At what point does y=ln(x) have the maximum curvature?

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[denki]

solved for argmax_y curvature(y=e^x) instead since they are the same curve, just flipped on the line x=y. Got the same result.

[karelvanderwalt]

Nice application and integration of Calc I knowledge

[Anonymous-zphb]

I wikipedia-cheated to get the curvature formula, but was then able to solve it :)
I'll be sure to watch some more vids on the subject to better understand the concept.

[allenaxp]

Excellent video on calculating curvature of a function!

[StuartSimon]

This is a great introduction to the concept of curvature in general. I had a problem in Algebra I in high school because I did not understand how a curve such as a parabola could gradually straighten out without ever becoming completely straight or approaching an asymptote. Of course I became familiar with the shape as time went on.

[thumper]

Hay I got this! I derived a formula for the turning radius from the 1st & 2nd derivatives on a circle. Then I found the local minimum. Then I got x = sqrt(2) because I multiplied by 2 instead of dividing.

No points for me 😞

[ianfowler]

I don't know if I posted something similar on a previous video. At any rate, I generalized by taking the derivative of k(x) and setting the numerator = 0. I did drop the absolute value sign.
k(x) = [ y '' ] / [ 1 + (y ')^2 ] ^(3/2)
After using the quotient rule, multiplying top and bottom by sqrt[ 1 + ( y ')^2 ] and setting the numerator = 0 we arrive at the following differential equation:

y ' ' ' [1 + (y')^2] - 3 y ' (y ' ' )^2 = 0

This DE must be satisfied for k(x) to have a stationary point. You can start with any y = f(x) and substitute into the DE.
I tried y = x^3 - 3x and found the exact x values to be +/- sqrt[ ( 6 +sqrt(86) ) / 15 ] The - sign just indicates a minimum with the same value as f is symmetrical about it's inflection point (0, 0).

Here's what's pretty cool. These values are so, so close the the x values for the stationary points of f(x). Ever so slightly to the right of the relative minimum at x = 1. Only out by .009. Tells us a lot about how the shape of f behaves close the the SPs.

k = 6 at x = 1 (the min) but max k is 6.027.

Great video. Try it for ln(x) - it works!

[pnachtwey]

I find the equation can be helpful for calculating the acceleration during a turn but I prefer, or need to use, more brute force methods because I don't have a nice function to differentiate. I must use finite differences to calculate the derivative. Sometimes the odeint() will return the velocity and acceleration as well as the position as a function of time.

[Rozenkrantzz]

I used a substitution to simplify and it worked very nicely:
With x^{-2}/(1 + x^{-2})^{3/2} let u^2 = 1 + x^{-2}. Then 2u du = -2x^{-3} dx and du/dx > 0 everywhere. The formula for curvature becomes
k = (u^{2} - 1)/(u^{3}) = u^{-1} - u^{-3}.
Hence
dk/dx = (-u^{-2} + 3u^{-4})du/dx. Maximum curvature occurs when dk/dx = 0. Because du/dx > 0 it must be that -u^{-2} + 3u^{-4} = 0, and in particular, u^2 = 3. Hence 3 = 1 + x^{-2} => x = sqrt(1/2) => (sqrt(1/2), ln(sqrt(1/2)) is the point of maximum curvature

[theartisticactuary]

Glad you demonstrated that turning point of curvature was a maximum rather than a minimum.

But I'm docking a mark for leaving x in the answer as one over root two rather than half root two. I have standards!

[DeJay]

The thumbnail for the maximum curvature of e^x is better, this one I was surprised because I had never considered such a question for stuff like lnx.

[Samir-zbxk]

Another interesting question you could do on this topic is to find the average curvature of ln(x) on the interval x∈(0, 1)

[dan-florinchereches]

Am i the only one disliking the derivative of function division formula? I don't find it particularly useful as you still have calculations to do and simplifications most of the time. I find it nicer to consider it product of the top and bottom to power -1 and then do calculations

[Anmol_Sinha]

Can we get a derivation for the curvature formula?

[naterthan]

Why didn't you use the second derivative or sign test to check if the critical point was really a maximum? Also, why didn't you check for when the first derivative was undefined for critical points?

[disnagamage]

Please can any one tell me how to find the maximum or minimum of a curve when the first derivative does not exits;i am not talking about vertical asymptotes

[Tony]

I do not recall this part of multi variable calc....

[howeworth]

You should check with the second derivative that the critical number does indeed yield the maximum of κ.

[cyrusyeung]

Faster way to take derivative, is to divide top and bottom by x first

curvature
= x/(x² + 1)^(3/2)
= [x^(4/3) + x^(-2/3)]^(-3/2)

d/dx(curvature)
= (-3/2)[x^(4/3) + x^(-2/3)]^(-5/2)
[(4/3)x^(⅓) - (2/3)x^(-⅓)]

derivative = 0 if and only if
[(4/3)x^(⅓) - 2/3x^(-⅓)] = 0
(4/3)x^(⅓) = (2/3)x^(-⅓)
x^(2/3) = ½
x = 1/sqrt(2)

[worldnotworld]

Very nice! It might be fun to look at graphs of the center of radius of various functions too, as a new function of (x, f(x)).

(I've lately been wondering, taking strolls through the woods and watching the trees and branches and the features of path and sky "revolve" around me, around what center point they mathematically describe...)