%% Math 273 - Section 2
%% Octave session in Lecture 14, October 09, 2014.

%% Additional comments are included using "% "

octave:2> v0=25;T0=5;
octave:3> W = 35.74 + 0.6215*T - 35.75*v^(0.16) + 0.4275*T*v^(0.16);
error: `T' undefined near line 3 column 20

% W evaluated using the original expression
octave:3> v=v0; T=T0; W = 35.74 + 0.6215*T - 35.75*v^(0.16) + 0.4275*T*v^(0.16)
W = -17.409

% partial derivatives
octave:4> W_v = -(35.75)*(0.16)*v^(0.16-1) + 0.4275*0.16*T*v^(0.16-1)
W_v = -0.36004
octave:5> W_T = 0.6215 + 0.4275*v^(0.16)
W_T =  1.3370

% The constant factor in L(v,T)
octave:6> -17.409 - 0.36004*(-25) + 1.337*(-5)
ans = -15.093

% Notice that L(v0,T0) = W(v0,T0)
octave:7> L = -15.093 - 0.36004*v + 1.337*T
L = -17.409

% second choice of v and T
octave:8> v=24;T=6;
octave:9> L = -15.093 - 0.36004*v + 1.337*T
L = -15.712
octave:10>  W = 35.74 + 0.6215*T - 35.75*v^(0.16) + 0.4275*T*v^(0.16)
W = -15.710

% Notice that the linearization is quite accurate here!

% another point
octave:11> v=27;T=2;
octave:12> L = -15.093 - 0.36004*v + 1.337*T
L = -22.140
octave:13>  W = 35.74 + 0.6215*T - 35.75*v^(0.16) + 0.4275*T*v^(0.16)
W = -22.143

% Again, the linearization is very accurate.

% third point
octave:14> v=5;T=-10;
octave:15> L = -15.093 - 0.36004*v + 1.337*T
L = -30.263
octave:16>  W = 35.74 + 0.6215*T - 35.75*v^(0.16) + 0.4275*T*v^(0.16)
W = -22.256

% Now, the linearization is way off from the correct value. Why?

% It has nothing to do with the negative temperature. 
% For instance, try a different point now.
octave:17> v=5;T=20;
octave:18> L = -15.093 - 0.36004*v + 1.337*T
L =  9.8468
octave:19>  W = 35.74 + 0.6215*T - 35.75*v^(0.16) + 0.4275*T*v^(0.16)
W =  12.981
octave:20>