Mathematics lesson note for SS3 First Term is now available for free. The State and Federal Ministry of Education has recommended unified lesson notes for all secondary schools in Nigeria, in other words, all private secondary schools in Nigeria must operate with the same lesson notes based on the scheme of work for Mathematics.
Mathematics lesson note for SS3 First Term has been provided in detail here on schoolgist.ng
For prospective school owners, teachers, and assistant teachers, Mathematics lesson note is defined as a guideline that defines the contents and structure of Mathematics as a subject offered at SS level. The lesson note for Mathematics for SS stage maps out in clear terms, how the topics and subtopics for a particular subject, group works and practical, discussions and assessment strategies, tests, and homework ought to be structured in order to fit in perfectly, the approved academic activities for the session.
To further emphasize the importance of this document, the curriculum for Mathematics spells out the complete guide on all academic subjects in theory and practical. It is used to ensure that the learning purposes, aims, and objectives of the subject meant for that class are successfully achieved.
Mathematics Lesson note for SS3 carries the same aims and objectives but might be portrayed differently based on how it is written or based on how you structure your lesson note. Check how to write lesson notes as this would help make yours unique.
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This post is quite a lengthy one as it provides in full detail, the Mathematics-approved lesson note for all topics and sub-topics in Mathematics as a subject offered in SS3.
Please note that Mathematics lesson note for SS3 provided here for First Term is approved by the Ministry of Education based on the scheme of work.
I made it free for tutors, parents, guardians, and students who want to read ahead of what is being taught in class.
SS3 Mathematics Lesson Note (First Term) 2024
FIRST TERM: E-LEARNING NOTES
SCHEME FIRST TERM
| WEEK | TOPIC | CONTENT |
|
1 |
GRAPHS OF TRIGONOMETRIC RATIOS | (a) Graphs of: (i) Sine 0 < x < 360 (ii) Cosine 0 < x < 360. (b) Graphical solution of simultaneous linear and trigonometric equations. |
|
2 |
DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 1 | (a) Meaning of differentiation/derived function. (b) Differentiation from the first principle. (c) Standard derivatives of some basic functions. |
|
3 |
DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 2 | (a) Rules of differentiation such as: (i) sum and difference (ii) chain rule (iii) product rule (iv) quotient rule. (b) Application to real life situation such as maxima and minima, velocity, acceleration and rate of change etc. |
|
4 |
INTEGRATION OF SIMPLE ALGEBRAIC FUNCTIONS | (a) Integration and evaluation of definite simple Algebraic functions. (b) Application of integration in calculating area under the curve. (c) Use of Simpson’s rule to find area under the curve. |
| 5 | REVISION | |
| 6 | REVISION | |
| 7 | REVISION | |
| 8 | REVISION | |
| 9 | REVISION | |
| 10 | REVISION |
WEEK 1:
Date:……………….
Subject: Mathematics
Class: SS 3
TOPIC: Trigonometry Graphs of Trigonometric Ratios
Content:
- Graphs of: (i) Sine 00< x < 3600 (ii) Cosine 00 < x < 3600
- Graphical solution of simultaneous linear and trigonometric equations.
THE GRAPH OF Y = SINq FOR 00 < q < 3600
The graph of y = sinq is drawn by considering the table of values for sin q from q = 00 to q = 3600 at intervals of 900 as shown in the table below.
| q | 00 | 900 | 1800 | 2700 | 3600 |
| y = sin q | 0 | 1 | 0 | -1 | 0 |
THE GRAPH OF Y= cos q for 00 < q0 < 3600
The graph of y = cos q is also drawn by considering the table of values for cos q from q = 00 to q = 3600 at intervals of 900 as shown in the table below.
| q | 00 | 900 | 1800 | 2700 | 3600 |
| Y = cos q | 1 | 0 | -1 | 0 | 1 |
THE GRAPH OF Y = tanq for 00 < q < 3600
The graph of y = tanq is drawn by considering the table of values for tanq from q = 00 to q = 3600 at intervals of 450 as shown below.
| q | 00 | 450 | 900 | 1350 | 1800 | 2250 | 2700 | 3150 | 3600 |
| y=tanq | 0 | 1 | udf | -1 | 0 | 1 | udf | -1 | 0 |
udf. Þ Undefined
CLASS ACTIVITY
- Draw the graph of each of the following functions
- Y= – sinx (b) y = – cosx
- Draw the graph of each of the following functions
(c)y = -tanx (d) y = sinx + cosx (e) y = 1+ sinx
GRAPHICAL SOLUTION OF SIMULTANEOUS LINEAR AND TRIGONOMETRIC GRAPH
Example 1:
(a) Copy and complete the table of values for the function y = 2 cos2x – 1
| X | 00 | 300 | 600 | 900 | 1200 | 1500 | 1800 |
| y=2cos2x-1 | 1.0 | 0.0 | 1.0 |
(b) Using a scale of 2cm to 300 on the x-axis and 2cm to 1 unit on the y-axis draw the graph of y = 2 cos 2x – 1 for 00 £ x £ 1800
(c) On the same axes draw the graph of
(d) Use your graph to find the
(i) Values of x for which 2 cos 2x + ½ = 0
(ii) Roots of the equation
2 cos 2x – + 1 = 0 (WAEC).
Solution:
y = 2cos2x – 1
| x | 00 | 300 | 600 | 900 | 1200 | 1500 | 1800 |
| y=2cos2x-1 | 1.0 | 0.0 | -2.0 | -3.0 | -2.0 | 0.0 | 1.0 |
For x = 600
y = 2 cos 2 x 600 – 1
= 2 cos 120 – 1
= -2 cos (180 – 120) – 1
= -2 cos 600 – 1
= -2 x 0.5 – 1
= -1 – 1
= -2
For x = 90
y = 2 cos 2 x 900 – 1
= 2 cos 180 – 1
= -2 – 1
= -3
For x = 120
y = 2 cos 2 x 1200 – 1
= 2 cos 240 – 1
= -2 cos (240 – 180) – 1
= -2 cos 60 – 1
= -2 x 0.5 – 1
= -1 – 1
= – 2
For x = 1500
y = 2 cos 2x – 1
= 2 cos 2 x 150 – 1
= 2 cos 300 – 1
= 2 cos (360 – 300) – 1
= 2 cos 60 – 1
= 2 x 0.5 – 1
= 1 – 1
= 0
(b) Turn to the next page for graph
(c) To draw the graph of
select any three values from the x-axis of the table above
| x | 00 | 900 | 1800 |
| y | -2.0 | -1.5 | -1.0 |
(d) (i)
2 cos 2x + ½ = 0
2 cos 2x = -½
2 cos 2x – 1 = -½ – 1
2 cos 2x – 1 = -1½
The values of x for which 2 cos 2x + ½ = 0 can be obtained at the point where y = -1½ (point A and B on the graph)
i.e. x = 520 or x = 1290
(ii) The roots of 2cos2x – x + 1 = 0
180
2cos2x – x + 1 = 0
180
2cos2x = x – 1
180
2cos2x – 1 = x – 1 -1
180
2 cos 2x – 1 = x – 2
180
{where x – 2 = 1 (x – 3600)}
180 180
The roots are found at the point where the two graphs y = 2 cos 2x – 1 and y = 1/180 (x – 3600) meet.(points C and D on the graph)
i.e. x = 550 or x = 1320
Example 2:
(a) Copy and complete the table of values for the function y = 2 cos 2q + sin q
| x | -1200 | -900 | -600 | -300 | 300 | 600 | 900 | 1200 | 00 |
| y | -1.87 | -0.13 | -1 | 0.13 | 2 |
(b) Using a scale of 2cm to 300 on q-axis and 2cm to 1 unit on y-axis draw the graph of y = 2 cos 2q + sin q for –1200 £ q £ 1200
(c) Using the same scale and axes draw the graph of
(d) From your graph, find the roots of the following equations
(i) 2 cos 2q + sin q = 0
(ii) 2 cos 2q + sin q + ½ = 0
(iii) 2 cos 2q + sin q =
Solution:
y = 2cos 2q + sin q
| x | -1200 | -900 | -600 | -300 | 300 | 600 | 900 | 1200 |
| y | -1.87 | -3.00 | -1.87 | 0.50 | 1.50 | -0.13 | -1.00 | -0.13 |
For q = -1200
-1200 ≡ 2400
y = 2 cos 2 x 240 + sin 240
= 2 cos 480 + sin 240
= 2 (-0.5) + (-0.8660)
= -1 –0.8660
= -1.87
For q = -30
-300 ≡ 3300
y = 2 cos 2 x 3300 + sin 3300
= 2 cos 6600 + sin 3300
= 2 (0.5) + (-0.5)
= 1 – 0.5
= 0.5
For q = -90
-900 ≡ 2700
y = 2 cos 2 x 2700 + sin 2700
= 2 cos 540 + sin 270
= 2 (-1) + (-1)
= -2 – 1
= -3
For q = 30
y = 2 cos 2 x 30 + sin 300
= 2 cos 600 + sin 300
= 2 (0.5) + (0.5)
= 1 + 0.5
= 1.5
(c)
| q | -1200 | 00 | 90 |
| y | -3.05 | -1 | 0.54 |
For q = -120
y = -7 x 120 – 1
410
y = -2.05 – 1
y = -3.05
For q = 0
y = 7 x 0 – 1
410
y = -1
For q = 900
y = 7 x 90 – 1
410
y = 630 – 1
410
y = 1.54 – 1
y = 0.54
(d) (i) The roots of 2 cos q + sin q = 0 is at the points where y = 0 i.e. where the graph crosses the q-axis.
q = -350 or q = 590
(ii) 2 cos 2q + sin q + ½ = 0
2 cos 2q + sin q = -1/2
The roots are the values of q where y = -1/2 i.e. q = -420 or q = 690 or q = 1120
(iii) 2 cos 2q + sin q =
The roots are at the points where the two graphs y = 2 cos 2q + sin q and y = meet. i.e. q = -1030 or q = -640 or q = 600
CLASS ACTIVITY
(1) (a) Copy and complete the following table of values for the function
y = sin q – cos 2q.
| q | 00 | 300 | 600 | 900 | 1200 | 1500 | 1800 |
| y | -1.00 | 0.00 | 1.37 | -1.00 |
(b) Using a scale of 2cm to 300 on q-axis and 2cm to 1 unit on y-axis draw the graph of y = sin q – cos 2q for 00 £ q £ 1800
(c) Using the same scale and axes draw the graph of
(d) From your graph, find the roots of the following equation.
(i) sin q – cos 2q = 0
(ii) sin q – cos 2q = 1
(iii) sin q – cos 2q =
(2) (a) Copy and complete the following table of values for the function
y = 1 – 3 sin 2x
| x | 00 | 150 | 300 | 450 | 600 | 900 | 1050 | 1200 | 1350 | 1500 | 1650 | 1800 | 750 |
| y | 1 | -0.5 | -1.6 | 1 | 2.5 | 3.6 | 2.5 | 1 | -0.5 |
(b) Using a scale of 1cm to 150 on the x-axis and 2cm to 1 unit on the y-axis draw the graph of y = 1 – 3 sin 2x for 00 £ x £ 1800
(c) Using the same scale and axes, draw the graph of
(d) From your graph, find the roots of the following equation
(i) 1 – 3 sin 2x = 0
(ii) 1 – 3 sin 2x = 2
(iii) 1 – 3 sin 2x =
PRACTICE EXERCISE
(1) (a) Copy and complete the following table of values for the function
y = 2 sin 2x + 1
| x | -450 | -300 | -150 | 00 | 150 | 300 | 450 | 600 | 750 | 900 | 1050 | 1200 |
| y | -0.7 | 0.0 | 1.0 | 2.7 | 2.0 | 0.0 | -0.7 |
(b) Using a scale of 1cm to 150 on the x-axis and 2cm to 1 unit on the y-axis draw the graph of y = 2 sin 2x + 1 for –45 £ x £ 1200
(c) From your graph, find the roots of the equations
(i) 2 sin 2x + 1 = 0
(ii) 2 sin 2x – 1 = 0
(2) (a) Copy and complete the following table of values for the function
y = cos 2x + 2 sin x
| x | 00 | 300 | 600 | 900 | 1200 | 1500 | 1800 | 2100 | 2400 | 2700 | 3000 | 3300 | 3600 |
| y | 1 | 1.5 | 1 | 1.24 | 1 | -0.5 | 1 |
(b) Using a scale of 1cm to 300 on the x-axis and 2cm to 1 unit on y-axis draw the graph of
y = cos 2x + 2 sin x for 00 £ x £ 3600
(c) From your graph, find the roots of the following equations
(i) cos 2x + 2sin x = 0
(ii) cos2x + 2sin x + 2 = 0
(3)(a) Copy and complete the following table of values for y = 3 sin 2 – Cos
| 0o | 30o | 60o | 90o | 120o | 150o | 180o | |
| Y | -1.0 | 0.0 | 1.0 |
(b) Using a scale of 2cm to 30o on theaxis and 2cm to 1 unit on the y axis, draw the graph of y=3sin 2-cos for 0o≤≤180o
(c) Use your graph to find the:
(i) solution of the equation 3 sin2 – cos=0, correct to the nearest degree;
(ii) maximum value of y, correct to one decimal place.
- (a)Copy and complete the following table of values for y = 9 cos x + 5 sin x to one decimal place.
| x | 0o | 30o | 60o | 90o | 120o | 150o | 180o | 210o |
| Y | 10.3 | -0.2 | -5.3 | 10.3 |
(b) Using a scale of 2cm to 30o on the x-axis and 2cm to 1 unit on the y-axis, draw the graph of y = 9 cos x + 5 sin x, for 0 ≤ x ≤ 210o
(c) Use your graph to solve the equation;
(i) 9 cos x + 5 sin x = 0
(ii) 9 cos x + 5 sin x = 3.5, correct to the nearest degree
(d) Find the maximum value of y, correct to one decimal place.
ASSIGNMENT
- (a)Copy and complete the table of values for y = 3 sin x + 2 cos x for 0o ≤ x ≤ 360o.
| x | 0o | 60o | 120o | 180o | 240o | 300o | 360o |
| y | 2.00 | – | – | – | – | – | 2.00 |
(b) Using a scale of 2cm to 60o on x-axis and 2cm to 1 unit on y-axis, draw the graph of y = 3 sin x + 2 cos x for 0o ≤ x ≤ 360o.
(c) Use your graph to solve the equation 3 sin x + 2 cos x = 1.5.
(d) Find the range of values of x for which 3 sin x + 2 cos x < – 1.
- (a) Copy and complete the table of values for y= sin x + 2 cos x, correct to one decimal place.
| x | 0o | 30o | 60o | 90o | 120o | 150o | 180o | 210o | 240o |
| y | 2.2 | -1.2 | -2.0 | -1.9 |
(b) Using a scale of 2cm to 30o on the x-axis and 2cm to 0.5 units on the y-axis, draw the graph of y = sin x + 2 cos x for 0o ≤ x ≤ 240o
(c) Use your graph to solve the equation:
(i) sin x + 2 cos x = 0;
(ii) sin x – 2.1 – 2 cos x.
(d) From the graph, find y when x = 171o.
- (a) Copy and complete the table of the relation y – 2sin x – cos 2x
| x | 0o | 30o | 60o | 90o | 120o | 150o | 180o |
| y | 0.5 | -1.0 |
Using a scale of 2 cm to 30o on the x-axis and 2cm to 0.5 unit on the y-axis, draw the graph of y = 2 sin x – cos 2x, for 0o ≤ x ≤ 180o.
(b) Using the same axes, draw the graph of y = 1.25
(c) Use your graphs to find the:
(i) values of x for which 2 sin x – cos 2x = 0.
(ii) roots of the equations 2 sin x – cos 2x = 1.25.
- Copy and complete the table of values for y = 1 – 4cosx
| x | 00 | 300 | 600 | 900 | 1200 | 1500 | 1800 | 2100 | 2400 | 2700 | 3000 |
| y | -3.0 | 1.0 | 4.5 | -1.0 |
(B) using a scale of 2cm to 300 on the x-axis and 2cm to 1 unit on the y-axis , draw the graph of y= 1- 4cosx for 00≤x≤3000.
(C)use the graph to:
(i)solve the equation 1- 4cosx=0;
(ii) find the value of y when x=1050
(iii)find x when y=1.5
- (a) Copy and complete the table for the relation y=2 cos 2x – 1.
| x | 0o | 30o | 60o | 90o | 120o | 150o | 180o |
| Y=2cos 2x-1 | 1.0 | 0.0 | 1.0 |
(b) Using a scale of 2cm=30o on the x -axis and 2cm =1 unit on the y –axis draw the graph of y = 2 cos 2x +1/2 = 0
(c) On the same axis draw the graph of y =1/180 (x – 360).
(d) Use your graphs to find the:
(i) values of x for which 2 cos 2x+ ½ = 0
(ii) roots of equation 2 cos 2x – x/180 + 1 = 0
KEYWORDS: roots, y-axis, x-axis, coordinate, graph, table of values, scale etc.
WEEK 2:
Date:……………….
Subject: Mathematics
Class: SS 3
TOPIC: Differentiation of Algebraic functions 1
Content:
- Meaning of differentiation/derived function.
- Differentiation from the first principle.
- Standard derivatives of some basic functions.
MEANING OF DIFFERENTIATION/DERIVED FUNCTION#
The process of finding the differential coefficient of a function is called differentiation.Differentiation deals with the measure of the rate of change in a particular function when some quantities in the function is either increased or decreased. For example, given the function y = f(x), a change in x will produce a corresponding change in y. When y is increased, x is bound to increase in proportion and vice versa. Note: The reverse of differentiation is integration.
DIFFERENTIATION FROM THE FIRST PRINCIPLE
The method of finding the derivative of a function from definition is called differentiation from the first principle. Note: A change in x to produces a corresponding change in y to .
Example 1:
Differentiate the following from the first principle
SOLUTION
Take increment in both x and y
Divide both sides by
Take limits of both sides as
Take increment in both x and y
EXAMPLE 2:
Differentiate the following from the first principle
SOLUTION
Take increment in both x and y
CLASS ACTIVITY
Differentiate the following from the first principle
*STANDARD DERIVATIVES OF SOME BASIC FUNCTIONS
(1) if y = a , where ‘a’ is constant , then
(2) if y = ax , then
(3) if y = , then
(4) if y = , then
(5) if y = , then
(6) if y = sin ax , then
(7) if y = cos ax , then
(8) if y = then
EXAMPLE 1:
Use the standard derivatives given above to find of the following functions
- Differentiate the following functions with respect to x
CLASS ACTIVITY
Find the derivative of the following function
PRACTICE EXERCISE
- Find from first principle, the derivative, with respect to of 2
- By first principle .
- Find the derivative of the following functions:
- -7 1/2c. 4
- Differentiate the following functions with respect to x.
- 2x2-x-1
- 6+5x-x2
ASSIGNMENT
- For each of the following functions , use the method of differentiation from the first principles to find
- 3x2
- -5x2
- Use the standard derivatives to find the derivatives of the following functions
- 2x2-3x+3
- y=2ex
- Differentiate the following:
a.y=-cosx
b.y=2sinx
- Differentiate with respect to x.
a.8
- 5x-2/3
- Differentiate with respect to x.
b.
KEYWORDS: derivative, differentiate , rate of change ,increase, increment, first principle, derived function etc.
WEEK 3:
Date:……………….
Subject: Mathematics
Class: SS 3
TOPIC: Differentiation of Algebraic functions 2
Content:
- Rules of differentiation such as: (i) sum and difference (ii) chain rule (iii) product rule (iv)quotient rule.
- Application to real life situation such as maxima and minima, velocity, acceleration and rate of change etc
RULES OF DIFFERENTIATION
SUM AND DIFFERENCE RULE
- Y =
- Y =
SOLUTION
- Y =
SOLUTION
Example 1:
Differentiate
(a)
CLASS ACTIVITY
.
Let and
EXAMPLE 2:
Find the derivative of the function:
SOLUTION
Let
CLASS ACTIVITY
- Differentiate with respect to x:
b.Find the derivative of the following function:
a.(2x+1)3(x2+1)
- x3(2x2-1)
HIGHER DERIVATIVE
If
Then
Also,
etc.
EXAMPLE 1:
Find the second derivative of y=3x3-5x2
SOLUTION
Y=3x3-5x2
EXAMPLE 2:
Find the
SOLUTION
CLASS ACTIVITY
Find the third derivative of the functions:
- Y =
- Y =
APPLICATION TO REAL LIFE SITUATION SUCH AS MAXIMA AND MINIMA, VELOCITY, ACCELERATION AND RATE OF CHANGE ETC.
GRADIENT
If y or f is a function of x , then the first derivative or f ‘(x) is called the gradient function. The gradient of a curve at any point P(x1,y1) is obtained by substituting the values of x1 and y1 into the expression for This is the same as the gradient of the tangent at that point.
EXAMPLE 1:
Find the gradient of the curve y=x2+7x-2 at the point (2,16)
SOLUTION
y=x2+7x-2
Thus at (2,16)
EXAMPLE 2:
If f(x)=(x2+3)3, find the gradient of f(x) at x= ½
SOLUTION
Let y=(x2+3)2
Therefore, the gradient at x= ½ is
CLASS ACTIVITY
- Find the gradient of the curve y=x2+3x-2 at the point x=3.
- Find the coordinate of the point on the given curve, y=x2-x+3 whose gradient is 1.
VELOCITY AND ACCELERATION
Suppose that a particle’s distance, s metres, after t seconds is given by s=t2+3t+5
The velocity is the rate of change of s compared with t, i.e. Since s= t2+3t+5, then
Hence the velocity after t seconds is given by 2t+3.
Acceleration is the rate of change of velocity compared with time. If velocity is vm/s then the acceleration is
EXAMPLE 1:
A particle moves in a straight line specified by the equation x=3t2-4t3. Find the velocity and acceleration after 2 seconds.
SOLUTION
X=3t2-4t3
At t=2, we have 6(2)-12(2)2
= 12- 48
=-36m/s
EXAMPLE 2:
An object projected vertically upwards satisfies the relation h=27t-3t2, where hm is the height after t seconds.
- Find the time it takes to reach the highest point.
- How high does it go?
SOLUTION
- h=27t-3t2
i.e. 27-6t=0
6t=27
t=27/6=4.5seconds
b.to find the highest point, substitute t=4.5s into the expression for h.
h=27(4.5)-3(4.5)2
h=60.75m
CLASS ACTIVITY
- A particle moves along a straight line in such a way that after t seconds it has gone s metres, where s=t2+2t.
Find the velocity of the particle after
- 1 second
- 3 seconds
- A stone is thrown vertically into the air , and its height is s metres after t seconds, where s =29.4t-4.9t2
- after how many seconds does it reach its greatest height ?
- what is the greatest height?
- What is its initial velocity?
INCREASING AND DECREASING FUNCTIONS
A function y is increasing if while a function is decreasing if
Example1:
Find the range of values of x for which x2-x is increasing.
SOLUTION
Let y=x2-x
x2-x is increasing if
2x>1
x> ½
EXAMPLE 2:
Find the range of values of x for which x2-x is decreasing?
SOLUTION
Let y=x2-2x
x2-x is decreasing if
2x<2
X< 1
CLASS ACTIVITY
- Find the range of values of x for which x2-5x is increasing.
- Find the range of values for which x2-4x is decreasing.
RATE OF CHANGE
EXAMPLE 1:
- Find the approximate increase in the area of a circle if the radius increases from 2cm to 2.02cm.
SOLUTION
Let A denote the area of the circle of radius r.
Then, A=
=0.2514cm2
EXAMPLE 2:
If the side of a square is increasing by 0.2%, find the approximate percentage increase in the area.
SOLUTION
A=x2
CLASS ACTIVITY
- The radius of a circle is increasing at the rate of 0.001m/s. find the rate at which the area is increasing when the radius of the circle is 10cm.
- Find the approximate change in the surface area of a cube of side x metres caused by decreasing its side by 1%.
MAXIMA AND MINIMA
A turning point/stationary point of a curve is a point at which the gradient is zero. The turning point is either maximum point(highest point ) or the minimum point(lowest point) or the point of inflexion.
PROCEDURE FOR TESTING AND DISTINGUISHING BETWEEN STATIONARY POINTS
- Given y=f(x),determine
- Put and solve for x
- Substitute x into equation to obtain the y, i.e.(x,y) of turning point.
NATURE OF TURNING POINT
Using
- If
- If >0
- If
EXAMPLE 1:
A curve is defined by the function y=x3-6x2-15x-1,find the maximum and minimum point.
SOLUTION
First, we find and equate to zero.
To test for maximum or minimum,
We differentiate the second time
=
Put x=5, 6(5)-12=18>0…………..minimum point at x=5
Put x=-1, 6(-1)-12=-6-12=-18<0……………maximum point at x=-1
To find the corresponding y put x=5 to the first equation i.e. y=x3-6x2-15x-1
Y=125-150-75-1=-101, minimum point (5,-101)
Maximum point(-1,7), minimum point(5,-101)
CLASS ACTIVITY
- Find the maximum and minimum points of the curve y=x3-3x+5
- Find the turning point of y=x3-6x2+12x-11
EXAMPLE 2:
Find the maximum or minimum value of the curve y=x2-6x+5
SOLUTION
2x-6=0
2x=6
X=3
, 2>0 therefore, it is minimum
To obtain y,
y=32-6(3)+5
y=9-18+5
y=-4
minimum point is (3,-4)
PRACTICE EXERCISE
1.Find the maximum and minimum values of y for the function
y=2x3+3x2-36x-6
2 . Differentiate the following with respect to x.
- (4x+9)3
- (3x-2)3(x2+4)2 NECO
3.If y= 2x3-6x2—15x+19, find the coordinates of the points on the graph at which the gradient is 3.
- Differentiate
- A moving body has gone s metres in t seconds, where s=3t2-4t+5. Find its velocity after 3 seconds. Show that the acceleration is constant, and find its value.
ASSIGNMENT
- After t seconds a particle has gone s metres where s=t3-6t2+9t-5. Find the time (in seconds) for its velocity and acceleration to be zero. Calculate also the velocity and acceleration initially and after 5 seconds.
- Differentiate
- Find dy/dx of (3-2x)-1/2
- Find the equation of the tangent to the curve y=x2-2x+3 at the point (2,3)
- If the radius of a circle is increased from 5cm to 5.1cm, find the approximate increase in area.
KEYWORDS: derivative, gradient, differentiate , rate of change ,increase, maximum, minimum, velocity, acceleration, derived function etc.
WEEK 4:
Date:……………….
Subject: Mathematics
Class: SS 3
TOPIC: Integration of Simple Algebraic functions:
Content:
- Integration and evaluation of definite simple Algebraic functions.
- Application of integration in calculating area under the curve.
- Use of Simpson’s rule to find area under the curve.
INTEGRATION AND EVALUATION OF DEFINITE SIMPLE ALGEBRAIC FUNCTIONS
Integration is the opposite of Differentiation. It is the process of obtaining a function from its derivative. A function F (x) is an anti derivative of a given function F (x) if d = f (x).
In general, if F (x) is any anti derivative of f (x), then the most general anti derivative of f(x) is specified by f (x) + c and we write: + c
The symbol is called an integral sign and is called the indefinite integral. The arbitrary constant c is called the constant of integration, and the function f (x) is called the integral.
For example , F(x) = x4 + c is an anti derivative of f(x) = 4x3 because F’(x) = = 4x3 = f(x).
In general, if n -1, then an anti derivative of f(x) = is F (x) = + C
To integrate a power of x ( apart from power n = – 1, increase the power of x by 1 ( one) and divide by the new power.
EXAMPLE 1:
- = + c
= + c
( + 1 ) = = x5
- 2 ) = x5
( + 3 ) = x5
- Similarly, + C ] =
:. dx = + C ; ( n ≠ -1 )
- Integrate the following:
(i). (ii). (iii). dx
(iv). If = 4 and y =2 when x = -1, find y in terms of x.
SOLUTION :
(I). =2
= 2[ ] + C
= + C
= + C
(ii). This can be done term by term.
= –
= + –
= + – 10 + C
= + – 10x + C
Notice that instead of giving three different constants of integration, the three can be combined and written as one.
(iii). dx = + C
= + C
= 2 + C
(iv). = 4 , so dy = 4dx
= ie y = 4x + C. When y = 2, x = -1 2 = 4(-1) + C
C = 6 Hence, y = 4x + 6
- The integral
Let u = ax + b
= a ,
du = adx , so dx = 1/a du
=
= 1/a
EXAMPLE 2 :a. Integrate (i). (ii).
Solution : For
Let u = 3x +2, = 3
So 3dx = du , hence dx = 1/3du
:. =
=
= ( ) + C
= + C
(ii). , Let U= 2x – 1
= 2
2dx = du
dx= ½ du
=
=
= du
= () + C
= + C
= . + C
= . + C
=
b.Integrate (i).) dx (ii). 2x( dx
dx = ( + – )dx
= + – 4 dx
= + + + C
(ii). 2x ( – 1 )dx = (2 – 2x )dx
= 2 –
= – + C
CLASS ACTIVITY
1) (x 2 + 3x – 2)dx
2) + ) dx
3). dx
APPLICATION OF INTEGRATION IN CALCULATING AREA UNDER THE CURVE
y-axis
y = f(x)
a b x-axis
The integral is called the definite integral of the function f(x) with ‘a’ and ‘b’ the lower and upper limits of the integral respectively.
geometrically represents the area bounded by the curve y = f(x), the lines x = a , x =b and the x-axis.
Example 1:
Evaluate
Solution:
Now substitute the value of the upper limit for x minus when you substitute the lower limit.
= = 64 + c – 1 – c = 63
Now we shall examine some properties of the definite integral,
y-axis
y = f(x)
a c b x-axis
If the area is below the x-axis it will have a negative sign attached to it. Negating such an area will make it positive.
It is very essential to sketch the curve y = f(x) if the definite integral, is to be used in finding the area bounded by the curve y = f(x), the lines x= a , x = b and the x – axis.
Example 2:
Find the area bounded by the curve the lines x = 2, x = 3 and the x –axis.
Solution:
Y-axis
2 3 x-axis
Let the shaded area be the required area. i.e
The area =
=
=
=
= sq. units
- units
Example 3:
Find the area of the finite region bounded by the curve y = the line y = 1, y = 9 and the y – axis.
Solution: y-axis y = x2
9
1
x-axis
The area = as y = x2 then x = , so the area =
=
=
=
=
= 17 sq. units
CLASS ACTIVITY
- Find the area of finite region between the axis and the curve (i) y = x(x – 2)(x – 3)[ans: 3] (ii) y = x(x – 1)(x – 3)
- Find the area bounded by the following curves and lines and the axis
- Y = x2 for x = -1 , x = 2
- Y = x2+ 1 for x = 1 , x = 3
- Y = for x = 2 , x = 4
EQUATION OF CURVE GIVEN GRADIENT
EXAMPLE 1:
A curve passes through the point (0,1) and its gradient at any point P(x,y)=3x2-5. Find the equation of the curve.
SOLUTION
Let
dy=(3x2-5)dx
-5)dx
y=x3-5x+c
at (0,1)
1=0-5(0)+c
C=1
The equation is y=x3-5x+1
EXAMPLE 2:
A particle moves in a straight line in such a way that its velocity after t seconds is (3t+4)m/s. find the distance travelled in the first 3 seconds.
SOLUTION
V=3t+4
3
S=
0
S=27/2 +12
S=25.5m
VELOCITY AND ACCELERATION
EXAMPLE 1:
A particle moves in a straight line with a constant acceleration of 2cm/s2. If its velocity after t seconds is vcm/s, find u in terms of t, given that the velocity after 3 seconds is 12cm/s.
SOLUTION
a=
When t=3 and v=12
12=2
12=6+c
C=6
EXAMPLE 2:
The velocity, Vms-1 of a body after time t seconds is given by V=3t2-2t-3. Find the distance covered during the 4th second.
SOLUTION
Let S be the distance covered
S=
4
S=
3
S=(64-16-12)-(27-9-9)
S=27m
SIMPSON’S RULE
Another rule for numerical integration is attributed to Thomas Simpson (1710-1761) an English Mathematician.
a=xo,x1,x2,x4,…xn=b
and their corresponding ordinates at y0,y1,y2…,yn, Simpson showed that
this can also be written as
EXAMPLE:
Using Simpson’s rule with 8 strips, evaluate
Correct to 2 decimal places.
h=
The working is set in a tabular form as follows:
| x | y | First last ordinates | Odd ordinates | Remaining ordinates |
| 1 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 | Y0 Y1 Y2 Y3 Y4 Y5 Y6 Y7 Y8 | 1
0.2 | 0.67
0.40
0.29
0.22 |
0.50
0.33
0.25 |
| totals | 1.2 | 1.58 | 1.08 |
≃1.613
Hence,
EXAMPLE 2:
Using Simpson’s rule with 4 strips, evaluate
Correct to 2 decimal places.
| X | Y | First last ordinates | Odd ordinates | Remaining ordinates |
| 2 3 4 5 6 | Y0 Y1 Y2 Y3 Y4 | 4 64 | 8 32 | 16 |
| totals | 68 | 40 | 16 |
=
=86.67 (2 d.p.)
CLASS ACTIVITY
Using Simpson’s rule with six strips, evaluate
- Find the area of the finite region bounded by y = x2 – 2x – 3 , x = -1 , x = 3 and the x – axis
- Find the area of the finite region bounded by y = x2 – 2x – 3 , x = 0 , x = 5 and the x – axis
- Find the area of the finite region bounded by the curve y = 9x2the line y = 1 , y = 9 and the y – axis.
INTEGRATE THE FOLLOWING
- dx
- ( dx
ASSIGNMENT
INTEGRATE THE FOLLOWING
- dx
- Find the equation of a curve with gradient given by 2x-3 and passes through the point (3,2)
- A particle moves along a straight line in such a way that its acceleration after t seconds is (2t-1)cm/s2.if its velocity after t seconds is vcm/s, find v in terms of t, given that v=9 and t=2.
- Find the area enclosed by the curve y=4+3x-x2and the x-axis.
- Evaluate
- Evaluate
KEYWORDS: integrate, integral, tangent, gradient, differentiate , rate of change ,increase, velocity, acceleration, derived function etc.
WEEK 5 REVISION
WEEK 6 EXAMINATION
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