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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.
 one year ago
 one year ago
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.
 one year ago
 one year ago

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vf321Best ResponseYou've already chosen the best response.2
By kappa do you mean curvature?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
kappa is curvature and what is tau...
 one year ago

vf321Best ResponseYou've already chosen the best response.2
Does he maybe mean T, the unit tangent?
 one year ago

thez25Best ResponseYou've already chosen the best response.0
Yes to both sorry for the confusion
 one year ago

vf321Best ResponseYou've already chosen the best response.2
Really? I never heard of that use of tau before...
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
me neither, I've only seen big T
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
anyway, it's very straightforward with the formulas
 one year ago

vf321Best ResponseYou've already chosen the best response.2
Well anyway ill handle the tangent u can do kappa, Turing.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
you need tangent for kappa, so...
 one year ago

thez25Best ResponseYou've already chosen the best response.0
its a T in like greek format i believe it is the unit tangent though
 one year ago

vf321Best ResponseYou've already chosen the best response.2
lol ur right i was thinking about 2D formula for curvature
 one year ago

vf321Best ResponseYou've already chosen the best response.2
BTW how do you do a vector overscript in this?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
ok you don't need T for kappa, but it helps
 one year ago

TuringTestBest 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
 one year ago

vf321Best ResponseYou've already chosen the best response.2
\[r(t)=(acost)i+(asint)j+btk\]\[r'(t)=v(t)=D_t(acost)i+D_t(asint)j+D_t(bt)k\]
 one year ago

vf321Best ResponseYou've already chosen the best response.2
yeah howd u get the arrow and hat superscripts?
 one year ago

TuringTestBest 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
 one year ago

vf321Best ResponseYou've already chosen the best response.2
sweet thx was always too lazy to learn latex lol
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
me too I just pick it up as I go along on here...
 one year ago

vf321Best ResponseYou've already chosen the best response.2
\[\vec v(t) = <a \sin(t), a \cos(t), b>\]
 one year ago

vf321Best ResponseYou've already chosen the best response.2
By what i showed u earlier, @OP
 one year ago

thez25Best ResponseYou've already chosen the best response.0
thanks im following you just fine thus far!
 one year ago

vf321Best ResponseYou've already chosen the best response.2
\[\vec T(t) = \frac{\vec v(t)}{\left \vec v(t) \right}\]
 one year ago

vf321Best ResponseYou've already chosen the best response.2
I trust you can handle all that fun algebra yourself...
 one year ago

vf321Best 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 \[\kappa = \left \frac{d\vec T}{dt}\right\]
 one year ago

vf321Best 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?
 one year ago

thez25Best ResponseYou've already chosen the best response.0
no just for what the largest value of T is Thanks!
 one year ago

vf321Best ResponseYou've already chosen the best response.2
\kappa = \left \frac{d\vec T}{ds}\right
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
I usually get curvature with\[\kappa=\frac{\\vec T'(t)\}{\\vec r'(t)\}\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
keeps arc length out of it
 one year ago

vf321Best ResponseYou've already chosen the best response.2
\[\kappa = \left \frac{d\vec T}{ds}\right\]
 one year ago

vf321Best ResponseYou've already chosen the best response.2
I suppose, if you want to avoid chain rule, that TT's method's better.
 one year ago

vf321Best 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...
 one year ago

thez25Best ResponseYou've already chosen the best response.0
Lol I liked calculus till i got into some of this stuff....
 one year ago

TuringTestBest 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
 one year ago

vf321Best 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: \[\kappa = \left \left \frac{d\vec T}{ds} \right \right=\left \left \frac{d\vec T/dt}{ds/dt} \right \right=\left \left \frac{d\vec T/dt}{v(t)} \right \right\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
ah, thanks for bridging the gap :)
 one year ago

vf321Best ResponseYou've already chosen the best response.2
but @thez25 the most important thing here is understanding the formulas. Do you get em?
 one year ago

thez25Best ResponseYou've already chosen the best response.0
yeah im familiar with those its just really the end part thats getting me lol
 one year ago

vf321Best 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
 one year ago

vf321Best 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
 one year ago

vf321Best ResponseYou've already chosen the best response.2
scratch that last statement tho its only valid in some cases.
 one year ago

thez25Best ResponseYou've already chosen the best response.0
So how do you maximize T for a given value of a?
 one year ago

vf321Best ResponseYou've already chosen the best response.2
what do you mean 'maximize'? T is a unit vector.
 one year ago

thez25Best 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?
 one year ago

vf321Best ResponseYou've already chosen the best response.2
Strange. T = 1 always...
 one year ago

vf321Best 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!
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
well a and b are susceptible to change, so perhaps we need the partial wrt b ??
 one year ago

vf321Best 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
 one year ago

vf321Best ResponseYou've already chosen the best response.2
for 2D circles. the kcomponent just adds a constant, orthogonal component to motion.
 one year ago

thez25Best 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!
 one year ago

vf321Best ResponseYou've already chosen the best response.2
haha yeah! i forgot. It probably was torsion!
 one year ago

thez25Best ResponseYou've already chosen the best response.0
Torsion is b/(a^2+b^2) hmm
 one year ago

vf321Best 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... \[\tau = \left \left \frac{d\vec B(t)}{ds} \right \right = \left \left \frac{d\vec B(t)/dt}{v(t)} \right \right\]
 one year ago

thez25Best ResponseYou've already chosen the best response.0
Lets just do it your way i can follow those formulas better
 one year ago

vf321Best ResponseYou've already chosen the best response.2
\[\vec v(t)=<−a sin(t),a cos(t),b>\] \[\vec B(t) = \vec T(t) \times \vec N(t)\] \[\vec N(t) = \frac{d\vec T(t)}{ds}\]
 one year ago

vf321Best 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
 one year ago

thez25Best ResponseYou've already chosen the best response.0
so we still have to find T lol
 one year ago

vf321Best ResponseYou've already chosen the best response.2
Hopefully you already know the unit binormal formula and the unit normal formula
 one year ago

thez25Best ResponseYou've already chosen the best response.0
The T in those formulas is the unit tangent right not torsion?
 one year ago

thez25Best ResponseYou've already chosen the best response.0
Yeah ive seen all of those before!
 one year ago

vf321Best ResponseYou've already chosen the best response.2
Yes. and BTW my formula for N should be divided by magnitude, forgot that.
 one year ago

vf321Best 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... \[\vec T(t) = \frac{\vec v(t)}{V} = \frac{<−a sin(t),a cos(t),b>}{V} \]
 one year ago

thez25Best ResponseYou've already chosen the best response.0
just to make sure its the magnitude of V right?
 one year ago

vf321Best ResponseYou've already chosen the best response.2
V IS the magnitude  v(t) = V \[\vec N(t) = \frac{d\vec T(t)/dt}{ds/dt} =\frac{ <a cos(t), a sin(t), 0>/V+\vec v(t) *V'/V^2}{\vec v(t)}\]
 one year ago

thez25Best 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!
 one year ago

vf321Best 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>/
 one year ago

vf321Best 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...
 one year ago

thez25Best 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!
 one year ago

vf321Best ResponseYou've already chosen the best response.2
Okay. Either way, some way or another, you get: \[\vec N = <cos(t), sin(t) ,0>\]\[\vec T = \frac{<−asin(t),acos(t),b>}{\sqrt{a^2+b^2}}\]
 one year ago

thez25Best ResponseYou've already chosen the best response.0
I got that earlier when i was finding them :p
 one year ago

thez25Best ResponseYou've already chosen the best response.0
is V^2 in the previous equation equal to 1 somehow?
 one year ago

vf321Best ResponseYou've already chosen the best response.2
When we cross them, the work for which I won't bother to show, we get \[\vec B = \frac{<a sin(t) cos(t), a sin(t)cos(t), 0>}{\sqrt{a^2+b^2}}\]
 one year ago

thez25Best ResponseYou've already chosen the best response.0
sorry didnt see that you explained that at the end of that paragraph
 one year ago

vf321Best ResponseYou've already chosen the best response.2
wait i messed my cross sry.. it shouldnt have no z value.
 one year ago

thez25Best ResponseYou've already chosen the best response.0
isnt z the 0 there? so no value right
 one year ago

vf321Best ResponseYou've already chosen the best response.2
Yes, but that's wrong. This is the right cross product (I checked with mma) \[\vec B = \frac{<b sin(t), bcos(t), a>}{\sqrt{a^2+b^2}}]\]
 one year ago

vf321Best ResponseYou've already chosen the best response.2
Now that we have B, we can find the torsion!
 one year ago

vf321Best 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
 one year ago

vf321Best ResponseYou've already chosen the best response.2
I think that you can handle the torsion from here, right?
 one year ago

thez25Best ResponseYou've already chosen the best response.0
its alright I got what you meant by it!
 one year ago

vf321Best 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)
 one year ago

vf321Best ResponseYou've already chosen the best response.2
k i gotta go now im sure u can handle it from there.
 one year ago

thez25Best 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!
 one year ago

TuringTestBest 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\[\frac{d\tau}{dt}=0\]
 one year ago

TuringTestBest 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\[\frac{\partial\tau}{\partial b}\]?
 one year ago

thez25Best ResponseYou've already chosen the best response.0
sorry i left but they want it with respect to any given value of a...
 one year ago

TuringTestBest 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
 one year ago

thez25Best 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?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
since tau only depends on a and b, yes
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
according to the earlier work, which I'm not going to independently verify
 one year ago

thez25Best ResponseYou've already chosen the best response.0
okay thank you for your help!
 one year ago
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