Embakasi East MP Babu Owino during his Maths lesson on Facebook.
Read all the form 4 notes here. You can also download a copy of the pdf notes on this link; MATH FORM FOUR NOTES
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 non-isometric 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 non-isometric 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 x-axis. When multiplying by this matrix, the x co-ordinate remains unchanged, but the y co-ordinate changes sign.
This transformation matrix creates a reflection in the y-axis. When multiplying by this matrix, the y co-ordinate remains unchanged, but the x co-ordinate 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 co-ordinates.
This transformation matrix creates a reflection in the line y=x. When multiplying by this matrix, the x co-ordinate becomes the y co-ordinate and the y-ordinate becomes the x co-ordinate.
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 anti-clockwise. When multiplying by this matrix, the point matrix is rotated 90 degrees anti-clockwise 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 co-ordinates 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
Solution
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;
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;
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 y-axis 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.
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.
clockwise through 900 about the origin. Find the coordinates of this image.
(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”
Transformation M = a b
c d
(ii) Coordinates of C1
Find a single matrix that maps T and T2
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
(a) Given that A (-6, 5) is mapped onto A (6,-4) by a shear with y- axis invariant
(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”
1 -1
Hence find the coordinates to the point at which the two lines
x + 2y = 7 and x – y =1
3 2 2 -4
Find the value of x if
(i) A- 2x = 2B
(ii) 3x – 2A = 3B
(iii) 2A – 3B = 2x
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 | 112-115 | 116-119 | 120-123 |
| Frequency | 1 | 15 | 42 | 31 | 8 | 3 |
Find the mean mass
Solution
Let use a working mean of 109.5.
| class | Mid-point x | t= x – 109.5 | f | f t |
| 100-103 | 101.5 | -8 | 1 | – 8 |
| 104-107 | 105.5 | -4 | 15 | – 60 |
| 108-111 | 109.5 | 0 | 42 | 0 |
| 112-115 | 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;
Example
The example above to be used to demonstrate the steps
| class | Mid-point x | t= | f | f t |
| 100-103 | 101.5 | -2 | 1 | – 2 |
| 104-107 | 105.5 | -1 | 15 | – 15 |
| 108-111 | 109.5 | 0 | 42 | 0 |
| 112-115 | 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 | 150-154 | 155-159 | 160- 164 | 165-169 | 170-174 | 175-179 |
| frquency | 2 | 5 | 16 | 9 | 5 | 2 | 1 |
Calculate the ;
Solution
| class | frequency | Cumulative frequency |
| 145-149 | 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 20thand 21ststudents 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 20th student = 154.5 +
= 154.5 + 4.0625
=158.5625
Height of the 21st = 154.5 +
= 154.5 + 4.375
=158.875
Therefore median height =
= 158.7 cm
The 10th student fall in the in 155 – 159 class
= 154.5 +
5 + 0.9375
4375
(ii) Upper quartile= L +
The 10th student fall in the in 155 – 159 class
= 159.5 +
5 + 3.888
3889
Note;
The median corresponds to the middle quartile or the 50th percentile
= 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
| 1-10 | 11-20 | 21-30 | 31-40 | 41-50 | 51-60 | 61-70 | 71-80 | 81-90 | 91-100 |
| 4 | 9 | 16 | 24 | 18 | 12 | 8 | 5 | 3 | 1 |
Marks
FRCY
(I) The middle 50 % of the students
(ii) The middle 80% of the students
Solution
| 1-10 | 11-20 | 21-30 | 31-40 | 41-50 | 51-60 | 61-70 | 71-80 | 81-90 | 91-100 |
| 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
The median = 39.5
The Lower quartile
The upper quartile
(II) The middle 80% include the marks between the first decile and the 9th 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 |
| fi | 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
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 |
Example
The table below shows the distribution of marks of 40 candidates in a test
| Marks | 1-10 | 11-20 | 21-30 | 31-40 | 41-50 | 51-60 | 61-70 | 71-80 | 81-90 | 91-100 |
| 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 | |
| 1-10 | 5.5 | 2 | 11.0 | – 39.5 | 1560.25 | 3120.5 |
| 11-20 | 15.5 | 2 | 31.0 | -29.5 | 870.25 | 1740.5 |
| 21-30 | 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 |
| 51-60 | 55.5 | 5 | 277.5 | 10.5 | 110.25 | 551.25 |
| 61- 70 71-80 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 | 152-156 | 157-161 | 162-166 | 167-171 | 172- 176 | 177-181 |
| frequency | 12 | 14 | 24 | 15 | 8 | 7 |
Calculate the mean and the standard deviation
Solution
Let A = 169
| Length | Mid-point x | x-169 | 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 |
| 172-176 | 174 | 5 | 1 | 8 | 8 | 8 |
| 177-181 | 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
| 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
| Wind speed ( knots) | 0 – 19 | 20 – 39 | 40 – 59 | 60-79 | 80- 99 | 100- 119 | 120-139 | 140-159 | 160-179 |
| 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
| Pupil | Mark x | x – a | ( x-a)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
| 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
(ii) The standard deviation
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.
Commodity Weight Price Relatives
X 3 125
Y 4 164
Z 2 140
Calculate the retail index for the group of commodities.
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.
| Mass (g) | 25- 34 | 35-44 | 45 – 54 | 55- 64 | 65 – 74 | 75-84 | 85-94 |
| 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 60th percentile mass
(ii) The percentage of potatoes whose masses lie in the range 53g to 68g
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.
| Marks | 0-10 | 10-30 | 30-60 | 60-70 | 70-100 |
| 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
| 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
| 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
| 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 3-dimensional geometry
(d) The angle between
iii) A plane and a plane
Introduction
Geometrical properties of common solids
Examples of three dimensional figures
Rectangular Prism
A three-dimensional figure having 6 faces, 8 vertices, and 12 edges
Triangular Prism
A three-dimensional 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 three-dimensional 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;
Solution
cm
= cm
Similar triangle MQR is right angled at M
cm
= cm
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.
(a) Calculate the
(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
Sketch and label the net of the solid.
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
(a) Sketch and label the pyramid
(b) Find the angle between a slanting edge and the base
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.
therefore=
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;
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;
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:
Solution
Sin 2x = sin (90 – x)
Therefore 2 x = 90 – x
X =
For the given range, x =.
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
| x | 00 | 100 | 200 | 300 | 400 | 500 | 600 | 700 | 800 | 900 | 1000 | 1100 | 1200 |
| 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 00 ≤ x ≤ 1200 on the grid provided
(ii) Hence solve the equation 2 sin 3x = -1.5
| X0 | 00 | 300 | 600 | 900 | 1200 | 1500 | 1800 | 2100 | 2400 | 2700 | 3000 | 3300 | 3600 |
| Cos x0 | 1.00 | 0.50 | -0.87 | -0.87 | |||||||||
| 2 cos ½ x0 | 2.00 | 1.93 | 0.52 | -1.00 | -2.00 |
Using the scale 1 cm to represent 300 on the horizontal axis and 4 cm to represent 1 unit on the vertical axis draw, on the grid provided, the graphs of y = cosx0 and y = 2 cos ½ x0 on the same axis.
(a) Find the period and the amplitude of y = 2 cos ½ x0
(b) Describe the transformation that maps the graph of y = cos x0 on the graph of y = 2 cos 1/2 x0
| X | 00 | 300 | 450 | 600 | 900 | 1200 | 1350 | 1500 | 1800 | 2250 | 2700 | 3150 | 3600 |
| 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 00. Take 1cm represent 300 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
| 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 + 300) | 0.50 | 1 |
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 + 300 ) = 0.
| X0 | 0 | 30 | 60 | 90 | 120 | 150 | 180 |
| 2 sin x0 | 0 | 1 | 2 | 1 | |||
| 1 – cos x0 | 0.5 | 1 |
(b) On the grid provided, using the same scale and axes, draw the graphs of
y = sin x0 and y = 1 – cos x0 ≤ x ≤ 1800
Take the scale: 2 cm for 300 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 xo + cos x0 = 1
values of y, correct to 1 decimal place.
| X | 00 | 150 | 300 | 450 | 600 | 750 | 900 | 1050 | 1200 |
| 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
00 ≤ x ≤ 1200
Take the scale 2 cm for 150 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
Sin2 x – 2 tan x = 0
Cos x
y
y= -3 cos 2x0 and y = 2 sin (3x/20 + 30) for 0 ≤ x ≤ 1800
| X0 | 00 | 200 | 400 | 600 | 800 | 1000 | 1200 | 1400 | 1600 | 1800 |
| -3cos 2x0 | -3.00 | -2.30 | -0.52 | 1.50 | 2.82 | 2.82 | 1.50 | -0.52 | -2.30 | -3.00 |
| 2 sin (3 x0 + 300) | 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 2x0 and y = 2 sin (3x/20 + 300)
(a) On the same axis. Take 2 cm to represent 200 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 x0 + 2 sin (3x/20 + 300) = 0
| x0 | 00 | 300 | 600 | 90 | 10 | 1500 | 180 | 210 | 240 | 270 | 300 | 330 | 360 |
| Cosx0 | 1.00 | 0.50 | -0.87 | -0.87 | |||||||||
| 2cos ½ x0 | 2.00 | 1.93 | 0.5 |
Using the scale 1 cm to represent 300 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 x0 and y = 2 cos ½ x0 on the same axis
(a) Find the period and the amplitude of y =2 cos ½ x0
Ans. Period = 7200. Amplitude = 2
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
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
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;
Solution
Is subtended by 60 nm
Is subtended by; 60 x 60.5 = 3630 nm
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 ;
Solution
Angle between points A and B is (
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 :
Solution
The positions of P and Q on earths surface are as shown below
PQ
– 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;
Solution
=
=
Therefore speed is
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.
(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?
Find
(a) The distance between the two towns in
(b) The local time at X when the local time at Y is 2.00 pm.
(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) 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 (300 N, 130E) at the same speed.
Calculate the total time to complete the journey from airport C, though airport B.
When its 8 am at A, the time at B is 11.00 am.
(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)
Find, to the nearest degree, the latitude on which A and B lie
1200 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 500S with long longitudes and 200 W. (take the radius of the earth to be 6370 km)
(i) Over the North Pole
(ii) Along the parallel of latitude 300 N
the earth to be a sphere with a circumference of 4 x 104 km, calculate in km the distance traveled by the ship.
(b) If a ship sails due west from San Francisco (370 47’N, 1220 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) two-variable linear case.
The general process for solving linear-programming exercises is to graph the inequalities (called the “constraints”) to form a walled-off area on the x,y-plane (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 co-ordinates ( 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;
Co-ordinates 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
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: x1 = number of red gadgets to produce (and sell)
x2 = number of blue gadgets to produce (and sell)
Step 2. Define the objective function.
Maximize R = 30 x1+ 50 x2 (total revenue in ksh)
Step 3. Define all constraints.
(1) x1 £ 50 (steel supply constraint in kilograms)
(2) 2 x2 £120 (wood supply constraint in board meters)
(3) 3 x1 + 2 x2 £ 180 (labor supply constraint in man hours)
x1 , x2³ 0 (non-negativity requirement)
Step 4. Graph all constraints.
Then determine area of feasible study
Note;
Optimization
List all corners (identify the corresponding coordinates), and pick the best in terms of the resulting value of the objective function.
(1) x1 = 0 x2 = 0 R = 30 (0) + 50 (0) = 0
(2) x1 = 50 x2 = 0 R = 30 (50) + 50 (0) = 1500
(3) x1 = 0 x2 = 60 R = 30 (0) + 50 (60) = 3000
(4) x1 = 20 x2 = 60 R = 30 (20) + 50 (60) = 3600 (the optimal solution)
(5) x1 = 50 x2 = 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) 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.
(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 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.
Type A: Kshs 600 per shirt
Type B: Kshs 400 per shirt
If one bag of the mixture contains x kg of N and y kg of S
(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.
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.
| Machine | Number of operators | Floor space | Daily profit |
| A | 2 | 5m2 | Kshs 1,500 |
| B | 5 | 8m2 | 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
(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;
iii) Loci involving points under given conditions;
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 mid-point 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
Intersecting Loci
Solution
In the figure below
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
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.
YZ = 8cm and ÐXYZ = 75o
(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
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 = 30o
(d) (i) Locus of P and locus of Q meet at X. Mark x
(ii) Construct locus R in which angle BRC 120o
(iii) Show the locus S inside triangle ABC such that XS ³ SR
Mark the required region as P
(ii) Locate the position of P such that triangle APB has a maximum area and calculate this area.
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,
iii) Is it possible for the foundation to be 3.4m from AE and in the region?
(a) Construct triangle ABC in which AB = 7 cm, BC = 8 cm and ∠ABC = 600.
(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.
(a) Construct triangle ABC in which BC is 6.7cm, angle ABC is 60o and ÐBAC is 90o.
(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 75o)
(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 ½o 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 co-ordinates
[(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;
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 = 5x3 + 9x2 + 7x + 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:
Lower each exponent by one.
To do this, simply subtract 1 from each exponent in each variable term. Here’s how you do it:
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
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;
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:
Solution
S =
The gradient function is given by;
V =
=
v =
= 24 + 16 – 8
=32m/s
v =
= 54 + 24 – 8
=70m/s
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
Solution
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=x2. 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
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
X = 6 gives
x = 6 gives
(162 + c) – (
We can summarize the steps in short form as follows:
=
=
=150
Example
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 co-ordinates 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
Solution
Therefore, v =-10 t + c
When t = 0, v = 40 m/s
Therefore, 40 = 0 + c
40 = c
When t = 0 , S = 0
C = 0
The expression for displacement is ;
v = 40 – 10 (2)
= 40 – 20
= 20 m/s
S =40t –
= 40 (2) – 5 (
= 80 – 20
= 60 m
V = 40 – 10 (5)
= -10 m/s
S
= 75 m
V = 40
S
= 320 – 320
= 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:
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 mid-ordinate rule;
(f) Apply mid-ordinate rule to approximate area under curves.
Content
(a) Area by counting techniques
(b) Trapezium rule
(c) Area using trapezium rule
(d) Mid-ordinate
(e) Area by the mid-ordinate rule
Introduction
Estimation of areas of irregular shapes such as lakes, oceans etc. using counting method. The following steps are followed
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
This the mid –ordinate rule.
Note:
The mid-ordinate 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 semi-circle 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 mid-ordinates 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 mid-ordinate 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.
| 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= 3x2 – 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
(b) Find ò(x2 – 2x – 3) dx
(c) Find the area bounded by the curve y = x2 – 2x – 3, the axis and the lines x= 2 and x = 4.
(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.
S = t3 – 5t2 + 2t + 5
2t2
Find its:
(a) Acceleration after 1 second
(b) Velocity when acceleration is Zero
The area bounded by the axis x = -2/3 and x = 2 is shown by the sketch below.
Find:
(a) (2x + 3 x2) dx
(b) The area bounded by the curve x – axis, x = – 2/3 and x =2
| 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.
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 = 2t3 + t2 – 1. Find the distance covered by the particle in the interval 1 ≤ t ≤ 3
At the points P and Q. The line also cuts x-axis at (7, 0) and y axis at (0, 7)
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) Find: dy
dx
(b) Determine the values of y at the turning points of the curve
y = 1/3x3 + x2 – 3x + 2
(c) In the space provided below, sketch the curve of y = 1/3 x3 + x2 – 3x + 2
(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
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
(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 = 4-x2, the line
y – 3x and the y-axis.
Calculate the values of a and b.
2007
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, y2) 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 |
| Y1 | 0 | 4 | 5.7 | 6.9 | 8 | 9 | 9.8 | 10.6 | 11.3 | 12 |
| Y2 | 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
(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
Y = 3x2 – 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 = 3x2 – 8x + 10 and the lines y – 0, x = 0 and x = 10
(b) Find (x2 – 2 x – 3) dx
(c) Find the area bounded by the curve Y = x2 – 2x – 3. The x – axis and the
lines x = 2 and x = 4
Find its
(a) Acceleration after t seconds
(b) Velocity when acceleration is zero
(a) Find ò(2x + 3x2) dx
(b) The area bounded by the curve, x axis x = –2/3 and x = 2
Find the
(a) Gradient of the curve at x = 1
(b) Equation of the tangent to the curve at the point (1, 3)
(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
(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
Get the latest Mwalimu National Sacco BOSA Loans Application Form {Free Download}, here. PDF Latest…
Here is the revised and latest Mwalimu National Sacco FOSA Salary Advance Application Form {Free…
Here is the latest Mwalimu National Sacco FOSA Instant Loan Application Form {Free Download}. Download…
The landscape of digital gambling has shifted toward high-octane mechanics that prioritize volatility and massive…
The Teachers Service Commission (TSC) has advertised 170 job vacancies across, covering senior, mid‑level, and…
Health Cabinet Secretary Aden Duale has ordered immediate changes to the Social Health Authority (SHA)…
