As an aside, since @Cris LaPierre has already suggested the use of LevelList to do the job. And that arguably is the correct answer, at least A correct answer. But there are always multiple ways to solve a problem. In that vein, we could have used fimplicit instead.
% no need to use clc and clear in an answers script
L = 100;
v_origin = 10;
v_opposite = 20;
% There is NO need to predefine t as a random number,
% since t is just the argument to a function handle. As importantly, this
% problem does not even use x_0 and y_0. So I can just comment them out and
% still make this work. I assume you use those functions later on though.
% x_0 = @(t) 2*t + 2;
% y_0 = @(t) t + 3;
k_ad = 10^(0.3591 * log10(v_origin) + 0.0952);
k_dt = 10^(0.5441 * log10(v_opposite) - 0.0795);
r_fort = (1 + 1.34 * sqrt(k_ad^2 + k_dt^2)) * L;
r_aft = (1 + 0.67 * sqrt(k_ad^2 + k_dt^2)) * L;
r_starb = (0.2 + k_ad) * L;
r_port = (0.2 + 0.75 * k_ad) * L;
syms x y;
segmentedFunction = @(m) (m < 0) * -1 + (m >= 0) * 1;
parameter = @(x,y) power(2*x ./ ((1 + segmentedFunction(x)) .* r_starb - (1 - segmentedFunction(x)) .* r_port), 2) + ...
power(2*y ./ ((1 + segmentedFunction(y)) .* r_fort - (1 - segmentedFunction(y)) .* r_aft), 2);
So parameter is an implicit function of x and y. A contour plot fixs that function at some value or set of values, and then plot the set of points in the (x,y) plane where the function takes on exactly those given values. That is what LevelList does for us. But again, parameter defines an implicit function of x and y. And that means we can use fimplicit to do the work, to identify exactly that desired contour.
fimplicit(@(x,y) parameter(x,y) - 0.5,[-400,400,-400,400],'linewidth',2)
So we can achieve exactly the same thing as countour does, merely by recognizing what a contour is, and how it relates to an implicit function.
Contour is still a good tool in general to use though, since by showing a full set of contours, we get a much better feeling for the general shape of the surface.
