Mathematics Lesson Note SS3 Second Term
Mathematics Notes SSS3 – Edudelight.com
MATHEMATICS SS3
SECOND TERM
SCHEME OF WORK MATHEMATICS
WEEK(S) TOPICS
- Review of first term work: (i) Bonds and debentures (ii) Shares (iii) Rates (iv) Income tax and (v) Value added tax.
- CO-ORDINATE GEOMETRY OF STRAIGHT LINE: Cartesian coordinate (ii) plotting the linear graph (iii) determine the distance between two coordinate points. (iv) Finding the mid-point of the line joining two point (v) practical application of coordinate geometry.(vi) Gradient and intercept of a straight line.
- COORDINATE GEOMETRY OF A STRAIGHT LINE CONTINUES: (I) Define gradient and intercepts of a line. (ii) Find the angle between two intersecting straight lines (iii) Application of linear graphs to real life student.
- DIFFERENTIATION OF ALGEBRAIC FUNCTION: (I) Meaning of differentiation/ derived function (ii) differentiation from first principle (iii) standard derivative of some basic functions.
- DIFFERENTIATION OF ALGEBRAIC FUNCTION CONTINUES: Rules of differentiation such as: (a) Sum and difference (b) Product rule (c) Quotient rule. (d) Application of real situation such as Maximal, Minima velocity, Acceleration and rate of change.
- INTEGRATION AND EVALUATION SIMPLE ALGEBRAIC FUNCTION: (i) definition (ii) Method of integration: (a) substitution method (b) partial fraction method (c) part. (iii) Application of integration in calculating area under the curve (iv) Use of Simpson’s rule to find the area under the curve.
7-12. Revision and Mock Examination.
WEEK 1
REVIEW OF FIRST TERM WORK
BONDS: A bond is a documentary obligation to pay a sum of money or to perform a contract.
DEBENTURE: Is a certificate that certifies an amount of money owed to someone or promises to pay of the issuer a specific amount of money.
SHARES: is a portion of something given or allotted to someone from investment (Dividends).
RATE: a rate can be defined as a payment or levy paid to an authority or individual on the use of property (ratable value).
INCOME TAX: This is the amount taken from salaries of workers for the services provided by the government. Such as security, education, health roads etc.
VALUE ADDED TAX (VAT): this is a certain amount taken any goods sold or purchased by a customer.
EXAMPLE RATE
Find the rate at 85kobo in the # on the house of ratable value of #216.
SOLUTION
Value of the house #216 or 21600kobo
Amount payable on every # to the government =
= #216 x = #183.60
ASSESSMENT: (a) A man with an annual salary of #4200 has allowances of #1400. How much does he pay each year in income tax?
(b) to raise an income of #4176000 a town declares a rate of 87 kobo in the naira. What is the ratable value of the town?
(c) the annual rates at 73 cents in the dollar on a house are $262.80. What is the ratable value?
(d) A certain doctor has a salary of #11500. His allowances, which include the expenses of running the practice, are #4800. Find the total amount he pays each year in tax.
Mathematics Notes SSS3 – Edudelight.com
WEEK 2
CO-ORDINATE GEOMETRY
Coordinate geometry is an aspect of geometry that deals with points and lines joining them.
GRADIENT OF A LINE: This is the sloping degree of a line joining two points and it is measured by the ratio of the increment in vertical axis to that of the horizontal axis.
B
A N
X1 X2
Gradient or slope (M) = . From the above diagram.
=
. Therefore, the slope or gradient can be regarded as tangent of the inclination angle to the Horizontal.
DISTANCE BETWEEN TWO POINTS:
This is defined as the length of the line segment joining any two points.
From the diagram above, (by Pythagoras theorem)
AB =
MID-POINT OF A LINE SEGMENT:
This is defined as the coordinates of the middle points of the line joining two points.
=
, ѝ =
Mid-point = (
GRADIENT AND INTERCEPT FORM:
This is the form that includes the gradient (M) and the intercept (C) on Y-axis
Y = MX + C. where M is the gradient of the line and C is the intercept on Y-axis (Equation of the straight line).
EXAMPLE DISTANCE
Find the distance between the points A(3,-2) and B(8,10).
SOLUTION
Distance (AB) =
= =
= =
Distance (AB) = 13
EXAMPLE MID-POINT
Find the mid- point of the line joining the points P(4,2) and Q(-5,0).
SOLUTION
Mid –point = ,
= = (
EXAMPLE GRADIENT AND INTERCEPT FORM
Determine the gradient and intercept from the equation 2x + 3y = 5.
SOLUTION
The equation of a straight line Y = mx + c
i.e. 2x +3y =5 3y = 5 – 2x
y =
Therefore, gradient (M) = , Intercept =
ASSESSMENT: Determine the solution of the following:
- Find the distance between the following pairs of points: (i) x(3,-9) and y(-4,15) (ii) p(-2,-7)and q(8,-9).
- Find the gradient and intercept of the following: (i) 6y-8x+3 = 0 (ii) x + 6y – 10 = 0 (iii) 3x = 15 – 5y
- Find the mid-points of the line joining the points: (i) (-6,-12) and (4,3) (ii) (p,q) and (q,p) (iii) (0,11) and (12,-9).
PARALLELISM OF A STRAIGHT LINE: For any lines to be parallel, the gradient of the two lines must be equal. i.e. .
PERPENDICULARITY OF LINES: For lines to be perpendicular or normal to each other, the product of the two gradients must be equal to -1. i.e. or
vise-visa.
EXAMPLE PARALLELISM
Find the equation of the line which pass through the point ( 1,2) and parallel to the line 4x – y = 2.
SOLUTION
4X –Y = 2 , then 4x -2 = y m = 4
Parallelism
Y – = m(x –
)
y – 2 = 4(x – 1)
Y -2 = 4x – 4 y = 4x – 2 or y – 4x + 2 =0 (the equation).
EXAMPLE PERPENDICULARITY
Find the equation of the line passing through the point (1,1)which is perpendicular to 2x – 3y = 4
SOLUTION
2x – 3y = 4. Then 2x -4 = 3y divide through by 3
= y. therefore, m =
.
For line to be perpendicular m = i.e. m =
M = .
Equation of the line y –
Y – 1 = (x – 1)
y – 1 =
, (cross multiply)
2y -2 = -3x = 3 2y + 3x -5 = 0 ( the equation).
EX AMPLE INTERSECTION OF LINE.
Find the coordinate of the points at which the lines
4y =3x +2 and 8y = 9x – 5 intersect.
SOLUTION
Solving the equation simultaneously,
4y = 3x + 2 ………….. (1) X 3
8y = 9x – 5 …………… (ii) X 1
12Y = 9X + 6
8Y = 9X – 5 (-)
4Y = 11 Y =
Or
Substitute the value of y in equation (1)
4) = 3x + 2
11 = 3x + 2
3x = 9 x = 3
Therefore, the coordinate is ( 3, )
EXAMPLE ANGLE BETWEEN TWO STRAIGHT LINE.
Find the acute angle between the lines 3x +2y =1 and 4x -2y + 6 = 0.
SOLUTION
3X + 2Y = 1 2Y = 1 – 3X
Y = therefore,
=
4x – 2y + 6 = o 4x + 6 = 2y
Y = = 2x + 3 therefore,
= 2
=
=
=
2
=
=
=
ASSESSMENT: Determine the solution set of the following:
- Find the equation of the which passes through the point (5,7) parallel to the line 7x +5y = 15.
- Find the equation of the straight line passing through the point (-2, 1) and perpendicular to the line 4 = 2x -7y.
- Find the angle between the two intersecting lines 3x +2y – 10 = 0 and x +6y – 10 = 0
- Find the coordinate of the intersecting lines 5x +12y + 13 =0 and 6y – 8x + 3 = 0.
- Find the acute angles between the following intersecting lines: (i) y =3x + 4 and y = 2x – 1 (ii) 3y = x + 4 and y = -3x – 4 (iii) 4y +3x = 2 and 2y –x = -3.
- Find the equation of the lines which passes through the following pairs of points: (i) (2,4) and (5,6) (ii) (
, 0) and (-1,2) (iii) (1,-1) and (-2,3).
Subtract the original equation from (i)
= (
+ 2x
+ x -1 – (
+ x -1)
= (
+ 2x
+
(divide through by
=
+ 2x + 1 i.e.
= 0 +2x + 1 = 2x + 1
STANDARD DERIVATIVE
This is the general rule for derivative function. i.e.
= na
, where n is the power of x, a is the coefficient of x.
NOTE: when differentiating any function the constant value becomes zero since the derivative is with respect to x.
EXAMPLE
Differentiate 3 – 7
SOLUTION
= (3X4
– (7×3)
+ (1×2)
+ (7×1)
– 0
= 12
– 21
+ 2x +7
ASSESSMENT: work the following question:
- Using first principle, find the derivative of the following: (i) y = x +25 (ii) y = 5
– 7 (iii) y =
– 3x + 65.
- Differentiate each of the following with respect to x: (i) 6
+ 4
-7
+ 3x -8. (ii) (3x + 2)(6x +5) (iii)
.
- From first principle, differentiate 5x –
- Differentiate with respect to x,
.
Let v = x + 5 and = 1
= (
-6)(1) + (x + 5)(4x)
=
– 6 +
+ 20x =
+ 20x – 6.
3. QUOTIENT RULE ( =
)
Find the derivative of y =
SOLUTION
Let u = 5x + 2 and = 5, v = 3x – 4 and
= 3
=
=
.
4. MAXIMA AND MINIMA RULES
The function can only be maxima if the following is satisfied: (i) f`(a) = 0 (ii) f`(a + h) < 0 (iii) f`(a-h) > 0
MINIMA if: (i) f`(b) = 0 (ii) f`(b+h) < 0 (iii) f`(b-h) > 0
EXAMPLE
Find the highest product of two numbers whose sum is 10.
SOLUTION
Let one of the number be x and the other be 10 – x
P(product) = x( 10 – x) = 10x –
= 10 – 2x
f`(a) = 0, 10 – 2x = 0, X = 5
Since = -2 which is < 0,
= -2(10 –(-2)) = -24
= x(10 – x) = 5 (10 – 5) = 25.
5. VELOCITY AND ACCELERATION
Acceleration is the rate of change of velocity with respect to time.
Velocity = i.e. the rate of change of displacement with respect to time.
Acceleration = (
=
.
EXAMPLE
The equation of motion of a particle along a straight line is specified by the equation x = 5 – 3
+ 6t. find the velocity and acceleration of the particle after 2 seconds if x is in meters.
SOLUTION
Velocity = =
+ 6 (after 2 seconds)
Velocity = 30(2– 12(2
+ 6 = 870
.
Acceleration = =
– 36
(after 2s)
Acceleration = 150(2– 36(2
= 2256
.
6. RATE OF CHANGE OF FUNCTION
Given that y = f(x), f`(x) can be interpreted as the rate of change of y with respect to x. if y increases when x increases the rate of change with respect to x is positive but if y decreases when x increases then the rate of change is nagetive.
EXAMPLE
The side of a square is increasing at the rate of 1cms. Find the rate of change of its area when the side is 10cm long.
SOLUTION
Let the area of the square = . i.e. A =
= 2x = 2×10 = 20cm
The rate of change of the side of the square = 1
=
= 20cm x1cms
= 20
.
ASSESSMENT: Find the solution of the following:
- A body moves along a straight line so that its distance (S) meters after t seconds is given by s =
-2
+ 3t + 4. Find its acceleration when t = 2seconds.
- Find the maximum and minimum points on the curve y =
- The displacement xm, of a particle from a fixed point after time ts is given by x =
.
- The times when the particle is momentarily at rest.
- In
, the acceleration of the particle at t = 4.
- The side of a square changes from 10cm to 10.01cm, find approximate increase in the area of the square.
WEEK 6
INTEGRATION OF ALGEBRAIC FUNCTION
INTEGRATION: Is defined as the anti-differentiation or is the process from the derivative form to the anti-derivative form.
Y = + c, where a is the coefficient of x. n is the power of x and c is the constant value through differentiation.
NOTE: when integrating = ln x + c.
EXAMPLE
= 3
– x + 5, find y?
SOLUTION
Y = + c.
y =
+ c
Y =
7+
+ c =
+ 5x + c.
STANDARD INTEGRAL FORMS
= G(x) + c
EXAMPLE
SOLUTION
=
+ C
= + C
+ C
INTEGRATION BY ALGEBRAIC SUBSTITUTION
EXAMPLE
SOLUTION
Let u = ,
=
=
.
=
=
+ c =
+ c.
AREA UNDER THE CURVE
The integral is called the definite integral of the function f(x) with a and b the lower and upper limits of the integral.
EXAMPLE
Evaluate.
SOLUTION
=[
+ c
=
=
= 913.5
ASSESSMENT: Integrate the following:
)
- Using substitution method, evaluate
.
WEEK 7 – 12
MOCK AND EXAMINATION
Mathematics Notes SSS3 – Edudelight.com
