recall that the curvature is defined as
K = |y"|/(1 + y' ^2)^(3/2)
So, with your function,
y' = 5/x
y" = -5/x^2
K = 5/(x^2 (1 + 25/x^2)^(3/2)) = 5x/(x^2 + 25)^(3/2)
So, max K occurs when K' = 0, at x = 5/√2
At what point does the curve have maximum curvature?
y = 5 ln(x)
1 answer