find a point on a graph

Hello,

How can I find a point on this curve with y=1000, so what is the x-value when y=1000?

Thanks in advance

Answer 1

[Star Strider]

Try somethiing like this with your data:

x = 10:50;                                  % Create Data
y = 4000 - 80*x;                            % Create Data

x1k = interp1(y, x, 1000);                  % Interpolate

figure
plot(x, y,    x1k, 1000, 'rs')              % Plot Result
xline(x1k, '--k')
yline(1000,'--k')

Use interp1 to find the x-value for the y-value at 1000.

[Jens Petit-jean]

then how does it work for a system of equations?

function [dy] = FunctieOpdracht10_2_1(t,y)
    dy = zeros(2,1); 
    dy(1) = y(2);
    dy(2) = -9.81 + (9.81.*(y(2)).^2)/10000 
end

this is what I have

[t,h]=ode23(@FunctieOpdracht10_2_1, [0 60], [4000 0])
plot(t,h)
ylim([0 4000])

Thanks in advance

[Star Strider]

You would need to do the same calculation for each curve. That is not a problem if the curves are monotonically increasing or decreasing, however if that is not the situation, it would be necessary to find the index of a point close to the one you want to interpolate, and then select a few points on either side of that index to do the interpolation with. That region would have the curve monotonically increasing or decreasing, satisfying the requirements for interpolation.

Answer 2

[John D'Errico]

You COULD approximate the curve with some function. Best to choose one that can be intelligently extrapolate. Then since you need ot find a value of x where y == 1000, you would either invert the function, perhaps using an analytical inverse if one exists, or using fzero.

Or, since this is fairly well behaved looking curve, you could approximate the data in that plot using a model of the form x(y). Now all you need do is evaluate the functino at y==1000.

We don't have your data, nor any clue as to what a reasonable model might be for all of this. So go and do one of the above.