Solve with multiple values

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Daniel
Daniel 2011-10-22
Below is the equation I'm trying to work with. g is a single value. vbprime, newtheta and y1 are 1x51 matrices. I just want to solve for t1, for use in c, but I can't seem to get it to solve. I keep getting an error about MUPAD engine in the solve line. I have declared t1 to by sym also. I expect c to S to have multiple values, as well as c. Any thoughts?
S=-0.5*g.*t1^2+vbprime.*sind(newtheta).*t1+y1
solve(S)
c=(100-vbprime.*cosd(newtheta).*t1)

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Walter Roberson
Walter Roberson 2011-10-22
syms t1 vbprime_ newtheta_ y1_
S=-0.5*g.*t1^2+vbprime_.*sind(newtheta_).*t1+y1_;
T1 = solve(S,t1);
T1Fun = matlabFunction(T1, vbprime_, newtheta_, y1_);
c = arrayfun(@(IDX) 100-vbprime(IDX).*cosd(newtheta(IDX)) .* T1Fun(vbprime(IDX), newtheta(IDX), y1(IDX)), 1:length(vbprime));
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Walter Roberson
Walter Roberson 2011-10-23
Hmmm... the function is a quadratic, so there are two solutions coming out of solve(). Possibly matlabFunction is interpreting that as requiring two output arguments.
Unfortunately as I do not have the toolbox, this is not something I can test for myself.
I would suggest that you would be better off solving the quadratic algebraically once (using the quadratic formula) on paper, and either choosing one of the two roots based upon additional information, or else creating both formula and handling them semi-independently.
For both together, I get
t3 = sin(newtheta .* pi / 180);
t4 = vbprime .* t3;
t5 = vbprime .^ 2;
t6 = t3 .^ 2;
t11 = sqrt(t5 .* t6 + (2 .* g .* y1));
t13 = 1 ./ g;
s1 = (t4 + t11) .* t13;
s2 = (t4 - t11) .* t13;
Daniel
Daniel 2011-10-25
Holy cow I got it. Not exactly sure why it works, but it works. I included the code below for future questions. Thanks so much for all your help. Quadratic equation!!! A little different then what you suggested, but same idea. I should have thought of this. Thanks so much again!
a=-0.5*g
b=vbprime.*sind(newtheta)
c=y
t=(-b-sqrt(b.^2-4.*a*c))/(2*a)

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