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

The aim of this question is to find the point in a curve where the curvature is maximum.

The question is based on the concept of differential calculus which is used to find the maximum value of curvature. In addition to that, if we want to calculate the value of curvature as $(x)$ tends to infinity, it will be derived by first finding the limit of curvature at $(x)$ tending to infinity.

$K=\frac{\left| f^{\prime\prime} \left(x\right)\right|} {\left[1+\left(f^\prime\left(x\right) \right)^2\right]^\frac{3}{2}}$

The function is given as:

$f\left(x\right) = \ln{x}$

$f^\prime\left(x\right) = \frac{1}{x}$

$f^{\prime\prime}\left(x\right) = -\frac{1}{x^2}$

Now putting it in the formula of curvature, we get:

$k\left(x\right) = \dfrac{\left| f^{\prime\prime} \left(x\right)\right|} {\ \left[1+\left(f^\prime \left(x\right)\right)^2 \right]^\frac{3}{2}}$

$k\left(x\right) = \dfrac{ \left|-\dfrac{1}{x^2} \right|} {\ \left[1+{(\dfrac{1}{x})}^2\right]^ \frac{3}{2}}$

$k\left(x\right) = \frac{1}{x^2\ \left[1+\dfrac{1}{x^2} \right]^\frac{3}{2}}$

Now taking derivative of $k\left(x\right)$, we have:

$k\left(x\right) = \frac{1}{x^2\ \left[1+\dfrac{1} {x^2}\right]^ \frac{3}{2}}$

$k\left(x\right)\ =\ x^{-2}\ \left[1 + \frac{1}{x^2}\right]^ \frac{-3}{2}$

$k^\prime\left(x\right)\ =\ -2\ x^{-3}\ \left[1+\frac{1}{x^2}\right]^\frac{3}{2}\ +\ x^{-2}.\ \frac{-3}{2}\ \left[1 +\frac{1}{x^2}\right]^\frac{-5}{2}\ (-2\ x^{-3})$

$k^\prime\left(x\right)\ =\ \frac{-2}{x^3\ \left[1+\dfrac{1} {x^2}\right]^\frac{3}{2}}\ +\ \frac{3}{x^5\ \left[1+\dfrac{1} {x^2}\right]^\frac{5}{2}}$

$k^\prime\left(x\right)\ =\ \ \frac{-2\ x^2\ (1+\dfrac{1}{x^2})+\ 3}{x^5\ \left[1+\dfrac{1}{x^2}\right]^\frac{5}{2}}$

$k^\prime\left(x\right)\ =\ \ \frac{-2\ x^2\ -2+\ 3}{x^5\ \left[1+\dfrac{1}{x^2}\right]^\frac{5}{2}}$

$k^\prime\left(x\right)\ =\ \ \frac{-2\ x^2\ +\ 1}{x^5\ \left[1+ \dfrac{1}{x^2}\right]^\frac{5}{2}}$

$k^\prime\left(x\right)\ =\ \ \frac{1\ -\ 2\ x^2\ }{x^5\ \left[1 +\dfrac{1}{x^2}\right]^\frac{5}{2}}$

Putting $k^\prime\left(x\right)\ =0$, we get:

$0\ =\ \ \frac{1\ -\ 2\ x^2\ }{x^5\ \left[1+\dfrac{1}{x^2}\right]^\frac{5}{2}}$

$0\ =\ \ 1\ -\ 2\ x^2$

Solving for $x$ we have the equation:

$2 x^2 = 1$

$x^2=\frac{1}{2}$

$x=\frac{1}{\sqrt2}\approx\ 0.7071$

We know that the domain of $\ln{x}$ does not include any negative roots, so the maximum interval can be:

$\left(0,0,7\right):\ \ \ K^\prime\left(0,1\right)\ \approx\ 0.96$

$\left(0,7,\infty\right):\ \ \ K^\prime\left(1\right)\ \approx\ -0.18$

We can notice that $k$ is increasing and then decreasing, so it will be maximum at infinity:

$\lim_{x\rightarrow\infty}{\frac{1}{x^2\ \left[1+\dfrac{1}{x^2}\right]^\frac{3}{2}}}$

$\lim_{x\rightarrow\infty}{\frac{1}{\infty\ \left[1+\dfrac{1}{\infty}\right]^\frac{3}{2}}}$

$\lim_{x\rightarrow\infty}{\frac{1}{\infty\ \left[1+0\right]^\frac{3}{2}}}=\ 0$

Thus, the curvature approaches $0$.

## Numerical Results

$k$ will be maximum at infinity

$\lim_{x\rightarrow\infty}{\frac{1}{x^2\ \left[1+\dfrac{1}{x^2}\right]^\frac{3}{2}}}$

$\lim_{x\rightarrow\infty}{\frac{1}{\infty\ \left[1+0\right]^\frac{3}{2}}}=\ 0$

Thus, the curvature approaches $0$.

## Example

For the given function $y = \sqrt x$, find the curvature and radius of curvature at $x=1$ value.

The function is given as:

$y = \sqrt x$

First derivative of the function will be:

$y^\prime = (\sqrt x)^\prime$

$y^\prime = \frac{1}{2\sqrt x}$

The second derivative of the given function will be:

$y^{\prime\prime} = (\frac{1}{2\sqrt x})^\prime$

$y^{\prime\prime} = (\frac{1}{2}x^{\frac{-1}{2}})^\prime$

$y^{\prime\prime} = \frac{-1}{4}x^{\frac{-3}{2}}$

$y^{\prime\prime} = \frac{-1}{4\sqrt {x^{3}}}$

Now putting it in the formula of curvature, we get:

$k\left(x\right) = \frac{\left|f^{\prime\prime} \left(x\right)\right| }{\ \left[1+\left(f^\prime\left(x\right)\right)^2\right]^\frac{3}{2}}$

$k\left(x\right) = \frac{\left|y^{\prime\prime}\right|}{\ \left[1+ \left(y^\prime\right)^2\right]^\frac{3}{2} }$

$k \left(x\right) = \frac{\left|\dfrac{-1}{4\sqrt {x^{3}}}\right|}{\ \left[1+\left(\dfrac{1}{2\sqrt x}\right)^2\right]^\frac{3}{2}}$

$k\left(x\right) = \frac{\dfrac{1}{4\sqrt {x^{3}}}}{\ \left(1+ \dfrac{1}{4 x}\right)^\frac{3}{2}}$

$k\left(x\right) = \frac{\dfrac{1}{4\sqrt {x^{3}}}}{\ \left(\dfrac{4x+1}{4 x}\right)^\frac{3}{2}}$

$k \left(x\right) = \frac{2} {\left(4 x +1\right)^\frac{3}{2}}$

Now putting $x=1$ in the curvature of the curve formula:

$k\left(1\right) =\frac{2} {\left(4 (1) +1\right)^\frac{3}{2}}$

$k\left(1\right) =\frac{2} {5 \sqrt 5}$

We know that the radius of curvature is reciprocal to the curvature:

$R =\frac{1}{K}$

Put the value of curvature and calculate above at $x=1$ in the formula of radius of curvature, which will result in:

$R = \frac{1}{\dfrac{2} {5 \sqrt 5}}$

$R = \frac {5 \sqrt 5}{2}$