求曲线｛x=tcost ,y=tsint｝在t=π/2处的法线方程

x'(t)=cost-tsint,x'(π/2)=-π/2
y'(t)=sint+tcost,y'(π/2)=1
y'(x)=y'(t)/x'(t)
y'(π/2)=-π/2
而x(π/2)=0,y(π/2)=π/2
因此该点的法线斜率为-1/y'(π/2)=2/π
法线方程为：y=2x/π+π/2