Find kappa and tau for the helix r(t)=(acost)i+(asint)j+btk, a,b>0 a^2+b^2 does not = 0. What is the largest value tau can have for a given value of a? give reasons for answer please.

- anonymous

- jamiebookeater

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- anonymous

By kappa do you mean curvature?

- TuringTest

kappa is curvature and what is tau...

- anonymous

Does he maybe mean T, the unit tangent?

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## More answers

- anonymous

Yes to both sorry for the confusion

- anonymous

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

- TuringTest

me neither, I've only seen big T

- TuringTest

anyway, it's very straightforward with the formulas

- anonymous

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

- TuringTest

you need tangent for kappa, so...

- anonymous

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

- anonymous

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

- anonymous

BTW how do you do a vector overscript in this?

- TuringTest

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

- anonymous

superscript*

- TuringTest

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

- anonymous

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

- anonymous

yeah howd u get the arrow and hat superscripts?

- TuringTest

\vec r\[\vec r\]

- TuringTest

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

- anonymous

sweet thx was always too lazy to learn latex lol

- TuringTest

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

- anonymous

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

- anonymous

By what i showed u earlier, @OP

- anonymous

thanks im following you just fine thus far!

- anonymous

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

- anonymous

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

- anonymous

haha yes!

- anonymous

And then curvature is pretty easy too. I forgot that it's torsion that's hard.
It's just
\[\kappa = \left| \frac{d\vec T}{dt}\right|\]

- anonymous

And I assume you can take a derivative as well. Do you need explanations for why the formulas for kappa and T are what they are?

- anonymous

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

- anonymous

ds lol soz

- anonymous

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

- TuringTest

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

- TuringTest

keeps arc length out of it

- anonymous

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

- anonymous

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

- anonymous

Thats a good idea :p

- anonymous

Yeah in fact I'm all for @TuringTest 's method. I was going to do all that messy chain rule stuff.... It's been a while since MVC...

- anonymous

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

- TuringTest

this site keeps me fresh on it
it's very easy to forget all the formulas
there's another way to where you use the cross product and the first and second derivatives also

- TuringTest

product of*...

- anonymous

Lol wait. I see what happened. We could have gone with mine from the start! I remember doing this in class:
\[\kappa = \left| \left| \frac{d\vec T}{ds} \right| \right|=\left| \left| \frac{d\vec T/dt}{ds/dt} \right| \right|=\left| \left| \frac{d\vec T/dt}{v(t)} \right| \right|\]

- anonymous

Man its been a while!

- anonymous

yea thats right!

- TuringTest

ah, thanks for bridging the gap :)

- anonymous

but @thez25 the most important thing here is understanding the formulas. Do you get em?

- anonymous

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

- anonymous

Well the T should be easy... But as for curvature, the best way I can explain that is the following. Curvature, in very lay-terms, is how "curvy" a function is. Thus, the turn of a very elongated ellipse, if you could imagine, should have a high curvature. On the other hand, if you imagine a straight line, we would have zero curvature. The value has to be scalar. If we need to look at something as it travels through space, the best measure of its difference from a straight line path (read: its TANGENT) over how far it travels (read: its ARC LENGTH) is, well, ||dT/ds||

- anonymous

Note we can't use time because curvature only cares about shape, and time is simply a parameter we invented to more easily describe a curve. It does not contain info about the curve itself. Also, dT/dt = c*N anyway

- anonymous

I got ya thanks :p

- anonymous

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

- anonymous

okay got ya

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

Strange. |T| = 1 always...

- anonymous

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

- TuringTest

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

- anonymous

By physics I mean a is the radius, and v is the velocity. And we know from circular motion that mv^2/(a) = acc

- anonymous

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

- anonymous

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

- anonymous

haha yeah! i forgot. It probably was torsion!

- anonymous

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

- anonymous

Well it might be i dunno. Let's solve for it, unless your book told you that's it...
\[\tau = \left| \left| \frac{d\vec B(t)}{ds} \right| \right| = \left| \left| \frac{d\vec B(t)/dt}{v(t)} \right| \right|\]

- anonymous

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

- anonymous

\[\vec v(t)=<âˆ’a sin(t),a cos(t),b>\]
\[\vec B(t) = \vec T(t) \times \vec N(t)\]
\[\vec N(t) = \frac{d\vec T(t)}{ds}\]

- anonymous

To explain the torsion formula: Basically the same as my curvature explanation, but torsion is how much your curve "wants" to move out of the 2D plane that it is currently "on" - the 2D plane that is made by the two vectors N and T

- anonymous

so we still have to find T lol

- anonymous

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

- anonymous

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

- anonymous

Yeah ive seen all of those before!

- anonymous

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

- anonymous

right right!

- anonymous

So let's start crackin (but b4 i do that, I'll first let |v(t)| be V, since I use it a lot, and it will start cancelling out...
\[\vec T(t) = \frac{\vec v(t)}{V} = \frac{<âˆ’a sin(t),a cos(t),b>}{V} \]

- anonymous

just to make sure its the magnitude of V right?

- anonymous

V IS the magnitude - ||v(t)|| = V
\[\vec N(t) = \frac{d\vec T(t)/dt}{ds/dt} =\frac{ <-a cos(t), -a sin(t), 0>/V+\vec v(t) *V'/V^2}{\vec v(t)}\]

- anonymous

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

- anonymous

Well, wait.
Remember, V is the magnitude, which depends on time, so I have to factor in the product rule! HOWEVER, in this case, we are dealing with helical movement, which, because of our physics, we know has CONSTANT velocity magnitude ( you could actually prove this on your own). Thus, V is a CONSTANT and V' = 0. We get:
\[\vec N(t)=\frac{<-a cos(t), -a sin(t) ,0>}{V ^2}\]
But if you remember I clumsily forgot to make it a unit vector by forgetting to divide by the magnitude. Thus, at the very end, you need something in the direction of <-a cos(t), -a sin(t) ,0> with magnitude 1. We get: <-cos(t), -sin(t) ,0>/

- anonymous

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

- anonymous

I sort of follow but i can figure out what u mean more when i read over it a few times. U can keep going though dont want to waste your time!

- anonymous

Okay. Either way, some way or another, you get:
\[\vec N = <-cos(t), -sin(t) ,0>\]\[\vec T = \frac{<âˆ’asin(t),acos(t),b>}{\sqrt{a^2+b^2}}\]

- anonymous

I got that earlier when i was finding them :p

- anonymous

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

- anonymous

When we cross them, the work for which I won't bother to show, we get
\[\vec B = \frac{}{\sqrt{a^2+b^2}}\]

- anonymous

sorry didnt see that you explained that at the end of that paragraph

- anonymous

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

- anonymous

isnt z the 0 there? so no value right

- anonymous

Yes, but that's wrong. This is the right cross product (I checked with mma)
\[\vec B = \frac{

**}{\sqrt{a^2+b^2}}]\]**

**
**

- anonymous

okay thanks!

- anonymous

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

- anonymous

\[\tau = \left|\left|\frac{d\vec B(t)/dt}{\vec v(t)}\right|\right|\]
Also, minor note - i was kind of using improp notation whenever I did that - I should have double bars around both the vector in the num and denom

- anonymous

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

- anonymous

its alright I got what you meant by it!

- anonymous

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

- anonymous

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

- anonymous

well if you get a chance later could u explain how to maximize it lol thanks for all your help though!

- TuringTest

maximize it with respect to what?
if you want to maximize it with respect to t find\[\frac{d\tau}{dt}=0\]

- TuringTest

but\[\tau=\frac b{\sqrt{a^2+b^2}}\]and you said you are "given a value of a" so perhaps they want\[\frac{\partial\tau}{\partial b}\]?

- anonymous

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

- TuringTest

so the only thing to take the derivative with respect to is b. so set\[\frac{\partial\tau}{\partial b}=0\]solve for b

- anonymous

alright thanks so that would give the largest value torsion can have for a given value of a then correct?

- TuringTest

since tau only depends on a and b, yes

- TuringTest

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

- anonymous

okay thank you for your help!

**
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