Question

Matlab 曲线到定点距离最短问题

已知x(t)=5t-10, y(t)=25t^2-120t+144.
t在[0, 4]之间，求曲线到(0, 0)的最短距离。

Answer 1

You can use Matlab's symbolic package to get the minimum distance, you might need to use a little bit calculus

t = sym('t','real') % define a real symbol variable

x=5*t-10;  

y=25*t.^2-120*t+144;

d = (x^2+y^2)^.5;  % distance function

d1 = diff(d); % the differentiation

t1s = solve(d1); % let the differentiation equal to 0 to get the maxmin point

t1v = eval(t1s); % get the numerical evaluation

min_d = subs(d,t,t1v)

min_d =

    1.3577

You can use the following code to generate a figure of illustrating the curve and the distance:

t = [0:0.001:4]; 

x=5*t-10;

y=25*t.^2-120*t+144;

figure(1);

hold on;

plot(x,y,'b');

axis([0 4 0 4],'square');

x=5*t-10;

d=(x.^2+y.^2).^.5;

figure(1);

hold on;

plot(x,d,'g');

t=t1v;

x=5*t-10;

y=25*t.^2-120*t+144;

plot(x,y,'r.');

plot([0 x],[0 y],'r');

plot([x x],[0 4],'--');

追问

And one more question, how to determine x, y at which the curve is the closest to the origin(0,0)? Thank you~

追答

when you have the parameter t, you can simply substitute it into the formula to obtain x and y, in my code:
t=t1v;
x=5*t-10;
y=25*t.^2-120*t+144;

And the point is a red point on the figure.

Answer 2

可这样：
t=0:0.01:4;x=5*t-10;y=25*t.^2-120*t+144;plot(x,y)
figure
plot(x,y)
hold on
syms x(t) y(t)
x=5*t-10;
y=25*t.^2-120*t+144;
kf=-diff(x)/diff(y);%法线的斜率
%设曲线上(x,y)点及(0,0)在法线上，则法线的斜率也等于下式
kk=(y-0)/(x-0);
%解方程求交点的t值
t1=solve(kf==kk,'MaxDegree',3,'Real',true);
t2=double(t1);
x1=double(subs(x,t,t2));
y1=double(subs(y,t,t2));
d=hypot(x1,y1);
fprintf('原点到曲线的最短距离是 %f\n',d)
plot(x1,y1,'r*')
plot(0,0,'r*')
plot([0,x1],[0,y1])
hold off
axis equal
axis([-1,+5,-1,10])

Answer 3

最容易懂的就自己编程把；
matlab有个直接求两点间的距离的函数；你在循环比较；
或者你把他看成线性规划问题，你应该懂得；
再则 用lingo 也行；实在不懂的话就写c语言把 干脆点也行 。具体函数还是你自己来熟练操作把