Second Term Lesson Note Mathematics SS 3

SECOND TERM E-LEARNING NOTE

SUBJECT: MATHEMATICS CLASS: SS 3

SCHEME OF WORK

WEEK TOPIC

• Calculation on interest on bonds and debentures using logarithm table and problems on taxes and value added tax.

• Coordinate Geometry of straight line: Cartesian coordinate graphs, distance between two points, midpoint of the line joining two points.

• Coordinate Geometry of straight lines: Gradient and Intercepts of a line, angle between two intersecting straight lines and application.

• Differentiation of algebraic functions: meaning of differentiation, differentiation from first principle and standard derivatives of some basic functions.

• Differentiation of algebraic functions: Basic rules of differentiation such as sum and difference, product rule, quotient rule and maximal and minimum application.

• Integration and evaluation of simple algebraic functions: Definition, method of integration: substitution, partial fraction and integration by parts, area under the curve and use of Simpson’s rule.

• – 12.         Revision and mock examination.

REFERENCE TEXT

• New General Mathematics for SS book 3 by J.B Channon

• Essential Mathematics for SS book 3

• Mathematics Exam Focus

• Waec and Jamb past Questions

WEEK ONE

• Calculation on interest on bonds and debentures using logarithm table 
• Problems on taxes and value added tax.

WEEK TWO

• Coordinate Geometry of straight line: Cartesian coordinate graphs
• distance between two points
• midpoint of the line joining two points
• Coordinate Geometry of Straight line:
• Cartesian coordinate graph:

Distance between two lines:

In the figure below, the coordinates of the points A and B are (x1, y1) and (x2, y2), respectively. Let the length of AB be l.

y

                                                                                B(x2, y2)

l

                                                                                              y2 – y1

                                                A(x1, y1)      x2 – x2          C

                                                                                                       X

Using Pythagoras theorem:

AB2 = AC2 + BC2

 l2 =(x2 – x1)2 + (y2 – y1)2

l  = √(x2 – x1)2 + (y2 – y1)2

Example:

Find the distance between the each pair of points: a. (3, 4) and (1, 2)   b. (3, – 3) and (- 2, 5)

Solution:

Using l =√(x2 – x1)2 + (y2 – y1)2

1. l = √(3 – 1)2 + (4 – 2)2

l = √22 + 22

l = √8 = 2√2 units

1. l = √(3 – (-2)2 + (-3 – 5)2

=  √52 + (-8)2

=  √25 + 64 = √89 = 9.43 units

Evaluation: Find the distance between the points in each of the following pairs leaving your answers in surd form:  1. (-2, – 5) and (3, – 6)     2. (- 3, 4) and (- 1, 2)

Mid-point of a line:

The mid-point of the line joining two points: 

y

                                                                                                         B(x2, y2)

                                                                                                              y2 – y

                                                                       M(x, y)                         D   

x2 –x     

                                                                                     y – y1

                                                A(x1, y1)   x – x1      N                      C

                                                                                                                              X

 Triangle MAN and BMD are congruent, so AM = MD and BD = MN

                      x – x1 = x2 – x                                          y – y1 = y2 – y

                        x + x = x2 + x1                                          y + y = y2 + y1

                            2x = x2 + x1                                          2y      = y2 + y1

x= x2 + x1                                                  y = y2 + y1

                                          2                                                                    2

Hence, the mid-point of a straight line joining two is    x2 + x1  ,y2 + y1

                                                                                                               2                 2

Example: Find the coordinates of the mid-point of the line joining the following pairs of points.

1. (3, 4) and (1, 2)         b. (2, 5) and ( – 3, 6)

Solution:

    Mid-point =         x2 + x1  ,y2 + y1

                                      2                2

1. Mid-point =  1 + 3  ,     4 + 2       =  (2, 3)

                            2               2

1. Mid-point =  – 3 + 2  ,   6+ 5       =   – 1  ,   11

                                                         2             2                  2       2

Evaluation: Find the coordinates of the mid-point of the line joining the following pairs of points.

1. (- 2 , – 5) and (3, – 6)         b. (3, 4) and ( – 1, – 2)

General Evaluation

1. Find the distance between the points in each of the following pairs leaving your answers     in surd form:  1. (7, 2) and (1, 6) 
2. What is the value of r if the distance between the points (4, 2) and (1, r) is 3 units?
3. Find the coordinates of the mid-point (-3, -2) and (-7, – 4)

Reading Assignment: NGM for SS 3 Chapter 9 page 77 – 78, 

Weekend Assignment:

1. Find the value of α2 + β2 if α + β = 2 and the distance between the points (1, α) and (β, 1) is 3 units.
2. The vertices of the triangle ABC are A (7, 7), B (- 4, 3) and C (2, – 5). Calculate the length of the longest side of triangle ABC.
3. Using the information in ‘2’ above, calculate the line AM, where M is the mid-point of the side opposite A.

WEEK THREE

• Coordinate Geometry of straight lines: 
• Gradient and Intercepts of a line
• Angle between two intersecting straight lines and application

Gradient and Intercepts of a line

Gradient of a line of the form y = mx + c, is the coefficient of x, which is represented by m and c is the intercept on the y axis.

Example

1. Find the equation of the line with gradient 4 and y-intercept -7.

Solution

m = 4, c = – 7,

Hence, the equation is; y =4x – 7.

Evaluation: 

1.What is the gradient and y intercept of the line equation 3x -5y +10=0 ?

2. Find the equation of the line with gradient – 9 and y-intercept 4.

Gradient and One Point Form

The equation of the line can be calculated given one point (x, y) and gradient (m) by using the formula; y – y1= m(x – x1)

Example

Find the equation of the line with gradient -8 and point(3, 7).

Solution

m = – 8, (x1, y1) =(3,7)

Equation: y – 7 = – 8(x – 3)

                 y = -8x + 24 +7

                 y = -8x + 31

Evaluation: 

1. Find the equation of the line with gradient 5 and point(-2, -7).
2. Find the equation of the line with gradient -12and point (3, -5).

Two Point Form:

Given two points (x1, y1) and (x2, y2), the equation can be obtained using the formula:

 y2 – y1 = y – y1

x2 – x1      x – x1

Example: Find the equation of the line passing through (2,-5) and (3,6).

Solution

6 – (-5)/3 – 2 = y – (-5)/x – 2

11 = y + 5/x – 2

11(x – 2) = y + 5

11x – 22 = y + 5

y – 11x + 27 = 0

Evaluation: 

1.Find the equation of the line passing through (3, 4) and (-1, -2).

2.Find the equation of the line passing through (-8, 5) and (-6, 2).

Angles between Lines

Parallel lines:

The angle between parallel lines is 00 because they have the same gradient

Perpendicular Lines:

Angle between two perpendicular lines is 900 and the product of their gradients is – 1. Hence, m1m2 = – 1

Examples: 

1. Show that the lines y = -3x + 2 and y + 3x = 7 are parallel.

solution:

         Equation 1: y = -3x + 2,   m1 = -3

         Equation 2:  y + 3x = 7,   

                                 y = -3x + 7, m2 = – 3

since; m1 = m2 = – 3, then the lines are parallel

1. Given the line equations x = 3y + 5 and y + 3x = 2, show that the lines are perpendicular.

solutions:

     Equation 1:     x = 3y + 5,   make y the subject of the equation.

                              3y = x + 5

                                y = x/3 + 5/3

                            m1 = 1/3 

  Equation 2:    y + 3x = 2,

                              y = – 3x + 2,   m2 = -3 

hence: m1 x m2 = 1/3 x – 3 = – 1 

since: m1m2 = – 1, then the lines are perpendicular.

Evaluation: State which of the following pairs of lines are: (i) perpendicular   (ii) parallel

             (1)    y = x + 5 and y = – x + 5      (2). 2y – 6 = 5x and 3 – 5y = 2x    (3) y = 2x – 1 and 2y – 4x = 8 

Angles between Intersecting Lines:

y

                                  y = mx + c               

                    θ                                                x  

                        0   

The gradient of y = mx + c is tan θ.    Hence m = tan θcan be used to calculate angles between two intersecting lines. Generally the angle between two lines can be obtained using: tan 0 = m2 -m1

                                                                                                                                                                 1 + m1m2

Example: Calculate the acute angle between the lines y=4x -7 and y = x/2 + 0.5.

Solution: 

Y=4x -7, m1= 4, y=x/2+0.2, m2 =1/2.

 Tan O= 0.5 – 4.       = -3.5/3

                    1 + (0.5*4)

Tan O =- 1.1667 

O=tan-1(-1.1667) = 49.4

Evaluation:Calculate the acute angle between the lines y=3x -4 and x – 4y +8 = 0.

General Evaluation:

1.Calculate the acute angle between the lines y=2x -1 and  2y + x = 2.

2.If the lines 3y=4x -1 and qy= x + 3 are parallel to each other, find the value of q.

3.Find the equation of the line passing through (2,-1) and gradient 3.

Reading Assignment: NGM for SS 3 Chapter 9 page 75 – 81 

Weekend Assignment

1.Find the equation of the line passing through (5,0) and gradient 3.

2.Find the equation of the line passing through (2,-1) and (1, -2).

3. Two lines y=3x – 4 and x – 4y + 8=0 are drawn on the same axes.

(a) Find the gradients and intercepts on the axes of each line.

(b) Find the equation parralel to x -4y + 8=0 at the point (3, -5)

WEEK FOUR

• Differentiation of algebraic functions: meaning of differentiation
• Differentiation from first principle 
• Standard derivatives of some basic functions.

Consider the curve whose equation is given by   y = f(x)  Recall that m = y2 – y1= f(x+x)-f(x) 

x2– x1x      

As point B moves close to A, dx becomes smaller and tends to zero.

The limiting value is written on  Lim   f(x +x) – f(x) and is denoted by as x –> 0

dx

fl(x) is called the derivative of f(x) and the gradient function of the curve

The process of finding the derivative of f(x) is called differentiation. The rotations which are commonly used for the derivative of a function are f1(x) read as f – prime of x,  df/dx read as  dee x of f

df/dx    read  dee – f  dee- x,    dy/dx read  dee – y  dee- x

If   y = f(x) , this dy/dx = f1(x) (it is called the differential coefficient of y with respect to x.

Differentiation from first principle: The process of finding the derivative of a function from the consideration of the limiting value is called differentiation from first principle.

Example 1 

Find from first principle, the derivative of   y = x2

Solution

y = x2

      y + y = (x + x)2

y + y = x2 + 2xx + (x)2

y =  x2 + 2xx+ (x)2  –  y

y = x2  +  2xx   +  (x)2  –  x2

y  =  2xx  +  (x)2

y  = (2x   +  x)x

y =  2x   +   x

x

Lim  x  =  0

dy =  2x

dx

Example 2:

Find from first principle, the derivative of 1/x

Solution

Let y    =    1

x

y + y =      1

               x + x

y  =       1           –  y

                x + x

y =     1       –  1

            x + x       x

y  =  x – (x +  x)

               (x  +x)x

y =  x  –  x  –  x

x2  +  xx

dy  =    -x

x2+ x

y  =   -1                                                                                                                                                  

x     x2 + x   

Lim  x = 0

dy =    -1

dx         x2

Evaluation: Find from first principle, the derivatives of y with respect to x:

1. Y = 3x3                      2. Y = 7x2     3. Y = 3x2 – 5x

Rules of Differentiation:     Let y = xn

y + dy = (x + dx)n

= xn + nxn-1dx + n(n -1) xn-2(dx)2 + … (dx)n

    2!

                            = xn + n xn-1dx + n(n-1) xn-2  (dx)2+ — + (dx)n – xn

        2!

                            = nxn-1dx + n (n – 1) xn–1 (dx)2

2! 

dy/dx = n xn-1 + n (n –1) xn-1 dx

Lim dy/dx = nxn-1

dx = 0

Hence;   dy/dx  =  nxn-1      if y = xn

Example 3:

Find the derivative of the following with respect to x:   (a) x7 (b) x½ (c) 5x2 – 3x (d) – 3x2 (e) y = 2x3 – 3x + 8

Solution

1. Let  y = x7

dy/dx = 7 x7-1 = 7x6

1. Let  y = x ½

dy/dx = ½ x½ -1 = ½ x– ½  =   1

                                                                  2√x

1.           Let y = 5x2 – 3x 

dy/dx = 10x – 3 

1. Let y = – 3x2

dy/dx =2× – 3x2-1 = – 6x

1.         Let y = 2x3 – 3x + 8

dy/dx= 3 x 2x3-1 – 3 + 0

= 6x2 – 3 

Evaluation:

• If  y=5x4 ,find  dy/dx        2.Given that y= 4x-1 find y1

General Evaluation

1. Find, from first principles, the derivative of  4x2 – 2  with  respect  to  x.
2. Find the derivative of the following       a.3x3 – 7x2 – 9x + 4   b. 2x3   c. 3/x
3. Using idea of difference of two square; simplify 243x2 – 48y2
4. Expand (2x -5)( 3x-4)
5. If the gradient of y=2x2-5 is -12 find the value of y.

Reading Assignment: NGM for SS 3 Chapter 10 page 82 -88, 

Weekend Assignment

Objective

1. Find the derivative of 5x3(a) 10x2     (b) 15x2     (c) 10x     (d) 15x3
2. Find dy/dx, if y = 1/x3(a) –3/x4 (b) 3/x4 (c) 4/x3 (e) –4/x3
3. Find f1(x), if f(x) = x3 (a) 3x (b) 3x2 (c) ½ x3 (d) 2x3
4. Find the derivative of   1/x(a) 1/x2   (b) –1/x2   (c) – x (d) –x2
5. If      y = – 2/3 x3. Find dy/dx (a) 4/3 x2      (b) 2x2   (c) – 2x2        (d) –2x

Theory

1. Find from first principle, the derivative of   y = x + 1/x
2. Find the derivative of 2x2 – 2/x3

WEEK FIVE

• Differentiation of algebraic functions: 
• Basic rules of differentiation such as sum and difference, product rule, quotient rule
• Maximal and minimum application.

Derivative of algebraic functions

Let f, u, v be functions such that 

f(x) = u(x) + v(x)

f(x +x ) = u(x +x ) + v(x + x)

f(x + x) – f(x) = {u(x+ x) + v(x+ x) – v(x + x) – u(x) – v(x)}

= u (x +x ) – u(x) + v(x +x ) – v(x)

f(x + x) – f(x) = u(x +x)-u(x) + v(x +x ) – v(x)

Lim  f(x + x) – f(x) = U1(x) + V1(x)     

if y = u + v and u and v are functions of x, then dy/dx = du/dx + dv/dx

Examples:Find the derivative of the following

1) 2x3 – 5 x2 + 2          2)3x2 + 1/x             3)2x3 + 2x2 +1

Solution

1. Let y = 2x3 – 5x2 + 2

dy/dx = 6x2 – 10x

2. Let y = 3x2+ 1/x = 3x2 + x-1

dy/dx = 6x – x-2 = 6x – 1

x2

3.       Let y = 2x3 + 2x2 + 1

dy/dx=6x2  +  4x

Evaluation:   1. If y = 3x4 – 2x3 – 7x + 5. Find dy/dx

2.Findd (8x3 – 5x2 + 6)

Dx

Function of a function (chain rule)

Suppose that we know that y is a function of u and that u is a function of x, how do we find the derivative of y with respect to x?

Given that y = f(x) and u = h(x), what is dy/dx?

dy/dx = ,   this is called the chain rule

Examples

Find the derivative of the following.(a)y = (3x2 – 2)3   (b)  y = (1 – 2x3)  (c) 5/(6-x2)3

Solution

1. y = (3x2 – 2)3

Let u = 3x2 – 2

y = (3x2 – 2)3 => y = u3

y = u3

dy/du = 3u2

du/dx = 6x

dy/dx =  = 3u2 x 6x

= 18xu2 = 18x(3x2 – 2)2

2. y = (1 – 2x3)1/2 => (1 – 2x3) 1/2

Let u =1 – 2x3,     hence  y = u1/2 

dy/dx = = ½ u-1/2  x( –6x2)

= -3x2 u – ½=  -3x2

u1/2

    -3x2 =        -3x2

    √ u   √(1 – 2x3)

3. y =      5 = 5(6 – x2)-3

        (6 – x2)3

Let u = 6 – x2

y = 5(u)–3

dy/du = -15u –4

du/dx = -2x

dy/dx = dy/du X du/dx = -15u-4 x (-2x) = 30x u-4 = 30x (6 – x2)-4

            =          30x_

                     (6 – x2)4

Evaluation:

1. Given that y =       1                find dy/dx

              (2x + 3)4

2. If y = (3x2 + 1)3 , Find dy/dx

Product Rule

We shall consider the derivative of y = uv where u and v are function of x.

Let y = uv

Then y + y = (u +u )(v + v)

= uv + uv + vu +uv

y = uv + uv + vu+ uv – uv

y= uv + vu + uv

y=  uv  +    vu   +    uv

x x

As x =>0 ,u=> 0 , v=> 0

          Lim       y   =     Lim       uv +    Lim        vu   +     Lim        uv

x=>0   x          x=>0    x        x=>0      x           x=>0      x

Hence   dy/dx= U dv + Vdu

dxdx

Examples

Find the derivatives of the following.

(a) y = (3 + 2x) (1 – x)               (b) y = (1 – 2x + 3x2) (4 – 5x2)

Solution

1. y = (3 + 2x) (1 – x)

Let u = 3 + 2x and v = (1 – x)

du/dx = 2 and dv/dx = -1

dv/dx = u dv + vdu

dx dx

=  (1 – x) 2 + (3 + 2x) (-1)     = 2 – 2x – 3 – 2x

dy/dx = – 1 – 4x

2. y = (1 – 2x + 3x2) (4 – 5x2)

Let u = (1 – 2x + 3x2)         and v = (4 – 5x2)

du/dx = -2 + 6x                      and dv/dx = – 10x

dy/dx = udv +   vdu

dxdx

= (1 – 2x + 3x2) (-10x) + (4 – 5x2) (- 2 + 6x)

= – 10x + 20x2 – 30x3 + (- 8 + 10x2 + 24x – 30x3)

= – 10x + 20x2 – 30x3 – 8 + 10x2 + 24x – 30x3

= 14x + 30x2 – 60x3 – 8

Evaluation 

Given that (i) y = (5+3x)(2-x)    (ii)  y = (1+x)(x+2)3/2 ,find dy/dx

Quotient Rule:

If y = u

v

then; dy =  vdu– udv

dxdxdx

                          v2

Examples:

Differentiate the following with respect to x.  (a)x2 + 1     (b)(x – 1)2

                                                                                              1 – x2               √x

Solution:

1. y = x2 + 1

                      1 – x2

             Let u = x2 + 1      du/dx = 2x

                   v= 1 – x2          dv/dx = – 2x

dy =  vdu–  udv

dxdxdx

                v2

dy/dx = (1 – x2)(2x) – (x2 + 1)(-2x)

                                             (1 – x2)2

=  2x – 2x3 + 2x3 + 2x

                 (1 – x2)2

dy/dx        =                        4x

1. – x2)2

1. y = (x – 1)2

          √x

Let u = (x – 1)2        du/dx = 2(x – 1)

        v = √x                    dv/dx = 1/2√x

dy/dx = √x 2(x – 1) -(x – 1)2 1/2√x       

                                  (√x)2

dy/dx = √x 2(x – 1) – (x – 1)2 1/2√x       

x

Evaluation:  Differentiate with respect to x: (1)  (2x + 3)3    (2)         √x

                                                                                (x3– 4)2                 √(x + 1)

Applications of differentiation:

There are many applications of differential calculus.

Examples:

1. Find the gradient of the curve y = x3 – 5x2 + 6x – 3 at the point where x = 3.

      Solution:

 Y = x3 – 5x2 + 6x – 3

dy/dx = 3x2 – 10x + 6

where x = 3; dy/dx = 3(32) – 10(3) + 6

                                      = 27 – 30 + 6

                                      = 3.

1. Find the coordinates of the point on the graph of y = 5x2 + 8x – 1 at which the gradient is – 2 

Solution:

   Y = 5x2 + 8x – 1

dy/dx = 10x + 8

replace; dy/dx by – 2 

                              10x + 8 = – 2 

                               10x = – 2 – 8

                           x = -10/10   = – 1 

1. Find the point at which the tangent to the curve y = x2  – 4x + 1 at the point (2, -3)

Solution:

                 Y =x2  – 4x + 1

dy/dx = 2x – 4 

at point (2, -3): dy/dx = 2(2) – 4 

dy/dx = 0

tangent to the curve:  y – y1 = dy/dx(x – x1)

                                                                  y – (-3) = 0 ( x- 2)

                                                                  y + 3 = 0

Evaluation:  

1. Find the coordinates of the point on the graph of y = x2 + 2x – 10 at which the gradient is 8.
2. Find the point on the curve y = x3 + 3x2 – 9x + 3 at which the gradient is 15.

Velocity and Acceleration

Velocity: The velocity after t seconds is the rate of change of displacement with respect to time.

               Suppose; s = distance and t = time,

Then;  Velocity = ds/dt

Acceleration:  This is the rate of change of velocity compared with time.

Acceleration = dv/dt

 Example:

A moving body goes s metres in t seconds, where s = 4t2 – 3t + 5. Find its velocity after 4 seconds. Show that the acceleration is constant and find its value.

Solution:

                     S = 4t2 – 3t + 5

ds/dt = 8t – 3 

velocity = ds/dt = 8(4) – 3

                                                      = 32 – 3

                                                     = 29

                      Acceleration: dv/dt = 8.

Maxima and Minimal

1. Find the maximum and minimum value of y on the curve 6x – x2.

Solution:

                     y = 6x – x2

dy/dx = 6 – 2x 

equatedy/dx = 0

                     6 – 2x = 0

                            6 = 2x

                            X = 3

The  turning point is (3, 9)

1. Find the maximum and minimum of the function x3 – 12x + 2.

Solution:

                      Y =x3 – 12x + 2

dy/dx = 3x2 – 12 

                           3x2 – 12 = 0

 3x2 = 12 

x2 = 12/3

x2 = 4                  x = ± 2

minimum point occur when d2y/dx2> 0

maximum point occurs when d2y/dx2< 0 

d2y/dx2= 6x 

substitute x = 2;   d2y/dx2 = 6 x 2 = 12

therefore: the function is minimum at point  x = 2 and y = – 14

substitute x = – 2; d2y/dx2 = 6(-2) = -12 

therefore:  the function is maximum at point x = – 2 and y = 18

Evaluation: 

1. A particle moves in such a way that after t seconds it has gone s metres, where s = 5t + 15t2 – t3 
2. Find the maximum and minimum value of y on the curve 4 –12x – 3x2.

General Evaluation

Use product rule to find the derivative of

1.   y = x2 (1 + x)½ 
2.   y = √x (x2 + 3x – 2)2
3.   Find  the  derivative   of   y =(7x2 -5)3
4. Using completing the square method find t if s=ut+1at2

                                                2

5. If 3 is a root of the equation x2 – kx +42=0 find the value of k and the other root of the equation

READING ASSIGNMENT: NGM for SS 3 Chapter 10 page 90 -101, 

WEEKEND ASSIGNMENT

OBJECTIVE

1.Differentiate the function  4x4 + x3 – 5 (a)4x3 +3x2   (b)16x2 +3x2 (c)16x3 +3x2 (d)16x4 +  3x2

2.Find d2y/dx2   of the function y = 3x5wrt x. (a) 15x3 (b) 45 x4 (c) 60x3 (d) 3x5 (e) 12x3

3.If f(x) = 3x2 + 2/x find f1(x)   (a) 6x + 2    (b) 6x + 2/x2    (c) 6x – 2/x2  (d)6x -2

4.Find the derivative of 2x3 – 6x2    (a) 6x2 – 12x    (b) 6x2 – 12x  (c) 2x2 – 6x  (d) 8x2 – 3x

5.Find the derivative of x3 – 7x2 + 15x   (a) x2 – 7x + 15 (b) 3x2 – 14x + 15 (c) 3x2 + 7x + 15 (d) 3x2 – 7x + 15

THEORY

1. Differentiate with respect to x. y2 + x2 – 3xy = 4
2. Find the derivative of 3x3(x2 + 4)2