x=cost+tsint,y=sint-tcost,t=(6/π.3/π),求弧长

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x=cost+tsint,y=sint-tcost,t=(6/π.3/π),求弧长
x=cost+tsint,y=sint-tcost,t=(6/π.3/π),求弧长

x=cost+tsint,y=sint-tcost,t=(6/π.3/π),求弧长
dx/dt=tcost,dy/dt=tsint,所以ds/dt=根号(前面两式平方和)=t,所以弧长是
int_(6/π.3/π)tdt=27/(2π的平方)

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