t1=input('t1=') w=50/3*pi a=2/3*pi for t=0:0.0001:t1 b=2*w*(pi-a)*t/pi c=cos(2/3*pi+b) d=sin(2/3*pit1=input('t1=')w=50/3*pia=2/3*pi for t=0:0.0001:t1b=2*w*(pi-a)*t/pic=cos(2/3*pi+b)d=sin(2/3*pi)h=cos(2/3*pi)[x,y,z]=solve('x^2+y^2+z^2=100^2+170^2','x^

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t1=input(''t1='')w=50/3*pia=2/3*pifort=0:0.0001:t1b=2*w*(pi-a)*t/pic=cos(2/3*pi+b)d=sin(2/3*pit1=input

t1=input('t1=') w=50/3*pi a=2/3*pi for t=0:0.0001:t1 b=2*w*(pi-a)*t/pi c=cos(2/3*pi+b) d=sin(2/3*pit1=input('t1=')w=50/3*pia=2/3*pi for t=0:0.0001:t1b=2*w*(pi-a)*t/pic=cos(2/3*pi+b)d=sin(2/3*pi)h=cos(2/3*pi)[x,y,z]=solve('x^2+y^2+z^2=100^2+170^2','x^
t1=input('t1=') w=50/3*pi a=2/3*pi for t=0:0.0001:t1 b=2*w*(pi-a)*t/pi c=cos(2/3*pi+b) d=sin(2/3*pi
t1=input('t1=')
w=50/3*pi
a=2/3*pi
for t=0:0.0001:t1
b=2*w*(pi-a)*t/pi
c=cos(2/3*pi+b)
d=sin(2/3*pi)
h=cos(2/3*pi)
[x,y,z]=solve('x^2+y^2+z^2=100^2+170^2','x^2+(z+170*c)
^2=100^2*d^2','y^2+z^2=170^2+100^2*h^2/c^2')
x=subs(x)
y=subs(y)
z=subs(z)
plot3(x,y,z)
end
就是求解这个三元二次方程,然后画出轨迹来,且Y>0的

t1=input('t1=') w=50/3*pi a=2/3*pi for t=0:0.0001:t1 b=2*w*(pi-a)*t/pi c=cos(2/3*pi+b) d=sin(2/3*pit1=input('t1=')w=50/3*pia=2/3*pi for t=0:0.0001:t1b=2*w*(pi-a)*t/pic=cos(2/3*pi+b)d=sin(2/3*pi)h=cos(2/3*pi)[x,y,z]=solve('x^2+y^2+z^2=100^2+170^2','x^
t1=input('t1=') w=50/3*pi a=2/3*pi for t=0:0.0001:t1 b=2*w*(pi-a)*t/pi c=cos(2/3*pi+b) d=sin(2/3*pi
20 - 离问题结束还有 14 天 8 小时
问题补充:t1=input('t1=')
w=50/3*pi
a=2/3*pi
for t=0:0.0001:t1
b=2*w*(pi-a)*t/pi
c=cos(2/3*pi+b)
d=sin(2/3*pi)
h=cos(2/3*pi)
[x,y,z]=solve('x^2+y^2+z^2=100^2+170^2','x^2+(z+170*c)
^2=100^2*d^2','y^2+z^2=170^2+100^2*h^2/c^2')
x=subs(x)
y=subs(y)
z=subs(z)
plot3(x,y,z)

matlab中的t1=double(t1)是什么意思? t1=input('t1=') w=50/3*pi a=2/3*pi for t=0:0.0001:t1 b=2*w*(pi-a)*t/pi c=cos(2/3*pi+b) d=sin(2/3*pit1=input('t1=')w=50/3*pia=2/3*pi for t=0:0.0001:t1b=2*w*(pi-a)*t/pic=cos(2/3*pi+b)d=sin(2/3*pi)h=cos(2/3*pi)[x,y,z]=solve('x^2+y^2+z^2=100^2+170^2','x^ t1=s/v1- matlab solve问题,同样使用solve,在使用input的情况下,少了一个解同一个方程15.24*ln(d)+15.24*d1/d-t1-35.54=0,用第一种方法算syms t1 d1d=solve('15.24*ln(d)+15.24*d1/d-t1-35.54=0','d');subs(d,{t1,d1},{input('t1='),input('d1=')} t2=T2-T1=[(根号2)-1]t1 是什么意思 为什么(s1/t1)/(s2/t2)=(s1/s2)/(t1/t2) 求证三角函数.2cos(wt1+θ)*s(wt2+θ)=cosw(t2-t1)+cos[w(t1+t2)+2θ] 关于lingo 求解目标函数为非线性规划的问题model:max=9*t1*w+9*t1*x+9*t1*z+9*y1+15*t2*w+15*t2*x+15*t2*z+15*y2-6*w-16*x-10*y1-10*y2-15*z;0.005*t1*w-0.015*t1*x-0.015*t1*z-0.025*y1200;z0;x>0;y1>0;y2>0;z>0;end W=△E=cm(t2-t1)中的△E是什么意思? T1*0.5+T2*0.707=600 T1*0.866=T2*0.707 怎样求T1和T2最好详细点 利用MATLAB如何求解如下非线性方程组,fai=pi/2; f=36.2*10^3 w=2*pi*f a=2*10^-6;b=2*10^-6;v=2.5*10^-3% m/s[t1,t4]=solve('a*cos(w*t1)-a*cos(w*t4)+(t4-t1)*v=2*pi*v/w','b*cos(w*t1+fai)-b*cos(w*t4+fai)=0','t1','t4');出现以下问题:警告: 利用MATLAB如何求解如下非线性方程组,(续)fai=pi/2; f=36.2*10^3 w=2*pi*f a=2*10^-6;b=2*10^-6;v=2.5*10^-3% m/s[t1,t4]=solve('a*cos(w*t1)-a*cos(w*t4)+(t4-t1)*v=2*pi*v/w','b*cos(w*t1+fai)-b*cos(w*t4+fai)=0','t1','t4');出现以下问题 为什么a1/a2=t2/t1 已知函数f(x)=-x4+2x3,对于任意t1,t1=[-1/2,2](t1,t2),那么|f(t1)-f(t2)| void delay(uint z) { uint t1,y; for(t1=z;t1>0;t1--) for(y=110;y>0;y--); } c51程序这两句不懂谢谢! int t1=5,t2=6,t3=7,t4,t5;t4 = t1 执行这段程序:int t1=1,t2=3,t3; t3=t1 若t1/|t1|+t2/|t2|+t3/|t3|=1,则|t1t2t3|/t1t2t3=