How to graph values of a function above and below 0 as separate functions?

I need to create a graph of y=sin(x). Where y is greater than zero, have a separate function of y^2, and where y is less than zero, have a function y^3. These should all exist on the same plot.

I think I have to define y2 and y3 ouside of the if statement. If I set y2=y; ouside 'if', I have y throughout the whole graph. Either way, the if statement doesn't seem to work.

And on top of that, I don't know how to graph this, because x and y2 will be different lengths if y2 only exists where y>0

Here is what I came up with, but it totally doesn't work. The photo included is a more complex function, but the idea is exactly the same. I'm supposed to use if and elseif. Thank you! Stay safe and healthy everyone!

t= 0:0.2:3*pi;
y=sin(t);

figure(1)
plot(t,y,'-.b')
grid on
hold on

if y>0
    y2=y.^2;
elseif y<0;            
    y3=y.^3;
end

figure(1)
plot(t,y2,'om')

figure(1)
plot(t,y3,'dk')

Answer 1

[Adam Danz]

The conditional statements in your code aren't doing what you think they are doing. They return true only when the entire vector meets the criteria. Instead, use indexing.

x = 0 : pi/12 : 4*pi; 
y = sin(x); 

y2(y>0) = y(y>0).^2; 
y2(y<0) = y(y<0); 

y3(y<0) = y(y<0).^3; 
y3(y>0) = y(y>0); 

hold on
plot(y, '-b', 'LineWidth', 4)
plot(y2, '--m', 'LineWidth', 2)
plot(y3, '--c', 'LineWidth', 2)

[Konrad Krusche]

Thank you. I knew that something about the if wasn't working because it's a vector, but didn't know exactly how it works.

Answer 2

[Image Analyst]

Try this:

t = 0:0.2:3*pi;
y = sin(t);
y2 = y .^ 2;
y3 = y .^ 3;

hFig = figure;
plot(t,y,'-.b')
grid on
hold on

positiveYIndexes = y > 0;
negativeYIndexes = y < 0;
plot(t(positiveYIndexes), y2(positiveYIndexes), 'om-', 'LineWidth', 2')
plot(t(negativeYIndexes), y3(negativeYIndexes), 'dk-', 'LineWidth', 2')
xlabel('t', 'FontSize', 20);
ylabel('y', 'FontSize', 20);
legend('y', 'y squared', 'y cubed', 'Location', 'north');
hFig.WindowState = 'maximized';

Answer 3

[Star Strider]

Subscript the elements and provide corresponding time vectors for each one:

t= 0:0.2:3*pi;
y=sin(t);

for k = 1:numel(t)
    if y(k)>0
        t2(k) = t(k);
        y2(k)=y(k).^2;
    elseif y(k)<0
        t3(k) = t(k);
        y3(k)=y(k).^3;
    end
end

figure(1)
plot(t,y,'-.b')
grid on
hold on

figure(1)
plot(t2,y2,'om')
figure(1)
plot(t3,y3,'dk')

Answer 4

[Image Analyst]

If you absolutely MUST do it with an if/else construct, and throw away most of the benefits of the vectorization capabilities of MATLAB, then you can do this:

% Initialization steps.
clc;    % Clear the command window.
fprintf('Beginning to run %s.m ...\n', mfilename);
close all;  % Close all figures (except those of imtool.)
clear;  % Erase all existing variables. Or clearvars if you want.
workspace;  % Make sure the workspace panel is showing.
format short g;
format compact;
fontSize = 15;

t= 0:0.2:3*pi;
y=sin(t);

hFig = figure;
plot(t,y,'-.b')
grid on
hold on

y2 = nan(1, length(y)); % Nan points will not get plotted.
y3 = nan(1, length(y));
for k = 1 : length(y)
	if y(k) >= 0
		y2(k) = y(k) ^ 2;
	else
		y3(k) = y(k) ^ 3;
	end
end
plot(t, y2, 'om-', 'LineWidth', 2')
plot(t, y3, 'dk-', 'LineWidth', 2')
xlabel('t', 'FontSize', 20);
ylabel('y', 'FontSize', 20);
legend('y', 'y squared', 'y cubed', 'Location', 'north');
hFig.WindowState = 'maximized';

[Konrad Krusche]

I'm supposed to do it with if / ifelse, which seems odd the more I read about this question, but if that's how the prof wants it, that's the way I need to do it.Thanks so much!

[Image Analyst]

So how are you going to do it now? You didn't tag this as homework, which means (if you had) we'd have given you hints on how to do it, not total solutions. If you turn in our solutions as your own work, and your professor finds out, you could be in serious trouble.

[Konrad Krusche]

My bad, I should have tagged it appropriately. He sent us additional materials on how to use if (which I only saw after this post), and the only main difference from what you posted above, is he defined y2 as an empty vector (y2=[]). I of course posted a different function and variables here, as it was not my intention to copy. The only part of my solution that is the same as above is the definition of y2, as a vector of non-numbers. I should probably inform him that I asked for help. Again, thank you, and I apologize for my mistake, I'll make sure to tag correctly if I ask for help for again.