Question
3.
(a)
Find the positive square
root of a given positive number
d where d = 25 using
(i)
Method of successive
Bisection with beginning number
x1 = 6
(ii)
Newton - Raphson Method.
(i)
let
x=square root of 25
so that x2-25=0 ....(1)
taking f(x)=x2-25
,
substituting
the value of x=1,2,3,4,5,6
we
get
f(1)=-24
, f(2) = -21 , f(3) = -16 ,
f(4) = -9 , f(5) =0 , f(6) =
11
since
f(4) = -ve and f(6) = +ve ,
a root of (1) lies between 4
and 6 therefore , first approximation
to the root is
x0
= ½ (4+6)=5
then f(x0)=25-25=0
i.e. +ve
therefore , a root of (1) lies
between x0 and 6
also
x1=6 (given )
therefore according to bisection
method ,
x2=(x0
+ x1 )/2
=5.5 and f(x2)=+ve
therefore , a root of (1) lies
between x0 and x2
x3=(x0
+ x2 )/2
=5.25 and f(x3)=+ve
therefore , a root of (1) lies
between x0 and x3
x4=(x0
+ x3 )/2
=5.125 and f(x4)=+ve
therefore , a root of (1) lies
between x0 and x4
x5=(x0
+ x4 )/2
=5.0625 and f(x5)=+ve
therefore , a root of (1) lies
between x0 and x5
x6=(x0
+ x5 )/2
=5.03125 and f(x6)=+ve
therefore , a root of (1) lies
between x0 and x6
note
:- like calculate x7
to x25 , you will
find that value of x24
and x25 is going
to be same equal to 5.0000001
. as x24 and x25
are same , we take square root
of 25 = 5.0000001
(ii)
let
x=square root of 25
so that x2-25=0 ....(1)
taking f(x)=x2-25
, newton raphson method gives
xn+1
= xn - [ f(xn)
/f '(xn) ]
=xn - [ x2n
- 25 /2xn ]
= ½
[ xn + (25
/ xn ) ] .....(2)
now since f(4) = -9 , f(6) =
11 ,
a root of (1) lies between 4
and 6
therefore , taking x0
= 4.5 , equ (2) gives
x1= ½ [ x0
+ (25 /x0 ) ]
=
½ [ 4.5 +(25/4.5)]
=
5.0277778
x2= ½ [ x1
+ (25 /x1 ) ]
=
½ [5.0277778
+(25/5.0277778 )]
=
5.0000767
x3= ½ [ x2
+ (25 /x2 ) ]
=
½ [5.0000767
+(25/5.0000767 )]
=
5
x4=½ [ x3
+ (25 /x3 ) ]
=
½ [5 +(25/5 )]
=
5
since x3=x4
we take square root of 25 =
5
Que.
3
(c)
A
rocket is launched from the
ground.
Its acceleration measured
every 5 seconds is tabulated
below.
Find the velocity and
position of rocket at t = 40
seconds.
Use Trapezoidal as well
as Simpson's rules.
Compare the answers.
t |
0 |
5 |
10 |
15 |
20 |
25 |
30 |
35 |
40 |
a(t) |
40.0 |
45.25 |
48.50 |
51.25 |
54.35 |
59.48 |
61.5 |
64.3 |
68.7 |
Trapezoidal
rule
:-
If
a be the acceleration in time
t sec at s position , then velocity
V,
V=
h/2 [ (a0+a8)
+ 2(a1+a2+a3+a4+a5+a6+a7)]
Hence the required velocity
=5/2 [(40.0 +68.7) + 2 (45.25+
48.50 +51.25+
54.35+59.48+61.5+
64.3)]
=5/2 [108.7 + 2(384.63)]
=5/2 [877.96]
=2194.9 cm/sec = 21.949 m/sec
and the position of the rocket
at t=40 having the velocity
21.949 m/sec
=21.949 * 40
=877.96 m
Simpson's
rule :-
v = h/3 [ X + 4.O
+2.E ]
Here h=5 , a0=40.0
,a1=45.25 , a2=48.50
, a3=51.25 , a4=54.35
, a5=59.48 , a6=
61.5 ,
a7= 64.3 , a8=
68.7
X=a0+a8 =
40.0 +68.7 =108.7
O=a1+a3+a5+a7
= 45.25+51.25+59.48+64.3=220.28
E=a2+a4+a6
= 48.50 + 54.35 +61.5=164.35
Hence the required velocity
= 5/3 [ (108.7) +4(220.28) +2
(164.35)]
= 5/3 [108.7+881.12+328.7]
=5/3 * 1318.52 = 2197.53 cm/sec
= 21.9753 m/sec
and the position of the rocket
at t=40 having the velocity
21.9753 m/sec
=21.9753 * 40
=879.012 m