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. 

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