Further Mathematics Lesson Note SS3 First Term

1 hours read

Further Mathematics Elearning Note – Edudelight.com

SS3 FURTHER MATHEMATICS

FIRST TERM SCHEME OF WORK

Review of SS2 examination questions. Quadratic inequalities and inequalities in two dimensions.
Matrices and Determinants (2×2, 3×3):

basic definitions
Matrices as linear transformations.
determinants
Solution of 2×3 simultaneous equations.

3. Partial fractions:

Basic definitions
Proper and rational functions with denominators as linear functions (distinct and repeated) and others.

4. Integration:

Understand integration as the reverse of process of differentiation.
Integration of algebraic polynomials including ¹/x, logarithmic functions
Definite integrals and application to kinematics, velocity- time and speed-time graph
Area under the curve, trapezoidal rule, volume of solids of revolution

5. Conic sections

Equations of parabola, hyperbola, ellipse in rectangular Cartesian coordinates
Parametric equations

6. Correlation:

Concept of correlation as measure of relationships
Scatter diagrams
Rank correlation
Tied ranks

7. Mid-term test

8. Probability distributions and approximations

Binomial distribution
Poisson distribution
Normal distribution
Binomial approximations by Poisson distribution

9. Probability distribution and variance

Normal approximations by binomial distribution
Mean
Variance
Coefficient of variance of binomial, Poisson and normal.

10. Static forces

Forces in equilibrium
Resultant of parallel forces (in the same direction and in opposite direction) acting on a rigid body

11. Revision

12. Examination

13.Vacation

WEEK 1

LINEAR INEQUALITIES

CONTENT

Linear & Analytical Solutions of Linear Inequalities in One Variable
Quadratic Inequalities in One Variables
Absolute Values

LINEAR INEQUALITIES IN ONE VARIABLE

Most of the rules for solving linear inequalities in one variable are similar to those for solving a linear equation in one variable with exception of the rules on multiplication and division by negative number which reverses the sense of the inequality

EXAMPLE: Find the solution set of each of the following inequalities and represent them graphically

(a) 2x – 3 < x + 7

(b) 3x + 4 > 1 – 2x

Solution

2x – 3 < x + 7

Adding 3 to both sides

2x < x + 10

Subtracting x from both sides

X < 10

3x + 4 > 1 – 2x

Subtracting 4 from both sides

3x > – 3 – 2x

Adding 2x to both sides

5x > -3

Dividing both sides by 5

x > -3/5

QUADRATIC INEQUALITIES IN ONE VARIABLE

To find the solution sets, of the quadratic inequalities of the form,

ax² + bx + c ≥ 0 or ax² + bx + c ≤ 0.

Note the following

1) If a>0 and b>0 then a.b>0

or a<0 and b<0 then a.b>0

2) If a<0 and b>0 then a.b<0

Or a>0 and b<0 then a.b<0

Worked examples

1) Find the solution set of x² + x – 6 > 0

Solution

x² + x – 6 > 0

(x – 2)( x + 3} > 0

x – 2> 0 or x + 3<0

x >2 or x < -3

x – 2 < 0 or x +3>0

x < 2 or x > -3

-3 < x < 2

–_{30 2}

2) Show graphically the solution Set of the inequality x² + 3x – 4 ≤ 0

Solution

X² + 3x – 4 ≤ 0

X² + 3x – 4 = 0

. (x – 1)(x + 4) ≤ 0

.x – 1 ≤ 0 or x + 4 ≥ 0

.x ≤ 1 or x ≥ -4

X – 1 ≥ 0 or x + 4 ≤ 0

X ≥ 1 or x ≤ -4

X ≤ 1 or x ≥ -4

– 4 ≤ x ≤ 1

-4 0 1

EVALUATION

Find the solution set of the inequalities

a) x² + 5x – 14 < 0

b) 2 – 3x – 9x²> 0

c) 1 – x² ≤ 0

Answers to evaluation questions

.a) x² + 5x – 14 < 0

(x + 7)(x – 2) < 0

.x + 7 < 0 or x – 2 > 0

.x < – 7 or x > 2

And x + 7 > 0 or x – 2 < 0

. x > – 7 or x < 2

Solution -7 < x < 2

-7 < x < 2

_{-7 0 2}

b) 2 – 3x – 9x²> 0

-9x² – 3x + 2 > 0

-9x² – 6x + 3x + 2 > 0

-3x (3x + 2) +1(3x + 2) > 0

(3x + 2) (1- 3x) > 0

3x + 2 > 0 or 1- 3x >0

3x > -2 or 1 > 3x

. x > -⅔ x < ⅓

3x + 2 < 0 or 1- 3x < 0

3x < -2 or 1 < 3x

.x < -⅔ or x > ⅓

Solution -⅔ < x < ⅓

-⅔ < x < ⅓

_–⅔ ₀ ⅓

.c) 1- x²< 0

(1 – x) (1 + x) < 0

1 – x < 0 or 1 + x > 0

X > 1 or x > – 1

1 – x > 0 or 1 + x < 0

. x < 1 or x < -1

Solution – 1 < x < 1

-1 0 1

Quadratic Inequality curve

We recall that th graph of f(x) = ax^² + bx + c is a parabola if D ≥ 0, the parabola crosses the axis at two distinct points, this fact can be used to solve the inequality ax² + bx + c ≥ 0 or ax² + bx + c ≤ 0

Worked examples

1) Determine the solution set of the inequality x² – x – 10 < 2

X² – x – 10 – 2 < 0

X² – x – 12 < 0

(x + 3)(x – 4) < 0

.x + 3 < 0 or x – 4 > 0

.x < -3 or x > 4

.x + 3 > 0 or x – 4 < 0

.x > -3 or x < 4

– 3 < x < 4

-3 0 4

USING PARABOLIC CURVE

Coordination of points at which the curve cuts the axis (x + 3)(x – 4) = 0

X = -3 , x = 4

.y = x² + x – 12

-3 4

-12

2) Find the solution of the inequality x² – 2x – 3 ≥ 0

Solution

x² – 2x – 3 ≥ 0

.(x + 1)(x – 3) ≥ 0

.x + 1 ≥ 0 or x – 3 ≥ 0

.x ≥ -1 or x ≥ 3

(x + 1) ≤ 0 or (x – 3) ≤ 0

.x ≤ -1 or x ≤ 3

Solution set -1 ≤ x ≤ 3

-1 3

-3

b) 9 – x² ≥ 0

3² – x² ≥ 0

(3 – x)(3 + x) ≥ 0

3 – x ≥ 0 or 3 + x ≥ 0 y = 9 – x²

– x ≥ -3 or x ≥ -3

.x ≤ 3 or x ≥ -3 -3 3

(3 – x) ≤ 0 or (3 + x) ≤ 0

-x ≤ -3 or x ≤ – 3

.x ≥ 3 or x ≤ – 3

Solution set -3 ≤ x ≤ 3

ABSOLUTE VALUES

If a number x is positive or negative the absolute value of x is denoted as │x│. The absolute value of a number is the magnitude of the number regardless of the sign.

Worked examples

1) │2x – 3│≥ 4

2x – 3 ≥ 4

2x ≥ 4 + 3

2x ≥ 7

.x ≥ 7/2

.x ≥ 3½

– (2x – 3) ≥ 4

-2 x + 3 ≥ 4

– 2x ≥ 4 – 3

-2x ≥ 1

.x ≤ -½

-½ 0 3½

2) Find the solution set of the inequality │x – 2│<│x + 3│

Solution

│x – 2│<│x + 3│

(x – 2)²< (x + 3)² ≡ x² – 4x + 4 < x² + 6x + 9

– 4x – 6x < 9 – 4

– 10x < 5

.x > – 5/10

.x > -½

-½ 0 1

EVALUATION

Find the solution set of the inequality

a) │2x – 1│>3

b) │x – 3│ – │x – 1│< 0

c) │x – 3│ ≤│x – 2│

Answers to evaluation questions

a) │2x – 1│>3

2x – 1 > 3

2x > 3 + 1

2x > 4

.x > 4/2

.x > 2

-(2x -1) . 3

-2x + 1 > 3

-2x > 3 -1

-2x > 2

.x < -2/2

.x < -1

.x<-1 x > 2

-1 0 2

.b) │x – 3│-│x -1│ < 0

(x – 3)² – (x – 1)²< 0

(x – 3) (x – 3) – (x – 1) (x – 1) < 0

.x² – 6x + 9 – (x² – 2x + 1) < 0

X² – 6x + 9 – x² + 2x – 1 < 0

– 4x + 8 < 0

– 4x < – 8

X > 8/4 = 2

.x > 2

0 1 2

c) │x – 3│ ≤│x – 2│

(x – 3)² ≤ (x – 2)²

. x² – 6x + 9 ≤ x² – 4x + 4

– 6x + 9 ≤ – 4x + 4

– 6x + 4x ≤ 4 – 9

– 2x ≤ -5

.x ≥ – 5/- 2

.x ≥ 2½

_{1 2½}

Reading Assignment : F/maths Project 1 pg 104 – 111

WEEKEND ASSIGNMENT

1) Find the range of x for which │2x – 1│> 3

.a) 1<x<3/2 b) -3/2 < x < -1 c) -3/2 < x < 1 d) x > 3/2 and x < -1

2) Find the range of the value that satisfies the inequality x² + 3x – 18 < 0

. a) -3 < x < 6 (b)-3 > x <6 (c)-6 >x >3 (d)-6 >x < 3 (e)-6 < x <3

3) Find the range of values of x for which 2x² – 5x + 2 ≥ 0

(a) -2<x<-½ (b) ½ <x<2 (c) x < -½ or x ≥ -2 (d) x ≤½ or x ≥ 2

4) Find the range of values of y which satisfies the inequality 2y – 1 < 3 and 2 – y ≤ 5

(a) – 3 ≤ y ≤ 1 (b) – 2 ≤y ≤ 3 (c) -3≤ y ≤ 4 (d) -3 ≤ y ≤ 2

5) Find the range of values of x for which 1/x + 3 < 2x is satisfy

(a) – 3 < x < 5/2 (b) x < -3 and x > -5/2 (c) x < 1 and x < ½

THEORY

1) Find the range of values of x for which 7x – 12 ≥ x²

2) For what values of x is 2x² – 11x + 12 positive?

LINEAR INEQUALITY (PART TWO)

CONTENT

Linear Inequalities in Two Variables by Graphical Method.
Graphical Solution of Simultaneous Linear Inequalities in Two Variables.
Linear Programming

GRAPHICAL SOLUTION OF INEQUALIIES IN TWO VARIABLES

A straight line has the general equation ax+by+c=0, where a,b and c are real numbers.

The line ax + by + c =0 partitions the x-y plane into two regions

The following graph show the region represented by the inequality

Y ≤ 0, y ≥ 0, x ≤ 0 and x ≥ 0

Worked examples

1.Show the region representing 2x + y + 1 > 0

Solution

2x + y + 1 > 0

Steps

make y the subject of the inequality
convert the inequality into a line equation
obtain x and y co ordinates of the line
draw the line and shade the required by the inequality

2x + y + 1 > 0

. y > – 2x -1

When x = 0 , y = -2(0) -1

. y = -1 (0, 1)

When y = 0, 0 > -2x -1

1 > -2x

X = – ½ (-½ ,0)

2. Show the region represented by x- 2y + 3 ≤ 0

Solution

x- 2y + 3 ≤ 0

2y = -3 – x . y = -3/-2 – x/-2 . y = 3/2 + x/2 or y = 3 + x

When x = 0

. y = 3 + 0 = 3 (0, 3/2)

2 2

When y =0

0 = 3 + x

. x =-3 (-3, 0)

EVALUATION

Show the region which represents the following inequality

a) 2x – 3y + 1 ≤ 0 b) x – 4y + 7 ≥0

Answers to evaluation question

2x – 3y + 1 ≤ 0

3y = -2x – 1

. y = -2x – 1

3 -3

. y = 2x + 1 , y = 2x + 1

3 3 3

When x = 0, y = 2(0) + 1 = 1

3 3 ( 0, 1/3)

When y = 0 , 0 = 2x + 1

2x = -1, x = -1/2 (-1/2, 0)

b.) x- 4y + 7 ≥0

4y = -7 – x → y = 7 + x = 7 + x

4 4 4

When x = 0 y = 7 + 0 = 7 (0, 7/4)

4 4

When y = 0, 0 = 7 + x , x = -7 (-7, 0)

Reading Assignment : Further Mathematics Project 1 pg 113 – 119

WEEKEND ASSIGNMENT

Find the range of x for which │2x – 1│> 3

.a) 1<x<3/2 b) -3/2 < x < -1 c) -3/2 < x < 1 d) x > 3/2 and x < -1

2) Find the range of the value that satisfies the inequality x² + 3x – 18 < 0

. a) -3 < x < 6 (b)-3 > x <6 (c)-6 >x >3 (d)-6 >x < 3 (e)-6 < x <3

3) Find the range of values of x for which 2x² – 5x + 2 ≥ 0

(a) -2<x<-½ (b) ½ <x<2 (c) x < -½ or x ≥ -2 (d) x ≤½ or x ≥ 2

4) Find the range of values of y which satisfies the inequality 2y – 1 < 3 and 2 – y ≤ 5

(a) – 3 ≤ y ≤ 1 (b) – 2 ≤y ≤ 3 (c) -3≤ y ≤ 4 (d) -3 ≤ y ≤ 2

5) Find the range of values of x for which 1/x + 3 < 2x is satisfy

(a) – 3 < x < 5/2 (b) x < -3 and x > -5/2 (c) x < 1 and x < ½

WEEK 7 DATE…………………………….

MATRICES

*Definition of matrix and uses

*Examples and types of matrix

*Matrix addition and subtraction

*Multiplication of matrices

Matrix

A matrix is an ordered set of numbers listed rectangular form. A matrix is, by definition, a rectangular array of numeric or algebraic quantities which are subject to mathematical operations. Matrices can be defined in terms of their dimensions (number of rows and columns). Let us take a look at a matrix with 4 rows and 3 columns (we denote it as a 4×3 matrix and call it A):

Each individual item in a matrix is called a cell, and can be denoted by the particular row and column it resides in. For instance, in matrix A, element a₃₂ can be found where the 3rd row and the 2nd column intersect.Matrices and Determinants were discovered and developed in the eighteenth and nineteenth centuries. Initially, their development dealt with transformation of geometric objects and solution of systems of linear equations. Historically, the early emphasis was on the determinant, not the matrix. In modern treatments of linear algebra, matrices are considered first. We will not speculate much on this issue.
Matrices provide a theoretically and practically useful way of approaching many types of problems including:

Solution of Systems of Linear Equations,
Equilibrium of Rigid Bodies (in physics),
Graph Theory,
Theory of Games,
Leontief Economics Model,
Forest Management,
Computer Graphics, and Computed Tomography,
Genetics,
Cryptography,
Electrical Networks,
Fractals

Here are a couple of examples of different types of matrices:

And a fully expanded m×n matrix A, would look like this:

… or in a more compact form:

Example. Let A denote the matrix

        [2  5  7  8]

        [5  6  8  9]

        [3  9  0  1]

This matrix A has three rows and four columns. We say it is a 3 x 4 matrix.We denote the element on the second row and fourth column with a_2,4.

Square matrix

If a matrix A has n rows and n columns then we say it’s a square matrix. In a square matrix the elements a_i,i , with i = 1,2,3,… , are called diagonal elements.
Remark. There is no difference between a 1 x 1 matrix and an ordinary number.

Diagonal matrix

A diagonal matrix is a square matrix with all de non-diagonal elements 0.
The diagonal matrix is completely defined by the diagonal elements.
Example.

        [7  0  0]

        [0  5  0]

        [0  0  6] The matrix is denoted by diag(7 , 5 , 6)

Row matrix

A matrix with one row is called a row matrix. [2 5 -1 5]

Column matrix

A matrix with one column is called a column matrix.

[2]

[4]

[3]

[0]

Matrices of the same kind

Matrix A and B are of the same kind if and only if
A has as many rows as B and A has as many columns as B

        [7  1  2]       [4  0  3]

        [0  5  6]  and  [1  1  4]

        [3  4  6]       [8  6  2]

The transposed matrix of a matrix

The n x m matrix B is the transposed matrix of the m x n matrix A if and only if
The ith row of A = the ith column of B for (i = 1,2,3,..m)
So a_i,j = b_j,I The transposed matrix of A is denoted T(A) or A^T

        [7  1 ]       [7  0  3]

        [0  5 ]    =  [1  5  4]

        [3  4 ]

0-matrix

When all the elements of a matrix A are 0, we call A a 0-matrix.We write shortly 0 for a 0-matrix.

An identity matrix I

An identity matrix I is a diagonal matrix with all the diagonal elements = 1.

A scalar matrix S

A scalar matrix S is a diagonal matrix whose diagonal elements all contain the same scalar value.
a_1,1 = a_i,i for (i = 1,2,3,..n)

        [7  0  0]

        [0  7  0]

        [0  0  7]

The opposite matrix of a matrix

If we change the sign of all the elements of a matrix A, we have the opposite matrix -A.
If A’ is the opposite of A then a_i,j‘ = -a_i,j, for all i and j.

A symmetric matrix

A square matrix is called symmetric if it is equal to its transpose.
Then a_i,j = a_j,i , for all i and j.

        [7  1  5]

        [1  3  0]

        [5  0  7]

A skew-symmetric matrix

A square matrix is called skew-symmetric if it is equal to the opposite of its transpose.
Then a_i,j = -a_j,i , for all i and j.

        [ 0  1 -5]

        [-1  0  0]

        [ 5  0  0]

Matrix Addition and Subtraction

DEFINITION: Two matrices A and B can be added or subtracted if and only if their dimensions are the same (i.e. both matrices have the same number of rows and columns. Take:

Addition

Addition and subtraction operations can easily be performed on matrices, provided the matrices have the same dimensions. All that is required is to add or subtract the corresponding cells of each matrix involved in the operation. Let us take a look at the addition of two 2×3 matrices, A and B:

If A and B above are matrices of the same type then the sum is found by adding the corresponding elements a_ij + b_ij.

Here is an example of adding A and B together.

Example:

             _     _     _    _

1. Problem: | -5 0  |   | 6 -3 |

            |       | + |      |

            |_ 4 1 _|   |_2  3_|

  Solution: Add the corresponding members.

 _                  _

| (-5 + 6)   (0 - 3) |

|                    |

    =       |_( 4 + 2)   (1 + 3)_|

 _      _

| 1   -3 |

|        |

    =       |_6    4_|

Subtraction of matrices is done in the same manner as addition. Always be aware of the negative signs and remember that a double negative is a positive!

SUBTRACTION

If A and B are matrices of the same type then the subtraction is found by subtracting the corresponding elements a_ij − b_ij.

Here is an example of subtracting matrices.

Example. Consider the three matrices J, F, and M from above. Evaluate

Answer.

We have

and since

we get

To compute J–M, we note first that $\begin{displaymath}(-1)M = \left(\begin{array}{cccc} -580&-270&-350\\ -625&-350&-410\\ \end{array}\right)\;.\end{displaymath}$

Since J–M = J + (-1)M, we get

And finally, for J–F+2M, we have a choice. Here we would like to emphasize the fact that addition of matrices may involve more than one matrix. In this case, you may perform the calculations in any order. This is called associativity of the operations. So first we will take care of –F and 2M to get

Since J–F+2M = J + (-1)F + 2M, we get

So first we will evaluate J–F to get

to which we add 2M, to finally obtain

MULTIPLICATION

Performing the operation product involves multiplying the cells of a particular rows in the first matrix by the cells of a particular column in the second matrix, adding the products, and storing the result in the cell of the resultant matrix whose coordinates correspond to the row of the first matrix and the column of the second matrix. For instance, in AB = C, if we want to find the value of c12, we must multiply the cells of row 1 in the first matrix by the cells of column 2 in the second matrix and sum the results.

Example 1

Multiply:

Answer

This is 2×3 times 3×2, which will give us a 2×2 answer.

Our answer is a 2×2 matrix

Multiplying 2 × 2 Matrices

The process is the same for any size matrix. We multiply across rows of the first matrix and down columns of the second matrix, element by element. We then add the products:

In this case, we multiply a 2 × 2 matrix by a 2 × 2 matrix and we get a 2 × 2 matrix as the result.

Example 2:

Multiply:

Answer

Note 2 – Commutativity of Matrix Multiplication

Does AB = BA?

Let’s see if it is true using an example.

Example 3:

math expression and

find AB and BA.

Answer

We performed AB above, and the answer was:

Now BA is (3 × 2)(2 × 3) which will give 3 × 3:

So in this case, AB does NOT equal BA

In general, when multiplying matrices, the commutative law doesn’t hold, i.e. AB ≠ BA. There are two common exceptions to this:

The identity matrix: IA = AI = A.
The inverse of a matrix: A^-1A = AA^-1 = I.

Example 4 – Multiplying by the Identity Matrix

Given that

find AI.

Answer

We see that multiplying by the identity matrix does not change the value of the original matrix.That is, AI = A

Further Examples

Example1. If possible, find BA and AB.

Answer

AB is not possible. (3 × 3) × (1 × 3).

Example2. Determine if B = A^-1.

Answer

If B = A^-1, then AB = I.

So B is NOT the inverse of A.

Example3. In studying the motion of electrons, one of the Pauli spin matrices is

where

Show that s² = I. [If you have never seen j before, it is on complex numbers].

Answer

Example4. Evaluate the following matrix multiplication:

Answer

The interpretation of this is that the robot arm moves from position (2, 4, 0) to position (-2.46, 3.73, 0). That is, it moves in the x-y plane, but its height remains at z = 0. The 3 × 3 matrix containing sin and cos values tells it how many degrees to move.

EVALUATION

Given that and

2A + 4B = …………………………..
BA = ………………………..
If I = identity matrix, find 7AB + 3I = ………………..
Solve for x and y , ……..x= ,y=
If X = and Y = , Find XY …………

THEORY

Determine X + Y if
If , find u and v if x = 3, y = 1 and A =30⁰

WEEK 8 DATE…………………..

TRANSPOSE AND INVERSE OF MATRICES, DETERMINANT OF MATRICES, APPLICATION OF DETERMINANTS.

Content:

*Transpose of matrices

*Determinant

*Application of determinants(Cramer’s rule)

*Inverse of a matrix

TRANSPOSE OF MATRICES

DEFINITION: The transpose of a matrix is found by exchanging rows for columns i.e. Matrix A = (a_ij) and the transpose of A is:

A^T = (a_ji) where j is the column number and i is the row number of matrix A.

For example, the transpose of a matrix would be:

In the case of a square matrix (m = n), the transpose can be used to check if a matrix is symmetric. For a symmetric matrix A = A^T.

The Determinant of a Matrix

DEFINITION: Determinants play an important role in finding the inverse of a matrix and also in solving systems of linear equations. In the following we assume we have a square matrix (m = n). The determinant of a matrix A will be denoted by det(A) or |A|. Firstly the determinant of a 2×2 and 3×3 matrix will be introduced, then the n×n case will be shown.

Determinant of a 2×2 matrix

Assuming A is an arbitrary 2×2 matrix A, where the elements are given by:

then the determinant of a this matrix is as follows:

Determinant of a 3×3 matrix

The determinant of a 3×3 matrix is a little more tricky and is found as follows (for this case assume A is an arbitrary 3×3 matrix A, where the elements are given below).

then the determinant of a this matrix is as follows:

Determinant of a n×n matrix

For the general case, where A is an n×n matrix the determinant is given by:

Where the coefficients α_ij are given by the relation:

where β_ij is the determinant of the (n-1) × (n-1) matrix that is obtained by deleting row i and column j. This coefficient α_ij is also called the cofactor of a_ij.

Calculating a 2 × 2 Determinant

In general, we find the value of a 2 × 2 determinant with elements a, b, c, das follows:

We multiply the diagonals (top left × bottom right first), then subtract.

Example 1.

The final result is a single number.

Application of Determinants in Solve Systems of Equations

We can solve a system of equations using determinants, but it becomes very tedious for large systems. We will only do 2 × 2 and 3 × 3 systems using determinants.

Cramer’s Rule.

The solution (x, y) of the system

can be found using determinants:

Example 2

Solve the system using Cramer’s Rule:

x − 3y =6 ; 2x + 3y = 3

Answer

First we determne the values we will need for Cramer’s Rule:

a₁ = 1 b₁ = -3 c₁ = 6

a₂ = 2 b₂ = 3 c₂ = 3

3 × 3 Determinants

A 3 × 3 determinant

can be evaluated in various ways.

We will use the method called “expansion by minors”. But first, we need a definition.

Cofactors

The 2 × 2 determinant

is called the cofactor of a₁for the 3 × 3 determinant:

The cofactor is formed from the elements that are not in the same row as a₁ and not in the same column as a₁.

Similarly, the determinant

is called the cofactor of a₂. It is formed from the elements not in the same row as a₂and not in the same column as a₂.

We continue the pattern for the cofactor of a₃.

Expansion by Minors

We evaluate our 3 × 3 determinant using expansion by minors. This involves multiplying the elements in the first column of the determinant by the cofactors of those elements. We subtract the middle product and add the final product.

Note that we are working down the first column and multiplying by the cofactor of each element.

Example 3

Evaluate

Answer

= -2[(-1)(2) − (-8)(4)] − 5[(3)(2) − (-8)(-1)] + 4[(3)(4) − (-1)(-1)]

= -2(30) − 5(-2) + 4(11)

= -60 + 10 + 44

= -6

Here, we are expanding by the first column. We can do the expansion by using the first row and we will get the same result

Cramer’s Rule to Solve 3 × 3 Systems of Linear Equations

We can solve the general system of equations,

a₁x + b₁y + c₁z = d₁

a₂x + b₂y + c₂z = d₂

a₃x + b₃y + c₃z = d₃

By using the determinants:

where

Example 4

Solve, using Cramer’s Rule:

2x + 3y+ z = 2

−x + 2y+ 3z = −1

−3x− 3y+ z = 0

Answer

where

Determinant Exercises

EVALUATION

1. Evaluate by expansion of minors:

2. Solve the system by use of determinants:

x + 3y+ z = 4

2x − 6y − 3z = 10

4x− 9y+ 3z = 4

The Inverse of a Matrix

DEFINITION: Assuming we have a square matrix A, which is non-singular (i.e. det(A) does not equal zero), then there exists an n×n matrix A^-1 which is called the inverse of A, such that this property holds:

AA^-1 = A^-1A = I, where I is the identity matrix.

The inverse of a 2×2 matrix

Take for example an arbitrary 2×2 Matrix A whose determinant (ad − bc) is not equal to zero.

where a,b,c,d are numbers, The inverse is:

Now try finding the inverse of your own 2×2 matrices.

The inverse of a n×n matrix

The inverse of a general n×n matrix A can be found by using the following equation.

Where the adj(A) denotes the adjoint (or adjugate) of a matrix. It can be calculated by the following method:

Given the n×n matrix A, define
B = b_ij
to be the matrix whose coefficients are found by taking the determinant of the (n-1) × (n-1) matrix obtained by deleting the i^th row and j^th column of A. The terms of B (i.e. B = b_ij) are known as the cofactors of A.
Define the matrix C, where
c_ij = (−1)^i+j b_ij.
The transpose of C (i.e. C^T) is called the adjoint of matrix A.

Lastly to find the inverse of A divide the matrix C^T by the determinant of A to give its inverse.

We’ll find the inverse of a matrix using 2 different methods. You can decide which one to use depending on the situation.

The first method is limited to finding the inverse of 2 × 2 matrices. It involves the use of the determinant of a matrix which we saw earlier.

Reminder: We can only find the determinant of a square matrix. For example, if A is the square matrix

Then we can find the determinant of A:

= 10 + 3 = 13.

For convenience, we could have written the determinant of A as |A| and so our final answer would be:

|A| = 13

Another way of writing the same thing is to use “det” for “determinant”. So for example, in this case we would write: det(A) = 13

Method 1 – Transposing and Determinants

This method is only good for finding the inverse of a 2 × 2 matrix.

Example.

Find the inverse, A^-1, of math expression using Method 1.

Answer

Method 1 is as follows. [1] Interchange leading diagonal elements:

-7 → 2; 2 → -7

[2] Change signs of the other 2 elements:

-3 → 3; 4 → -4

[3] Find the determinant |A|

math expression = -14 + 12 = -2

[4] Multiply result of [2] by

So we have found the inverse, as required.

Is it correct?

We check by multiplying our inverse by the original matrix. If we get the identity matrix (I) for our answer, then we must have the correct answer.

Method 2: Adjoint matrix

Method 2 uses the adjoint matrix method.

Answer

The inverse of a 3×3 matrix is given by:

“adjA” is short for “the adjoint of A“. We use cofactors (that we met earlier) to determine the adjoint of a matrix.

Cofactors Recall: The cofactor of an element in a matrix is the value obtained by evaluating the determinant formed by the elements not in that particular row or column.

Example: Consider the matrix

The cofactor of 6 is

The cofactor of -3 is

We find the adjoint matrix by replacing each element in the matrix with its cofactor and applying a + or – sign as follows:

and then finding the transpose of the resulting matrix. The transpose means the 1^stcolumn becomes the 1^st row; 2^nd column becomes 2^nd row, etc.

Example 1:
Find the inverse of the following by using the adjoint matrix method:

A = matrix

Solution:

Step 1:

Replace elements with cofactors and apply + and –

Step 2

Transpose the matrix:

adjA = matrix

Before we can find the inverse of matrix A, we need det A:

Now we have what we need to apply the formula

So, matrix

A^-1 = matrix

Example 2.
Find the inverse of

using method 2.

Answer

Interchange rows and columns:

DetA = So

OBJECTIVE

Find the non-zero positive value of x which satisfies the equation
2 B C. D. 1
Find the value of k. , A 1 B 2 C 3 D 4
Find the matrix T if ST = I where S = and I is the identity matrix. ………
Given that Q = , evaluate
A matrix P = is such that P^T = -P , where P^T is the transpose of P . I f b = I, then P is.

THEORY

Find the inverse of the matrix
Find the values of t for which the determinant of the matrix below will give zero.

WEEK 4

INTEGRATION:

Integration: This is defined as anti- differentiation. Suppose, y = x³ + 2x, the first derivative is

3x² + 2. (dy/dx = 3x²+ 2)

then the anti – derivative of 3x² + 2 = x³ + 2x

Thus, integration is the reverse process of differentiation and denoted by the symbol ∫.

If dy/dx = xⁿ , then ∫ dy/dx = xⁿ⁺¹ + C

n + 1 (n ≠ -1)

where c is the arbitrary constant.

Indefinite integral concepts:

General Concept:

Example: Evaluate the following integrals:
1. ∫x² dx 2. ∫x^5/2 dx 3. ∫4/ x5 dx 4. ∫√x⁸ 5. ∫(7x⁴ + 2) dx 6. ∫(x⁵ + 2x⁴ – x³ + 6) dx

Solution;

1. x³+ C 2. x ^{5/2 + 1}= 2x^7/2+ c 3. ∫4x^-5 dx = 4x^-5+1= – 4^{x – 4}+ C

3 5/2 + 1 7 -5 + 1

4. ∫ x⁴ = x⁴⁺¹= x⁵ + C 5. 7x^{4 +1}+ 2x⁰⁺¹= 7x⁵ + 2x + C

4 + 1 5

6. x⁶ + 2x⁵ – x⁴+ 6x + C

6 5 4

NB: Integral of a constant is not zero but the variable in the question.

Evaluation: Evaluate the following integrals; 1. ∫ (12×3 – x⁶ +1/x²) dx 2. ∫x²(3x² + 4x) dx

Trigonometric integral:

The trigonometric integrals can be summarized in the following table. Remember that this is the reverse process of differentiation.

F(x) ∫ f(x)dx

Sin x – cos x + c

Cos x sin x + c

Sec²x tan x + c

Cosec²x – cot x + c

Sec x tan x sec x + c

e^x e^x + c

1/x ln x + c

Example: Evaluate each of the following integrals.

1. ∫sin x – 5 cos x )dx 2. ∫ (5sinx + 3x²)dx

Solution:

1. . ∫sin x – 5 cos x )dx = ∫sin x dx – ∫5 cos x dx

= – cos x – 5 ( sin x ) + c

= – cos x – 5sin x + c

2.∫ (5sinx + 3x²)dx = ∫5 sin x dx + ∫3×2 dx

= -5cos x + x³ + c

3. ∫ e^2x dx = e^2x/2 + c

Evaluation: Evaluate the integrals:

1. ∫ (3 cos x + 2 sin x) dx 2. ∫e ^{2×2 + 5x} dx

INTEGRATION BY ALGEBRAIC SUBSTITUTION

Sometimes integral are not given in the standard form, such integral are then reduced to standard form format before evaluation by algebraic substitution.

Suppose, an integral is given in the form ∫ f(ax + b)ⁿ dx

Then, the algebraic substitution is to represent the function in the bracket by any letter.

Let u = (ax + b) du/dx = a, dx = du/ a

∫uⁿ dx = ∫ uⁿ du/a

= 1/a ∫ uⁿ du

Example:

Evaluate the following integrals.

1. ∫(2x² – 5x )⁴ dx 2. ∫ 7 dx 3. ∫x cos 2x² dx 4. ∫ x² √(x³ + 5)dx

(5x – 4 )

Solution:

1. . ∫(2x² – 5x )⁴ dx let u = 2x² – 5x , du/dx = 4x – 5 , dx = du/4x -5

. ∫u 4 du/ 4x -5 = u⁵+ c

5(4x – 5)

= (2x² – 5x )⁴+ c

20x – 25

2. ∫ 7 dx let u = ( 5x – 4) , du/dx = 5, dx = du/5

(5x – 4 )⁵

∫7 u ^-5 du/5 = 7u^-4+ c

5 x – 4

= 7(5x – 4 )^-4

-20

3. ∫x cos 2x² dx let u = 2x², du/dx = 4x, dx = du/4x

then; ∫x cos 2x² dx = ∫x cos u du/4x = 1 x ( sin u ) = 1 sin 2x² + c

4 4

4. ∫ x² √(x³ + 5)dx let u = x³ + 5, du/dx = 3x², dx = du/3x²

∫ x² √(x³ + 5)dx = ∫x² u^1/2du/3x² = 1 x u^3/2=2 (x³ + 5)^3/2+ c

3/2 x 3 9

Evaluation: Evaluate the following integrals:

1. ∫(5x – 7 )^7/2dx 2.∫cos 9x dx 3. ∫ x cos 2x dx.

INTEGRATION BY PARTS

This technique is uniquely useful in evaluating integrals that are not in the standard form. Such integrals can not be solved by algebraic substitution.

From the product rule of differentiation, it can be generalized thus;

∫ vdu = uv – ∫ udv.

Example: Evaluate the following integral by parts.

∫ 2x sin x dx 2. ∫ e^2x cos 2x dx

solution:

1. ∫ 2x sin x dx , let v = 2x, dv/dx = 2, dv = 2dx

∫du = sin x dx

u = – cos x

∫ vdu = uv – ∫ udv.

∫ x² sin x = – x²cos x – ∫- cos x x 2 dx

= – x² cos x + 2∫ cos x

= – x² cos x + 2 sin x + c

2. ∫ x²e^x dx , let v = x², dv/dx = 2x, dv = 2xdx

du = e^x u = e^x dx

∫ vdu = uv – ∫ udv.

∫ x²e^x = e^x x² – ∫ e^x2xdx

= x² e^x – 2∫e^x x dx

the integral part in the RHS will have to be evaluated using integration by parts;

thus, v = x, dv/dx = 1, dv = dx , du =e^x, u = e^x

∫ vdu = uv – ∫ udv.

∫x e^x = e^x. x – ∫e^xdx

= e^x.x – e^x

finally, ∫ x²e^x = x² e^x – 2(e^x.x – e^x)

= x²e^x – 2xe^x + 2e^x + c

Evaluation: Evaluate 1. ∫x² cos x dx 2.∫x³ e^-x dx

INTEGRATION BY PARTIAL FRACTION

Sometimes rational functions are not expressed in the proper standard form; such function can be evaluated by transforming them into standard form through partial fractions. The knowledge of partial fractions is needed here to evaluate the functions.

Example; Integrate each of the following with respect to x;

1. 2x + 32. x + 8

(2x + 1) (x – 1) ( x² + 3x + 2)

solution:

1. resolve into partial fraction; 2x + 3 = A + B

(2x + 1)(x – 1) ( 2x + 1) (x – 1)

2x + 3 = A(x-1) + B(2x + 1)

when x = 1, 2(1) + 3 = B(2 + 1)

5 = 3B, B= 5/3

when x = -1/2, 2(-1/2) + 3 = A(-1/2 – 1 )

2 = – 3/2 A A = – 4/ 3

; 2x + 3 =∫-4+∫ 5dx

(2x + 1)(x – 1) 3( 2x + 1) 3(x – 1)

= -4 ln (2x + 1)+5 ln (x – 1)

2 x 3 3

= – 2/3 ln (2x + 1) + 5/3 ln ( x – 1) + c

2. x + 8 =x + 8 = A+ B

(x² + 3x + 2) (x + 1)(x + 2) x + 1 x + 2

x + 8 = A(x+2) + B(x+1)

when x = -2,

– 2 + 8 = B(-2+1)

6 = -B, B = – 6

when, x = -1, – 1 + 8 = A(-1 + 2)

7 = A.

thus, x + 8 = ∫7 + ∫ – 6

x + 1 x + 2

= 7ln (x+ 1) – 6 ln(x+2) + c

Evaluation: Integrate by partial fraction.

1. 4x + 3 2. 1

(x – 3)(x+2) (x²+ 3x + 2)

WEEKEND ASSINGMENT:

1. Evaluate ∫ (x⁵ + 3)dx . A. x⁶/6 + 3x + c B. x⁵/6 + c C. x⁶/6 + c

2. Evaluate ∫ cos 7x dx A. 7sin 7x B. 1/7 sin 7x + c C. 7sin 7x + c

3. Integrate the function; (3x + 5)⁵ wrt x .A 12(2x+3)⁶ + c B. (2x +3)⁶+ c C. (3x + 5)⁶ + c

4. Find ∫(x+1)(x²– 2)dx A. x⁴ + x³ – x² – 2x + c B. x⁴ – x³ + x² + c C. x + x⁴ – x³ – x² + c

4 3 3 4

5. Integrate 1/x⁵ wrt x. A. x ⁶ + c B . x ^-4+ c C. x^-4+ c

6 – 4

Theory:

1. Find ∫x sin2xdx 2. Evaluate ∫ dx

x ( x + 2)

Definite integral

The process of differentiation can be used to determine areas under curve. The definite integral is used to determine area under curve.

Consider the curve y = f(x) in the range of x = q and x = r. The area under the curve can be determine by; f(q) – f(r)

Thus^,^∫^q^{f(x) dx is called the definite integral}.

The integral constant “c” is not needed here as it will be eliminated during calculation.

Example:

₁

Evaluate ∫₀dx = [ x ] = 1 – 0 = 1.

_-2

Evaluate ∫_-1(x² – x )dx = x³ – x²

3 2

= (-2)³ – (-2)² – [ (-1)³ + (-1)²]

3 2 3 2

= -8/3 – 2 +1/3 – ½ = -16 – 12 + 2 – 3

₃ = – 29/6.

3. Evaluate ∫₁(x-1)(3-x)dx = [2×2 – x³– 3x ]³₁ = 0 – (-4/3) = 4/3

4 Given that ∫_1.25 (4x – 5)⁴dx = 51.2, find the value of p.

Solution;

Let u = 4x – 5, du/dx = 4, dx = du/4

∫(4x – 5)⁴dx = ∫u⁴du/4 = u ⁵

p5 – 1.25⁵= 51.2 , p⁵ – 3.052 = 1024

20 20

p⁵ = 1024 + 3.052

p = 4 . 002

_{3 5}

Evaluation: Evaluate the following integrals; 1. ∫₁(x + 2)^-1dx 2.∫₁ (2x – 1)³dx

AREA UNDER CURVE

The definite integral is a useful tool in determining the area under curve.

I Curve and x – axis

Suppose we have a function y = f(x), the ordinates corresponding to x = a and x = b and the x – axis.

If ∫_af(x)dx , then the area bounded by the curve and the x – axis is;

_{c b c}

∫_af(x)dx = ∫_a f(x)dx + ∫_bf(x)dx

Example: Find the area bounded by the curve y = 3x², the ordinates x= 2, x = 5 and the x – axis.

Solution;

F(x) = 3x²dx

∫₂ 3x²dx = 3x³= 5³ – 2³=125 – 8 = 117.

Evaluation; Find the area bounded by the curve y = x² the ordinates x = 1, x = 3 and the x – axis.

II CURVE y – axis and two abscissa.

This is when the the curve makes with the y – axis is required and the y ordinates given.

WEEK5

THE CIRCLE:

A&B. Definition, General Equation, examples and evaluation.

Definition:

A circle is defined as the locus of point equidistant from a fixed point. A circle is completely specified by the centre and the radius.

Equation of a circle with centre (a, b) and radius r.

Q (x,y)

y – b

ppppp

x – a

a——–→

PR = x – a.

QR = y – b

Since D PQR is a right angle triangle, we have:

PQ ² = PR ²+ QR ²

r² = (x-a)² + (y-b)²

Hence, the equation of a circle with centre (a, b) and radius r is

(x- a )²+ (y – b)² = r²

If the centre of the circle is the origin (0, 0), the equation become x² + y² = r²

GENERAL EQUATION OF A CIRCLE

From (x – a)² + (y – b)² = r²

r² – 2ax + a² +y²– 2by+b² –r² = 0

x²+ y² -2ax -2by +a² + b²– r² = 0

The above equation can be written as x² +y² +2gx + 2fy +c = 0

Where a = – g; b= – f; c= a²+b² – r²

Hence: x² + y² + 2gx + 2fy + c = 0 is called the general equation of a circle. Observe the following about the general equation.

i It is a second degree equation in x and y

ii The co- efficient of x² and y² are equal

iii It has no xy term.

Examples:

1 Find the equation of a circle of centre (3, -2) radius 4 units.

Solution:

a = 3; b = -2 and r = 2

(x – a)² + (y – b )² = r²

(x- 3)² +(y +2)² = 4²

x²-6x + 9 +y² + 4y + 4 = 16

x²+y² – 6x + 4y + 9 +4 – 16= 0

x² + y² – 6x + 4y-3= 0

Find the centre and radius of a circle whose equation is x² + y² – 6x + 4y – 3 = 0

Solution:

X² + y² – 6x + 4y – 3 = 0

X²– 6x+ y² + 4y = + 3

Complete the square for x and y

X² – 6x + 9 +y² 4y + 4 = 3 + 9 + 4

(x – 3)² + (y + 2)² = 16

Compare with (x- a)² + (y – b) = r²

a = 3, b = – 2, r² = 16, \ r = 16 = 4.

Hence the centre is (3, – 2) and the radius is 4 units.

Evaluation:

Find the equation of the circle with center (-1, – 1 ) and radius 3.
Find the centre and radius of the circle x² + y² -6x+ 14y +49 = 0

C. EQUATION OF TANGENT TO A CIRCLE AT POINT (x₁,y₁)

Let the equation of the circle be x²+ y² + 2gx +2fy + c = 0

At x₁, y₁

x₁² + y₁²+ 2gx₁ + 2fy₁ +c = 0

C = – (x₁² + y₁² +2gx₁ +2fy₁)…………… (i)

Differentiating the equation of circle above

2x +2y dy/dx + 2g + 2f dy/dx = 0

Divide through by 2

x + y dy/dx + g +f dy/dx = 0

(y + f)dy/dx = – (x+g)

The equation of the tangent at x₁,y₁

(y-y₁) (y+f) = -(x-x₁) (x+g)

yy₁ + yf – y₁² -y₁f = – (xx₁ + xg – x₁²-x₁g)

yy₁+ yf – y₁²-y₁f = – xx₁ – xg +x₁² + x₁g

yy₁+ yf +xx₁ + xg = x₁²+ x₁g +y₁²+y₁f

yy₁ + xx₁ + yf + xg = x² + y² +x₁g +y₁f

Adding gx₁+ y₁f to both sides

yy₁ + xx₁ +y₁f + yf+xg +gx₁ = x₁² + y₁² + x₁g +x₁g + y₁f+y₁f

yy₁ + xx₁+ (y₁ +y) f + (x+x₁) g = x₁² + y₁²+2x₁g+2y₁f

but x₁²+y₁² +2x₁g +2y₁f = – C

yy₁ +xx₁ + (y₁ + y) f +(x + x₁) g + C = 0

Hence the equation of the tangent to the circle x² + y² + 2gx +2fy+c= 0 at (x₁,y₁) on the circle is xx₁+ yy₁ + (x+x₁)g + (y + y₁)f + c = 0

Example:

Show that the point (2,3) lies on the circle x² + y² – 3x + 4y – 19 = 0. Hence or other wise, determine the equation of the tangent to the circle at the point (2, 3).

Solution:

x² +y²– 3x + 4y – 19 = 0

At (2, 3)

2² + 3²– 3(2) +4(3) – 19 = 0

4 + 9 – 6 + 12 – 19 = 0

R. H. S = L. H. S, hence the point (2,3) lies on the circle.

x²+ y² -3x +4y – 19 = 0

Compare with:

x²+ y² + 2gx +2fy + c= 0

Þ 2g = – 3, 2f = 4

g = ^-3/₂, f = ⁴/₂ = 2 c = – 19

Equation of Tangent:

yy₁ + xx₁ + (x +x₁) g + (y + y₁) f + c = 0

3y + 2x + (x + 2) (-³/₂) + (y + 3)2 – 19 = 0

3y + 2x – (^3x/₂) – 3 + 2y + 6 – 19 = 0

6y + 4x – 3x – 6 + 4y + 12 -38 = 0

10y + x – 32 = 0

Alternatively:

x² + y² – 3x + 4y – 19 = 0

2x + 2y dy/dx – 3 +dy/dx = 0

(2y + 4) = 3 – 2x

at 2,3

= =

y – y₁ = m(x –x₁)

y – 3 = -1/10 (x-2)

10(y- 3) = -1 (x – 2)

10y – 30 = -(x + 2)

10y + x – 30 – 2 = 0

10y + x – 32 = 0

EVALUATION:

Find the equation of the tangent to the circle

x² + y² + 4x – 10y – 12= 0 at (3,1)
x² + y² – 6x – 3y = 16 at (-2, 0 )

READING ASSIGNMENT

Read equation of a circle , Further Mathematics Project II, page 205 -210

WEEKEND ASSIGNMENT

1. What is the radius of the circle whose equation is x² +y² – 6 x – 7 = 0

a) 2 b) 3 c) 4 d) 9

2. Which of the following is not an equation of a circle?

(a) x² + y² = 4 (b) x² + y² – 2x – 3 = 0 (c) x² + y² – 2xy + 4x -6y+ 1 = 0 (d) 2x² + 2y² – 6x + 4 y + 3 = 0

3. The equation of a circle with centre (-2,5) and radius 3 units is

a) x² + y² + 4x – 10y + 20 = 0 b) x² + y² + 4x – 10y + 26 = 0 c) x² + y² + 4x – 10y – 38 = 0 d) x² + y² + 4x – 10y + 39 = 0

4. The coordinates of the centre of the circle

2x² + 2y² – 4x + 12y -7 = 0 is

a) (-1, 3) b) (1,-3) c) (2, -6) d) (-2, 6)

5. The equation of a circle of radius 3 is x² + y² + 10x – 8y + k = 0. Find the value of the constant k.

a) -50 b) 18 c) 32 d) 41 e) 10

THEORY

1. The equation of a circle is x² + y² – 10x + 8y = 0 find

(i) its radius (ii) its area.

2. A circle passes through the points (0, 3) and (4,1), if the centre of the circle is on the x –axis, find the equation of the circle.

THE PARABOLA

D. Definition, Parabola equation, Examples and evaluation.

Definition: A parabola is the locus of points equidistant from a given point called the focus and from a given line called the directrix.

The line AB is a distance a from the y-axis, it is called the directrix. Line AF is called axis of symmetry.

Since BP = FP

BP ² = FP ²

(x+ a)² = (x – a)² + (y – 0)²

x² +2ax + a² = x² – 2ax + a² + y²

2ax = -2ax + y²

2ax + 2ax = y²

4ax = y²

Hence the equation of a parabola is y² = 4ax.

The line segment through the focus and perpendicular to the axis of symmetry and with end points R and Q on the parabola is called latus rectum. The point V is called the vertex of the parabola.

Other cases of parabola.

1. 2.

y²= -4ax x² = 4by

x² = – 4by

If the vertex of a the parabola y² = 4ax is translated to the point (x₁, y₁) the equation of the corresponding parabola becomes (y-y₁)2 = 4a (x – x₁)

The above equation is said to be in the standard or canonical form.

Example:

Find the focus and directrix of the parabola y² = 16x

Solution:

Compare y² = 16x with y² = 4ax

Þ 4a = 16

a =¹⁶/₄ = 4.

Hence the focus is (4, 0) and the directrix is -4

2. Find the equation of the parabola whose vertex is the origin and whose focus is the point

F(5,0)

Solution:

1 a = 5, y2= 4ax

y² = 4 x 5 x x

y² = 20x

2 If the canonical form of a parabola is given by y² – 4y – 12x + 40 = 0. Find

(i) The vertex (ii) the focus (iii) the directrix of the parabola.

Solution.

y² – 4y – 12x + 40 = 0

y² -4y + 4 – 12x + 36 =0

y² – 4y + 4 = 12x – 36

(y – 2)² = 12 (x – 3)

Vertex is (3, 2)

4a = 12 Þ a = ¹²/₄ = 3

Focus (3+3, 2) = (6, 2)

Directrix is x = – 3+3 = 0

Evaluation:

1 Find the foci and directrices of (i) y² = 32x (ii) x² = – 50 y

2. Write in its canonical form y² – 6y -2 x + 19 = 0 and hence determined the focus and the directrix

TANGENT AND NORMAL TO PARABOLA

TANGENT TO A PARABOLA

Equation of a tangent to y² = 4 ax at the point (x₁, y₁)

Recall y² = 4ax

Differentiating implicitly 2y dy/dx = 4a

dy/dx = 4a/2y

dy =2a

dx y

At point (x₁, y₁), dy = 2a

dx y1

y- y₁ = m (x -x₁)

y – y₁ = ^2a/y₁ (x – x₁ )

y₁ (y-y₁) = 2a (x-x₁)

yy₁ – y₁² = 2ay1x – 2ax₁

Since (x₁, y₁) is on y² = 4 ax it implies that y₁² = 4ax₁

Thus yy₁ – 4ax₁ =2ax – 2ax₁

yy₁ = 2ax – 2ax₁ + 4ax₁

yy₁ = 2ax + 2ax₁

yy₁ = 2a (x + x₁)

-y₁ 2a

Equation of the Normal to y² = 4ax at the point (x₁, y₁).

Recall, gradient of the tangent = 2a/y₁ gradient of normal will be =

Þ (y – y₁)_/(x – x₁) = -y₁/2a

2a (y- y₁) = – y₁ (x – x₁)

2ay – 2ay₁ = – xy₁ + x₁ y₁

2ay + xy₁ = 2ay₁ + x₁y₁ is the equation of the normal to the parabola y²= 4ax at the point

(x₁, y₁) on the parabola.

Example

Find the equation of the tangent and normal to the parabola y² = 12x at the point (3,6 )

Solution:

y² = 12x compare with y² = 4ax

4a = 12

a = 12/4 = 3

Equation of the tangent is given by:

y₁y – y₁²= 2ax – 2ax_1.

At (3, 6) i.e x₁ = and y₁ = 6, we have

6y- 6² = 2 x 3 x X – 2 x3x3

6 y – 36 = 6x – 18

6y – 6x + 18 – 36 = 0

6y – 6x – 18 = 0

y –x -3 = 0 the required equation

For normal,

2ay + xy₁ = 2ay₁ + x₁ y₁

(2) (3) y + x (6) = 2 x 3 x 6 + 3 x 6

= 36 + 18 = 54

6y + 6x – 54 = 0

y + x – 9 = 0

EVALUATION

Find the equation of (i) tangent (ii) normal to y² = 16x at (1, -4)

READING ASSIGNMENT:

Read Parabola equation, Further Mathematics Project II, page 210 -213

WEEKEND ASSIGNMENT:

1) Find the equation of a tangent at point (2, 2) to the circle x²+ y² = 8

(a) x – y -4 = 0 (b) x + y – 4 = 0 (c) x + y + 4 = 0 (d) x + y² – 4 = 0

(e) 2x + y² – 8 = 0

2. Find the coordinates of the point at which the gradient of the curve

y = x² – x + 4 is 3. (a) (1, 4) (b) (2, 6) (c) (1, 6) (d) (3, 5)

3, Find the equation of the normal to the line x + 2 y -5 = 0 at (1, 2)

(a) x – 2y + 3 = 0 (b) 2x – y = 0 (c)2x – y – 6 =0 (d) x – y = 0

4 The locus of a point equidistant from a given point and a given line is called a

(a) Circle (b) directrix (c) hyperbola (d) ellipse (e) parabola.

5. Find the equation of the normal to the curve y = x³ – 3x – 5 at the point (2, -3)

(a) x + 9y + 25 = 0 (b) x + 9y – 27 = 0 (c) x- 9y – 25 = 0 (d) x – 9y + 2 7 = 0

THEORY

A is the point (3, -1 ) and B is the point (5,3)
1. Show that the locus of a point P(x, y) which moves so that (PA)² + (PB)² = 28 is a circle.
1. Find the centre of the circle.
Find the equation of the normal to the curve y = x² -4x – 12, at the point where the curve cuts the y axis.

Definition, Ellipse equation, example and evaluation

Definition:

An ellipse is the locus of point P, moving in a plane such that the sum of its distances from two fixed points F₁ and F₂ called the foci is a constant.

The points V₁, V₂ V₃ and V₄are called the vertices of the ellipse. The line segment V₁ V₂ is called the major axis while the V₃ V₄ is called the minor axis. The point 0 is called the centre of the ellipse.

From above.

V₁V₂ = 2a and V₃V4 = 2b

V₁F₁ = V₂ F₂

F₂ P + PF₁= constant

Where P is on V₁

F₂P + PF₁ = F₂ V₁ + V₁F₁

= F₂V₁ + V₂ F₂

=V₁ V₂

= 2a

Thus F₂ P + PF₁ = 2a

x² + y² a² b²

b² = a² – c²

x² + y² a² b²

Equation (i) becomes = 1The above equation = 1 is the equation

x² + y² b² a²

of the ellipse when the major axis is in x – axis. If the major axis is on the y – axis, the equation becomes = 1

(x – x₁) + (y – y₁) a² b²

The canonical form of the equation of the ellipse is = 1 at (x₁, y₁)

Examples:

Find the four vertices and the foci of the ellipse x²/9 + y²/25 = 1

Solution:

Compare x²/9+ y²/25 = 1 with x² /b² + y² /a² = 1

Þ b = 9 = 3 and a = 25 = 5

v₁= (0,5 ) , v₂ = (0, – 5 ) v₃ = (3, 0 ) v₄ (-3, 0)

c² = a² – b²

= 25 – 9 = 16

\ c = 16 = 4

The foci are (0, 4 ) and (0, – 4)

2. Write the equation of the ellipse 25x² + 4y² – 50 – 16 y – 59 = 0 in its canonical form and hence determine

the coordinates of the center of the ellipse
- The two foci of the ellipse.

Solution

25x² + 4y² – 50x – 16y – 59 = 0

25x² – 50x + 4y² – 16y – 59 = 0

25(x² – x + 4 (y² – 4y) = 59

25( x² – x + 1 – 1) + 4 (y² – 4y + 4 – 4 ) -= 59

25( x² – x + 1 ) + 4 ( y² – 4y + 4) – 25 – 16 = 59

25( x – 1) ²+ 4 (y – 2 )² = 59 + 25 + 16 =100

25(x – 1 )²+ 4(y-2)² = 59 + 25 16 = 100

100 100 100

(x – 1)² + (y – 2) 4 25

= 1

Hence the coordinate of the centre is (1, 2)

x – x₁ + y – y₁ b² a²

(x – 1)² + (y – 2)² 4 25

Compare = 1 with

Þ b²= 4 and b = 2

a² = 25 and a = 5

Hence the v₁ = [0 + x₁, a + y] = 1, 7

v₂ = [ 0 + x₁, a – y] = 1, -3

v₃ = [b + x₁, 0 + y₁]

v₄ = -b + x₁, 0 + y₁] = -1, 2

c² = a² – b²

c² = 25 – 4 = 21

\c = 21

The foci are F₁ (0 + x₁, c + y₁) = 1 , 21 + 2

F₂ (0 + x₁, -c + y₁) = 1 – 21 + 2

EVALUATION

1. Find the foci and vertices of 9x² + 10y² = 90

2. Write the ellipse 4x² + 5y² – 24x – 20y + 36 = 0 in its canonical form and hence determine its foci and vertices..

C. Equation of tangent and normal to an ellipse at (x₁, y₁)

From x²/ a²+ y²/b² = 1

Differentiating explicitly:

2x + 2ydy a² b² dx

= 0

At the point (x₁, y₁), dy/dx = – b²x₁/a²y₁

y – y₁ = -b² x₁ x – x₁ a²y₁

The equation of the tangent becomes

At the point (x₁, y₁ ) is on the ellipse

xx₁ + yy₁ a² b²

The equation of the normal to the Ellipse

From above, gradient of tangent = – b² x₁/a²y₁

Gradient of normal = -1/( – b²x₁/a²y₁)= a²y₁/bx₁

The equation of normal becomes (y- y₁₎/(x – x₁₎ = a²y₁/b²x₁

b²x₁ (y –y₁) = a²y₁ (x- x₁)

b²x₁y – b²x₁y₁ = a²xy₁ –a² x₁ y₁

a²xy₁ –b²x₁y = a²x₁ y₁ –b² x₁ y₁

a²xy₁ –b²x₁y = (a² –b²) x₁y₁

Example

Find the equation of tangent and normal to the ellipse

4x²+ 25y²= 100 at (-3, 8/5)

Solution

4x² + 25y²= 100

4x²/100 + 25y²/100 = 100/100

x²/25 + y²/4 = compare with x²/a² + y²/b² = 1

a² = 25 and a = 5 b²= 4 and b = 4

xx₁ + yy₁ a² b²

Equation of tangent = 1

25 4

-3x + 8y = 25 20

= 1 11

-12x + 40y = 100

-3x + 10y = 25

10y – 3x = 25

Equation of normal is

a²xy₁ – b²x₁y = (a² – b²)x₁ y₁

25x (8/5) – 4(-3)y = (25 – 4) (-3) (8/5)

-504 5

40x + 12y =

200x + 60y = -504

200x + 60y + 504 = 0

EVALUATION

Find the equation of tangent and normal to the ellipse x² + 2y² = 2 at (-√2, 0)

D. THE HYPERBOLA

Definition, Hyperbola equation, Examples and evaluation.

Definition:

A hyperbola is the locus of a point P, moving in a plane such that its distances from two fixed points called foci have a constant differen

The locus condition is that

PF₁– PF₂ = constant

When P is at V₂

PF₁ = c + a

PF₂ = c- a

\PF₁ – PF₂ = (c + a) – (c – a)

PF₁ – PF₂ = 2a

Hence the equation of hyperbola is : x² – y² = 1

a² b²

Example:

Find the vertices and foci of the hyperbola 25x² – 4y² = 100.

Solution:

25x²/100 – 4y²/ 100 = 100 / 100

x²/4 – y²/25 = 1 compare with x²/a² – y²/ b² = 1

a² = 4 Þ a = + 2

b²= 25 Þ b = + 5

Hence the V₁= 2, 0 and v₂ = -2, 0

b² = c² – a²

c² = b² + a² = 25 + 4 = 29

c = 29 F₁ = 29, 0 F₂ = – 29 , 0

EVALUATION:

Find the vertices and foci of the hyperbola 9×2 – 4y² = 36

Equation of tangent at the point (x₁, y₁) to x²/a² – y²/ b² =1

If we differentiate x²/a² – y²/b² = 1 we have:

Equation of the normal at the point (x₁, y1) to the hyperbola

x²– y² = 1

a² b²

2x – 2ydy = 0

a² b^{2 dx}

2x= 2ydy

a²b² dx

dy= 2b²x =b²x

dx 2a²y a²y

at x₁,y₁, dy = b²x₁

dx a²y

let m be the gradient of the normal at the point (x₁, y₁), then

m = -a²y₁, thus the equation of the normal is y – y =-a²y₁

b²x₁ x – x₁b²x₁

b²x₁(y – y₁) = -a²y₁(x –x₁)

b²x₁y – b² x₁y₁= -a²xy₁ + a² x₁ y₁

b²x₁y + a²xy₁ = a²x₁y₁ + b²x₁y₁

a²xy₁ + b²x₁y = (a² + b² ) x₁ y₁

Examples

Find the equations of tangent and the normal to the hyperbola

x² – 2y² = 6 at (2 3, 3 )

SOLUTION:

x² – 2y²= 6

6 6 6

x² – y² = 1 compare with x²– y²= 1

6 3 a²b²

a² = 6, a è 6 b² = 3 è b = 3

Equation of tangent is x₁x – y₁y = 1

a² b²

2√3x – √3y = 1

6 3

2 √3x – 2 √ 3 y = 6

x – y = 6 = 3

2√3 √3

x-y = √3

y – x +√3 = 0

The equation of normal is

a²xy₁ + b²x₁y = (a² + b²) x1 y₁

6x (√3) + 3 (2 √3) y = (6 + 3) (√2) (√3)

6√3x + 6√3 y = 54

√3x +√3y = 9

x + y = 9 = 3√3

√3

x + y – 3√3 = 0

EVALUATION:

1. Find the equation of tangent and normal to the hyperbola 4x² – 9y² = 36 at

(3 √2, 2)

WEEKEND ASSIGNMENT

1. 16x² – 25y2= 400 is an equation of a / an

(a) Circle (b) curve (c) ellipse (d) hyperbola (e) parabola

2. Find the equation of the circle with centre (3, 7) and circumference 8π units.

(a) x²+ y² – 6x – 14y + 74 = 0 ( b) x² + y² – 6x -14y + 42 = 0

3. Which of the following shows the equation of an ellipse

(a) x² + y² = r² ( b) x² + y² =1 (c) x² +y² +2gx +2fy + c = 0 (d) y² = 4ax

a² b²

4. y² = 16x . This is the equation of a\an

(a) ellipse ( b) parabola (c) hyperbola (d) circle

5. If y² = 12x . Find the value of a

(a) 12 (b) 4 (c) 3 (d) 0

WEEK 5

REGRESSION LINE AND CORRELATION COEFFICIENT

SCATTER DIAGRAM

Definition: a scatter diagram is a graphic display of bivariate data. A bivariate data involves two variables

TYPES OF SCATTER DIAGRAM:

Linear positive correlation.

A positive correlation between two variables x any y means that in general, increase in x is accompanied by increase in y. The regression line has a positive slope.

Linear negative correlation

A negative correlation between x and y means that an increase in x is accompanied by a decrease in y, negative correlation has a negative slope.

3. Zero Correlation:

There is no apparent association between x and y.

Non Linear Correlation:

Most of the points lie on or near a curve which is parabolic in shape. The parabolic curve is called a regression curve.

REGRESSION LINE OR LINE OF BEST FIT OR THE LEAST SQUARES LINE

There are two variables where one is dependent and the other is independent variable. The regression line can be fit using scatter diagram method and the least squares method.

LEAST SQUARES METHOD: If x is independent variable and y dependent variable, that is y on x. then :The equation of the regression line is written as y = ax + b

Where a is the slope and b is the y – intercept. Given two sets of variables x and y it can be deduced that

a = n ∑ xy – ∑ x ∑ y

∑ x² – ( ∑ x)²

b = y a – ax

Where x = ∑ x

y = ∑ y

Example: use the least square method to fit a regression line of y on x for the following data

X	3	5	6	9	11	14	15	18
Y	2	3	5	7	10	12	13	17

Find value of y when x = 8

SOLUTION:

X	y	Xy	x²
3	2	6	9
5	3	15	25
6	5	30	36
9	7	63	81
11	10	110	121
14	12	168	196
15	13	195	225
18	17	306	324
∑ x = 81	∑ y = 69	∑ xy = 893	∑ x²= 1017

a = n ∑ xy – ∑x ∑ y = 8 (893) – 81x 69

n∑(x²) – ( ∑x)² 8 (1017) – (81)²

a = 7144 – 5589 = 1555

8136 – 6561 1575

a = 0. 9873

x = ∑ x = 81 = 10.125

n 8

y = ∑ y = 69 = 8. 625

n 8

b = y – ax

b = 8.625 — 0.9873 (10.125)

= 8.625 – 9.996

b = -1.37

y = ax + b

y = 0.9873x – 1.37 (regression line of y on x )

When x = 8

y = 0. 9873 (8) – 1.37

y = 6. 5284 ~ 6. 5

EVALUATION

Use the least square method to fit a regression line of y on x for the following data

X	1	4	5	7	8	10	12	16	19	20
Y	2	3	4	5	7	8	10	15	20	18

Use the line obtained to find the value of y when x = 9

CORRELATION COEFFICIENT

DEFINITION:

The correlation coefficient determines the amount or degree of linear relationship between two variables. The correlation coefficient is represented by r

The characteristics of r are as follows:

The value of r is the same irrespective of the variable labelled x or y.
the value of r satisfies the inequality -1< x < + 1
if r is close to +1, the variables are highly positively correlated. If r is close to -1 then, x and y are highly negatively correlated. If r is close to zero, the correlation between x and y is very low. There is no correlation between x and y when r = 0

There are two methods of obtaining the correlation coefficient.

Pearson’s coefficient of correlation or product moment correlation coefficient
Rank correlation coefficient .

RANK CORRELATION COEFFICIENT: It is also known as Spearman’s rank correlation coefficient and defined as :

r_k = 1 – 6 ∑ D²

n(n² -1)

As the name implies, the variables (if not ranked) can be ranked in ascending order or descending order. Where there are ties, the average is used as the rank.

Where D is the difference between the pairs of variables and n is the number of variables. D = Rx – Ry

Example:

The table below gives the examination marks of 10 students in mathematics and history.

Maths	51	25	33	55	65	38	35	53	61	44
History	20	65	25	36	51	50	77	31	60	5

A Calculate the rank correlation coefficient

b) Comment briefly on your result

SOLUTION:

MATHS (x)	HISTORY (y)	Rx	Ry	D	D²
51	20	5	9	-4	16
25	65	10	2	8	64
33	25	9	8	1	1
55	36	3	6	-3	9
65	51	1	4	-3	9
38	50	7	5	2	4
35	77	8	1	7	49
53	31	4	7	-3	9
61	60	2	3	-1	1
44	5	6	10	-4	16

∑D² = 178

r_k = 1- 6 ∑D²

n(n² – 1)

=1 – 6 x 178

10 (10² – 1)

1 – 1068/990

= 1-1.178= -0.078

There is a very low negative correlation between the marks obtained in mathematics and history.

EVALUATION:

The table below shows the marks obtained by ten students in both theory (x) and practical (y) examination.

X	50	70	85	35	60	65	75	40	45	80
Y	45	55	75	40	50	60	70	35	30	65

Calculate the rank correlation coefficient between x and y comment on your result.

PEARSON’S CORRELATION COEFFICIENT: It is fully called Pearson’s product moment correlation coefficient. It is simple to calculate and it does not recognise any of the variables as independent or dependent. It is obtained using the formula below.

r = n ∑ xy – ∑x∑ y

√ [n∑(x² ) – (∑x)² ][n∑(y²) – (∑y)²

Example:

Calculate the product moment correlation coefficient for the following data

X	2	4	7	9	11
Y	1	2	3	7	9

Comment on your result.

SOLUTION:

X	Y	XY	X²	Y²
2	1	2	4	1
4	2	8	16	4
7	3	21	49	9
9	7	63	81	49
11	9	99	121	81
∑x = 33	∑y = 22	∑xy = 193	∑x² = 271	∑y² = 144

r = 5 x 193 – 33 x 22

√[5(271) – ( 33)²][5(144) – ( 22)²]

r = 965 – 726

√266 x 236

r= 239

250.55

r = 0.9539. r= 0.95 (approximately to 2 s.f)

Comment: The relationship between x and y is highly positive.

EVALUATION: The following data are the marks obtained by five students in statistics (X) and mathematics(Y). Calculate the product moment correlation coefficient and comment on your result.

X	33	36	42	52	40
Y	42	46	38	62	52

WEEKEND ASSIGNMENT

Use the table below to answer questions 1 and 2.

Height	160	161	162	163	164	165
No of students	4	6	3	7	8	2

The mean of the distribution is

(a) 4875.1 cm ( b) 4001.2 (c) 3571.0cm (d) 162.2 cm (e) 129.2cm

2. The median of the distribution is

(a) 160 (b) 162 (c) 163 (d) 164 (e) 165

3. Calculate the standard deviation of 3,4, 5,6,7,8,9

(a) 2 (b) 2.4 (c) 3.6 (d) 4.0 (e) 4.2

4. Calculate the mean deviation of 6 , 8 , 4 , 0 , 4

(a) 4 .0 (b) 3.6 (c) 3.0 (d) 2. 8 (e) 2 . 1

5. The table below shows the rank Rx and Ry of marks scored by 10 candidates in an oral and

written tests respectively. Calculate the spearman’s rank correlation coefficient of the data.

R_x	1	2	3	4	5	6	7	8	9	10
R_y	2	3	4	1	6	5	8	7	10	9

(a)51/55 b) 6/55 c)49/55 d)54/55 e) 61/55

THEORY

1 The distribution of marks scored in statistics and mathematics by ten students is given in the table below:

Maths(x	11	20	23	42	48	50	57	64	80	90
Stat(y)	26	23	35	46	44	50	50	58	68	70

Plot a scatter diagram for the distribution
Draw an eye- fitted line of best fit
Use your line to estimate the students marks in statistics if his mark in maths is 40

2. The table below gives the marks obtained by members of a class in maths and physics examination

STUDENTS	A	B	C	D	E	F	G	H	I	J
Maths	85	75	59	43	74	69	62	80	54	63
Physic	92	72	62	48	85	73	46	74	58	50

Calculate the product moment correlation coefficient.
Comment on your result.

WEEK 7- MID-TERM TEST

WEEK 8

PROBABILITY DISTRIBUTION

-BINOMIAL PROBABILITY DISTRIBUTION

– POISSON PROBABILITY DISTRIBUTON,

Content.

Probability distribution deals with theoretical probability model based on the randomness of certain natural occurrences. The binomial and Poisson distribution are discrete distribution

BINOMIAL DISTRIBUTION.

This arises from a repeated random experiment which has two possible outcomes.

The two possible outcomes of the random experiment are usually called success and failure.

Prob( success) = P, Prob(failure) = q

Since the two events are complementary, hence p+ q = 1 or p = 1-q, q = 1 – p

The probability of success or failure of an event is the same for each trials and does not influence the probability of success or failure of another trial of the same event.

:. Binomial distribution of n trails and r required outcome(s) is defined as :

pr(x = r) = ⁿC_r Pr q^n-r

when ⁿC_r = n!

(n-r)! r!.

The binomial distribution is suitable when the number of trials is not too large.

Example: Find the probability that when two fair coins are tossed 5 times a head and a tail appear three times.

Solution:

Two fair coins = (HT,TH,TT, HH) = 4

Prob( a head and a tail) = ²/₄ = ½

i.e p = ½ , q = ½ (p + q = 1)

n = 5, r = 3.

:. P(x = r) = ⁿCr p^rq^n-r

p ( x = 3) = ⁵C₃ ( ½ )³ ( ½ ) ^5-3

p (x = 3) = 10 x ¹/₈ x ¼ = ¹⁰/₃₂ = ⁵/₁₆

p (x = 3) = 0.3125.

It is known that 2 out of every 5 cigarettes smokers is a village have cancer of the lungs. Find the probability that out of a random sample of 8 smokers from the village, 5 will have cancer of the lungs.

Solution.

Prob( a smoker has cancer) = ²/₅ i.e p = ²/₅

Prob( a smoker doesn’t have cancer) = 1 – ²/₅ = ³/₅

:. q = ³/5

n = 8, r = 5

Prob( x = -5) = ⁸C₅ (²/₅)⁵ (³/₅)³

= 56 x 32 x 27 = 48384

3125 125 390625

Prob (x = 5 ) = 0.124.

Evaluation.

Find the probability that when a fair six-faced die is tossed six times, a prime number appears exactly four times.

in an examination, 60% of the candidates passed. Use the binomial distribution to calculate the probabilities that a random sample of 10 candidates contain exactly 2 failures.

POISSON DISTRIBUTION: The Poisson distribution is more suitable when the number of trials is very large and probability of successes is small. It is defined as :

Pr(x) = / e^– , x = 0, 1,2,3,

Where ʌ = np, e = 2.718

P = probability of success, n = number of trials.

Example: If 8% of articles in a large consignment are defective, what is the chance that 30 articles selected at random will contain fewer than 3 defective articles?

Solution:

P = 8/100 = 0.08, n = 30

:. ʌ = np = 0.08 x 30 = 2.4.

Prob( fewer than 3 ) i.e prob( (0) + prob (1) + prob (2)

Prob ( x = 0 ) = 2.4^o x e^-2.4 = 1 x ^e-2.4

Prob( x = 1 ) = 2.4^o x e^-2.4 = 2.4 x e^-2.4

Prob(x = 2) = 2.4^o x e^-2.4 = 2.88 x e ^-2.4

Prob( x <3) = e ^-2.4 + 2.4 x e^-2.4 + 2.88 x e^-2.4

= e^-2.4 ( 1 + 2.4 + 2.88)

= e^-2.4x 6.28.

EVALUATION: The probability that a person gets a reaction from a new drug on the market is o.001. if 200 people are treated with this drug. Find approximately, the probability that:

exactly three persons will get a reaction
1. more than two person will get a reaction

Properties of Binomial and Poisson Distribution.

Binomial:

a. It assigns probability to non-occurrence of events i.e Prob( x = 0)

Mean µ = np

WEEK 9

PROBABILITY DISTRIBUTION

Standard deviation , r = √npq

Variance ð² = npq

Poisson

It assigns probability to non-occurrence of events i.e Prob(x = 0)

Mean µ = = np

Standard deviation, ð = √ = √np

Variance, ð² = = np

Example: In the probability of tossing a fair coin three times, a head shows up twice fine the mean and standard deviation .

Solution.

n = 3 Prob(a head) = ½ , i.e p= ½ , q = ½

I. Mean µ = np = 3 x ½ = 3/2

II. Standard deviation :r = √npq = 3 x ½ x ½ = ¾

Example 2. 0.2% of the cooks produced by a machine were found to be defective. If there are 1000 corks, find the mean and standard deviation?

Solution:

P = 0.2% = 0.002.

N = 1000

I. mean µ = ʌ = np = 1000 x 0.002 = 2

II. r = √np = √2

EVALUATION

1. In an examination, 60% of the candidates pass. If 10 candidates were sampled. Find the mean, standard deviation and variance of the candidates.

2. The probability that a person gets a reaction from a new drug in the market is 0.001. If 2000 people were treated with this drug, find the mean and standard deviation.

READING ASSIGNMENT: Read probability distribution, further maths. project 3 from page 198-201.

WEEKEND ASSIGNMENT.

1. What is the variance of a binomial distribution?

(a) np (b) √npq ( c ) npq (d)

2. The mean (µ) of a poisson distribution is the same as

(a) Standard deviation (b) variance ( c) poisson distribution (d )

3. If number of trials is 100 and probability of success is 0.0001, what is the variance of this distribution.?

(a) 0.00999 (b) 0.1 (c ) 0.01 (d)

4. If the birth of a male child and that of a female child are equiprobable. Find the probability that in a family of five children exactly 3 will be male.

(a) 16/5 (b) 5/16 ( c) 5/32

5. If an unbiased die is thrown repeatedly, what are the chances that the first, six to be thrown will be the third throw?

(a) 25/216 (b) 1/6 (c) 25/36

Theory.

1. 20% of the total production of transistors produced by a machine are below standard. If a random sample of 6 transistors produced by the machine is taken, what is the probability of getting (1) exactly 2 II exactly 1 II at least, 2 III at most 2 standard transistors?

2., A fair die is thrown five times. Calculate correct to 3 decimal places, the probability of obtaining

(a) at most two sixes

( b) exactly three sixes

Week 8-9

PROBABILITY DISTRIBUTION (CONTINUATION)

-Normal distribution

– properties and area

– z – scores application.

Normal Distribution: This is a continuous distribution and it takes the form

P(x) = 1

ð√2x e^{-½ (x – u)2}

^ð

where o is the standard deviation µ. U is the mean and e = 2.718.

The graphical representation of a normal distribution is a bell-shaped curve.

P(x)

PROPERTIES OF THE NORMAL DISTRIBUTION

It depends on the mean (u) and standard deviation
The shape is bell-shaped
The function is continuous, hence the range is from –ά to + ά
The curve is symmetrical about the vertical line through the mean.

A normal distribution function is a probability function, hence the total area under the curve is .

The normal distribution has a complicated equation, but it can be shown in shaded area under the shape

Values within 1 sd of the mean

µ-ð × µ+ð

pr( µ-r < x <µ + r) = 0.68.

2. Values within 2sd of the mean

Pr(u-2r<x < + 2R) = 0.955

µ– 2r µ µ+2r

3. Values within 3sd of the mean.

u– 2r µ µ+2r

Pr(u-3r<x < u + 3R) = 0.997

Example: A random variable X is normally distributed with mean 65 and standard deviation 5, find:

Pr(60 <x < 70 )
Pr(55 <x < 75 )
Pr(50 <x < 80 )

Solution: µ = 65, r = 5

1. Pr (60 < x < 70) = Pr(µ – r < x <µ + r)

= Pr(65-5 < x <65+5)

= Pr(60 < x < 70)

II. Pr (55 < x < 75) = Pr(µ -2r < x <µ + 2r)

= 0.95

III. Pr (50 < x < 80) = Pr(µ -35 < x <µ +3 r)

= 0.997

Evaluation:

1. A random variable x is normally distributed with mean 45 and standard deviation 12.

Find I Pr(9 < x <81)

II. Pr(36 < x < 57)

AREA UNDER NORMAL CURVE

The area under a normal curve can be defined by checking the probability value in the Normal distribution probabilities table.

Example:

Find the area between Z = 0 and Z = 2.13

Solution:

Pr( 0 < z < 2.13) check the value against 2.13.

Pr( 0 < z < 2.13) = 0.4834

2.13

2. Find the area z = -1.3 and z = 1.2

Solution:

Pr(-1.3 < z <1.2)

= Pr(0 < z <1.2) + Pr(0 < z <1.3)

= 0. 3849 + 0.4032.

= 0.7881.

3. Find the area between Z = 0.36 and Z = 1.89

Solution;

Pr(0.36 < z <1.89)

= Pr(0 < z <1.2) – Pr(0 < z <0.36)

= 0.4706 – 0.1406

= 0.33.

Evaluation: Using the standard deviation normal distribution table. Find the area under:

1.Pr(-1.5 < z < 2.0 )

2. Pr(z <1.5)

3. Pr(z >2.6)

Z Scores

The area under a normal distribution curve between two values depends on the number of standard deviations from the mean. Therefore, the standardize normal curve is obtained from the normal curve by the substitution.

Z = X – µ

:. Z is called the standardized score or Z score at mean zero (o) and standard deviation 1.

Example: A random variable whose distribution is normal has mean 25 and standard deviation 5. Find

1. Pr(22 < z < 27)

II. Pr(x <20)

III. Pr(x > 26.5)

Solution:

µ = 25, б = 5

I. Pr(22 < x < 27) Pr ( x₁-µ<z <x₂ – µ )

ð ð

= Pr(z₁< z < z₂)

= z₁= 22 – 25 = -3 = -.0.6

= z₂ = 27-25 = ²/₅ = 0.4

Pr(-1.3 < z <1.2) = Pr(-0.6 < z < 0.4)

= Pr(z < 0.4) + Pr( z < 0.6)

= 0.1554 + 0.2258

= 0.3812

II. Pr(x <20) = Pr( z < 20-25)

= Pr( z < -1) = Pr(-1 < z < 0)

= Pr( z < 0) – Pr( z <1)

= 0.5 – 0.3413.

:. Pr(x <20) = 0.1587

III. Pr(x > 26.5 ) = = Pr( z < 20-25)

= Pr(z > 0.3)

= Pr(0.3 < z < 0 )

= Pr( z < 0 ) – Pr(z < 0.3 )

= 0.5 – 0.1179

= 0.3821.

Evaluation: The weight of packets of sugar produced by a machine have a mean of 1kg and a standard deviation of 0.1kg. What is the probability that in a random sample of 50 packets the combined weight will exceed 52kg.

Reading Assignment: Read Z scores and Normal distribution . Further mathematics project III from pag2 202-210.

Weekend Assignment

Find the area between z = 0.36 and z= 1.89

(a) 0.33 (b) 0.6112 (c) 1.00

2. Use the information below to answer questions 2-4.

A distribution with mean 85 and standard deviation 10 is normally distributed. If X is a random variable of the distribution, find

2. Pr(80 < x < 8.9) (a) 0.9332 (b) 0.5 (c ) 0.3469

3. Pr(x >83) (a) 0.1587 (b) 0.789 (c ) 0.4207

4. Pr(x > 87) (a) 0.0047 (b) 0.35 (c) 0.4207

5. Find, with the usual notations, P (z < 1.810) from the table of normal distribution.

(a) 0.311 (b) 0.0288 (c ) 0.9649

Theory

The scores of some 500 candidates in an examination were found to be approximately normally distributed with mean 40 and standard deviation 5. Find the number of candidates who scored at least 48.
The length of nails produced in a factory are approximately normally distributed with mean 2cm and a standard deviation 0.01cm. Find the proportion of nails that will be shorter than 1.98cm.

Week 10

^STATICS

Definition of Concepts
Resultant of two forces
Components resolution of forces.

Definition of Concept.

Statics is the study of bodies which remain at rest under the action of given forces.

Mass : This is the quantity of matter contain in a body. Mass of a particular body does not change and the standard unit is kilogram.

Force: Force is that action which tends to change the state of rest or uniform motion of a body in a straight line. It’s a vector quantity sine it has magnitude and direction. The unit of force is Newton.

:. F = Ma

Where M= Mass, a = acceleration.

Composition of Forces: Two or more concurrent forces can be combined to obtain a single force. Therefore, resultant force is the force produced or obtained when two or more concurrent forces are combined.

A force can be resolve by

I Graphical Method II. Analytical Method.

Analytical Method: The parallelogram law of composition of two forces is used to find the resultant force of two or more forces. Hence, parallelogram law states that if two forces acting at a point are represented in magnitude and direction by two adjacent sides of a parallelogram, then the resultants of the two forces, is represented in magnitude and direction by the diagonal of the parallelogram, drawn from the point of action of the two forces.

180-θ

R is the resultant force and can be obtained using cosine rule:

R2 = P2 + Q2 – 2PQ cos (180 –θ )

The angle of inclination is ά and can be obtained using sine rule or tan.

:. Tan ά = CD

Tan ά= P Sin θ

Q + P Cos θ

Example 1: the angle between two forces of magnitude 8B and 5N IS 120⁰. Find in N, the magnitude of their resultant .

Solution

P 5N R

8N Q

Let R to the resultant vectorial force .

R2 = P2 + Q2 – 2PQ Cos R.

= 52 + 82 – 2 (5 x 8 )Cos 60o

= 25 + 64 – 80 x ½

= 89 – 40

R2 = 49

:. R = √49

R = 7N

2. Calculate, correct to one decimal place, the angle between two forces 20N and 30N if their resultant is 40N.

Let R be the resultant vectorial force

R2 = P2 + Q2 – 2PQ cos R

= 52 + 82 – 2 (5 x 8) Cos 60^o

= 25 + 64 – 80 x ½

= 89 – 40

R2 = 49 :. R = √49, R = 7N.

2. Calculate, correct to one decimal place, the angle between two forces 20N and 30N if their resultant is 40N.

R2 = P2 + Q2 – 2PQ Cos (180 – θ)

402 = 202 + 302 – 2 (20 x 30 ) Cos (180 – θ)

1600 = 400 + 900 – 1200 cos (180 – θ)

1600-1300 = 1200Cos (180 –θ)

300 = -1200 Cos (180 – θ)

= Cos ( 180 –θ )

-1200

-1/4 = Cos (180 – R )

– 180 –θ = Cos ^-1-1/₄ )

180 – θ = 104. 5

180 – 104.5 = θ

θ = 75.5

:. The angle between them is 75.5^o

Evaluation.

A vertical force of 6N and a horizontal force of 8B act on a body . Find the magnitude and the inclination of the resultant force to the horizontal .
The angle between two forces of magnitude 8N and 11N is 35^o. Find the magnitude and inclination to the 11N force of resultant force.

RESOLUTION OF FORCES

A given force can be resolved into two parts and each part is called the resolute. A given force can be resolve in two directions which are perpendicular to each other.

The component along the y axis is called the vertical component and the component alng the x- axis is the horizontal component.

py θ

Let Px the horizontal component

Let py the vertical component.

If p is inclined to the upward vertical

0 px

Px Sin θ = px

Px = P sin θ.

Cos θ = py

Py = p cos θ

Force Horizontal Component Vertical Component

P1 P1 Sin θ P1 Cos θ

P2 P2 cos θ -P2 Sin θ

P3 -P3 cos θ – P3 Sin θ

P4 – P4 cos θ P4 sin θ

Resulant of several concurrent forces :the resultant is obtained as :

R = √(EPx)² + (EPy)²

Where Epx = Sum of horizontal components

EPy = Sum of vertical components

Tan θ = Py

Θ = tan ^-1Py

Px Angle of inclination to the resultant.

Example 1:

The horizontal component of a force p which makes an angle of 50o with the horizontal is 30N. Find the force P

30 = P x 0.6 + 28

P = 30/0.6428

P = 46.67N

Three forces (2N, 060^o), ( 4.5n, 180 ^o) and (5n, 300^o) act on a body of mass 2kg which is initially at rest. Find I the resultant force on the body II the acceleration with which the body begins to move.

Solution:

Let P1 = (2N.060o) . P2 = (4.5N, 180 o) P3 = (5N,300o )

Horizontal Component Vertical Components

P1 = 2 Sin 60o = 1.7321 P1 = 2 Cos 60o = 1

P2 = 4.5 Cos 90 o = O P2 = -4.5 Sin 90o = -4.5

P3 = 5Cos 30o = -4.33 P3 = 5 Sin 30o = 2.5

Epx = 1.732 + 0 – 4.33 Epy = 1 – 4.5 + 2.5

= – 2.598 Epy = -1

(EPx)² = 6.7496 (Epy)² = 1

:. R =√ ( EPx )²+ (EPy)²= √6.7496 + 1 = √7.7496

R = 2.78N

Acceleration of the body ; F = ma

M = 2kg F = Resultant force 2.78

F=ma

= 2 x a

a = ^2.78/₂ a = 1.39ms^-2

Evaluation.

1. The vertical component of a force F which makes an angle of 35o with the horizontal is 45N. Find the force F.

2. The forces of magnitude 35N and 45N act on a particle in the directions 180o and 315o respectively. Find the resultant of these forces giving:

a. the magnitude correct to the nearest whole number

b. the direction correct to the nearest degree.

Reading Assignment .

Read Compositon and Resolution of Coplanar forces on pages 154 to 165 of further mathematics project III.

Weekend Assignment .

1. Two forces each of magnitude PN are inclined to each other at an angle of 120o. find the magnitude of their resultant.

(a) P√3 N (b) P2N ( c) PN

2. Find the angle between the two forces 5N and 6N if their resultant is 8N.

(a) 60o (b) 120o (c) 180 o

3. A force P of magnitude 60N makes an angle of 40o with the horizontal . use the information to answer questions 3 and 4

3. Find the horizontal component of ^oP

(a) 20N (b) 45.96N (c ) 38.57N

4. Find the vertical component of P

(a) 45.96N (b) 38.57 (d) 20N.

5. Find the resultant of forces 8N and 10N inclined at an angle 120o to each other .

(a) 2√61N (b)61√2N C 39N

Theory

Forces F1 = ( 10N, 090o), F2 = (20N, 210o) and F3 = (4N, 330o) act on a body at rest on a smooth table. Find , correct to one decimal place the magnitude of the resultant force.
Find the magnitude and direction of the resultant of the forces shown in the diagram below:

STATIC – CONTINUATION.

Definition of equilibrium
Condition of equilibrium of rigid body
Application of the condition to solve problems
Lami’s theorem and application .

Definition.

Equilibrium is when abody remains at rest under the action of given forces.

Translational Equilibrium : The state of equilibrium of bodies which remain at rest under the action of forces have tendency to cause translation.

Condition of Equilibrium

When a block is placed on a tab;e as shown below and force F1 and F2 are applied to the block.

The block remains in translational equilibrium if the magnitude of F1 and F2 are equal .

Since F1 and F2 are acting in opposite direction, but have equal magnitudes.

Then, F1 = -F2

F1 + F2 = 0

Also, the upward force N balances the downward force mg on the block.

:. N = -mg , N + mg = 0

hence, the sum of the vertical components and horizontal components of forces acting on a body in translational equilibrium is equal to zero.

Example 1: A particle of mass 5kg is supported by two light inelastic strings inclined at angles 30o and 45o respectively to the horizontal. If the system is in equilibrium, calculate the tension in each

Solution:

Let the two inelastic strings to be DP and OQ

Resolve each string vertically and horizontally:

Horizontal : Efx = T1Cos 30o + T2 cos 45o + 0

Vertical : Efy = T1 Sin 30o + T2 sin 45o – 50

:. Efx = 0.7061T2 – 0.866T1 = 0 eqa I

Efy = 0.7071T2 + 0.5T1 = 50 eqa II

Solving equation I and II simultaneously :

1.366T1 = -50

T1 = -50/-1.366

T1 = 36.6N

Substitute T1 in equation I or II using equation 1, 0.7071T1 – 0.866T1 = 0.

0.7071T2 = 0.866 x 36.6

T2 = 31.6956

0.7071

T2 = 44.82N

:. The tensions in the strings are T1 = 36.6N , T2 = 44.82N

2. A particle of mass 5kg is suspended by a light in extensive string which makes an angle of 30o with the downward vertical and horizontal forced F. If the system is in equilibrium ,calculate

The tension in the string
The magnitude of the force F (take g = 10ms^-2)

Solution.

I using Sin θ = AB

Sin 60o = 5 x 10

T = 50/sin 60o, T = 57.74N.

The tension in the string is 57.74 N

II. magnitude of F1 Tan 30o = OB

50 x Tan 30o = F :. F = 28.87N

Evaluation;

!. A street lamp of mass 10kg is suspended at a position ) by two wires OP and OQ across a rood such that each wire is inclined at an angle of 80o to the upward vertical, If the system is in equilibrium, calculate the tension in one of the two wires. (take g = 10ms^-2)

2. A particle of mass 98kg is suspended by two light inelastic strings of length 9m and 12m from two fixed point P and which are 15m apart. Calculate I the angles made by the strings with the upward vertical II the tension in the strings.

TRIANGLE OF FORCES

If three coplanar forces act on a body in such a way that the system is in equilibrium ,then the forces can be represented in magnitude and direction by the sides of a triangle taken in order

The triangle representing the three coplanar force is called a triangle of forces.

Example 1:

A body of mass 6.5kg is supported by two strings, One of the stings is inclined at an angle of 30o and the other 40o to the horizontal . Find the tension in each strings, if the system is in equilibrium (take g = 10ms^-2)

Solution

Using Sine rule;

T1 = 65

Sin 50o Sin 70o

T1 = 65 x sin 50o

Sin 50o Sin 70

T1 = 65 x 0.766

0.9397

T1 = 52.98N

Similarly:

T2 = W

Sin 60o Sin 70o T2, = 65 x sin 60o = 65 x 0.8660

Sin 70o 0.9397

T2 = 59.9N

Evaluation.

A body of mass s10kg is suspended by means of two light inextensible strings. AP and BP which are inclined at angles 60o and 30o respectively to the downward vertical . if T1 and T2 are the magnitude of the tension AP and BP respectively, Calculate the values of T1 and T2.

LAMI’S THEOREM: This theorem states that if three forces acting at a point are in equilibrium, then each force is proportional to the sine of the angle between the liens of action of the other two forces.

Consider the forces F1, F2 and F3 below

By Lami’s theorem: F1x Sin B

F2 x sin B

F3 x sin θ

Since the forces are proportional to the sine of the angle: then F1 = F2 = F3

Sin B sin8 sin θ

Example: A body of weight 91N is suspended by two inelastic string 5m and 12 m long attached to two points a the same horizontal level, whose distance apart is 13m. using Lami’s theorem or otherwise, find the tension along the strings .

Solution:

Le the tension in the strings be T1 nd T2 and x,B be the angles made with the horizontal.

= OA2 + OB2 = 52 +122

= 25 + 144 =169

but AB = 13

AB2 = 13 2 = 169

:. OA2 + OB2 + AB2

hence AOB is a right angle triangle

:. Sin B = ⁵/₁₃, sin x = ¹²/₁₃

B = sin -1 0.3845 = sin-1 12/13

B = 22.6 = 0.9231 = 67.4

Using Lami’s theorem:

T1 = W

Sin 157.4 sin 90o

T1 = 91 x sin 157.4o = 91 x 0.3843

Sin 90o 1

T1 = 34.97N

T2 = W

Sin 112.6 sin 90o = T2 = 91 x sin 112.6

Sin 90o

T2 = 91 x 0.9232

T2 = 84.01N.

Evaluation: A particle of mass 10kg is connected by two strings of length 3m and 4m to two points on the same horizontal level and 5m apart, find the tension in the strings.

Reading Assignment

Read Equilibrium page 170-177 of Further Mathematics project III.

Weekend Assignment.

1. Two forces (8N, 030o) and (10N, 120o) act on a body, find the magnitude of the force that would be applied to keep the system in equilibrium.

(a) 16.6N (b) 12.8N (c)11.1N (d) 9.2N

Week 6

Review of first half lessons.

Week 2

In a community, 10% of the people tested positive to the HIV virus. If 6 persons from the community

are selected at random, one after the other with replacement, calculate correct to four decimal places,

the probability that (i) exactly 5 (ii) none (iii) at most 2, tested positive to the virus.

Week 3

The mean score of 200 students in an examination is 40 and the standard deviation is 8. if the scores

are assumed to be normally distributed, find the :

Proportion of students obtaining more than 46
Number of students scoring between 32 and 48.

Week 4

Coplanar forcews 4N, 8N 6N,4N and 5N act at a point as shown in the diagram . if the 6Nforce act in the direction 090o calculate the:

Magnitude of the resultant force
Direction of the resultant force

Week 5

A uniform plank PQ of length 8m and mass 10kg is supported horizontally at the end P and at point R. 3 metres from Q. A boy of mass 20kg walks along the plank starting from P. If the plank is in equilibrium, calculate the

a. reaction at P and R when he has walked 1.5 metres

b. distance he had walked when the two reactions are equal. (take g = 10ms^-2)

STATICS – CONTINUATION.

Definition of Moment of a force

Principles of Moments

Application of the principle in solving problems.

Definition:

Moments of a force: The moment of a force about a reference point is defined as the product of the force and the force arm. Moment of a force is a vector quantity and its units is Nm.

The direction of movement or sense of movement about the point is duly considered.

The object can move in clockwise moment and ant-clockwise moments about the given point.

Suppose we have an object acted upon by two forces, F1 and F2 in the opposite direction, then

M₁ = F1 x di

M₂= F2 x d₂

Where M₁ = Magnitude of F₁

M₂ = Magnitude of F₂

di = Forced arm of d₁

d₂ = Force arm of d₂

PRINCIPLE OF MOMENTS

1. When a system of coplamar forces are in equilibrium, tehn the sum of the clockwise moment is equal to the sum of the anti-clockwise moments about the same point in the plane.

2. When the system is not in equilibrium, then the resultant of two coplanar forces F1 and F2 denote by R is represented by the relationship below:

M1 + M2 + MR

Where M1 = Moment f F1

M2 = Moments of F2

MR = Moments of T.

Centre of Gravity. The centre of gravity of a uniform plank or rod is the midpoint of the plant or rod

Examples 1 : A uniform rod PQ is 15m long and has mass 20kg. The rod rests on two supports at P and Q. An object of mass 5kg is suspended at a point R on the rod 5m from the end P. Calculate the reaction of the supports P and Q (take g = 10ms^-2)

Solution

Kp kq

7.5m

5m Q

5kg

20kg

Let the reaction at P be Kp and at Q be K2

NB: The weight of the rod acts downward through the midpoint of the rod.

Moments about the point P, MR = M1 + M2

But M1 = F1 x di and F = Mg.

Kq x 15 = (5 x 10 ) x 5 + ( 20 x 10 ) x 7.5

15kq = 50 x 5 + 200 x 7.5

K1 = 250 + 1500

15.

Kq = 116. 7N.

Moment about the point Q,

Kp x 15 = ( 5 x 10 ) x 10 + (20 x 10 ) x 7.5.

15Kp = 500 + 1500

Kp = 2000

15.

Kp = 133.3N

2. A uniform beam AB of length 6m and mass 20kg rests on support P and Q placed 1m from each end of the beam. Masses of 10kg and 8kg are placed at A and B respectively. Calculate the reactions at P and Q (g = 9.8ms^-2)

Solution :

8kg

10kg

3 3

2 2

20kg.

Let reaction at point P be Rp and at point Q be Rq.

:. Moment about the point Q

Rp x 4 = ( 10 x 9.8) x 5 + (20 x 9.8 ) x 2 – ( 8 x 9.8 x 1 )

4Rp = 490 + 392 – 78.4

Rp = 803.6

4. Rp = 200.9N

Moment about the point P:

Rq x 4 = ( 8 x 9.8 ) x 5 + (20 x 9.8 ) x 2 – ( 10 x 9.8) x 1

4Rq = 392 + 392 – 98

Rq = 689

Rq = 171.5N

Evaluation

1. A uniform rod PQ, is 20 m long and weighs 80N, has weights 20 N and 50N suspended at P and Q respectively . Find the distance from P where the rod must be supported so that it will rest horizontally.

2. A uniform rod PQ of length 10 m and mass 2kg rest on two supports at x and y. If PX = 2m and QY = 1m, find the reaction of X. ( take g = 10ms^-2)

Reading Assignment:

Read Rotational Equilibrium and Principle of Moments. Page 178-185 of Further Mathematics Project

III.

Week 8

Topic: Friction.

Basic Concept
Coefficient of friction
Forces acting on a body.

Basic Concept: When two bodies are in contact, each one is exerting a force on the other. Therefore, friction can be defined as the force which tends to oppose the relative sliding motion of two surfaces in contact. Frictional force is the opposing force between two forces in contact.

The direction of friction is opposite to the direction in which the motion will occur. The frictional force for any two surfaces in contact has a value given by”

F = UR

Where R is the normal reaction between the bodies and obtained as ; R = m x g

U = Coefficient of friction and the value depends only on the nature of the surfaces in contact.

Example 1: A mass of 4kg rests on a rough horizontal table, with U = 0.4. find the least force sufficient to move the mass: (take g = 10ms^-2)

Solution

Sufficient force; F = UR.

R = m x g = 4 x 10 = 40

F = 0.4 x 40

F = 16N

2. If a force of 10N is just sufficient to move a mass of 2kg resting on a rough horizontal table, find the coefficient of friction ( g = 10ms^-2)

Solution.

F = 10 N , m = 2kg.

F = UR

10 = U x 2 x 10

10 = 20U

u = 10/20

U = 0.5

Evaluation :

A body of mass 8kg rests on a horizontal surfacr. If the coefficient of friction between the body and the horizontal surfacr is 0.65, calculate the minimum horizontal force enough to just move the body.

FORCES ACTING ON A BODY PLACED ON A ROUGH INCLINED PLANE

Rough Inclined Plane: There are two basic forces acting on an object placed on an inclined plane; the applied forces P and the frictional force F

P= Mg Sin Ө

F = Mg Sin Ө

R = Mg Cos Ө

Recal F = UR

:. F = UMgCos Ө where F is the limiting

Friction:

The least force required in P to make the body move up the inclined plane r is P = UR + Mg Sin Ө

:. P = F + MgSinӨ

P = UR + MgSinӨ

The least force required to make the body slide down the plane is thus

P + f = mgSin Ө

P = mg Sin Ө- Ur.

Where m = mass and Ө= angle of friction.

The value of the coefficient of fricton from F = UR and

MgSinӨ = UMgCos Ө is defined by

U = mg sin Ө

Mg cos Ө

:. U = tan Ө

Example :

1.A block is placed on an inclined plane at an angle of 30^o to the horizontal and just remains at rest. Find the coefficient of friction.

Solution:

Ө = 30^o, U = tan Ө

U = tan 30^o

U = 1/√ or 0.5774.

2. A body of mass 4.5kg rests on a smotth plane inclined at an angle of 47^o to the horizontal. Calculate the magnitude of the force P parallel to the plane just enough to prevent the body from sliding down the plane (g= 9.8ms^-2).

Solution.

P= mg Sin Ө

P = 4.5 x 9.8 x sin 47^o

= 44.1 x 0.7314.

P = 32.25N.

Evaluation.

A particle of mass 25kg slides down a rough plane inclined at angle 30^o to the horizontal. If the coefficient of friction is 0.2, find, in ms^-2 the acceleration of the particle correct to 3 significant figures (take g = 10ms^-2)

Reading Assignment:

Read Friction page 191 -195

Weekend Assignment .

1. A body is in limiting equilibrium on a plane inclined at an angle to the horizontal . if Cos = 0.8, calculate the coefficient of friction.

2. A big stone is of mass 13kg. A boy whose weight is 59N sits on the stone . Find the minimum horizontal force P required to move the stone on the ground . If the coefficient of friction is 0.25 (g = 9.8ms^-2).

(a) 46.6N (b) 59N ( c) 186.4N

3. A mass of 8kg tests on a rough table with U = 0.6. find the least force which will make the mass move (g = 10ms^-2).

4. A mass of 10kg slides down a rough plane inclined at Ө to the horizontal where sin Ө = 0.6. if U = 0.3, find the acceleration of the box.

5. A body can just rest in equilibrium on a slope inclined at Ө to the horizontal where sin Ө = ⁵/₁₃ find U

1 hours read

Symmetric	Diagonal	Upper Triangular	Lower Triangular	Zero	Identity

Further Mathematics Elearning Note – Edudelight.com

GRAPHICAL SOLUTION OF INEQUALIIES IN TWO VARIABLES

The line ax + by + c =0 partitions the x-y plane into two regions

EVALUATION

. y = -2x – 1

Matrix Addition and Subtraction

DEFINITION: Two matrices A and B can be added or subtracted if and only if their dimensions are the same (i.e. both matrices have the same number of rows and columns. Take:

Example 1

Example 4 – Multiplying by the Identity Matrix

Further Examples

AT = (aji) where j is the column number and i is the row number of matrix A.

For example, the transpose of a matrix would be:

The Determinant of a Matrix

Determinant of a 2×2 matrix

Determinant of a 3×3 matrix

Determinant of a n×n matrix

Calculating a 2 × 2 Determinant

Application of Determinants in Solve Systems of Equations

Cramer’s Rule.

Example 2

3 × 3 Determinants

Cofactors

Expansion by Minors

Example 3

Example 4

The Inverse of a Matrix

The inverse of a 2×2 matrix

The inverse of a n×n matrix

Method 1 – Transposing and Determinants

Is it correct?

Cofactors Recall: The cofactor of an element in a matrix is the value obtained by evaluating the determinant formed by the elements not in that particular row or column.

Example: Consider the matrix

Read Next

Christian Religious Studies (CRS) Lesson Note SS1 Second Term

CCA Lesson Note JSS 3 Third Term

English Studies Lesson Note for SS 1 Second Term

Lesson note on English Language Jss1 First Term

Read Also

Leave a Reply Cancel reply

ACEOndo Admission List (NCE) 2018/2019 | 1st, 2nd, 3rd Batch

Physics Lesson Notes SS1 First Term

Lesson Note on French SS1 Second Term

A^T = (a_ji) where j is the column number and i is the row number of matrix A.