I am trying to perform the second derivative test of this function I came up with
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Oliver Ries
le 29 Fév 2024
Commenté : Oliver Ries
le 29 Fév 2024
Trying to do the second derivative test to find the minimum W, but none of this code is working. How do I get the critical points and then the second derivative to test if those points are a minimum. Also how do I go back and plug them into the equation for R and W when I set them up as syms.
E = 9.9*10^6;
p = .098;
F = 1500;
L = 20;
g = 32.2*12;
I = (F*L^2)/(E*pi^2);
syms R1 t
x = I == (pi/4)*(((R1+t)^4)-(R1^4));
R = solve(x, R1, 'MaxDegree', 4);
W0 = 2.*pi.*p.*L.*g.*t.*R;
disp(W0);
W = inline(W0,'t');
disp(W);
dW = diff(W(t),t)==0;
disp(dW);
cp = vpasolve(dW,t);
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Walter Roberson
le 29 Fév 2024
Modifié(e) : Walter Roberson
le 29 Fév 2024
Q = @(v) sym(v);
Pi = Q(pi);
E = Q(9.9)*10^6;
p = Q(.098);
F = Q(1500);
L = Q(20);
g = Q(32.2)*12;
I = (F*L^2)/(E*Pi^2);
syms R1 t
x = I == (Pi/4)*(((R1+t)^4)-(R1^4));
R = solve(x, R1, 'MaxDegree', 4);
W0 = 2.*Pi.*p.*L.*g.*t.*R;
disp(W0);
dW = diff(W0,t)==0;
disp(dW);
cp = arrayfun(@(F) solve(F,t), dW(1), 'uniform', 0)
cp{1}
vpa(cp{1})
Hmmm... those look suspiciously similar...
Note that the general solution involves
cp = arrayfun(@(F) solve(F,t), dW, 'uniform', 0)
I had to restrict it to dW(1) to fit within the time limits here.
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