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I have a function of one variable. In this graph, we can see that there are a couple of places where the graph "bends" a lot -- a local maximum of "bending", if you will. The ordinary second derivative measures "bending" in the $y$ direction, which will be greatest near the singularity (which is at $-1$). However, if we measure "bending" orthogonal to the tangent line to the graph at a point, then the "bending" is maximized at the two "elbows" near $x=-2$ and $x=0.75$ or so. Given a function $y=f(x)$, how might I go about finding these elbows? Presumably, I will set the third derivative of something to 0 and solve for $x$, but that something is not $f$...

enter image description here

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  • $\begingroup$ Don't mind the red text in the image. Those are other variables, and this is actually a level set.... I've got a dozen dozen dozen of these very-similar-but-not-quite graphs. :) $\endgroup$
    – Him
    Jul 13 at 15:56
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It seems like you are searching for the point where the curvature is greatest. In this case, if $y = f(x)$, use the formula $$\kappa = \frac{|y''|}{\left(1 + \left(y'\right)^{2}\right)^{3/2}}.$$ This will give you the curvature for every $x$ that exists in the domain. Now, to get the value of $x$ whose curvature is greatest, solve for $$\frac{d\kappa}{dx} = 0$$ to get the extreme values, and if necessary, solve for $$\frac{d^{2}\kappa}{dx^{2}} = 0$$ to check which ones are the maxima.

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What you're looking for is to maximize the curvature. A formula for the curvature of the graph of a function $y=f(x)$ can be found on the same page here. As you can see from that formula, the curvature does not involve the third derivative, it is instead proportional to the second derivative, with a proportionality constant that depends only on the first derivative.

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  • $\begingroup$ But if I'm maximizing the curvature, then I would take a derivative of that, which would involve the third derivative of $f$. :) $\endgroup$
    – Him
    Jul 13 at 16:05
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    $\begingroup$ That's a good point. But it's not the third derivative itself that you would set to zero. $\endgroup$
    – Lee Mosher
    Jul 13 at 16:06

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