Question

)10 12 32-5272-92130 180 230280330 temperature, T (°C) 15.3 15.2915.28 15.27 15.2615.22 15.1414.19 13.3112.9 knots index, i depth, z (m) 3 6i Table 1: Temperature of seawater in the upper layer of the ocean. Problem 1 (hand-calculation): Consider the seawater temperature data in table 1 to estimate the temperature at depth z --150 m. Use natural cubic splines to find the piecewise cubic polynomials that pass through the knots 5 through 8. You can use MATLAB to solve the linear system of equations for the coefficients of the polynomials / splines as discussed in class.

Answer 1

Here is Code:

function y=natural_cubic_spline_interp(x1,y1,x)
%x1=independent variable;x2=dependent variable;x=value at which we have to
%find the dependent variable;y=corresponding value of dependent variable at x;
y = spline(x1,y1,x);
end

clear all
close all
z=[-12 -32 -52 -72 -92 -130 -180 -230 -280 -330];
T=[15.31 15.31 15.32 15.32 15.32 15.22 15.14 14.19 13.31 12.9];

depth=[-330:2:-10];
for i=1:length(depth)
temp_LPI(i)=lagrange_interp(z,T,depth(i));
end
hold on
plot(z,T,'*')
plot(depth,temp_LPI,'b')

for i=1:length(depth)
temp_NDD(i)=newtondivdiff_interp(z,T,depth(i));
end
plot(depth,temp_NDD,'--g')

for i=1:length(depth)
temp_nCS(i)=natural_cubic_spline_interp(z,T,depth(i));
end
plot(depth,temp_nCS,'k')
title('Depth Vs. Temperature plot')
xlabel('Depth in meter')
ylabel('Temperature in degree c')
legend('Nodes','Lagrange','Newton','Cubic Spline','Location','best')
temp_LP=lagrange_interp(depth, temp_LPI,-150);
fprintf('\nThe temperature value at z=-150 m using Lagrange interpolation is %f \n',temp_LP)
temp_ND=newtondivdiff_interp(depth, temp_NDD,-150);
fprintf('\nThe temperature value at z=-150 m using Newton interpolation is %f \n',temp_ND)
temp_sp=natural_cubic_spline_interp(depth, temp_nCS,-150);
fprintf('\nThe temperature value at z=-150 m using Cubic Spline interpolation is %f \n',temp_sp)