At what point does the curve have maximum curvature? What happens to the curvature as x tends to infinity y= ln x?

Solution:

The point of maximum curvature of any curve (defined by the function f(x) is determined from the following expression

K  =   | f’’(x) | / [1 + f’(x)²]^3/2

Given f(x) = ln x

f’(x) = -x^-2 = 1/x

f’’(x) = -1/x³

Therefore,

K = ( 1/x²)/[1+ (1/x)²]^3/2

   =  1/x³[1+ 1/x²]^3/2

   =   x/[1 + x²]^3/2

Differentiating the above equation we get

dK/dx  = ([1 + x²]^3/2 dx/dx  - (x)(3/2)(2x)[1 +  x²]^1/2) / [1 + x²]³

          =  1/[1 + x²]^3/2 -  3x²/[1 + x²]^5/2

          =   ((1 + x²) -  3x² )/ [1 + x²]^5/2

          =   (1 - 2x² ) / [1 + x²]^5/2

For maximum curvature, dK/dx has to be zero. The above expression will be zero: 

When x = 1/√2 or 0.707

On either side of x = 1/√2 

K ≷ 0

Therefore the maximum curvature will be at x = 1/√2

At what point does the curve have maximum curvature? What happens to the curvature as x tends to infinity y= ln x?

Summary:

The point at which the curve y = lnx will have the maximum curvature will be at x = 1/√2. The y value will be ln(1/√2) = -0.347 at that point.