Formula for a curvature

[Mondo]

Hi,

I read this document https://www-users.cse.umn.edu/~mille003/1372curvature.pdf which I found while googling on curvature formula derivation. This is quite well described but I got stuck at one equation, right in the middle. The author says that curvature is essentially a derivative of an angle which position vector makes with a positive X axis, namely [imath]\frac{d\phi}{dt}[/imath]. That of course makes a perfect sense. Then author says that we can also get a measure of how fast the curve is turning by focusing on the curve alone and shows equation [math]k = \frac{d\phi}{ds} = \frac{d\phi}{dt}\frac{dt}{ds}[/math] with a note that in the last equality he used a chain rule. Here is my problem - how can a derivative of angle in respect to path become a derivative of angle in respect to time, multiplied by a derivative of time in respect to path? Especially the last part is confusing and I have a hard time relating it to physical world - a derivative of time in respect to path? What is it? The converse would be a speed.

 

[Dr.Peterson]

Hi,

I read this document https://www-users.cse.umn.edu/~mille003/1372curvature.pdf which I found while googling on curvature formula derivation. This is quite well described but I got stuck at one equation, right in the middle. The author says that curvature is essentially a derivative of an angle which position vector makes with a positive X axis, namely [imath]\frac{d\phi}{dt}[/imath]. That of course makes a perfect sense. Then author says that we can also get a measure of how fast the curve is turning by focusing on the curve alone and shows equation [math]k = \frac{d\phi}{ds} = \frac{d\phi}{dt}\frac{dt}{ds}[/math] with a note that in the last equality he used a chain rule. Here is my problem - how can a derivative of angle in respect to path become a derivative of angle in respect to time, multiplied by a derivative of time in respect to path? Especially the last part is confusing and I have a hard time relating it to physical world - a derivative of time in respect to path? What is it? The converse would be a speed.

By "converse", I presume you mean "reciprocal".

If t is in seconds, and s is in meters, then [imath]\frac{d\phi}{dt}[/imath] is in radians per second, [imath]\frac{ds}{dt}[/imath] is in meters per second, and [imath]\frac{dt}{ds}[/imath] is in seconds per meter: reciprocal speed. The latter is how many seconds it takes to go one meter along the path.

It might make more sense to you if you write [imath]k = \frac{d\phi}{dt}\frac{dt}{ds}[/imath] as [math]k = \frac{\frac{d\phi}{dt}}{\frac{ds}{dt}},[/math] that is, the rate of change of angle in radians per second, divided by the speed in meters per second. This gives the rate of change of angle per meter along the path. This makes it independent of how fast you are driving along the path, correcting for that dependence in the initial (wrong) definition of curvature as mere rate of change of angle.

 

[Mondo]

@Dr.Peterson thank you for an answer.

By "converse", I presume you mean "reciprocal".

Yes, a reciprocal which is speed in this case.

This makes it independent of how fast you are driving along the path...

How is it independent of speed if there is derivative in respect of time which exactly says "How fast?"
To be honest I still don't see the physical sense of [imath]\frac{dt}{ds}[/imath]. How can time depend on path?
You also said:

.. correcting for that dependence in the initial (wrong) definition of curvature as mere rate of change of angle.

The definition of curvature as a speed in which an angle changes actually makes sense to me. Why do you think it is wrong or incomplete?

Thank you.

 

[Dr.Peterson]

The definition of curvature as a speed in which an angle changes actually makes sense to me. Why do you think it is wrong or incomplete?

Because if you drive faster, the angle will change faster! If curvature is to be a property of the curve itself, and not of how you move along it, then it must be independent of your speed (that is, of how you move along it).

When you said,

The author says that curvature is essentially a derivative of an angle which position vector makes with a positive X axis, namely [imath]\frac{d\phi}{dt}[/imath].

you were misstating what he said, which was actually

It should be clear from this that the derivative dφ/dt gives information about how fast the curve is turning, and whether it is
turning in a clockwise or counterclockwise direction

He is not saying that this is what curvature is. He then makes the adjustment I mentioned, defining curvature instead as dφ/ds:

To get a measure of how fast the curve is turning that depends on the curve alone, and not the specific parameterization, we fix on arc length s as a standard parameterization for the curve. Thus the curvature k at a point (x, y) on the curve is defined as the derivative k = dφ/ds = dφ/dt dt/ds, where we have used the chain rule in the last equality.

In effect, the curvature is the rate of change of angle if you drive along it at a unit rate (say, 1 meter per second). It's the change of angle per meter.

How is it independent of speed if there is derivative in respect of time which exactly says "How fast?"
To be honest I still don't see the physical sense of \(\frac{dt}{ds}\). How can time depend on path?

We've divided out the speed, to leave a rate that depends only on the curve, not on your speed along it.

Again, I suggested that you think of it not as a multiplication by [imath]\frac{dt}{ds}[/imath], but as division by [imath]\frac{ds}{dt}[/imath], which, again, cancels out the dependence on how you are "driving".

How can time depend on path?

Not on path (that's the curve itself), but on motion (which combines the path and how you move along it). Are you thinking that "s" means the path, and not "distance along the path"?

I suspect, too, that you may be missing the idea of a curve having different parameterizations. For example, if I have a curve, say x=t, y=t^2, and replace t with, say, u=t/2 so that t=2u, I get x=2u, y=4u^2, which is the same curve, but traversed twice as fast at each point. If I made a different substitution, the speed might be varying in a different way, not just proportional to the original. By changing the parameter to s, we are changing to a constant speed.

 

[Mondo]

When you said,

Mondo said:
The author says that curvature is essentially a derivative of an angle which position vector makes with a positive X axis, namely \frac{d\phi}{dt}dtdϕ.
Click to expand...

you were misstating what he said, which was actually

It should be clear from this that the derivative dφ/dt gives information about how fast the curve is turning, and whether it is
turning in a clockwise or counterclockwise direction

He is not saying that this is what curvature is. He then makes the adjustment I mentioned

Well he also said:

This information is, essentially, what we mean by the curvature of the curve at the point (x(t), y(t))

Yes I rephrased what he said but nonetheless this sentence suggests that [imath]\frac{d\phi}{dt}[/imath] is what we define by curvature. I think the essence is that this is a curvature in a point only. Author then clearly says that it depends on parametrization hence not a generic measure.

We've divided out the speed, to leave a rate that depends only on the curve, not on your speed along it.

Yes this makes sense from "physical" perspective. From a mathematical one, I have a problem with how this chain rule was applied. In general we can apply it in a situation where we have two functions like [imath]y = f(u)[/imath] and [imath]u = g(x)[/imath] then [imath]\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}[/imath]. But here if I define [imath]\phi = y(s)[/imath] and [imath]s = u(t)[/imath] then [imath]\frac{d\phi}{ds}[/imath] should just end here as [imath]\phi[/imath] does not depend on [imath]t[/imath] (time). On the other hand if I define [imath]\phi = y(t)[/imath] then [imath]\frac{d\phi}{ds} = 0[/imath]. The only arrangement where I can apply chain rule is when I set [imath]\phi = y(t)[/imath] so phi depends on time only and [imath]t = u(s)[/imath] time depends on arc length which is very awkward but this finally leads to: [imath]\frac{d\phi}{ds} = \frac{d\phi}{dt} \frac{dt}{ds}[/imath]. What is wrong here?

 

[Deleted member 4993]

so phi depends on time only and t=u(s)t = u(s)t=u(s) time depends on arc length

But that is how we measure "time" - using an old-fashioned pocket-watch!!!

 

[Mondo]

But that is how we measure "time" - using an old-fashioned pocket-watch!!!

@Subhotosh Khan but [imath]t = u(s)[/imath] means we measure time not by pocket watch but by distance we went

 

[Deleted member 4993]

Do you agree s = f(t)

Then if 'u' is inverse function of 'f', we have

t = f^-1(s) = u(s)

 

[Dr.Peterson]

The only arrangement where I can apply chain rule is when I set [imath]\phi = y(t)[/imath] so phi depends on time only and [imath]t = u(s)[/imath] time depends on arc length which is very awkward but this finally leads to: [imath]\frac{d\phi}{ds} = \frac{d\phi}{dt} \frac{dt}{ds}[/imath]. What is wrong here?

It is given that x and y are continuously differentiable functions of parameter t, so vector r is a function of t, and he shows how to define arc length s as a function of t. It is assumed that ds/dt > 0 ("the particle never stops"). This implies that, at least locally, s is an invertible function of t, so t is a function of s. Therefore dt/ds makes perfect sense.

Using your notation,

But here if I define [imath]\phi = y(s)[/imath] and [imath]s = u(t)[/imath] then [imath]\frac{d\phi}{ds}[/imath] should just end here as [imath]\phi[/imath] does not depend on [imath]t[/imath] (time). On the other hand if I define [imath]\phi = y(t)[/imath] then [imath]\frac{d\phi}{ds} = 0[/imath].

we have [imath]\phi = y(t)[/imath] and [imath]t = u^{-1}(s)[/imath], so [imath]\phi = y(u^{-1}(s))[/imath], and we have the proper setup for the chain rule: [imath]\phi[/imath] depends on t, which depends on s. It's silly to claim that [imath]\phi[/imath] is a constant function of s.

 

[Mondo]

It is very clever! I was confused by the unusual and weird looking [imath]\frac{dt}{ds}[/imath]. This is just a neat trick to express angle in terms of "s" so it is independent of speed or any other parameter.
The rest of the document is just a pure algebra and is rather straightforward. However they wrote [imath]\frac{dt}{ds} = \frac{1}{\sqrt{x^2 + y^2}dt}[/imath]. The one in numerator is because dt is understood as derivative of t in respect to t which is just one, right?

 

[Dr.Peterson]

Are you not aware of the derivative of an inverse function? None of this should surprise you.

 

[Mondo]

@Dr.Peterson, yes I missed that. Thank you!
To wrap this up I have a few more questions:
1. I try to make a note on the crucial part which is [imath]k = \frac{d\phi}{ds} = \frac{d\phi}{dt} \frac{dt}{ds}[/imath] and I write "...Since phi (angle function) and s (a distance) both depend on time we need to express t (time) using s, so we know how a change in s affects input argument of phi - a time...". Is my note correct here?

2. [imath]r(t) = x(t)i + y(t)j[/imath] hence [imath]tan_\phi(t) = \frac{y(t)}{x(t)}[/imath] but he has this tangent as [imath]\frac{\dot y}{\dot x}[/imath]. Yes I know that tangent is a derivative but it can't be both fractions in the same time, where is the discrepancy?

3. I noticed in many physical texts curvature is often simplified to just [imath]k = \frac{1}{R}[/imath] which is the result for a circle with radius R. It is a great simplification especially that we can always at least locally match it to some circle.

 

[Deleted member 4993]

2. r(t)=x(t)i+y(t)jr(t) = x(t)i + y(t)jr(t)=x(t)i+y(t)j hence tanϕ(t)=y(t)x(t)tan_\phi(t) = \frac{y(t)}{x(t)}tanϕ(t)=x(t)y(t) but he has this tangent as y˙x˙\frac{\dot y}{\dot x}x˙y˙. Yes I know that tangent is a derivative but it can't be both fractions in the same time, where is the discrepancy?

If x(t) = r(t) * cos[φ(t)] and y(t) = r(t) * sin[φ(t)] then what are the expressions for \(\displaystyle \frac{dx}{dt}\) and \(\displaystyle \frac{dy}{dt}\)?

 

[Dr.Peterson]

@Dr.Peterson
1. I try to make a note on the crucial part which is [imath]k = \frac{d\phi}{ds} = \frac{d\phi}{dt} \frac{dt}{ds}[/imath] and I write "...Since phi (angle function) and s (a distance) both depend on time we need to express t (time) using s, so we know how a change in s affects input argument of phi - a time...". Is my note correct here?

It's not how I'd say it, but not incorrect.

@Dr.Peterson
2. [imath]r(t) = x(t)i + y(t)j[/imath] hence [imath]tan_\phi(t) = \frac{y(t)}{x(t)}[/imath] but he has this tangent as [imath]\frac{\dot y}{\dot x}[/imath]. Yes I know that tangent is a derivative but it can't be both fractions in the same time, where is the discrepancy?

I presume you mean [imath]\tan\phi(t) = \frac{y(t)}{x(t)}[/imath], without the subscript. (You can use Preview to check your formatting.)

This would be true if [imath]\phi[/imath] were defined as the direction from the origin to the point (x,y); but it isn't. It's the direction of motion along the curve, which is defined by the derivative, not the point itself.

@Dr.Peterson
3. I noticed in many physical texts curvature is often simplified to just [imath]k = \frac{1}{R}[/imath] which is the result for a circle with radius R. It is a great simplification especially that we can always at least locally match it to some circle.

If you already knew the radius of curvature for a given curve, then obviously this would be a simplification. But you don't, do you?

In fact, the radius of curvature can be defined as R = 1/k !

 

[Mondo]

[imath]\phi[/imath]Mondo said:
@Dr.Peterson
1. I try to make a note on the crucial part which is [imath]k = \frac{d\phi}{ds} = \frac{d\phi}{dt} \frac{dt}{ds}[/imath] and I write "...Since phi (angle function) and s (a distance) both depend on time we need to express t (time) using s, so we know how a change in s affects input argument of phi - a time...". Is my note correct here?

It's not how I'd say it, but not incorrect.

Heh I can imagine that. How would you say it compactly?

I presume you mean [imath]\tan\phi(t) = \frac{y(t)}{x(t)}[/imath] without the subscript. (You can use Preview to check your formatting.)

Yes that's what I meant. BTW, I feel like the quoting mechanism here does not work well - it does not care about latex + does not quote nested responses like I want to quote your reply which quotes my earlier one.

Okay so I overlooked what angle is assumed by [imath]\phi[/imath]. Anyway I can NOT visualize it by plotting [imath]r(t)[/imath] vector from origin to some point on curve and then a perpendicular [imath]\dot r[/imath] together with the angle right? It is just taken from derivative definition.

If you already knew the radius of curvature for a given curve, then obviously this would be a simplification. But you don't, do you?

Yes I don't.


Thank you!

 

[Mondo]

Hi,

I would like to ask a follow up question on [imath]tan \phi = \frac{\dot y}{\dot x}[/imath] I still have a problem visualizing it. I created a crude sketch in pint to visualize the problem.


There is a curve [imath]k[/imath], a position vector [imath]r[/imath]and a speed vector [imath]\dot r[/imath]. Now, we know that a derivative of [imath]r[/imath] will be tangent to a position vector and that this tangent makes [imath]\phi[/imath] with positive x axis. (the last fact is given in the text, not some mathematical property). Also [imath]x[/imath] and [imath]y[/imath] are components of [imath]r[/imath]. But how can we tell what [imath]\dot X [/imath] and [imath]\dot Y[/imath] are?

PS: dotted lines are just to show the point of intersection with cartesian axes.

Thank you.

 

[Dr.Peterson]

There is a curve [imath]k[/imath], a position vector [imath]r[/imath]and a speed vector [imath]\dot r[/imath]. Now, we know that a derivative of [imath]r[/imath] will be tangent to a position vector and that this tangent makes [imath]\phi[/imath] with positive x axis. (the last fact is given in the text, not some mathematical property).

The "fact [that] is given" is just how they are defining [imath]\phi[/imath], so it doesn't have to be proved.

I wouldn't say "a derivative of [imath]r[/imath] will be tangent to a position vector"; you mean tangent to the curve traced by the position vector, right?

And technically, your [imath]\phi[/imath] is on the wrong side of the tangent, since you say it should be the angle with the positive x axis:

​

But those are just details, not your actual question.

Also [imath]x[/imath] and [imath]y[/imath] are components of [imath]r[/imath]. But how can we tell what [imath]\dot X [/imath] and [imath]\dot Y[/imath] are?

Do you mean [imath]\dot x [/imath] and [imath]\dot y[/imath], or are you using upper case with a different meaning?

And what are you actually asking? You can't see these two quantities on the picture, because they are rates of change, and the picture is static. But if you look at points P and Q here as point R moves,

​

then [imath]\dot x [/imath] and [imath]\dot y[/imath] are the speeds with which P and Q move, as indicated by arrows. And they are the components of the vector velocity of point R.

The angle "phi" should be the angle at the origin between the positive x-axis and the position vector 'r'

No, I think you're missing the dots, representing time derivatives.

 

[Mondo]

The angle "phi" should be the angle at the origin between the positive x-axis and the position vector 'r'

No it isn't. That's what I thought oryginally to but in the document I linked in the first post you san see that this is the angle a tangent makes wit positive X axis.

 

[Deleted member 4993]

No, I think you're missing the dots, representing time derivatives.

Yes ... so I did. I removed my response now.

 

[Deleted member 4993]

No it isn't. That's what I thought oryginally to but in the document I linked in the first post you san see that this is the angle a tangent makes wit positive X axis.

I removed my response.