Using matlab to find critical points?

2016 3 8
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更新时间：2016 3 9
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For Problem 1 in the attached document, I am struggling to get the right answers. I cant seem to get the correct code to find the critical points and angles.
This is what I have so far:
pos_trap = zeros(size(vel_meas));
pos_trap(1) = 0.35; % Initial position of the piston
for k = 2:length(vel_meas)
pos_trap(k) = pos_trap(k-1) + 0.5*(vel_meas(k-1)+vel_meas(k))*DeltaT;
end
% Check accuracy of Trapezoid estimate
pos_actual = r*cos(w*t)+sqrt(L^2-r^2*sin(w*t).^2);
Error_trap = abs(pos_actual - pos_trap);

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Geoff Hayes
Geoff Hayes 2016-3-8
Nick - how is the above code related to problem one in the document? How is the trapezoid estimate related to the problem since it is not mentioned at all?
Nick Haufler
Nick Haufler 2016-3-9
That was the way we were taught in class, but I think I figured it out now. Does this look right?
r=0.03
L=0.5
w = 15.708
vel_meas = sym('-0.03*15.708*sin(Th) - 0.03^2*15.708*sin(2*Th)/(2*0.5)')
der1=diff(vel_meas,'Th')
CriticalPoints = solve(der1==0,'Th')
der2 = diff(der1,'Th');
der2_test = subs(der2,'Th',CriticalPoints)
CriticalPoints = double(CriticalPoints)*180/pi
Geoff Hayes
Geoff Hayes 2016-3-9
Nick - I don't have the Symbolic Toolbox so can't comment on whether the above will give the correct answer or not. But it looks like you are using diff correctly to get the first and then second derivative of your function. Do you get similar results if you were to work through the problem by hand?
Walter Roberson
Walter Roberson 2016-3-9
Better would be
syms Th
vel_meas = -0.03*15.708*sin(Th) - 0.03^2*15.708*sin(2*Th)/(2*0.5);
der1 = diff(vel_meas, Th)
CriticalPoints = solve(der1==0, Th)
Note: if you are using a MATLAB version before about R2011b, then change the "derl==0" to just "der1"

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