I am trying to input my found critical point into the second derivative, but when I double it to get an actual value, I get an array of zeros.

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;

dW = diff(W0,t)==0;

cp = arrayfun(@(F) vpasolve(F,t), dW(1), 'uniform', 0);
cp{1};
var = double(vpa(vpa(cp{1})));

ddW = diff(dW,t);

ddW_t = subs(ddW, t, var); % Substitute t with the value 

number = double(ddW_t);

Answer 1

[Dyuman Joshi]

"... but when I double it to get an actual value, I get an array of zeros."

That is because dW is defined as an equation, not an expression.

You can change the definition to an expression and get the results accordingly -

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;

%define dW as an expression
dW = diff(W0,t);

cp = arrayfun(@(F) vpasolve(F,t), dW(1), 'uniform', 0)

cp = 1x1 cell array
    {[0.13882567791546955545640596714379]}

var = cp{1}

var = 

0.13882567791546955545640596714379

ddW = diff(dW,t)

ddW = 

ddW_t = subs(ddW, t, var) % Substitute t with the value 

ddW_t = 

number = double(ddW_t)

number = 
   1.0e+03 *

  -8.3670 + 0.0000i
  -2.9543 + 0.0541i
  -2.9543 - 0.0541i

[Dyuman Joshi]

Any updates, @Oliver Ries?

[Oliver Ries]

Yes, sorry I forgot to respond to this one. It ended up working although I went a completely different direction and instead of doing the second derivative test, I just returned the lowest weight and the corresponding values, because although this ended up working it still only returned one value every time rather than the 3x1 matrix you are showing.

[Dyuman Joshi]

I see.

Well, you can now implement the code above, accordingly.