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anonymous
 3 years 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.
anonymous
 3 years 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.

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0By kappa do you mean curvature?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1kappa is curvature and what is tau...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Does he maybe mean T, the unit tangent?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes to both sorry for the confusion

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Really? I never heard of that use of tau before...

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well anyway ill handle the tangent u can do kappa, Turing.

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0its a T in like greek format i believe it is the unit tangent though

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0lol ur right i was thinking about 2D formula for curvature

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0BTW how do you do a vector overscript in this?

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

TuringTest
 3 years ago
Best 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[r(t)=(acost)i+(asint)j+btk\]\[r'(t)=v(t)=D_t(acost)i+D_t(asint)j+D_t(bt)k\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah howd u get the arrow and hat superscripts?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sweet thx was always too lazy to learn latex lol

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\vec v(t) = <a \sin(t), a \cos(t), b>\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0By what i showed u earlier, @OP

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thanks im following you just fine thus far!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\vec T(t) = \frac{\vec v(t)}{\left \vec v(t) \right}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I trust you can handle all that fun algebra yourself...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0And 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
 3 years ago
Best ResponseYou've already chosen the best response.0And 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
 3 years ago
Best ResponseYou've already chosen the best response.0no just for what the largest value of T is Thanks!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\kappa = \left \frac{d\vec T}{ds}\right

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

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1keeps arc length out of it

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\kappa = \left \frac{d\vec T}{ds}\right\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I suppose, if you want to avoid chain rule, that TT's method's better.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah 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
 3 years ago
Best ResponseYou've already chosen the best response.0Lol I liked calculus till i got into some of this stuff....

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1this 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Lol 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
 3 years ago
Best ResponseYou've already chosen the best response.0Man its been a while!

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but @thez25 the most important thing here is understanding the formulas. Do you get em?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah im familiar with those its just really the end part thats getting me lol

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Note 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
 3 years ago
Best ResponseYou've already chosen the best response.0scratch that last statement tho its only valid in some cases.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So how do you maximize T for a given value of a?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what do you mean 'maximize'? T is a unit vector.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The question asks what is the largest value t can have for a given value of a?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Strange. T = 1 always...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if it was velocity, as in \[\vec v(t)\] Then I'd say look at the physics!

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0By physics I mean a is the radius, and v is the velocity. And we know from circular motion that mv^2/(a) = acc

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0for 2D circles. the kcomponent just adds a constant, orthogonal component to motion.

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0haha yeah! i forgot. It probably was torsion!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Torsion is b/(a^2+b^2) hmm

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well 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
 3 years ago
Best ResponseYou've already chosen the best response.0Lets just do it your way i can follow those formulas better

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\vec v(t)=<−a sin(t),a cos(t),b>\] \[\vec B(t) = \vec T(t) \times \vec N(t)\] \[\vec N(t) = \frac{d\vec T(t)}{ds}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0To 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
 3 years ago
Best ResponseYou've already chosen the best response.0so we still have to find T lol

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hopefully you already know the unit binormal formula and the unit normal formula

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The T in those formulas is the unit tangent right not torsion?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah ive seen all of those before!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes. and BTW my formula for N should be divided by magnitude, forgot that.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So 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
 3 years ago
Best ResponseYou've already chosen the best response.0just to make sure its the magnitude of V right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0V 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
 3 years ago
Best ResponseYou've already chosen the best response.0ohh that makes it easier to just cancel rather than figure out T right away good work!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well, 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
 3 years ago
Best ResponseYou've already chosen the best response.0Sorry if this is hard to understand. If you don't get it, I can go back...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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
 3 years ago
Best ResponseYou've already chosen the best response.0Okay. 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
 3 years ago
Best ResponseYou've already chosen the best response.0I got that earlier when i was finding them :p

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sorry didnt see that you explained that at the end of that paragraph

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wait i messed my cross sry.. it shouldnt have no z value.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0isnt z the 0 there? so no value right

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes, but that's wrong. This is the right cross product (I checked with mma) \[\vec B = \frac{<b sin(t), bcos(t), a>}{\sqrt{a^2+b^2}}]\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Now that we have B, we can find the torsion!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\tau = \left\left\frac{d\vec B(t)/dt}{\vec v(t)}\right\right\] 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
 3 years ago
Best ResponseYou've already chosen the best response.0I think that you can handle the torsion from here, right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0its alright I got what you meant by it!

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0k i gotta go now im sure u can handle it from there.

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

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sorry i left but they want it with respect to any given value of a...

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

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

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay thank you for your help!
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