# 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/x^{3}

Therefore,

K = ( 1/x^{2})/[1+ (1/x)^{2}]^{3/2}

= 1/x^{3}[1+ 1/x^{2}]^{3/2}

= x/[1 + x^{2}]^{3/2}

Differentiating the above equation we get

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

= 1/[1 + x^{2}]^{3/2} - 3x^{2}/[1 + x^{2}]^{5/2}

= ((1 + x^{2}) - 3x^{2 })/ [1 + x^{2}]^{5/2}

= (1 - 2x^{2 }) / [1 + x^{2}]^{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.

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