finding the time closest to origin

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)

[vicente estrada]

why was 4 used for t

Answer 1

[Cedric]

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.

[Suman Koirala]

Thank you Cedric!