@Huy Ngo, to solve the system of equations (x^2 + y^2 = 0) and (y = \tan(x)) in MATLAB, we need to find the intersection points of these equations. The first equation implies that both (x) and (y) must be zero, since the sum of their squares is zero. The second equation, (y = \tan(x)), is a trigonometric function that repeats periodically.
Here's a step-by-step guide on how you can set up a MATLAB script to find and plot these solutions:
- Define the Equations: We will use anonymous functions to define the equations.
- Set Up a Loop: We will use a loop to evaluate the equations over a range of (x) values.
- Plot the Equations: Use MATLAB's plotting functions to visualize the solutions.
- Check for Solutions: Since (x^2 + y^2 = 0) implies (x = 0) and (y = 0), we will check these conditions.
Here's a MATLAB script that achieves this:
% Define the range for x
x = linspace(-2*pi, 2*pi, 1000); % Evaluate from -2*pi to 2*pi
% Define the equations
y1 = tan(x); % y = tan(x)
y2 = sqrt(0 - x.^2); % y = sqrt(0 - x^2) which is 0 since x^2 + y^2 = 0 implies y = 0
% Plot the equations
figure;
plot(x, y1, 'r', 'DisplayName', 'y = tan(x)');
hold on;
plot(x, y2, 'b', 'DisplayName', 'x^2 + y^2 = 0');
plot(x, -y2, 'b'); % Plot the negative part of y2 for completeness
hold off;
% Set plot properties
xlabel('x');
ylabel('y');
title('Plot of y = tan(x) and x^2 + y^2 = 0');
legend show;
grid on;
% Find solutions
tolerance = 1e-4;
solutions = [];
for i = 1:length(x)
if abs(x(i)) < tolerance && abs(y1(i)) < tolerance
solutions = [solutions; x(i), y1(i)];
end
end
% Display solutions
disp('Solutions found:');
disp(solutions);
