φ(t)=acost,ψ(t)=bsint,φ'(t)=-asint,ψ'(t)=bcost,φ"(t)=-acost,ψ"(t)=-bsint,φ'3(t)=asint
故d^2y/dx^2=(-abcost*cost-absint*sint)/asint=-b/sint
先求一阶:dy/dx=dy/dt除以dx/dt=bcost/-asint=-bcott/a 再求二阶=d(dy/dx)/dx=d(-bcott/a)/dx=d(-bcott/a)/dt除以dx/dt=bcsc^2t/-asint=结果,楼主自己化简吧,用手机打得太费劲了。
dy/dx=(dy/dt)/(dx/dt)=bcost/(-asint)=-bcott/a
d²y/dx²=d(dy/dx)/dx=[d(dy/dx)/dt]/(dx/dt)=(b/asin²t)/(-asint)
=-b/a²sin³t