Question

求正弦曲线y=sinx在点（π/2,1）处的曲率

Answer 1

解:y'(π/2)=cos(π/2)=0,y''(π/2)=-sin(π/2)=-1K(π/2)=|y''(π/2)|/{1+[y'(π/2)]^2}^(3/2)=1/(1+0)^(3/2)=1故正弦曲线y=sinx在点(π/2,1)处的曲率为1