curvature of y=sin(x) at x=pi/4
So if y = f(x) = sin x, then f'(x) = cos x and f''(x) = -sin x. This gives k = | -sin x | / [ 1 + (cos x)^2 ]^(3/2) for the curvature at the point x. For x = pi/4, we have that sin pi/4 = cos pi/4 = 1/sqrt(2). So the curvature at the point x=pi/4 is given by: k = (1/sqrt(2) ) / [ 1 + 1/2 ]^(3/2) = 2/ ( 3*sqrt(3) )
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