Week 9 – SS2 Third Term Further Mathematics Notes

WEEK NINE
TOPIC : APPLICATION OF INTEGRATION II : SOLID REVOLUTION AND TRAPEZOIDAL RULE
A solid whichb has a central axis of symmetry is a solid of revolution.Forexample, a cone, a cylinder , a vase etc.

 
y

 
 
 
 
 
 
 
 
 
 
 
 Consider the area under a portion AB of the curve y = f(x) revolved about the x axis through four right angle or 3600, each point of the curve describes a circle centered on the x axis. A solid revolution can be thought of as created in this way with the circular plane ends cutting the x – axis at x = a and x = b.

 Let v be the volume of the solid for x = a up to an arbitrary value of x between a and b. Given abincreament dx in x , and y takes an increamentdy and v increases by dv.

 
 
 
 The figure shows a section through the x axis, from this it is seen that the slice dv of bthickness dx is enclosed between two cylinders of outer radius y +dy and inner radius y .
Then ,πy2dx< dv < π (y + dy)2dx;
With appropriate modification, if the curve is falling at this point.
π y2 < dv/dx < π (y + dy)2
if dx 0, dy 0 as dv/dx dv/dx
: . dv/dx = πy2 or V =

 
 
 
 
 
 Where y = f(x) and v = volume of solid revolution of the curve where y = f(x) is rotation completely and x – axis between limits x = a and x =b.

 Examples;The portion ofthe curve y = x² between x = 0 and x = 2 is rotated complrtely around the x axis, find the volume of the solid generated?

V =
=
V =
V = =
Put x = 2 and x = 0 then substitute into the expression above
V =

 THE TRAPEZOIDAL RULE
There are many definite integrals which can’t be evaluated and thus required advance techniques e.g
etc
We can find an approximate value for such integralsbyb finding the area approximately. There are many methods methods of doing this and such methods include the Trapezium rule.
Y

 
 
 
  Y = f(x)

  y₁ y₂y₃ y_n-1y_n
x₁ x₂ x₃ x_n-1x_n

 
 
  = ½ (y₁ +y₂)h + ½ (y₂ +y₃)h + ½ (y_n-1 +y_n)h
½ h(y1+2y2+2y3+……….+2yn-1 + yn)
½ (width of each trap. ) × (first ordinate + last ordinate )+ 2( sum of all other ord.)

 F(x)dx = ½ h{y1 + yn} +2{y2 +y3 + …Yn}.

 Example
Find the approximate value of at interval 0.5

Applying the rule;
{ ½ . ½ { (1 +0.33) + 2 ( 0.67 + 0.5 + 0.4)}
¼ {(1.33) +2(1.57)}
¼ (4.47) = 1.12.

 
 
 
 
  1 0.67 0.5 0.4 0.33

 Ex (2). Make a table of value of y for which y = for which x =2 t0 x = 3 at interval of 0.2.

 Using the rule.
= ½ *0.2 (0.5774 +0.3536 + 3.5354)
0.44664 *2
0.89 correct to 2 dp

 APPLICATION OF INTEGRATION TO KINEMATICS
If the ve;locityis given as a function of time, the displacement is the integral of the velocity function with respect to the time .
ds/dt = f(t)
then S =
= f(t) + C
Similarly, if the acceleration is a function of time, the velocity is the integral of the acceleration function.
Ex. A particle is projected in a straight line from O until a speed of 6m/s is attained. At time t secs.Later,its acceleration is (1 + 2t) m/s² for the value of t = 4. Calculate for the particle (i) its velocity (ii) its distance from O

 dv/dt = 1+ 2t
v = = t + t² + C
when t = 0
v = 6m/sand c = 6.
V = (t² + t + 6) m/s
When t = 4, v = 16 + 4 + 6 = 26m/s.
(ii) distance (s) = ds/dt = t2 + t + 6
S =
S = {t3/3 + 16/2 6t}⁴
= 160/3
m.

 Evaluation
1. Find the area enclosed byb y = x2 – x -2 and the x axis
2. find the area under the curve y = x2/3 between x = 2 and x = k is 8 times the area under the same curve between x = 1 and x =2, hence find the value of k.

 GENERAL EVALUATION

1. A particle moves in astraight line from O until the initial velocity was 2m/s. its acceleration is given by (2t -3)m/s2. Calc. (i) its velocity after 3 secs. (ii) the distance from O when it is momentarily at rest.
2. Find the volume of solid revolution when a is the region bounded by the cuerve y = 2x. and the ordinate at x = 2,and x = 4 and the x axis is revolved by 2π.

    Reading Assignment :F/Matrhs Project, pg 47 – 63

    WEEKEND ASSIGNMENT
   1. Integrate 2√x A. B.4x^3/2 + C C. + C D.
   2. Integrate A. + C B. x^3/2 + C C. x² + C D. + C
   3. The gradient of a curve is 6x + 2 and it passes through the point (1,3), find its equation A. 3x² – 2x + 2 B. 3x² -2x -2 + 2x + 2 C. 3x² – 2x + 2 D. 3x² – 2x -2
   4. Evaluate A. – + x2 – 2x + C B. x3/3 – x3/2 + x2 + 2x +C C. x3/3 + x2/2 +x3-2x + C D. x4/4+x3/3-x2+2x+C
   5. Eval. A. 9 +C B. 8/3 + C C. 24 D. 18 + C
   THEORY
   

3. Evaluate
4. Using trapezoidal rule, with ordinate x = -3,-2,-1,0,1,2,3 and 4. Calc correct to 3 dp an approximate value of

    
    WEEK TEN
   TOPIC: 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.
   

    
    
    X

    
    
    
    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.
   

    
    
    x

    
    
    
    Zero Correlation:
   There is no apparent association between x and y.
   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.

    
    
    x

    
    
    
    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
   n

    y = ∑ y
   n
   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:

5. The value of r is the same irrespective of the variable labelled x or y.
6. the value of r satisfies the inequality -1< x < + 1
7. 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.

8. Pearson’s coefficient of correlation or product moment correlation coefficient
9. 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

    GENERAL EVALUATION/REVISIONAL QUESTIONS
   

10. If Cos A = 24/25 and Sin B= 3/5, where A is acute and B is obtuse, find without using tables, the values of (a) Sin 2A (b) Cos 2B (c) Sin (A-B)
11. Use the addition formula to find the values of the following

    (a)Sin 75⁰ (b) cos 75⁰ (c) tan 45⁰
    

12. Calculate the Product moment correlation coefficient and the Spearman’s rank correlation coefficient.

    X	50	45	43	30	30	43	23	43	25
    Y	12	13.5	14	11	12	15	13.5	12	14

     READING ASSIGNMENT: Read correlation and regression.Page313–320. Further Mathematics project 2.
    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

     

13. 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.

     
     

    Rx	1	2	3	4	5	6	7	8	9	10
    Ry	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

     

14. Plot a scatter diagram for the distribution
15. Draw an eye- fitted line of best fit
16. 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

     

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