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By kappa do you mean curvature?

kappa is curvature and what is tau...

Does he maybe mean T, the unit tangent?

Yes to both sorry for the confusion

Really? I never heard of that use of tau before...

me neither, I've only seen big T

anyway, it's very straightforward with the formulas

Well anyway ill handle the tangent
u can do kappa, Turing.

you need tangent for kappa, so...

its a T in like greek format i believe it is the unit tangent though

lol ur right i was thinking about 2D formula for curvature

BTW how do you do a vector overscript in this?

ok you don't need T for kappa, but it helps

superscript*

\[\vec r(t)=(a\cos t)\hat i+(a\sin t)\hat j+bt\hat k;~~~ a,b>0; ~~~a^2+b^2 \neq0\]most likely

\[r(t)=(acost)i+(asint)j+btk\]\[r'(t)=v(t)=D_t(acost)i+D_t(asint)j+D_t(bt)k\]

yeah howd u get the arrow and hat superscripts?

\vec r\[\vec r\]

right-click any LaTeX you don't know and you can do
show as ->tex commands
then see the source code

sweet thx was always too lazy to learn latex lol

me too I just pick it up as I go along on here...

\[\vec v(t) = <-a \sin(t), a \cos(t), b>\]

thanks im following you just fine thus far!

\[\vec T(t) = \frac{\vec v(t)}{\left| \vec v(t) \right|}\]

I trust you can handle all that fun algebra yourself...

haha yes!

no just for what the largest value of T is Thanks!

ds lol soz

\kappa = \left| \frac{d\vec T}{ds}\right|

I usually get curvature with\[\kappa=\frac{\|\vec T'(t)\|}{\|\vec r'(t)\|}\]

keeps arc length out of it

\[\kappa = \left| \frac{d\vec T}{ds}\right|\]

I suppose, if you want to avoid chain rule, that TT's method's better.

Thats a good idea :p

Lol I liked calculus till i got into some of this stuff....

product of*...

Man its been a while!

yea thats right!

ah, thanks for bridging the gap :)

yeah im familiar with those its just really the end part thats getting me lol

I got ya thanks :p

scratch that last statement tho its only valid in some cases.

okay got ya

So how do you maximize T for a given value of a?

what do you mean 'maximize'? T is a unit vector.

The question asks what is the largest value t can have for a given value of a?

Strange. |T| = 1 always...

if it was velocity, as in \[\vec v(t)\]
Then I'd say look at the physics!

well a and b are susceptible to change, so perhaps we need the partial wrt b ??

for 2D circles. the k-component just adds a constant, orthogonal component to motion.

I think since were dealing with a helix T could actually be torsion not unit tangent sorry!

haha yeah! i forgot. It probably was torsion!

Torsion is b/(a^2+b^2) hmm

Lets just do it your way i can follow those formulas better

so we still have to find T lol

Hopefully you already know the unit binormal formula and the unit normal formula

The T in those formulas is the unit tangent right not torsion?

Yeah ive seen all of those before!

Yes.
and BTW my formula for N should be divided by magnitude, forgot that.

right right!

just to make sure its the magnitude of V right?

ohh that makes it easier to just cancel rather than figure out T right away good work!

Sorry if this is hard to understand. If you don't get it, I can go back...

I got that earlier when i was finding them :p

is V^2 in the previous equation equal to 1 somehow?

wait i messed my cross sry.. it shouldnt have no z value.

isnt z the 0 there? so no value right

okay thanks!

Now that we have B, we can find the torsion!

I think that you can handle the torsion from here, right?

its alright I got what you meant by it!

dB/dt = {b Cos[t], b Sin[t], 0}/V
||v(t)|| = V
V = sqrt(a^2+b^2)

k i gotta go now im sure u can handle it from there.

sorry i left but they want it with respect to any given value of a...

since tau only depends on a and b, yes

according to the earlier work, which I'm not going to independently verify

okay thank you for your help!