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
x_n+1      = x_n - [ f(x_n) /f '(x_n) ]
            =x_n - [ x²_n - 25 /2x_n ]
            = ½  [ x_n + (25 / x_n ) ] .....(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