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thez25
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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.
 2 years ago
 2 years ago
thez25 Group Title
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.
 2 years ago
 2 years ago

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vf321 Group TitleBest ResponseYou've already chosen the best response.2
By kappa do you mean curvature?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
kappa is curvature and what is tau...
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
Does he maybe mean T, the unit tangent?
 2 years ago

thez25 Group TitleBest ResponseYou've already chosen the best response.0
Yes to both sorry for the confusion
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
Really? I never heard of that use of tau before...
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
me neither, I've only seen big T
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
anyway, it's very straightforward with the formulas
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
Well anyway ill handle the tangent u can do kappa, Turing.
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
you need tangent for kappa, so...
 2 years ago

thez25 Group TitleBest ResponseYou've already chosen the best response.0
its a T in like greek format i believe it is the unit tangent though
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
lol ur right i was thinking about 2D formula for curvature
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
BTW how do you do a vector overscript in this?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
ok you don't need T for kappa, but it helps
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
\[\vec r(t)=(a\cos t)\hat i+(a\sin t)\hat j+bt\hat k;~~~ a,b>0; ~~~a^2+b^2 \neq0\]most likely
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
\[r(t)=(acost)i+(asint)j+btk\]\[r'(t)=v(t)=D_t(acost)i+D_t(asint)j+D_t(bt)k\]
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
yeah howd u get the arrow and hat superscripts?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
\vec r\[\vec r\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
rightclick any LaTeX you don't know and you can do show as >tex commands then see the source code
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
sweet thx was always too lazy to learn latex lol
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
me too I just pick it up as I go along on here...
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
\[\vec v(t) = <a \sin(t), a \cos(t), b>\]
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
By what i showed u earlier, @OP
 2 years ago

thez25 Group TitleBest ResponseYou've already chosen the best response.0
thanks im following you just fine thus far!
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
\[\vec T(t) = \frac{\vec v(t)}{\left \vec v(t) \right}\]
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
I trust you can handle all that fun algebra yourself...
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
And then curvature is pretty easy too. I forgot that it's torsion that's hard. It's just \[\kappa = \left \frac{d\vec T}{dt}\right\]
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
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?
 2 years ago

thez25 Group TitleBest ResponseYou've already chosen the best response.0
no just for what the largest value of T is Thanks!
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
\kappa = \left \frac{d\vec T}{ds}\right
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
I usually get curvature with\[\kappa=\frac{\\vec T'(t)\}{\\vec r'(t)\}\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
keeps arc length out of it
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
\[\kappa = \left \frac{d\vec T}{ds}\right\]
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
I suppose, if you want to avoid chain rule, that TT's method's better.
 2 years ago

thez25 Group TitleBest ResponseYou've already chosen the best response.0
Thats a good idea :p
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
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...
 2 years ago

thez25 Group TitleBest ResponseYou've already chosen the best response.0
Lol I liked calculus till i got into some of this stuff....
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
product of*...
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
Lol wait. I see what happened. We could have gone with mine from the start! I remember doing this in class: \[\kappa = \left \left \frac{d\vec T}{ds} \right \right=\left \left \frac{d\vec T/dt}{ds/dt} \right \right=\left \left \frac{d\vec T/dt}{v(t)} \right \right\]
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
Man its been a while!
 2 years ago

thez25 Group TitleBest ResponseYou've already chosen the best response.0
yea thats right!
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
ah, thanks for bridging the gap :)
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
but @thez25 the most important thing here is understanding the formulas. Do you get em?
 2 years ago

thez25 Group TitleBest ResponseYou've already chosen the best response.0
yeah im familiar with those its just really the end part thats getting me lol
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
Well the T should be easy... But as for curvature, the best way I can explain that is the following. Curvature, in very layterms, 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
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
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
 2 years ago

thez25 Group TitleBest ResponseYou've already chosen the best response.0
I got ya thanks :p
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
scratch that last statement tho its only valid in some cases.
 2 years ago

thez25 Group TitleBest ResponseYou've already chosen the best response.0
So how do you maximize T for a given value of a?
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
what do you mean 'maximize'? T is a unit vector.
 2 years ago

thez25 Group TitleBest ResponseYou've already chosen the best response.0
The question asks what is the largest value t can have for a given value of a?
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
Strange. T = 1 always...
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
if it was velocity, as in \[\vec v(t)\] Then I'd say look at the physics!
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
well a and b are susceptible to change, so perhaps we need the partial wrt b ??
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
By physics I mean a is the radius, and v is the velocity. And we know from circular motion that mv^2/(a) = acc
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
for 2D circles. the kcomponent just adds a constant, orthogonal component to motion.
 2 years ago

thez25 Group TitleBest ResponseYou've already chosen the best response.0
I think since were dealing with a helix T could actually be torsion not unit tangent sorry!
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
haha yeah! i forgot. It probably was torsion!
 2 years ago

thez25 Group TitleBest ResponseYou've already chosen the best response.0
Torsion is b/(a^2+b^2) hmm
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
Well it might be i dunno. Let's solve for it, unless your book told you that's it... \[\tau = \left \left \frac{d\vec B(t)}{ds} \right \right = \left \left \frac{d\vec B(t)/dt}{v(t)} \right \right\]
 2 years ago

thez25 Group TitleBest ResponseYou've already chosen the best response.0
Lets just do it your way i can follow those formulas better
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
\[\vec v(t)=<−a sin(t),a cos(t),b>\] \[\vec B(t) = \vec T(t) \times \vec N(t)\] \[\vec N(t) = \frac{d\vec T(t)}{ds}\]
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
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
 2 years ago

thez25 Group TitleBest ResponseYou've already chosen the best response.0
so we still have to find T lol
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
Hopefully you already know the unit binormal formula and the unit normal formula
 2 years ago

thez25 Group TitleBest ResponseYou've already chosen the best response.0
The T in those formulas is the unit tangent right not torsion?
 2 years ago

thez25 Group TitleBest ResponseYou've already chosen the best response.0
Yeah ive seen all of those before!
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
Yes. and BTW my formula for N should be divided by magnitude, forgot that.
 2 years ago

thez25 Group TitleBest ResponseYou've already chosen the best response.0
right right!
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
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... \[\vec T(t) = \frac{\vec v(t)}{V} = \frac{<−a sin(t),a cos(t),b>}{V} \]
 2 years ago

thez25 Group TitleBest ResponseYou've already chosen the best response.0
just to make sure its the magnitude of V right?
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
V IS the magnitude  v(t) = V \[\vec N(t) = \frac{d\vec T(t)/dt}{ds/dt} =\frac{ <a cos(t), a sin(t), 0>/V+\vec v(t) *V'/V^2}{\vec v(t)}\]
 2 years ago

thez25 Group TitleBest ResponseYou've already chosen the best response.0
ohh that makes it easier to just cancel rather than figure out T right away good work!
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
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>/
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
Sorry if this is hard to understand. If you don't get it, I can go back...
 2 years ago

thez25 Group TitleBest ResponseYou've already chosen the best response.0
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!
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
Okay. Either way, some way or another, you get: \[\vec N = <cos(t), sin(t) ,0>\]\[\vec T = \frac{<−asin(t),acos(t),b>}{\sqrt{a^2+b^2}}\]
 2 years ago

thez25 Group TitleBest ResponseYou've already chosen the best response.0
I got that earlier when i was finding them :p
 2 years ago

thez25 Group TitleBest ResponseYou've already chosen the best response.0
is V^2 in the previous equation equal to 1 somehow?
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
When we cross them, the work for which I won't bother to show, we get \[\vec B = \frac{<a sin(t) cos(t), a sin(t)cos(t), 0>}{\sqrt{a^2+b^2}}\]
 2 years ago

thez25 Group TitleBest ResponseYou've already chosen the best response.0
sorry didnt see that you explained that at the end of that paragraph
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
wait i messed my cross sry.. it shouldnt have no z value.
 2 years ago

thez25 Group TitleBest ResponseYou've already chosen the best response.0
isnt z the 0 there? so no value right
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
Yes, but that's wrong. This is the right cross product (I checked with mma) \[\vec B = \frac{<b sin(t), bcos(t), a>}{\sqrt{a^2+b^2}}]\]
 2 years ago

thez25 Group TitleBest ResponseYou've already chosen the best response.0
okay thanks!
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
Now that we have B, we can find the torsion!
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
\[\tau = \left\left\frac{d\vec B(t)/dt}{\vec v(t)}\right\right\] 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
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
I think that you can handle the torsion from here, right?
 2 years ago

thez25 Group TitleBest ResponseYou've already chosen the best response.0
its alright I got what you meant by it!
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
dB/dt = {b Cos[t], b Sin[t], 0}/V v(t) = V V = sqrt(a^2+b^2)
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.2
k i gotta go now im sure u can handle it from there.
 2 years ago

thez25 Group TitleBest ResponseYou've already chosen the best response.0
well if you get a chance later could u explain how to maximize it lol thanks for all your help though!
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
maximize it with respect to what? if you want to maximize it with respect to t find\[\frac{d\tau}{dt}=0\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
but\[\tau=\frac b{\sqrt{a^2+b^2}}\]and you said you are "given a value of a" so perhaps they want\[\frac{\partial\tau}{\partial b}\]?
 2 years ago

thez25 Group TitleBest ResponseYou've already chosen the best response.0
sorry i left but they want it with respect to any given value of a...
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
so the only thing to take the derivative with respect to is b. so set\[\frac{\partial\tau}{\partial b}=0\]solve for b
 2 years ago

thez25 Group TitleBest ResponseYou've already chosen the best response.0
alright thanks so that would give the largest value torsion can have for a given value of a then correct?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
since tau only depends on a and b, yes
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
according to the earlier work, which I'm not going to independently verify
 2 years ago

thez25 Group TitleBest ResponseYou've already chosen the best response.0
okay thank you for your help!
 2 years ago
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