Question

How can you show that ||T'(t)|| = ||r'(t) x r''(t)|| / ||r'(t)||^2 ?

Answer 1

Those are vectors by the way.

Answer 2

Apply matrix theory.

Answer 3

Is there a method of showing that using just vectors?

Answer 4

Try using eigenvectors

Answer 5

Eliminate the lurking variables by analytically applying the Riemann Hypothesis throughout the manifold.

Answer 6

@agd : Not helpful.

What's the definition of T?

Answer 7

It's a vector-valued function. T(t).

Answer 8

Yes, that is clear, but what is the set-up here? Uniform circular motion with a rope of force and tension T?

Answer 9

Nothing in particular, the above equation is supposed to be a general equation.

Answer 10

Anything which is a capital letter is a vector in this proof, if it is lower case it will be a scalar. Ok, here we go...
The definition of curvature is \[k(t)=\left| dT/ds \right|=\left| \frac{dT/dt}{ds/dt} \right|=\frac{|T'(t)|}{|R'(t)|}\]Another way to express curvature is:
\[k(t)=\frac{ |R'(t) X R''(t)|}{|R'(t)^3|}\] The big X is supposed to be a cross product. Now, your expression follows immediately from the two expressions for curvature. If you want me to prove the second expression for curvature, let me know. The first expression takes the definition of curvature, then re-written in an equivalent form.

Answer 11

Yes, please, I'm interested in the proof for the second expression.

Answer 12

By definition, we have\[T=R'/|R'|\] and \[|R'|=ds/dt\]So,\[R'=|R'|T=\frac{ds}{dt}T\]Differentiating, using the product rule gives:\[R''=\frac{d^2s}{dt^2}T+\frac{ds}{dt}T'\]Using the fact that T X T=0, we have:\[R' X R''=(ds/dt)^2(T X T')\]Now |T(t)|=1 for all t (it is a unit vector) so T and T' are orthogonal (if you're not sure why this is true, ask!). Thus,\[|R' X R''|=(ds/dt)^2|T X T'|=(ds/dt)^2|T||T'|\]\[=(ds/dt)^2|T'|\]Thus,\[|T'|=\frac{|R'XR''|}{(ds/dt)^2}=\frac{|R'XR''|}{|R'|^2}\]and,\[k=\frac{|T'|}{|R'|}=\frac{|R'XR''|}{|R'|^3}\]
Whew!

Answer 13

I don't quite understand how you jumped from the product rule derivation part to the next one (the one where you said T x T' = 0). Also, I don't quite understand how T and T' are orthogonal (I understand how R and R' are orthogonal, however.)

Answer 14

no prob. Ok starting at the product rule expression:\[R''=\frac{d^2s}{dt^2}T+(ds/dt)T'\]We are going to take the cross product of R' and R''( as expressed above).\[R'XR''=R X(\frac{d^2s}{dt^2}T+\frac{ds}{dt}T')\]Remember that we can use the distributive property with cross-products, so they just multiply through as usual. We have,\[R'X \frac{d^2s}{dt^2}T + R' X (ds/dt)T'\]Ignore the expressions for ds/dt above, they are just scalars. What is R'XT? Well, going back to the first line of my proof (the definition of the T vector) we see R'=T|R'|. Thus, R'XT=|R'|TXT=0. So the first term is zero and goes away. I hope this answers your first question.

Answer 15

sorry I forgot to put the ' in the second equation. typo

Answer 16

Now to show why T and T' are orthogonal...lets take the dot product of T with itself: \[T*T=|T|^2\]But |T|=1 so we have T*T=1
If we differentiate T*T, we get \[\frac{d}{dt}(T*T)=T' * T + T* T'=2T'*T=0\]Because (d/dt)1=0. Two vectors are orthogonal if and only if there dot product is 0. Thus T and T' are orthogonal.

Answer 17

No. It only works for constant length vectors. If |R(t)|=c, where c is a constant, this holds.

Answer 18

Sorry, should have thought before asking.

Answer 19

no, these concepts can be real mind-benders hehe

Answer 20

go through the proof, line-by-line and make sure you got it. there are some fancy tricks used but all the steps are based in concepts and rules you should know.

Answer 21

Thanks so much! I wish I could give you more medals. I'll try to get this some other time; I'm quite tired.

Answer 22

no worries, ill check back with you another day :)

Answer 23

:)

Answer 24

Thanks so much, I get it now!

Answer 25

So mind-boggling how you can also extend that to show:
\[a_{N} = \frac{||v(t) \times a(t)||}{||v(t)||}\]

Answer 26

Yeah, vectors are cool:) Oh yeah, above you said that you understand why R(t) and R'(t) are orthogonal...this is not true in general, only in specific circumstances (uniform circular motion for example). Remember R is the position vector of an object and R' is the change in position vector of the object over time. The object can go any way it pleases, it is not constrained to only travel in a circle (unless of course this is the problem you are faced with lol). If an object is travelling in a straight line at constant velocity then R' always lies on this line but (if you choose the origin to also lie on the line the particle travels on) your position vector R also lies on the line. Here R and R' are parallel NOT orthogonal. Hope this helps! Let me know if you have trouble with any of this stuff :)