MATLAB: I dont know where im going wrong with the code

incorrect use

adj = 20; % ft

opp = 200; % ft
% pythagorus formula a^2+b^2=c^2
c^2 = adj^2+opp^2;
c^2 = sqrt(ans)

Best Answer

The left-hand side should be the variable alone. Change that to

c = sqrt(adj^2 + opp^2)

c^2 cannot be the variable name as it has mathematical operator, on the other hand, c2 is a valid variable name.

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MATLAB: Though the code is pretty simple, still the value of theta remains 0 every time I execute the results with different values of w,x,y,z,u and v. The value of theta doesn’t change anytime.

Without even trying to exceute your code...

You set theta to zero.

The value of theta will ONLY ever be changed if a test is satisfied for exact equality between two floating point numbers. So tests like this:

if theta1_1_a==theta1_2_a || theta1_1_a==theta1_2_b

Now, l'll look at the result of your code.

[theta1_1_a, theta1_2_a]
ans =
          64.4196851608345         -64.4196851608345
>> [theta1_1_a, theta1_2_b]
ans =
          64.4196851608345          244.419685160834

So that test must fail.

Then you have an elseif clause.

elseif (theta1_1_b==theta1_2_a) || (theta1_1_b==theta1_2_b)

Is it satisfied?

[theta1_1_b, theta1_2_a]
ans =
         -64.4196851608345         -64.4196851608345

Gosh, it looks like they are the same numbers. Are they really the same?

[theta1_1_b == theta1_2_a]
ans =
  logical
   0
   
theta1_1_b - theta1_2_a
ans =
       1.4210854715202e-14

As you can see, they are not in fact identical.

NEVER TEST FOR EXACT EQUALITY BETWEEN FLOATING POINT NUMBERS. At least not until you truly understand why I just told you not to do so, and you understand when you can expect such a test to be valid. And even then, be careful.

Learn to use tolerances.

MATLAB: Graphing 2 variables in one equation (can not be transposed)

Square both sides to get rid of the sqrt(). Then multiply both sides by the denominator of the right hand side, to get two polynomials equal to each other. Distribute the Y^2 (that used to be Y but you squared) over the (k^2+(c*w)^2) to get Y^2*k^2 + Y^2*c^2*w^2 . Now you have c^2 times coefficients on each side, so subtract the Y^2*c^2*w^2 from both sides so that you end up with all the c^2 on the left and nothing in c on the right. Now you have of the form something1 * c^2 + something2 = something3 and that is clearly a quadratic so proceed to solve it: re-arrange to something1 * c^2 = something3 - something2, then divide through by the something1 to get c^2 = (something3 - something2) / something1 . Square root both sides, remembering to take both negative and positive square roots, and you have c in terms of +/- sqrt(an expression). Remember to plot both roots.

c1 = sqrt(-((X.^2 - Y.^2) .* (m.^2 .* w.^4 .* X.^2 - 2 .* k .* m .* w .^2 .* X.^2 + k.^2 .* X.^2 - k.^2 .* Y.^2))) ./ (w.*(X.^2 - Y.^2));
c = [c1(:),-c1(:)];
plot(k, c);

Best Answer

Related Solutions

MATLAB: Though the code is pretty simple, still the value of theta remains 0 every time I execute the results with different values of w,x,y,z,u and v. The value of theta doesn’t change anytime.

MATLAB: Graphing 2 variables in one equation (can not be transposed)

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