For an assignment, I am supposed to use MatLab to plot the vector field. I've tried
[x,y]=meshgrid(-5:1:5,-5:1:5);
quiver(x,y,0.2*(x.^2+y.^2),0.2*(x-y),0);
which gives me the following picture:
vectorfield1.PNG
I have also tried this:
[x,y]=meshgrid(-5:1:5,-5:1:5);
quiver(x,y,0.2*(x^2+y^2),0.2*(x-y),0);
which gives me this picture:
vectorfield2.PNG
These are so different, and I'm not sure which one is correct because I don't understand why .^ is giving a different result than just ^ alone.
Any advice is greatly appreciated. Thank you!

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madhan ravi
madhan ravi 2018년 12월 4일

1 개 추천

when you have a vector the right usage is .^ which is element wise operation each element is raised to the power ^ is only for scalars so use .^

댓글 수: 4

Jay
Jay 2018년 12월 4일
So since x and y are not vectors, does that mean the second image is the correct image since it was generated with just the ^ notation?
madhan ravi
madhan ravi 2018년 12월 4일
편집: madhan ravi 2018년 12월 4일
I am not the one who says this is correct and that is wrong you should learn about Array vs Matrix operations. Basically what happens is if a matrix is square ^ performs matrix multiplication and .^ performs element wise operation so now you are the one to decide which one is apt for your need.
Jay
Jay 2018년 12월 4일
Thank you so much!
madhan ravi
madhan ravi 2018년 12월 4일
Anytime :)

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추가 답변 (1개)

Image Analyst
Image Analyst 2018년 12월 4일
편집: Image Analyst 2018년 12월 4일

4 개 추천

Well you're sort of close - at least you tried - but I don't think either of your attempts is correct. Madhan is correct in that using dot means that it's element-by-element raising to a power. But you need to look at what quiver wants, and that is vectors whereas your x and y are matrices. If you take a second look at it, I think you'll realize you need to use (:) to turn them into vectors when you compute u and v, and you'll get something like this:
[x, y] = meshgrid(-5:1:5, -5:1:5);
u = 0.2 * (x(:) .^ 2 + y(:) .^ 2);
v = 0.2 * (x(:) - y(:));
% Plot quiver:
subplot(2, 1, 1);
quiver(x(:), y(:), u, v, 'LineWidth', 1);
grid on;
axis square;
xlabel('x', 'FontSize', 20);
ylabel('y', 'FontSize', 20);
title('Quiver Plot', 'FontSize', 20);
% Plot surface of the magnitude
subplot(2, 1, 2);
z = reshape(sqrt(u .^ 2 + v .^ 2), [11,11]);
surf(-5:1:5, -5:1:5, z);
xlabel('x', 'FontSize', 20);
ylabel('y', 'FontSize', 20);
title('Magnitude Plot', 'FontSize', 20);
0000 Screenshot.png
Notice that when x = y (along the diagonal), the vectors are flat, meaning no vertical component, as you'd expect from the equation which involves x-y.

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Jay
Jay 2018년 12월 5일
That makes so much more sense. Thank you so much for your help and explanation!

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Jay
2018년 12월 4일

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Jay
2018년 12월 5일

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