|
#2
Genial2006-05-29 08:16
|
用matlab求出方程组得一个解
[ atan(20/(2190400*sin(b1)^2+938961)*((2190400*sin(b1)^2+938961)*(1480*cos(b1)+1769))^(1/2)*sin(b1),-1.0000000000000000000000000000000/(2190400*sin(b1)^2+938961)*((2190400*sin(b1)^2+938961)*(1480*cos(b1)+1769))^(1/2)*(-37.+20.*cos(b1)))]
[ atan(20/(2190400*sin(b1)^2+938961)*((2190400*sin(b1)^2+938961)*(1480*cos(b1)+1769))^(1/2)*sin(b1),-1.0000000000000000000000000000000/(2190400*sin(b1)^2+938961)*((2190400*sin(b1)^2+938961)*(1480*cos(b1)+1769))^(1/2)*(-37.+20.*cos(b1)))]
[ atan(20/(2190400*sin(b1)^2+938961)*((2190400*sin(b1)^2+938961)*(1480*cos(b1)+1769))^(1/2)*sin(b1),-1.0000000000000000000000000000000/(2190400*sin(b1)^2+938961)*((2190400*sin(b1)^2+938961)*(1480*cos(b1)+1769))^(1/2)*(-37.+20.*cos(b1)))]
[ atan(20/(2190400*sin(b1)^2+938961)*((2190400*sin(b1)^2+938961)*(1480*cos(b1)+1769))^(1/2)*sin(b1),-1.0000000000000000000000000000000/(2190400*sin(b1)^2+938961)*((2190400*sin(b1)^2+938961)*(1480*cos(b1)+1769))^(1/2)*(-37.+20.*cos(b1)))]
[ atan(-20/(2190400*sin(b1)^2+938961)*((2190400*sin(b1)^2+938961)*(1480*cos(b1)+1769))^(1/2)*sin(b1),1.0000000000000000000000000000000/(2190400*sin(b1)^2+938961)*((2190400*sin(b1)^2+938961)*(1480*cos(b1)+1769))^(1/2)*(-37.+20.*cos(b1)))]
[ atan(-20/(2190400*sin(b1)^2+938961)*((2190400*sin(b1)^2+938961)*(1480*cos(b1)+1769))^(1/2)*sin(b1),1.0000000000000000000000000000000/(2190400*sin(b1)^2+938961)*((2190400*sin(b1)^2+938961)*(1480*cos(b1)+1769))^(1/2)*(-37.+20.*cos(b1)))]
[ atan(-20/(2190400*sin(b1)^2+938961)*((2190400*sin(b1)^2+938961)*(1480*cos(b1)+1769))^(1/2)*sin(b1),1.0000000000000000000000000000000/(2190400*sin(b1)^2+938961)*((2190400*sin(b1)^2+938961)*(1480*cos(b1)+1769))^(1/2)*(-37.+20.*cos(b1)))]
[ atan(-20/(2190400*sin(b1)^2+938961)*((2190400*sin(b1)^2+938961)*(1480*cos(b1)+1769))^(1/2)*sin(b1),1.0000000000000000000000000000000/(2190400*sin(b1)^2+938961)*((2190400*sin(b1)^2+938961)*(1480*cos(b1)+1769))^(1/2)*(-37.+20.*cos(b1)))]
请高手指教一下怎样画出关于变量b1的图形，还用括号中出现的逗号是怎么回事？