Find the radius of curvature at x = pi/2 on the curve y = sin x?

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Answer 1 · 2018-07-12T15:17:10

Radius of curvature at $x=pi/2$ is $-1$ .

Radius of curvature at a point on function $y=f(x)$ is given by

$R=[1+y'^2]^(3/2)/(y'')$

For $y=sinx$ , $y'=cosx$ and $y''=-sinx$

Hence radius of curvature is

$[1+cos^2x]^(3/2)/(-sinx)$

and at $x=pi/2$ , $R=[1+0]^(3/2)/(-1)$

= $1/(-1)=-1$