Formula for a curvature

[Mondo]

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

Yes that is was I meant - they just assume this angle.

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

Yes, I should have said it would be perpendicular to position vector and tangent to the curve that position vector traces. A position vector is just a point so I think we can't speak on a tangency with a single point.

And technically, your ϕ\phiϕ is on the wrong side of the tangent, since you say it should be the angle with the positive x axis:

Good point, thanks for spotting it.

Do you mean x˙\dot x x˙ and y˙\dot yy˙, or are you using upper case with a different meaning?

I really don't know why I made them uppercase.. but yes I need to get [imath]\dot x \text{ and } \dot y[/imath]

then x˙\dot x x˙ and y˙\dot yy˙ 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.

Yes this is exactly what I wanted. The question also is if we can get the length of these velocity vectors, namely [imath]\dot x \text{ and } \dot y[/imath] geometrically?


PS: Pity math symbols don't work when quoting... I would need to fix them manually each time I quote somebody response with math expression which is error prone. Would be good to fix that.

 

[Dr.Peterson]

Yes, I should have said it would be perpendicular to position vector and tangent to the curve that position vector traces. A position vector is just a point so I think we can't speak on a tangency with a single point.

No, the tangent vector is only perpendicular to the position vector in the case of a circle centered at the origin. All you can say in general is that it is tangent to the curve (by definition).

Yes this is exactly what I wanted. The question also is if we can get the length of these velocity vectors, namely [imath]\dot x \text{ and } \dot y[/imath] geometrically?

Didn't I already answer that? The velocity vector depends on speed, which is not shown on the graph, so you can't derive it from what you see on the graph. The best you can do is to work backward: Given the tangent vector, you can find its components. If you draw the tangent line, you can arbitrarily choose a vector along it.

PS: Pity math symbols don't work when quoting... I would need to fix them manually each time I quote somebody response with math expression which is error prone. Would be good to fix that.

I agree; you have to quote the entire post (using the Reply button), which does get math right, and remove what you don't want. I just did that above.

 

[Mondo]

Didn't I already answer that? The velocity vector depends on speed, which is not shown on the graph, so you can't derive it from what you see on the graph. The best you can do is to work backward: Given the tangent vector, you can find its components. If you draw the tangent line, you can arbitrarily choose a vector along it.

"Given the tangent vector" - yes once I have it on a graph I can find its components. But the question is if I can get this tangent vector from the position vector. Say [imath]r(t) = x [/imath] so the path is really a infinite straight line from origin, that makes 45 degrees with positive X axis. The derivative is of course 1 and in this case the angle that path and its tangent make with positive X axis is the same. Now, how should I locate the velocity vector on the graph? I found it is equal to 1 but where should it start, end and what would be its angle in relation to path?

 

[Dr.Peterson]

"Given the tangent vector" - yes once I have it on a graph I can find its components. But the question is if I can get this tangent vector from the position vector.

No -- no more than a policeman can tell how fast you are going from seeing where you are right now - or even from seeing a map of where you have been going. The graph, or the map, contains no time information!!

Say [imath]r(t) = x [/imath] so the path is really a infinite straight line from origin, that makes 45 degrees with positive X axis.

This is meaningless. I think [imath]r(t) = x [/imath] is supposed to be the position vector [imath]\mathbf{r}(t) = x [/imath]. But x is not a vector! And it is not an expression in t, so it isn't a function of t!

It appears that you are confusing [imath]r(t)[/imath] with [imath]y[/imath]; [imath]y = x[/imath] would be the line you describe.

The derivative is of course 1 and in this case the angle that path and its tangent make with positive X axis is the same. Now, how should I locate the velocity vector on the graph? I found it is equal to 1 but where should it start, end and what would be its angle in relation to path?

If you meant [imath]\mathbf{r}(t) = \langle t,t\rangle [/imath], so that [imath]y = x[/imath], then the velocity vector would be [imath]\dot\mathbf{r}(t) = \langle 1,1\rangle [/imath].

Again, you appear to have differentiated the real-valued function [imath]f(x) = x[/imath] with respect to x; this is entirely different from differentiating the vector function [imath]\mathbf{r}(t)[/imath] with respect to t. The latter is the velocity vector.

But if you did have a particle moving along the line [imath]y = x[/imath], you couldn't tell from that how fast it was moving along that path. The velocity vector would be in the direction of [imath]\langle 1,1\rangle [/imath], but could be any multiple of that, depending on the function.

 

[Mondo]

It appears that you are confusing r(t)r(t)r(t) with yyy; y=xy = xy=x would be the line you describe.

Yes I did a lot of mistakes in my last post, sorry about that. What I really meant was a velocity vector obtained from a position vector [imath]r(t) = [t,t][/imath] that would be [math]\dot r(t) = [1,1][/math]. So now knowing this fact I can graph it. If I would do this on the same graph as position vector then they would overlap right?

Also, there are two things I would clarify:

1. The tangent angle is supposed to be an angle with positive X axis. Now if the slope is negative, say m = -1 then this angle should be [imath]> 90[/imath] but [imath]tan^1(-1) = 45 \degree [/imath] what is the conversion that I miss here?

2. I remember hearing that velocity shall always be perpendicular to path but this is probably only in regards to circular motion right? For example if path is defined as [imath]y(t) = t^2[/imath] then velocity would be [imath]\frac{dy}{dt} = 2t[/imath] and hence there is no way for it to be perpendicular to path nor position vector all the time.

 

[Mondo]

what is the conversion that I miss here?

Ok I think I got it - we need to find the corresponding acute angle and its sign. which in this case will be [imath]-45 \degree[/imath].

 

[Dr.Peterson]

So now knowing this fact I can graph it. If I would do this on the same graph as position vector then they would overlap right?

In this very special case, yes.

1. The tangent angle is supposed to be an angle with positive X axis. Now if the slope is negative, say m = -1 then this angle should be [imath]> 90[/imath] but [imath]tan^1(-1) = 45 \degree [/imath] what is the conversion that I miss here?

No, the inverse tangent [imath]\tan^{-1}(-1) = -45^\circ[/imath], not +45. and that's an appropriate answer.

Ok I think I got it - we need to find the corresponding acute angle and its sign. which in this case will be [imath]-45 \degree[/imath].

Yes, you have to take quadrant into account, in general (though it doesn't really matter in this case). You could also use [imath]+135^\circ[/imath].

2. I remember hearing that velocity shall always be perpendicular to path but this is probably only in regards to circular motion right? For example if path is defined as [imath]y(t) = t^2[/imath] then velocity would be [imath]\frac{dy}{dt} = 2t[/imath] and hence there is no way for it to be perpendicular to path nor position vector all the time.

Right. Things you learn in geometry (or pre-calculus physics) don't apply to all curves, only to the curves then under discussion.

Now try finding the velocity of a uniform circular motion by calculus and confirm what you learned then! (Then try it for non-uniform circular motion!)

 

[Mondo]

Yes, you have to take quadrant into account, in general (though it doesn't really matter in this case). You could also use +135

You mean use for [imath]tan^1(\theta)[/imath] function? If so then yes because algorithms that calculate it know tangent function range and can convert it. What I wanted to know if how to get that manually. I think I got it now, thank you.

Now try finding the velocity of a uniform circular motion by calculus and confirm what you learned then! (Then try it for non-uniform circular motion!)

Sounds like a good example, do you have an actual problem to solve? Or at least a function that represent the object path?

 

[blamocur]

...Sounds like a good example, do you have an actual problem to solve? Or at least a function that represent the object path?

E.g. [imath]x(t) = r \cos\omega t[/imath], [imath]y(t) = r\sin\omega t[/imath].

 

[Dr.Peterson]

You mean use for [imath]tan^1(\theta)[/imath] function?

Do you mean to say [imath]tan^{-1}(\theta)[/imath], the inverse tangent function? It's important to check what you type.

This function has a range from -90 to +90 degrees.

If so then yes because algorithms that calculate it know tangent function range and can convert it. What I wanted to know if how to get that manually. I think I got it now, thank you.

I don't understand what you are saying here.

Sounds like a good example, do you have an actual problem to solve? Or at least a function that represent the object path?

I suggested that you think about this yourself. What would uniform motion around a circle look like? It's very simple, but the exercise will help you think about the topic. I would start by putting the unit circle in polar form: [imath]r = 1, \theta=at[/imath], where a would be the speed. (I see @blamocur has taken it one step beyond that.)