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WEEK 2
Reference Materials: New Further Mathematics project 1, by Adigun et al. Page 178
Previous Knowledge: Students can identify calculating devices.
Instructional Materials: Charts showing flowcharts.
Content FLOWCHART
A flowchart is a diagrammatical representation of a solution to a problem.
Example: The perimeter of a rectangle is 2(l+b). Draw a flowchart to determine and represent the information.
Advantages of flowcharts
- A flowchart plays a very important role in computer programming.
- It facilitates the interpretation and solution of problems.
- It can be easily understood.
- It can help in planning and development of algorithm for solving problems.
WEEK 3
Topic: Gradients of straight lines and curves
Sub-topic:
Gradients of straight lines
Duration: 40 minutes
Learning Objectives: By the end of the lesson, students should be able to calculate the gradient of a straight line.
Reference Materials: i. New General Mathematics for SSS 2, by M.F Macrae et al. Pages 184 – 192.
Previous Knowledge: Students can draw the graph of a linear equation (straight-line graph).
Instructional Materials: Graph board and graph book.
Content: GRADIENT OF A STRAIGHT LINE
The gradient of a straight line is the rate of change of y compared with x.
For example, if the gradient is 2, then for any increase in x, y increases two times as much.
Gradient of AB = Increase in y from A to B = MB
Increase in x from A to B AM
Example
Find the gradient of the line joining P(7, -2) and Q(-1, 2)
Gradient of PQ = increase in y = – AQ
Increase in x PA
=
Example 2
Find the gradient of the line 7x + 4y – 8 = 0
Re-arrange the equation: 4y = – 7x + 8
y = + 2
Therefore, gradient (m) = , y – intercept (c) = 2
SKETCHING GRAPHS OF STRAIGHT LINES
Given the equation
y = 3x – 2 , gradient = 3, y – intercept(c) = -2
2x + 3y = 6, gradient = , y – intercept(c) = 2
Example
Sketch the graph of the line whose equation is 4x – 3y = 12
Solution
When x = 0 ,- 3y = 12
y = – 4
The line crosses the y – axis at (0, – 4).
When y = 0 , 4x = 12
x = 3
The line crosses the x – axis at (3, 0).
From the graph:
Gradient m =
= =
y – intercept = – 4
Lines parallel to axes
Any line parallel to the x – axis has a gradient of zero. The equation of such lines is always in the form
y = c, where c may be any number.
The figure below shows the graph of y = 5 and y = – 3.
Notice that the equation of the x – axis is y = 0
————————————
————————————-
The gradient of a line that is parallel to the y – axis is undefined. The equations of such a lines are always in the form x = a , where a may be any number.
The figure below shows the graph of line x = 2 and x = – 4.
Notice that the equation of the y – axis is x = 0
EQUATION OF A STRAIGHT LINE
Equation of a straight line is of the form y = mx + c, where m is the gradient and c is the y – intercept.
Example 1
Determine the equation of a straight line whose gradient is and passes through the point (- 3, 2).
Solution
Using the formula y – y_{1} = m(x – x_{1})
Where (x_{1}, y_{1}) = (- 3, 2) and m =
y – 2 = (x + 3)
3y – 6 = – x – 3
x + 3y = 3
Example 2
Find the equation of the straight line passing through the points (1, 4) and (- 2, 6).
Using the formula
=
Where = (x_{1}, y_{1}) = ( 1, 4) and (x_{2}, y_{2}) = (- 2, 6), the equation is
=
cross multiply
– 3y + 12 = 2x – 2
2x + 3y = 14
GRADIENT OF A CURVE
Example
Draw the graph of y = for values of x from –2 to 3. Find the gradient of the curve at the point where x has the value (a) (b) – 2
Solution
x | -2 | -1 | 0 | 1 | 2 | 3 |
y | 1 | ¼ | 0 | ¼ | 1 | 2¼ |
(a) Gradient of the curve where x = 3
= gradient of tangent PT
= = 2.25 = 2¼ = 9 = 1^{1}/_{2}
1.5 1.5 6
(b) Gradient of curve where x = – 2
= gradient of tangent QR
= – = – 1 = – 1
1
WEEK 4
Topic: Straight line
Sub-topic:
Angle of slope and angle between lines
Duration: 80 minutes
Learning Objectives: By the end of the lesson, students should be able to calculate the angle of slope and angle between two lines.
Reference Materials: New Further Mathematics Project 2 by M. R Tuttuh Adegun
Previous Knowledge: Students can draw the graph of a linear equation (straight-line graph).
Instructional Materials: Graph board and graph book.
Content: ANGLE OF SLOPE
Example: Find the gradient of the line joining (3, 2) and (7, 10) and the angle of slope of the line.
Solution
Let m be the gradient of the line, then
m =
Let be the angle of slope of the line; then:
ANGLE BETWEEN TWO LINES
Condition for Parallelism
If two lines are parallel, the angle between them is zero, hence
Example: Determine if AB is parallel to PQ in each of the following.
- A(3, 1); B(4, 3) and P(4,6); Q(5, 8)
- A(-1, -2); B(2, -3) and P(5, 4) Q(6, 7)
Solution
- Let be the gradient joining A and B and be the gradient joining P and Q.
Since AB||PQ
- Let be the gradient joining A and B and be the gradient joining P and Q.
Since AB is not parallel to PQ
CONDITION FOR PERPENDICULARITY
If the lines are perpendicular, and therefore:
1 +
Example: Determine if AB is parallel to PQ in each of the following.
- A(5, -1); B(3, 2) and P(2, 4); Q(5, 6)
- A(-1, -2); B(2, -3) and P(5, 4) Q(6, 7)
Solution
- Let be the gradient joining A and B and be the gradient joining P and Q.
Since AB is perpendicular to PQ
- Let be the gradient joining A and B and be the gradient joining P and Q.
Since AB is perpendicular to PQ
EQUATION OF A LINE
The equation of a straight line is given by: y =mx + c
Example: Find the gradient and intercept on the y-axis of the following lines:
- y = 3x – 4
- y = – ½x – 3
Solution:
- Compare y = 3x – 4 with y = mx + c Hence the gradient is 3, intercept on y-axis is -4
- Gradient is – ½ , intercept on y-axis
GRADIENT AND ONE POINT FORM
Example: Find the equation of a straight line of slope 2, if it passes through the point (3, -2)
y –
m = 2;
Hence the equation of the straight line is:
y – (-2) = 2(x – 3)
y + 2 = 2x – 6
y = 2x -6 -2 = 2x – 8
y = 2x – 8
WEEK 5
Topic: Vectors
Sub-topic:
Modulus of a vector
Duration: 80 minutes
Learning Objectives: By the end of the lesson, students should be able to perform simple operations on vectors.
Reference Materials: New Further Mathematics Project 2 by M. R Tuttuh Adegun
Previous Knowledge: Students can perform arithmetic operations on vectors
Instructional Materials: Mathematical set.
Content: MAGNITUDE OF A VECTOR
The magnitude of a vector a, sometimes called the modulus of the vector is represented by |a|.
Zero Vector: The zero vector is a vector with zero magnitude.
Unit Vector: The unit vector is the vector represented by a and is such that a = |a| a
Negative Vector: The negative vector of a is written as – a
Equality of vector: Two vectors are equal when they have same magnitude and direction.
Example: Find the modulus of each of the following vectors
- 3i + 4j
- -2i – 5j
Solution
- Let r = 3i + 4j then |r| =
- Let r = -2i – 5j then |r| =
- Let r = then |r| =
Example: If find the modulus and direction cosines of:
Solution
|r_{1} + r_{2}| =
Let cos be the direction cosines of
cos
|| =
Let cos be the direction cosines of
cos
UNIT VECTOR
Example: Find the unit vectors in the directions of the following vectors
- r = 21 + 3j
- q = 4i – 5j
- p = 7i + 2j – 3k
- t = 3i -5j -3k
Solution
- Let be the unit vector in the direction of r; then
- Let be the unit vector in the direction of q; then
- Let be the unit vector in the direction of p; then
- Let be the unit vector in the direction of t; then
ARITHMETIC OPERATIONS ON VECTORS
Example: If p = 2i – 3j; q = 3i + 5j and r = i + j; Find the values of
- 2p + q + 3r
- 3p – 2q
Solution
- 2p = 2(2i – 3j ) = 4i – 6j
3r = 3( i + j ) = 3i + 3j
Therefore; 2p + q + 3r = (4i – 6j) + (3i + 5j) + (3i + 3j)
= 10i + 2j
- 3p = 3(3i – 3j) = 9i – 9j
2q = 2(3i + 5j) = 6i + 10j
Therefore 3p – 2q = (9i – 9j) – (6i + 10j) =3i – 19j
Example: Given that = a – b and = 2a + 3b, where a = 2i + 3j and b = 3i – 2j, find
= (2a + 3b) – (a – b)
= 2a + 3b – a + b = a + 4b
= (2i + 3j) + 4(3i – 2j) = 14i – 5j
Evaluation:
New Further Mathematics Project 2, by M.R Tuttuh Adegun et al. Page 262, Exercise 14, no 5
Conclusion: Teacher summarizes the topic, marks the students’ notes, does correction and allows the students to copy.
Assignment:
New Further Mathematics Project 2, by M.R Tuttuh Adegun et al. Page 262, Exercise 14, no 6
WEEK 6
MAGNITUDE OF A VECTOR
The magnitude of a vector a, sometimes called the modulus of the vector is represented by |a|.
Zero Vector: The zero vector is a vector with zero magnitude.
Unit Vector: The unit vector is the vector represented by a and is such that a = |a| a
Negative Vector: The negative vector of a is written as – a
Equality of vector: Two vectors are equal when they have same magnitude and direction.
Example: Find the modulus of each of the following vectors
- 3i + 4j
- -2i – 5j
Solution
- Let r = 3i + 4j then |r| =
- Let r = -2i – 5j then |r| =
- Let r = then |r| =
Example: If find the modulus and direction cosines of:
Solution
|r_{1} + r_{2}| =
Let cos be the direction cosines of
cos
|| =
Let cos be the direction cosines of
cos
Evaluation:
New Further Mathematics Project 2, by M.R Tuttuh Adegun et al. Page 262, Exercise 14, no 10
Conclusion: Teacher summarizes the topic, marks the students’ notes, does correction and allows the students to copy.
Assignment:
New Further Mathematics Project 2, by M.R Tuttuh Adegun
UNIT VECTOR
Example: Find the unit vectors in the directions of the following vectors
- r = 21 + 3j
- q = 4i – 5j
- p = 7i + 2j – 3k
- t = 3i -5j -3k
Solution
- Let be the unit vector in the direction of r; then
- Let be the unit vector in the direction of q; then
- Let be the unit vector in the direction of p; then
- Let be the unit vector in the direction of t; then
Evaluation:
New Further Mathematics Project 2, by M.R Tuttuh Adegun et al. Page 262, Exercise 14, no 10
Conclusion: Teacher summarizes the topic, marks the students’ notes, does correction and allows the students to copy.
Assignment:
New Further Mathematics Project 2, by M.R Tuttuh Adegun et al. Page 262, Exercise 14, no 12
ARITHMETIC OPERATIONS ON VECTORS
Example: If p = 2i – 3j; q = 3i + 5j and r = i + j; Find the values of
- 2p + q + 3r
- 3p – 2q
Solution
- 2p = 2(2i – 3j ) = 4i – 6j
3r = 3( i + j ) = 3i + 3j
Therefore; 2p + q + 3r = (4i – 6j) + (3i + 5j) + (3i + 3j)
= 10i + 2j
- 3p = 3(3i – 3j) = 9i – 9j
2q = 2(3i + 5j) = 6i + 10j
Therefore 3p – 2q = (9i – 9j) – (6i + 10j) =3i – 19j
Example: Given that = a – b and = 2a + 3b, where a = 2i + 3j and b = 3i – 2j, find
= (2a + 3b) – (a – b)
= 2a + 3b – a + b = a + 4b
= (2i + 3j) + 4(3i – 2j) = 14i – 5j
Evaluation:
New Further Mathematics Project 2, by M.R Tuttuh Adegun et al. Page 262, Exercise 14, no 5
Conclusion: Teacher summarizes the topic, marks the students’ notes, does correction and allows the students to copy.
Assignment:
New Further Mathematics Project 2, by M.R Tuttuh Adegun et al. Page 262, Exercise 14, no 6
WEEK 8
Topic: Straight line
Sub-topic:
Angle between lines
Duration: 80 minutes
Learning Objectives: By the end of the lesson, students should be able to calculate the angle between two lines.
Reference Materials: New Further Mathematics Project 2 by M. R Tuttuh Adegun
Previous Knowledge: Students can draw the graph of a linear equation (straight-line graph).
Instructional Materials: Graph board and graph book.
Content: ANGLE BETWEEN TWO LINES
The acute angle between lines of gradient m_{1} and m_{2}
Example: Find the acute angle between the lines x + 4y = 12 and y – 2x + 6 =0.
Solution
The gradients are -1/4 and 2
= |4.5| =77.47^{o}
GRADIENT INTERCEPT FORM
The gradient intercept form of the equation of a line is y = mx + c
Example: Determine the equation of the line whose gradient is -2 and y-intercept is 3..
Solution
Let the equation of the line be y = mx + c, where m = -2 and c = 3
Hence, the equation of the line is y = -2x + 3
Presentation:
Step I: Teacher revises the last topic with the students and does necessary corrections.
Step II: Teacher introduces the new topic to the students and explains by giving illustrative examples.
Step III: Teacher welcomes and answers questions from the students.
Step IV: Teacher gives notes to the students and ensures they copy correctly.
Step V: Teacher evaluates the students on topic discussed.
Evaluation:
1. Determine the equation of the line whose gradient is 3 and y-intercept is -4.
2. Find the acute angle between the lines 2y = 3x – 8 and 5y = x + 7.
Conclusion: Teacher summarizes the topic, marks the students’ notes, does correction and allows the students to copy.
Assignment: 1. Determine the equation of the line whose gradient is 3¼ and y-intercept is -6.
2. Find the acute angle between the lines 2y = – 5x + 8 and y = 3x – 7.
PERIOD 3
Topic: Equation of a straight line
Sub-topic: Gradient and one point form and two point form
Duration: 40 minutes
Learning Objectives: By the end of the lesson, students should be able to determine the equation of a line in different forms.
Reference Materials: i. New Further Mathematics for SSS 2 Project 2.
Previous Knowledge: Students can calculate angle between two lines.
Instructional Materials: Graph book.
Content: GRADIENT AND ONE POINT FORM
Equation of a line through (x, y) with gradient m is y – y_{1} = m(x – x_{1})
Example: A straight line has a gradient of -3/2 and passes through the point (1, 4). Find its equation and its intercept on the y-axis
Solution
In this case (x, y) = (1, 4) and m = – 3/2
So the equation is
y – 4 = – (x – 1)
- 2y – 8 = -3(x – 1)
- 2y + 3x =11
So y = – Hence the intercept on y-axis is 5½
GRADIENT AND TWO POINT FORM
Equation of a line through the two points (x_{1}, y_{1}) and (x_{2}, y_{2})is =
Example: Find the equation of a line AB which passes through the points (1, -1) and (-2, -13)
Solution
=
Therefore y + 1 = 4x – 4
- y = 4x – 5.
Thus the gradient of AB is 4.
Evaluation:
Find the equation of a line AB which passes through the points (-2, -3) and (-2, -13)
Conclusion: Teacher summarizes the topic, marks the students’ notes, does correction and allows the students to copy.
Assignment:
Find the equation of a line AB which passes through the points (-1, 2) and (3, 0)
Assignment:
New General Mathematics for SSS 2, by M.F Macrae et al. Page 190, Exercise 16d, no 2a, 2c