I found this:
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I found it here:
http://www.informit.com/articles/article.aspx?p=675274&seqNum=3
Answer:
t = [0:0.01:4];
x = 5*t - 10;
y = 25442 - 120*t + 144;
d = x.^2 + y.^2;
min = 1e+14;
for k = 1:length(t)
if d(k) < min
min = da(k);
tmin = t(k);
end
end
disp('The minimum distance is: ')
disp(sqrt(min))
disp('and it occurs at t = ')
disp(tmin)Output:
>> withLoop
The minimum distance is:
2.5106e+04
and it occurs at t =
4
Explanation:
