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

This calculus tutorial goes over how to find the maximum curvature of the function f(x)=ln(x). We will also discuss what will happen to the curvature if x approaches infinity. This is a common Calculus 3 problem, which is included in the Multivariable Calculus textbook by James Stewart. Subscribe to @bprpcalculusbasics for more calculus tutorials. Here's a proof of the curvature formula: kzbin.info/www/bejne/oYq9dIdmjbSMq7s
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#calculus #bprpcalculus #apcalculus #tutorial #math

[@bprpcalculusbasics]

Maximum curvature of e^x: kzbin.info/www/bejne/pHPHl5mGi9-rjpI

[@appybane8481]

You can use that result and switch x and y to get this video's result instantly.

[@kennethgee2004]

so here is where things are not exactly as they appear. The equal sign there is not a true equality. As you said you do not feel the curvature, but it still is there. Your proof was given that there is no horizontal asymptote. It would be better to say that the value would be infinitesimal which i think the Greek letter epsilon in lower case is used. The way the formula sits with the equal zero it appears there is no curvature at all, but that is not true as the graph is forever increasing and between any two giving points there is still curvature.

[@Grecks75]

​@@kennethgee2004Still, the limit for the curvature is exactly zero as x grows to infinity. There is no need for any other words or interpretation. The very definition of "limit" tells you what this means: The curvature becomes smaller than any positive real value if you only consider large enough x.

[@kennethgee2004]

@@Grecks75 no false it is never zero as the function for ln x is ever increasing.

[@Grecks75]

@@kennethgee2004 You seem to have no clue what a limit is. Please return to the basics of Calculus and make yourself familiar with the concepts of "limit of a sequence" and "limit of a function". Those are very important concepts to understand in the beginning. Without profoundly understanding them you're gonna get nowhere in Calculus.

[@karelvanderwalt3625]

Nice application and integration of Calc I knowledge

[@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.

[@1001legoboy]

It was clear!

[@ianfowler9340]

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!

[@bprpcalculusbasics]

Hi Ian! You showed me how to prove the formula k(x) = [ y '' ] / [ 1 + (y ')^2 ] ^(3/2) previously but not this. This is cool that you just differentiated that formula to find a generalization of the max of k! I will investigate a bit more later between the point where k vs relative min/max happens. Thanks.

[@ianfowler9340]

@@bprpcalculusbasics You are very welcome. k ' (x) tells us how fast is the curvature changing per unit of arclength. It's almost like some strange kind of accelaeration of that angle theta in d(theta)/ds.

[@ianfowler9340]

@@bprpcalculusbasics Another thought. If y is differentiable then we can find the equation of the path of the center of curvature. I did it for y = x^2 and the equation of the path of the center of curvature is : y = 3(x/4)^(2/3) + 1/2. It all comes together so nicely. You can graph both curves and then plot the circle of curvature at x = 1.

[@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!

[@Samir-zb3xk]

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

[@bprpcalculusbasics]

Ah, I can see why!

[@Samir-zb3xk]

​@@bprpcalculusbasicsA simalar question like that would also work nicely if f(x)=cosh(x)

[@NotBroihon]

Spoiler: 1 - 1/√2 if I'm not mistaken

[@Samir-zb3xk]

@@NotBroihon yes

[@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...)

[@Anmol_Sinha]

Can we get a derivation for the curvature formula?

[@erenoguz3297]

kzbin.info/www/bejne/oYq9dIdmjbSMq7ssi=X3PUTCHPKsBglYu2

[@Ninja20704]

He has videos on it already on the calculus channel

[@frood1499]

It’s in the description too 😊

[@Anmol_Sinha]

​@frood1499 ohhh, i did look there but missed it 😅

[@Nerdwithoutglasses]

google it, don't ask for something you could find yourself easily

[@dan-florinchereches4892]

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

[@anghme28ang11]

Its generally the same to i use either one

[@unturnedd]

me too, i did not even bother memorizing the quotient rule i just use the product

[@carultch]

Here's how you can remember it: Low d high Minus high d low All over low squared down below

[@Grecks75]

You are not alone, I think exactly the same. Most of the time I just use the product rule plus the derivative of 1/x together with the chain rule. It doesn't make a huge difference but sometimes there is a smaller power in the denominator in intermediate steps which can assist in simplifications.

[@Nerdwithoutglasses]

Your thoughts are understandable: Let's say you want to find critical points of f/g, you need the know how the sign of (f/g)' change. If you use the product rule, you get (f/g)'=f'*(1/g)+f*(-1/g^2)*g' ,now to get the sign, you have to turn (f/g)' into its "product" form (f/g)'=(f'*g-f*g')/g^2, that's the quotient rule! You see the quotient rule is only helpful if you are dealing with simple critical point problems (since you can do 2 steps in 1). However, if the function (in fraction) is "not simple" and you need to find its next derivative, using quotient rule is nonsense (doing nothing but increasing expression messiness), instead, just go ahead and use product rule or its generalization-Leibniz product rule . Try asking yourself (or your fellows) what technique we should use in the following problem: Find second derivative of cos(x)*ln(x)/x^2

[@howeworth]

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

[@Samir-zb3xk]

Do you really want to take the derivative of that again? You can verify that the maximum of κ(x) occurs at x=1/√2 by showing that it is increasing before x=1/√2 and decreasing after x=1/√2 (through the first derivative only)

[@cyrusyeung8096]

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)

[@bprpcalculusbasics]

Ah! Very nice!

[@naterthan5569]

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?

[@Samir-zb3xk]

You can clearly see at a glance (x²+1)^(5/2) = 0 has no real solutions You're right, he should have verified that the critical number corresponds to a maximum.

[@disnagamage131]

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

[@dbliss314]

Places where the first derivative does not exist are just like the places where it equals zero. They are candidates for being a maximum or minimum. All local maxima and minima either have f'(x)=0 or f'(x) does not exist. To give you more I would need an example.

[@disnagamage131]

@@dbliss314 here's a one f(x)= |x| at x = o

[@disnagamage131]

I need an algebraic explanation

[@emilioturbay667]

I miss the BP....

[@joshuahillerup4290]

I was hoping he would write ln`x and ln``x