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The formula for the curvature of a $y = f(x)$ function is given by:

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


Given that $y = 3ln(x),$ we know:

$y'(x) = 3/x$ $y''(x) = -3/x^2$


Plugging these in:

$k(x) = \frac{3/x}{[1+\frac{9}{x^4}]^\frac{3}{2}}$

$k(x) = \frac{3}{x[1+\frac{9}{x^4}]^\frac{3}{2}}$

$k(x) = \frac{3}{x[\frac{x^4+9}{x^4}]^\frac{3}{2}}$

$k(x) = \frac{3}{x[\frac{(x^4+9)^\frac{3}{2}}{x^6}]}$

$k(x) = \frac{3}{\frac{x}{x^6}{(x^4+9)^\frac{3}{2}}}$

$k(x) = \frac{3}{\frac{1}{x^5}{(x^4+9)^\frac{3}{2}}}$

$k(x) = \frac{3x^5}{{(x^4+9)^\frac{3}{2}}}$

From here, I need the critical points, which I get by taking the derivative of k(x):

$k'(x) = \frac{15x^4(x^4+9)^\frac{3}{2} - (3x^5)(3/2)(x^4+9)^\frac{1}{2}(4x^3)}{(x^4+9)^3}$

$k'(x) = \frac{15x^4(x^4+9)^\frac{3}{2} - 18x^8(x^4+9)^\frac{1}{2}}{(x^4+9)^3}$

$k'(x) = \frac{3x^4(x^4+9)^\frac{1}{2}[5(x^4+9)^3-6x^4]}{(x^4+9)^3}$

Beyond this, I have no idea what to do. I can solve it graphically, but how can I do it algebraically?

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    Hint: where is $k'(x)$ zero? – 2017-01-20
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    @Wojowu Did you read the entire post? :/ – 2017-01-20
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    @MichaelMcGovern That doesn't have any relevance to $k'$ being equal to $0$, though – 2017-01-20
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    @AlexandrH You first need to find where the first and second derivatives of the function whose critical you want to find are. – 2017-01-20
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    Your formula for the curvature seems wrong. See https://en.wikipedia.org/wiki/Curvature#Curvature_of_a_graph – 2017-01-20
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    Urgh...it is wrong. Time to redo this whole thing... – 2017-01-20
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    Sorry I deleted my comment about finding where $y′$ is $0$ about a minute after I posted because I realized it wasn't relevant. – 2017-01-20

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The correct form of the (unsigned) curvature of a function $y(x)$ is $$\kappa(x) = \frac{|y''(x)|}{(1+(y'(x))^2)^{3/2}}.$$ For $y(x) = 3 \log x$, we then easily find $$y'(x) = \frac{3}{x}, \quad y''(x) = -\frac{3}{x^2},$$ hence $$\kappa(x) = \frac{3x}{(9+x^2)^{3/2}}, \quad x > 0.$$ It follows that $$\kappa'(x) = \frac{3(2x^2-9)}{(9+x^2)^{3/2}}$$ which has a critical point $x = 3/\sqrt{2}$. It remains an exercise to show that this value corresponds to a local maximum of $\kappa$ and therefore the $x$-value of the point on $y = 3 \log x$ that has maximum curvature.

As an additional exercise, what is the location of the center of the osculating circle that corresponds to this maximal curvature?

As a second additional exercise, what is the locus of the centers of the osculating circles for all points on $y = 3 \log x$? Choose a suitable parametrization.