Find minimum y value of vectors

The first part of the code I need to show the values of x and y vectors when t is between 0 and 10 for which I have the following code

for i=1:10;
    t=i;
    x = 5 + sqrt(t);
    y= 2 + sin(2*t)/(t+1);
disp(['The Value for t is: ',num2str(t) , ' which gives us x as: ',num2str(x),' and y as ',num2str(y)])

end

The next task is to identify the minimum value Y vector (and the corresponding t and x values at that point). Im feeling pretty stumped as to how to find a minimum value? Any help appreciated

Answer 1

[Star Strider]

Use the min function with both outputs.

I leave the rest to you.

Answer 2

[Andriy Kavetsky]

If you write for i=1:10; t=i, then you would'nt get vector because t=10, maybe you should write t(i)=i for creating an vector and y should be like y= 2 + sin(2.*t)./(t+1);.To find minimum y use [ymin,ind]=min(y); ymin-minimum value and ind its index, so display t(ind) and x(ind), they will correspond to y index.

Answer 3

[Image Analyst]

Rebecca, since t = i, which goes from 1 to 10, t will ALWAYS be between 0 and 10 with your code - how could it not be? So you need to display ALL x and y. To find the min of a current value and a previous min value, you can do

minValue = min([minValue, currentValue]);

Adapt as needed.