different smooth curves coordinates (x, y) and (x, y_{2}) in the table below, represents points on the boundaries as claimed by Kazungu Ndoe respectively.
X  0  1  2  3  4  5  6  7  8  9 
Y_{1}  0  4  5.7  6.9  8  9  9.8  10.6  11.3  12 
Y_{2}  0  0.2  0.6  1.3  2.4  3.7  5.3  7.3  9.5  12 
(a) On the grid provided above draw and label the boundaries as claimed by Kazungu and Ndoe.
(b) (i) Use the trapezium rule with 9 strips to estimate the area of the
section of the land in dispute
(ii) Express the area found in b (i) above, in hectares, given that 1 unit on each axis represents 20 metres
 The gradient function of a curve is given by the expression 2x + 1. If the curve passes through the point (4, 6);
(a) Find:
(i) The equation of the curve
(ii) The vales of x, at which the curve cuts the x axis
(b) Determine the area enclosed by the curve and the x axis
 A particle moves in a straight line through a point P. Its velocity v m/s is given by v= 2 t, where t is time in seconds, after passing P. The distance s of the particle from P when t = 2 is 5 metres. Find the expression for s in terms of t.
 Find the area bonded by the curve y=2x – 5 the xaxis and the lines x=2 and x = 4.
 Complete the table below for the function
Y = 3x^{2} – 8 x + 10
X  0  2  4  6  8  10 
Y  10  6  –  70  –  230 
Using the values in the table and the trapezoidal rule, estimate the area bounded by the curve y = 3x^{2} – 8x + 10 and the lines y – 0, x = 0 and x = 10
 (a) Find the values of x which the curve y = x^{2} – 2x – 3 crosses the axis
(b) Find (x^{2} – 2 x – 3) dx
(c) Find the area bounded by the curve Y = x^{2} – 2x – 3. The x – axis and the
lines x = 2 and x = 4
 Find the equation of the tangent to the curve y = (x + 1) (x 2) when x = 2
 The distance from a fixed point of a particle in motion at any time t seconds is given by s = t – ^{5}/_{2}t^{2} + 2t + s metres
Find its
(a) Acceleration after t seconds
(b) Velocity when acceleration is zero
 The curve of the equation y = 2x + 3x^{2}, has x = – ^{2}/_{3} and x = 0, as x intercepts. The area bounded by the curve, x – axis, x = –^{2}/_{3} and x = 2 is shown by the sketch below.
(a) Find ò(2x + 3x^{2}) dx
(b) The area bounded by the curve, x axis x = –^{2}/_{3} and x = 2
 A curve is given by the equation y = 5x^{3} – 7x^{2} + 3x + 2
Find the
(a) Gradient of the curve at x = 1
(b) Equation of the tangent to the curve at the point (1, 3)
 The displacement x metres of a particle after t seconds is given by x = t^{2} – 2t + 6, t> 0
(a) Calculate the velocity of the particle in m/s when t = 2s
(b) When the velocity of the particle is zero,
Calculate its
(i) Displacement
(ii) Acceleration
 The displacement s metres of a particle moving along a straight line after t seconds is given by s = 3t + ^{3}/_{2}t^{2} – 2t^{3}
(a) Find its initial acceleration
(b) Calculate
(i) The time when the particle was momentarily at rest.
(ii) Its displacement by the time it comes to rest momentarily when
t = 1 second, s = 1 ½ metres when t = ½ seconds
(c) Calculate the maximum speed attained