quick query on parametric differentiation

[abskebabs]

From my maths notes, I have come to a point in the section concerning paremtric differentiation on curves and have a query concerning it. I will label the change in x, y or z using a vector quantity [math]\vec{r}[/math] Basically the tangent of any curve can be represented using the following equation:

[math]T=d\vec{r}/dt[/math]

With which we can define a "unit tangent" whereby:

[math]\hat{T}=|d\vec{r}/dt[/math]

But another quantity is established:

[math]ds=|d\vec{r}|=(dx^2+dy^2+dz^2)[/math]

The following equality is then established:

[math]ds/dt=|d\vec{r}/dt|[/math]

Although it seems highly suggestible, the final statement still erks me a little. Can anyone show why it has to be true? In my mathematical cosncience I feel it may not necessarily be a trivial connection.

 

Thanks for the help

[the tree]

That change in distance (s) equals absolute change in position vector?

 

What do you understand 's' to mean, if not |r|?

[abskebabs]

That change in distance (s) equals absolute change in position vector?

 

What do you understand 's' to mean, if not |r|?

I already understand what s means, as I have already said, I understand it to equal |r|. What I guess I am struggling with is the idea that modulus of the rate of change in a vector necessarily equals the rate of change of the modulus of a vector(the same thing in words).

 

I guess its just the rate of change thing that pickles me a little. IFor example I have no problem with your above statement, basially that the modulus of change in a vector quantity equals s. Indeed this is what I have already stated too above. I guess, I just have to think a bit on this one, n maybe I'm being a bit slow...

[Tom Mattson]

The following equality is then established:

[math]ds/dt=|d\vec{r}/dt|[/math]

 

The statement is false. I can give you a simple counterexample from physics. Consider a mass on a rope that is swung in a circle at a constant speed. The velocity vector [math]\vec{v}[/math] constantly changes, so [math]\frac{d\vec{v}}{dt}\neq 0[/math], but the speed is constant so [math]\frac{d|\vec{v}|}{dt}=0[/math]

[abskebabs]

The statement is false. I can give you a simple counterexample from physics. Consider a mass on a rope that is swung in a circle at a constant speed. The velocity vector [math]\vec{v}[/math] constantly changes, so [math]\frac{d\vec{v}}{dt}\neq 0[/math], but the speed is constant so [math]\frac{d|\vec{v}|}{dt}=0[/math]

 

If the statement is false then why is it in my lecturer's online maths notes(I am doing a physics course btw)! Could it be true for certain cases, but not generally? I'm confused now...

[Tom Mattson]

The statement is true whenever the magnitude of a vector changes, but not its direction.

[Royston]

abskebabs, sorry for going off topic, just wondered what course you were studying...I know it's a physics course, but who's it through ?

[abskebabs]

abskebabs, sorry for going off topic, just wondered what course you were studying...I know it's a physics course, but who's it through ?

It's ok I never mind talking about myself(who does mind tho really? ). I'm studying theoretical physics at the University of Birmingham, but you could surely have found that out by looking at my profile? I guess I must be one of the few that ever looks at those...

 

Back on the subject, I'm still worried about the confusion I have over this and may consult either my lecturer or my tutor about this. The above identity which Tom mattson has refuted in the case of velocity is crucial for what follows, concerning the unit tangent, it's derivatives and the orthonomal triad of unit vectors is it is perpendicular to. I guess one thing I should clarify is t does not necessarily represent time, but is just a general parameter. I don't think this really chnages the situation much though.

 

I guess I'm much more likely to get questions on differential equations(these were like 1/3 or 1/2 of my 2nd semester maths) tho, than the more awkward apects of this part of the course.