Radius of curvature formula when the denominator = 0

[nomadreid]

TL;DR Summary

If f(x) = 0 , then the formula for the radius of curvature at x=a fails for f"(a)=0. But, e.g., it would seem that y=x^4 at x=0 should have just as much right to a radius of curvature as y=x^2 at x=0, and the latter has one (1/2). So is there a different formula when the standard formula fails [except for straight lines, points of inflection, etc. where the radius is infinity (i.e, curvature 0)]?

The formula for the radius of curvature.

which fails, for example, at y=x^n (n>2) at x=0, which makes sense for odd n, but not for even n.
That is, the curvature 1/R =0 for those examples, which seems wrong. What am I approaching incorrectly?
Thanks.
(formula copied from https://mathworld.wolfram.com/RadiusofCurvature.html )

 

[anuttarasammyak]

That is, the curvature 1/R =0 for those examples, which seems wrong.

Why wrong ? Does 1/R=0 cause any problem ?

 

[nomadreid]

Why wrong ? Does 1/R=0 cause any problem ?

Mathematically, no: after all, the curvature 1/R =0 for straight lines; no worries. However, whereas the interpretation of having no curvature for a straight line is OK, the interpretation of zero curvature at x=0 for y=x^4 seems odd. Put another way: why should y=x^2 be considered curved at x=0, yet y=x^4, which bears a strong resemblance to the parabola, be considered flat at that point?

 

[anuttarasammyak]

I do not observe the resemblance. Can you draw the circle R ?

 

[Ibix]

Isn't it to do with the fact you have four roots there?

 

[anuttarasammyak]

We may be able to regard it from Taylor expansion.
$$y=x^4$$
does not have x^2 term of expansion around zero. The circle R is quadratic function touching (0,0).
$$x^2+(y-R)^2=R^2$$
$$y=\frac{1}{2R}x^2+\frac{y^2}{2R}=\frac{1}{2R}x^2+o(\frac{x^4}{R^3})$$

No x^2 expansion term leads No circle R or infinite R.

 

[nomadreid]

Thanks for the replies, anuttarasammyak and Ibix. Before getting to details, I want to make my confusion a little bit more explicit. A radius of curvature, or more directly its curvature, is supposed to say two things: (a) "This graph is curved, i.e., not straight, in any small but finite neighborhood of this point" and (b) to give some measure to this fact. It is a safe bet to say that the graph of y=x⁴ is not a straight line at x=0, but that it is less curved at x=0 than the curve, say, of y=x² at that point. Alas, this formula does not seem to provide this measure.

Now,

I do not observe the resemblance. Can you draw the circle R ?

If I could, that would mean I would know how to compute R, but that is precisely my question. The rough definition of the radius of curvature given by Mathworld is the radius of the osculating circle. Fine. There are two definitions of the osculating circle given;
(1) a circle having the same tangent and the same curvature. (2) The circle best approximating the curve.
(1) becomes circular, and in (2), "best approximates" is vague. Perhaps
( purple is the curve y=x⁴)

this circle better approximates the curve in a small neighborhood of zero, even though in a larger neighborhood perhaps this one does

How to tell what "best" means?
So no, I cannot draw the circle.

Isn't it to do with the fact you have four roots there?

Could you be more explicit?

We may be able to regard it from Taylor expansion.
$$y=x^4$$
does not have x^2 term of expansion around zero. The circle R is quadratic function touching (0,0).
$$x^2+(y-R)^2=R^2$$
$$y=\frac{1}{2R}x^2+\frac{y^2}{2R}=\frac{1}{2R}x^2+o(\frac{x^4}{R^3})$$

No x^2 expansion term leads No circle R or infinite R.

(typo: leads to circle...?)
I am not sure I understand the reply: are you saying that this attempt does not lead us anywhere?

 

[Ibix]

Could you be more explicit?

Consider something like $y=(x-a)^2(x+a)^2$. It has a maximum of height $a^4$ between two zeros separated by $2a$. As $a\rightarrow 0$ the height of the bump drops faster than the width - it is getting flatter.

 

[anuttarasammyak]

y=x is not a curve. It has finite R(x) nowhere.

y=x^2 is a curve. It has finite R(x) everywhere.

y=x^3 has finite R(x) except x=0. It is a curve.

y=x^4 has finite R(x) except x=0. It is a curve.

You might have too strict sense that a curve should have finite R(x) i.e. non zero curvature, everywhere.

 

[anuttarasammyak]

There are two definitions of the osculating circle given;

Circle R is made from three points on it ; the point on the curve we are looking at, its right neighbor point and left neighbor point both on the curve with taking limit of their closeness.

 

[nomadreid]

Circle R is made from three points on it ; the point on the curve we are looking at, its right neighbor point and left neighbor point both on the curve with taking limit of their closeness.

View attachment 347528

Interesting, assuming that "closeness" refers to the distance between the point on the curve and the point on the circle of the same x-value, then those arrows would be pointing down until reaching an interval approximately (-0.5, 0.5), then would reverse direction as the radius of curvature increases
[y on the graph is the radius of curvature as calculated by the above formula.

Anyway, what you are saying, that y=x^4 simply doesn't have a radius of curvature at x=0, as the corresponding limit tends to infinity, makes perfect mathematical sense, as done defining

y=x is not a curve. It has finite R(x) nowhere.

y=x^2 is a curve. It has finite R(x) everywhere.

y=x^3 has finite R(x) except x=0. It is a curve.

y=x^4 has finite R(x) except x=0. It is a curve.

You might have too strict sense that a curve should have finite R(x) i.e. non zero curvature, everywhere.

I just rankles that this interpretation is counter-intuitive to the usual concept of curvature as a measure of the difference to a straight line. [Put another way: it makes a teacher's life trickier to explain this concept of curvature to a secondary school student who is being told that something that, to her, is clearly curved, is for the wacky mathematicians not curved.]

I am not sure, but perhaps Ibix's reply

Consider something like $y=(x-a)^2(x+a)^2$. It has a maximum of height $a^4$ between two zeros separated by $2a$. As $a\rightarrow 0$ the height of the bump drops faster than the width - it is getting flatter.

was proposing that one alters the definition a bit so that the curve near zero should be dampened to give some finite value at 0 (sort of like

)

[ Ibix: if this was not your intention, please correct me.]

 

[anuttarasammyak]

Interesting, assuming that "closeness" refers to the distance between the point on the curve and the point on the circle of the same x-value, t

y=x4
We are looking at $(0,0)$
Left neighbor point $(−\delta,\delta^4)$
Right neighbor point $(+\delta,\delta^4)$
Circle R : $$x^2+(y−R)^2=R^2$$
The three points are on Circle R. We get the relation between R and $/delta$
$$R=\frac{1+δ^6}{2δ^2}\rightarrow \infty$$
for
δ→0
We can perceive it by the Figure

As the right and the left points on the violet curve y=x4 approach to (0,0) , radius of the green circle on which the three points becomes larger and larger, and diverges in the limit above said.

$$y=x^2$$
We are looking at $(0,0)$
Left neighbor point $(−\delta,\delta^2)$
Right neighbor point $(+\delta,\delta^2)$
Circle R : $$x^2+(y−R)^2=R^2$$
The three points are on Circle R. We get the relation between R and δ
$$R=\frac{1+\delta^2}{2} \rightarrow \frac{1}{2}$$
for
δ→0

I just rankles that this interpretation is counter-intuitive to the usual concept of curvature as a measure of the difference to a straight line. [Put another way: it makes a teacher's life trickier to explain this concept of curvature to a secondary school student who is being told that something that, to her, is clearly curved, is for the wacky mathematicians not curved.]

In your intuition is the letter S a curve ? If yes, how much is the curvature in the middle point ? Mathematics says it is zero. Does this contradict your intuition ?

In your intuition is
$$y=\sin ⁡x$$ a curve ? If yes, how much is the curvature at x=0 ? Mathematics says it is zero. Does this contradict your intuition ?

 

[WWGD]

Notice there's essentially just one tangent line to $y=x^4$ at the two sides of $0$, while there are two for $y=x^2$.

 

[nomadreid]

In your intuition is the letter S a curve ? If yes, how much is the curvature in the middle point ? Mathematics says it is zero. Does this contradict your intuition ?

In your intuition is
$$y=\sin ⁡x$$ a curve ? If yes, how much is the curvature at x=0 ? Mathematics says it is zero. Does this contradict your intuition ?

That the radius of curvature becomes infinite as x approaches zero for y=x⁴ is immediate from the original formula; your calculation asking what open infinitesimal neighborhood of (0,0) the circle can coincide with the curve (answer: none) is an interesting strengthening of the definition I quoted earlier as the circle that "best approximates" the circle. Given this strengthening, of course the result becomes more reasonable.

It still goes against the more subjective intuition of curvature, in that there is no neighborhood, of the curve at that point which is a straight line. That such inflection points have no curvature does fit my intuition, because as the curve is concave on one side of the point and convex on the other, then labeling one of them a positive curvature and the other one a negative curvature, the idea of a zero curvature is natural. But looking at the graph of y=x⁴, it does not seem obvious that the concavity changes at x=0. To put it another way, one can draw a tangent, and the points corresponding to δ and -δ will both be onthe same side of the tangent.. (Not the case in the examples you gave.) So whereas a proper intuition will allow for the zero curvature of y=x³ at x=0, it struggles at y=x⁴

Notice there's essentially just one tangent line to $y=x^4$ at the two sides of $0$, while there are two for $y=x^2$.

I would appreciate an explanation in more detail why it is incorrect that a tangent at a point to the right has negative slope, and a tangent to the left has positive slope,

 

[anuttarasammyak]

To put it another way, one can draw a tangent, and the points corresponding to δ and -δ will both be onthe same side of the tangent.. (Not the case in the examples you gave.) So whereas a proper intuition will allow for the zero curvature of y=x3 at x=0, it struggles at y=x4

$$R(x)=\frac{(1+y'^2)^{3/2}}{y^"}$$
$$y=x^2\ : \ R(x)=\frac{(1+4x^2)^{3/2}}{2}$$
$$y=x^3\ : \ R(x)=\frac{(1+9x^4)^{3/2}}{6x}$$
$$y=x^4\ : \ R(x)=\frac{(1+16x^6)^{3/2}}{12x^2}$$
The graphs of their inverse, i.e. curvature are

I confirm your #11 that, though its curvature is zero at Origin, y=x^4 has peaks of curvature in the both sides. y=x^2 and y=x^4 have different features of curving. This twin peaks feature is more clear for higher n y=x^n

Now we may say that curvature zero at a point does not have to mean that the figure is not curved whether it is one side of tangential line or not.
Your intuition on curvature would work better on a segment than on a point.
$$\int_0^L k(s) ds$$
where s is length parameter, k(s) is curvature, L is length of segment, which results crossing angle of tangential lines of segment’s two ends. Zero curvature on some points do not prevent a segment to curve.

 

[WWGD]

That the radius of curvature becomes infinite as x approaches zero for y=x⁴ is immediate from the original formula; your calculation asking what open infinitesimal neighborhood of (0,0) the circle can coincide with the curve (answer: none) is an interesting strengthening of the definition I quoted earlier as the circle that "best approximates" the circle. Given this strengthening, of course the result becomes more reasonable.

It still goes against the more subjective intuition of curvature, in that there is no neighborhood, of the curve at that point which is a straight line. That such inflection points have no curvature does fit my intuition, because as the curve is concave on one side of the point and convex on the other, then labeling one of them a positive curvature and the other one a negative curvature, the idea of a zero curvature is natural. But looking at the graph of y=x⁴, it does not seem obvious that the concavity changes at x=0. To put it another way, one can draw a tangent, and the points corresponding to δ and -δ will both be onthe same side of the tangent.. (Not the case in the examples you gave.) So whereas a proper intuition will allow for the zero curvature of y=x³ at x=0, it struggles at y=x⁴


I would appreciate an explanation in more detail why it is incorrect that a tangent at a point to the right has negative slope, and a tangent to the left has positive slope,

Sorry, I was talking in mostly an informal sense, in that the two arcs of $y=x^4$ approached the x-axis enough that they are more line-like , than arc-like. I'll try to make the claim more formal, rigorous.

 

[nomadreid]

$$R(x)=\frac{(1+y'^2)^{3/2}}{y^"}$$
$$y=x^2\ : \ R(x)=\frac{(1+4x^2)^{3/2}}{2}$$
$$y=x^3\ : \ R(x)=\frac{(1+9x^4)^{3/2}}{6x}$$
$$y=x^4\ : \ R(x)=\frac{(1+16x^6)^{3/2}}{12x^2}$$
The graphs of their inverse, i.e. curvature areView attachment 347582
I confirm your #11 that, though its curvature is zero at Origin, y=x^4 has peaks of curvature in the both sides. y=x^2 and y=x^4 have different features of curving. This twin peaks feature is more clear for higher n y=x^n View attachment 347584
Now we may say that curvature zero at a point does not have to mean that the figure is not curved whether it is one side of tangential line or not.
Your intuition on curvature would work better on a segment than on a point.
$$\int_0^L k(s) ds$$
where s is length parameter, k(s) is curvature, L is length of segment, which results crossing angle of tangential lines of segment’s two ends. Zero curvature on some points do not prevent a segment to curve.

Thanks, anuttarasammyak The formula at the end of your post looks interesting, but I am not sure how to interpret everything due to the odd phrasing at the end: "which results crossing angle of tangential lines of segment’s two ends" Allow me to take an example and one can then correct me on the basis of the example. Let's take the simple parabola y=x² over the interval [-1,1].
s = x
[Edit: I had taken the formula for the radius of curvature rather than the curvature for k(s), resulting in an incorrect integration. The correct calculation follows.]
k(x) =2/(1+4x²)^3/2 (using the reciprocal of the formula in the original post)
L=2
Then

So the definite integral from 0 to 2 would be 4/√17.

The crossing of the tangents to the two points (-1,1) and (1,1) [corresponding to the endpoints of the interval] would be the point (0.-1), which I am not sure how to tie in with the above.

As this does not advance my intuition, I conclude that I am doing something wrong. I would much appreciate being corrected. Thanks.

 

[anuttarasammyak]

Let's take the simple parabola y=x2 over the interval [-1,1].

$$\int k ds = \int k \frac{ds}{dx}dx=\int \frac{y^"}{(1+y'^2)^{3/2}} (1+y'^2)^{1/2}dx=\int \frac{y^"}{1+y'^2}dx$$
y=x^2, x[0,X]
$$\int k ds = tan^{-1}2X$$
It is in accordance with y'(X)=2X.

 

[nomadreid]

Thanks, anuttarasammyak. I shall mull over this (correcting the obvious typos: missing integration signs, and y' =2x instead of y" = 2x) for a while. It looks good; having a curvature for an interval makes sense.