anonymous
  • anonymous
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.
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
By kappa do you mean curvature?
TuringTest
  • TuringTest
kappa is curvature and what is tau...
anonymous
  • anonymous
Does he maybe mean T, the unit tangent?

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anonymous
  • anonymous
Yes to both sorry for the confusion
anonymous
  • anonymous
Really? I never heard of that use of tau before...
TuringTest
  • TuringTest
me neither, I've only seen big T
TuringTest
  • TuringTest
anyway, it's very straightforward with the formulas
anonymous
  • anonymous
Well anyway ill handle the tangent u can do kappa, Turing.
TuringTest
  • TuringTest
you need tangent for kappa, so...
anonymous
  • anonymous
its a T in like greek format i believe it is the unit tangent though
anonymous
  • anonymous
lol ur right i was thinking about 2D formula for curvature
anonymous
  • anonymous
BTW how do you do a vector overscript in this?
TuringTest
  • TuringTest
ok you don't need T for kappa, but it helps
anonymous
  • anonymous
superscript*
TuringTest
  • 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
  • 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
  • anonymous
yeah howd u get the arrow and hat superscripts?
TuringTest
  • TuringTest
\vec r\[\vec r\]
TuringTest
  • TuringTest
right-click any LaTeX you don't know and you can do show as ->tex commands then see the source code
anonymous
  • anonymous
sweet thx was always too lazy to learn latex lol
TuringTest
  • TuringTest
me too I just pick it up as I go along on here...
anonymous
  • anonymous
\[\vec v(t) = <-a \sin(t), a \cos(t), b>\]
anonymous
  • anonymous
By what i showed u earlier, @OP
anonymous
  • anonymous
thanks im following you just fine thus far!
anonymous
  • anonymous
\[\vec T(t) = \frac{\vec v(t)}{\left| \vec v(t) \right|}\]
anonymous
  • anonymous
I trust you can handle all that fun algebra yourself...
anonymous
  • anonymous
haha yes!
anonymous
  • 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
  • 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
  • anonymous
no just for what the largest value of T is Thanks!
anonymous
  • anonymous
ds lol soz
anonymous
  • anonymous
\kappa = \left| \frac{d\vec T}{ds}\right|
TuringTest
  • TuringTest
I usually get curvature with\[\kappa=\frac{\|\vec T'(t)\|}{\|\vec r'(t)\|}\]
TuringTest
  • TuringTest
keeps arc length out of it
anonymous
  • anonymous
\[\kappa = \left| \frac{d\vec T}{ds}\right|\]
anonymous
  • anonymous
I suppose, if you want to avoid chain rule, that TT's method's better.
anonymous
  • anonymous
Thats a good idea :p
anonymous
  • 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
  • anonymous
Lol I liked calculus till i got into some of this stuff....
TuringTest
  • 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
  • TuringTest
product of*...
anonymous
  • 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
  • anonymous
Man its been a while!
anonymous
  • anonymous
yea thats right!
TuringTest
  • TuringTest
ah, thanks for bridging the gap :)
anonymous
  • anonymous
but @thez25 the most important thing here is understanding the formulas. Do you get em?
anonymous
  • anonymous
yeah im familiar with those its just really the end part thats getting me lol
anonymous
  • 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
  • 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
  • anonymous
I got ya thanks :p
anonymous
  • anonymous
scratch that last statement tho its only valid in some cases.
anonymous
  • anonymous
okay got ya
anonymous
  • anonymous
So how do you maximize T for a given value of a?
anonymous
  • anonymous
what do you mean 'maximize'? T is a unit vector.
anonymous
  • anonymous
The question asks what is the largest value t can have for a given value of a?
anonymous
  • anonymous
Strange. |T| = 1 always...
anonymous
  • anonymous
if it was velocity, as in \[\vec v(t)\] Then I'd say look at the physics!
TuringTest
  • TuringTest
well a and b are susceptible to change, so perhaps we need the partial wrt b ??
anonymous
  • 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
  • anonymous
for 2D circles. the k-component just adds a constant, orthogonal component to motion.
anonymous
  • anonymous
I think since were dealing with a helix T could actually be torsion not unit tangent sorry!
anonymous
  • anonymous
haha yeah! i forgot. It probably was torsion!
anonymous
  • anonymous
Torsion is b/(a^2+b^2) hmm
anonymous
  • 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
  • anonymous
Lets just do it your way i can follow those formulas better
anonymous
  • anonymous
\[\vec v(t)=<−a sin(t),a cos(t),b>\] \[\vec B(t) = \vec T(t) \times \vec N(t)\] \[\vec N(t) = \frac{d\vec T(t)}{ds}\]
anonymous
  • 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
  • anonymous
so we still have to find T lol
anonymous
  • anonymous
Hopefully you already know the unit binormal formula and the unit normal formula
anonymous
  • anonymous
The T in those formulas is the unit tangent right not torsion?
anonymous
  • anonymous
Yeah ive seen all of those before!
anonymous
  • anonymous
Yes. and BTW my formula for N should be divided by magnitude, forgot that.
anonymous
  • anonymous
right right!
anonymous
  • 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... \[\vec T(t) = \frac{\vec v(t)}{V} = \frac{<−a sin(t),a cos(t),b>}{V} \]
anonymous
  • anonymous
just to make sure its the magnitude of V right?
anonymous
  • 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
  • anonymous
ohh that makes it easier to just cancel rather than figure out T right away good work!
anonymous
  • 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
  • anonymous
Sorry if this is hard to understand. If you don't get it, I can go back...
anonymous
  • 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
  • anonymous
Okay. Either way, some way or another, you get: \[\vec N = <-cos(t), -sin(t) ,0>\]\[\vec T = \frac{<−asin(t),acos(t),b>}{\sqrt{a^2+b^2}}\]
anonymous
  • anonymous
I got that earlier when i was finding them :p
anonymous
  • anonymous
is V^2 in the previous equation equal to 1 somehow?
anonymous
  • 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
  • anonymous
sorry didnt see that you explained that at the end of that paragraph
anonymous
  • anonymous
wait i messed my cross sry.. it shouldnt have no z value.
anonymous
  • anonymous
isnt z the 0 there? so no value right
anonymous
  • 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
  • anonymous
okay thanks!
anonymous
  • anonymous
Now that we have B, we can find the torsion!
anonymous
  • 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
  • anonymous
I think that you can handle the torsion from here, right?
anonymous
  • anonymous
its alright I got what you meant by it!
anonymous
  • anonymous
dB/dt = {b Cos[t], b Sin[t], 0}/V ||v(t)|| = V V = sqrt(a^2+b^2)
anonymous
  • anonymous
k i gotta go now im sure u can handle it from there.
anonymous
  • anonymous
well if you get a chance later could u explain how to maximize it lol thanks for all your help though!
TuringTest
  • TuringTest
maximize it with respect to what? if you want to maximize it with respect to t find\[\frac{d\tau}{dt}=0\]
TuringTest
  • 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
  • anonymous
sorry i left but they want it with respect to any given value of a...
TuringTest
  • 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
  • anonymous
alright thanks so that would give the largest value torsion can have for a given value of a then correct?
TuringTest
  • TuringTest
since tau only depends on a and b, yes
TuringTest
  • TuringTest
according to the earlier work, which I'm not going to independently verify
anonymous
  • anonymous
okay thank you for your help!

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