写出用matlab绘制四叶玫瑰线P=sin2Θ(Θ∈(0,2π))的程序

2025-06-21 08:32:44
推荐回答(2个)
回答1:

theta=linspace(0,2*pi,400); rou=4*sin(2*theta); polar(theta,rou)

^>> x = -2:0.1:2;

>> y = exp(-(x.^bai2));

>> theta = linspace(0, 2*pi);

>> rho = sin(2*theta);

>> t1 = linspace(0,30,1000);

>> x3 = 3*t1./(1+t1.^3);

>> y3 = 3*t1.^2./(1+t1.^3);

>> t2 = linspace(0,2*pi);

>> x4 = t2 - sin(t2);

>> y4 = 1 - cos(t2);

>> subplot(2,2,1);plot(x,y);title('概率曲线 y = exp(-x^2)')

>> subplot(2,2,2);polar(theta,rho),title('四叶玫瑰线 p = sin(2*t)')

>> subplot(2,2,3);plot(x3,y3);title('叶形线');

>> subplot(2,2,4);plot(x4,y4);title('摆线');

扩展资料:

四叶玫瑰线(four-leaved的一种.定长线段AB =2a,它的两个端点在垂直两直线上滑动,从两直线的交点O向线段AB作垂线OM,垂足M的轨迹称一’为四叶玫瑰线(见图).其极坐标方程为P=a sin 2B。

^clc;clear

subplot(2,2,1)

ezplot('exp(-x^bai2)')

subplot(2,2,2)

ezpolar('sin(2*t)')

subplot(2,2,3)

t=0:pi/100:2*pi;

x=3*t./(1+t.^3);

y=3*t.^2./(1+t.^3);

参考资料来源:百度百科-四叶玫瑰线

回答2:

theta=linspace(0,2*pi,400); rou=4*sin(2*theta); polar(theta,rou)