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thez25
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.
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
\[\vec r(t)=(a\cos t)\hat i+(a\sin t)\hat j+bt\hat k;~~~ a,b>0; ~~~a^2+b^2 \neq0\]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?
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...
\[\vec v(t) = <-a \sin(t), a \cos(t), b>\]
By what i showed u earlier, @OP
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...
And then curvature is pretty easy too. I forgot that it's torsion that's hard. It's just \[\kappa = \left| \frac{d\vec T}{dt}\right|\]
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?
no just for what the largest value of T is Thanks!
\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.
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...
Lol I liked calculus till i got into some of this stuff....
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
Lol wait. I see what happened. We could have gone with mine from the start! I remember doing this in class: \[\kappa = \left| \left| \frac{d\vec T}{ds} \right| \right|=\left| \left| \frac{d\vec T/dt}{ds/dt} \right| \right|=\left| \left| \frac{d\vec T/dt}{v(t)} \right| \right|\]
ah, thanks for bridging the gap :)
but @thez25 the most important thing here is understanding the formulas. Do you get em?
yeah im familiar with those its just really the end part thats getting me lol
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||
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
scratch that last statement tho its only valid in some cases.
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 ??
By physics I mean a is the radius, and v is the velocity. And we know from circular motion that mv^2/(a) = acc
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
Well it might be i dunno. Let's solve for it, unless your book told you that's it... \[\tau = \left| \left| \frac{d\vec B(t)}{ds} \right| \right| = \left| \left| \frac{d\vec B(t)/dt}{v(t)} \right| \right|\]
Lets just do it your way i can follow those formulas better
\[\vec v(t)=<−a sin(t),a cos(t),b>\] \[\vec B(t) = \vec T(t) \times \vec N(t)\] \[\vec N(t) = \frac{d\vec T(t)}{ds}\]
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
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.
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... \[\vec T(t) = \frac{\vec v(t)}{V} = \frac{<−a sin(t),a cos(t),b>}{V} \]
just to make sure its the magnitude of V right?
V IS the magnitude - ||v(t)|| = V \[\vec N(t) = \frac{d\vec T(t)/dt}{ds/dt} =\frac{ <-a cos(t), -a sin(t), 0>/V+\vec v(t) *V'/V^2}{\vec v(t)}\]
ohh that makes it easier to just cancel rather than figure out T right away good work!
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>/
Sorry if this is hard to understand. If you don't get it, I can go back...
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!
Okay. Either way, some way or another, you get: \[\vec N = <-cos(t), -sin(t) ,0>\]\[\vec T = \frac{<−asin(t),acos(t),b>}{\sqrt{a^2+b^2}}\]
I got that earlier when i was finding them :p
is V^2 in the previous equation equal to 1 somehow?
When we cross them, the work for which I won't bother to show, we get \[\vec B = \frac{<a sin(t) cos(t), -a sin(t)cos(t), 0>}{\sqrt{a^2+b^2}}\]
sorry didnt see that you explained that at the end of that paragraph
wait i messed my cross sry.. it shouldnt have no z value.
isnt z the 0 there? so no value right
Yes, but that's wrong. This is the right cross product (I checked with mma) \[\vec B = \frac{<b sin(t),- bcos(t), a>}{\sqrt{a^2+b^2}}]\]
Now that we have B, we can find the torsion!
\[\tau = \left|\left|\frac{d\vec B(t)/dt}{\vec v(t)}\right|\right|\] 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
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.
well if you get a chance later could u explain how to maximize it lol thanks for all your help though!
maximize it with respect to what? if you want to maximize it with respect to t find\[\frac{d\tau}{dt}=0\]
but\[\tau=\frac b{\sqrt{a^2+b^2}}\]and you said you are "given a value of a" so perhaps they want\[\frac{\partial\tau}{\partial b}\]?
sorry i left but they want it with respect to any given value of a...
so the only thing to take the derivative with respect to is b. so set\[\frac{\partial\tau}{\partial b}=0\]solve for b
alright thanks so that would give the largest value torsion can have for a given value of a then correct?
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!