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#