I am trying to find the time and distance at which the object is closest to the origin where the x and y coordinates vary with time. THis is what I have. But, something seems to be not working. First I used a point at time t=4 and used for loop to check for all times from 0 to 4.
disp('Finding the closest point') t=4; x=5*t- 10; y= (25*t^2)- 120*t+144; point =sqrt(x^2+y^2);
for t=0:0.1:4; x= 5.*t-10; y= 25.*t.^2- 120.*t+144; point1= sqrt(x^2+y^2); if point1<point xmin=x; ymin=y; tmin=t; end end
a= 5*tmin- 10; b= (25 *tmin^2)- 120*tmin +144; dist=sqrt(a^2+b^2); disp(dist) disp(tmin)
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This could be solved analytically, but if you want to do it numerically using the approach that you tried to implement, you could do the following:
>> t = 0:0.1:4 ; % Vector of all time steps. >> x = 5.*t-10 ; % Vector of all x values. >> y = 25.*t.^2- 120.*t+144 ; % Vector of all y values. >> d2 = x.^2 + y.^2 ; % Vector of all squared distances to origin.
>> id = find(d2 == min(d2)) % Index of the min. distance. id = 23 >> t(id) % Corresponding time. ans = 2.2000 >> x(id) % Corresponding x. ans = 1 >> y(id) % Corresponding y. ans = 1
>> plot(x, y) ; % Plot trajectory. >> axis([-1, 4, -1, 4]) ; % Zoom on region of interest. >> hold on ; >> plot(0,0,'Gx') ; % Add origin as a green cross. >> plot(x(id),y(id), 'Rx') ; % Add loc of the min as a red cross.
You will realize, however, that your approach can be dangerous if you lower the time step just a little, as the segment OP (where P is the estimate of the point realizing the min) will really be far from orthogonal to the curve.
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