Question

is O1. The olume I of liquid in a hollow horizontal cylinder of radius r and length related to the depthh h of the liquid by r-h + (h-r)V2rh-h1 arccos V= With the asd of Newron-Raphson's method, determine the depthh of the liquid for which the 2m and /=5m, Take au initial guess of 1.5m and stop the iterations volume is 8m if r when tho rclstive error between two successive iterations is at most U.0001. 110 points

Answer 1

Matlab code for Newton Method clear all close all syms h function for which root have to find 1-5 fun(h)=(r^2* acos ( (r-h) /r ) + (h-r) *sqrt ( 2 *r*h-h^2 ) ) *1-8; gl (h) -diff ( fun, h) ; 81st Derivative of this function % initial guess] xx-1.5 %function for Newton method fprintf(Function for Newton method f(h)-\n') disp (vpa (fun, 3)) fprintf nDerivative of the function disp (vpa (gl,3) ) is df/dh=\n') fprintf\nUsing Newton iterations \n) %Loop for all intial guesses n 0.0001 Serror limit for close itteration for i 1:500 maximum iterations x2 double (xx-(fun (xx). /g1(xx)) ) ; SNewton Raphson Formula cc abs (x2-xx); Relative Error erri) cc xx-x2 fprintf\nAfter %d iteration\n', i) fprintfValue of h=%f \n ',xx) fprintfFunction value is %f\n',fun (xx) ) fprintfDerivative of the function %f\n\n',g1 (xx)) if cc<n break end end root-xx; fprintf n\tDepth h of the liquid is Sf m\n',root) 88888%88888888 8 8888 888$88$388 Function for Newton method f(h) = 20.0 acos (1.0- 0.5*h) 5.0 (h - 2.0) *(- 1.0*h^2 4.0*h) (1/2) - 8.0 symbolic function inputs: h Derivative of the function is df/dh 5.0 1.0*h ^2 + 4.0*h)^(1/2) + 10.0/(1.0 - 1.0* (0.5*h - 1.0)^2)^(1/2) - (2.5* (h - 2.0)* (2.0 *h 4.0))/(4.0*h - 1.0*h^2) ^(1/2) 1

symbolic function inpu ts: h iterations Using Newton After 1 iteration Value of h=0.801774 Function value is 0.974297 Derivative of the function 16.013288 After 2 iteration Value of h-0.740931 Function value is 0.01 4224 Derivative of the function 15.5394 49 After 3 iteration Value of h 0.740015 Function value is 0.000003 Derivative of the function 15.532028 After 4 iteration Value of h=0.740015 Function value is 0.000000 Derivative of the function 15.532026 Depth hof the liquid is 0.740015 m Published with MATLAB® R2018a 2

%Matlab code for Newton Method
clear all
close all

syms h
%function for which root have to find
r=2; l=5;
fun(h)=(r^2*acos((r-h)/r)+(h-r)*sqrt(2*r*h-h^2))*l-8;
g1(h) =diff(fun,h);   %1st Derivative of this function
xx=1.5;            %initial guess]
%function for Newton method
fprintf('Function for Newton method f(h)=\n')
disp(vpa(fun,3))

fprintf('\nDerivative of the function is df/dh=\n')
disp(vpa(g1,3))

fprintf('\nUsing Newton iterations\n')

%Loop for all intial guesses
    n=0.0001; %error limit for close itteration
    for i=1:500 %maximum iterations
      
        x2=double(xx-(fun(xx)./g1(xx))); %Newton Raphson Formula
        cc=abs(x2-xx);                   %Relative Error
        err(i)=cc;
        xx=x2;
      
        fprintf('\nAfter %d iteration\n',i)
        fprintf('Value of h=%f\n',xx)
        fprintf('Function value is %f\n',fun(xx))
        fprintf('Derivative of the function %f\n\n',g1(xx))
      
        if cc<=n
            break
        end
      
      
    end
    root=xx;

fprintf('\n\tDepth h of the liquid is %f m\n',root)

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