Question

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.

Answer 1

By kappa do you mean curvature?

Answer 2

kappa is curvature and what is tau...

Answer 3

Does he maybe mean T, the unit tangent?

Answer 4

Yes to both sorry for the confusion

Answer 5

Really? I never heard of that use of tau before...

Answer 6

me neither, I've only seen big T

Answer 7

anyway, it's very straightforward with the formulas

Answer 8

Well anyway ill handle the tangent
u can do kappa, Turing.

Answer 9

you need tangent for kappa, so...

Answer 10

its a T in like greek format i believe it is the unit tangent though

Answer 11

lol ur right i was thinking about 2D formula for curvature

Answer 12

BTW how do you do a vector overscript in this?

Answer 13

ok you don't need T for kappa, but it helps

Answer 14

superscript*

Answer 15

\[\vec r(t)=(a\cos t)\hat i+(a\sin t)\hat j+bt\hat k;~~~ a,b>0; ~~~a^2+b^2 \neq0\]most likely

Answer 16

\[r(t)=(acost)i+(asint)j+btk\]\[r'(t)=v(t)=D_t(acost)i+D_t(asint)j+D_t(bt)k\]

Answer 17

yeah howd u get the arrow and hat superscripts?

Answer 18

\vec r\[\vec r\]

Answer 19

right-click any LaTeX you don't know and you can do
show as ->tex commands
then see the source code

Answer 20

sweet thx was always too lazy to learn latex lol

Answer 21

me too I just pick it up as I go along on here...

Answer 22

\[\vec v(t) = <-a \sin(t), a \cos(t), b>\]

Answer 23

By what i showed u earlier, @OP

Answer 24

thanks im following you just fine thus far!

Answer 25

\[\vec T(t) = \frac{\vec v(t)}{\left| \vec v(t) \right|}\]

Answer 26

I trust you can handle all that fun algebra yourself...

Answer 27

haha yes!

Answer 28

And then curvature is pretty easy too. I forgot that it's torsion that's hard.
It's just
\[\kappa = \left| \frac{d\vec T}{dt}\right|\]

Answer 29

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?

Answer 30

no just for what the largest value of T is Thanks!

Answer 31

ds lol soz

Answer 32

\kappa = \left| \frac{d\vec T}{ds}\right|

Answer 33

I usually get curvature with\[\kappa=\frac{\|\vec T'(t)\|}{\|\vec r'(t)\|}\]

Answer 34

keeps arc length out of it

Answer 35

\[\kappa = \left| \frac{d\vec T}{ds}\right|\]

Answer 36

I suppose, if you want to avoid chain rule, that TT's method's better.

Answer 37

Thats a good idea :p

Answer 38

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...

Answer 39

Lol I liked calculus till i got into some of this stuff....

Answer 40

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

Answer 41

product of*...

Answer 42

Lol wait. I see what happened. We could have gone with mine from the start! I remember doing this in class:
\[\kappa = \left| \left| \frac{d\vec T}{ds} \right| \right|=\left| \left| \frac{d\vec T/dt}{ds/dt} \right| \right|=\left| \left| \frac{d\vec T/dt}{v(t)} \right| \right|\]

Answer 43

Man its been a while!

Answer 44

yea thats right!

Answer 45

ah, thanks for bridging the gap :)

Answer 46

but @thez25 the most important thing here is understanding the formulas. Do you get em?

Answer 47

yeah im familiar with those its just really the end part thats getting me lol

Answer 48

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||

Answer 49

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

Answer 50

I got ya thanks :p

Answer 51

scratch that last statement tho its only valid in some cases.

Answer 52

okay got ya

Answer 53

So how do you maximize T for a given value of a?

Answer 54

what do you mean 'maximize'? T is a unit vector.

Answer 55

The question asks what is the largest value t can have for a given value of a?

Answer 56

Strange. |T| = 1 always...

Answer 57

if it was velocity, as in \[\vec v(t)\]
Then I'd say look at the physics!

Answer 58

well a and b are susceptible to change, so perhaps we need the partial wrt b ??

Answer 59

By physics I mean a is the radius, and v is the velocity. And we know from circular motion that mv^2/(a) = acc

Answer 60

for 2D circles. the k-component just adds a constant, orthogonal component to motion.

Answer 61

I think since were dealing with a helix T could actually be torsion not unit tangent sorry!

Answer 62

haha yeah! i forgot. It probably was torsion!

Answer 63

Torsion is b/(a^2+b^2) hmm

Answer 64

Well it might be i dunno. Let's solve for it, unless your book told you that's it...
\[\tau = \left| \left| \frac{d\vec B(t)}{ds} \right| \right| = \left| \left| \frac{d\vec B(t)/dt}{v(t)} \right| \right|\]

Answer 65

Lets just do it your way i can follow those formulas better

Answer 66

\[\vec v(t)=<−a sin(t),a cos(t),b>\]
\[\vec B(t) = \vec T(t) \times \vec N(t)\]
\[\vec N(t) = \frac{d\vec T(t)}{ds}\]

Answer 67

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

Answer 68

so we still have to find T lol

Answer 69

Hopefully you already know the unit binormal formula and the unit normal formula

Answer 70

The T in those formulas is the unit tangent right not torsion?

Answer 71

Yeah ive seen all of those before!

Answer 72

Yes.

and BTW my formula for N should be divided by magnitude, forgot that.

Answer 73

right right!

Answer 74

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

Answer 75

just to make sure its the magnitude of V right?

Answer 76

V IS the magnitude - ||v(t)|| = V
\[\vec N(t) = \frac{d\vec T(t)/dt}{ds/dt} =\frac{ <-a cos(t), -a sin(t), 0>/V+\vec v(t) *V'/V^2}{\vec v(t)}\]

Answer 77

ohh that makes it easier to just cancel rather than figure out T right away good work!

Answer 78

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>/

Answer 79

Sorry if this is hard to understand. If you don't get it, I can go back...

Answer 80

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!

Answer 81

Okay. Either way, some way or another, you get:
\[\vec N = <-cos(t), -sin(t) ,0>\]\[\vec T = \frac{<−asin(t),acos(t),b>}{\sqrt{a^2+b^2}}\]

Answer 82

I got that earlier when i was finding them :p

Answer 83

is V^2 in the previous equation equal to 1 somehow?

Answer 84

When we cross them, the work for which I won't bother to show, we get
\[\vec B = \frac{<a sin(t) cos(t), -a sin(t)cos(t), 0>}{\sqrt{a^2+b^2}}\]

Answer 85

sorry didnt see that you explained that at the end of that paragraph

Answer 86

wait i messed my cross sry.. it shouldnt have no z value.

Answer 87

isnt z the 0 there? so no value right

Answer 88

Yes, but that's wrong. This is the right cross product (I checked with mma)
\[\vec B = \frac{<b sin(t),- bcos(t), a>}{\sqrt{a^2+b^2}}]\]

Answer 89

okay thanks!

Answer 90

Now that we have B, we can find the torsion!

Answer 91

\[\tau = \left|\left|\frac{d\vec B(t)/dt}{\vec v(t)}\right|\right|\]
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

Answer 92

I think that you can handle the torsion from here, right?

Answer 93

its alright I got what you meant by it!

Answer 94

dB/dt = {b Cos[t], b Sin[t], 0}/V
||v(t)|| = V
V = sqrt(a^2+b^2)

Answer 95

k i gotta go now im sure u can handle it from there.

Answer 96

well if you get a chance later could u explain how to maximize it lol thanks for all your help though!

Answer 97

maximize it with respect to what?
if you want to maximize it with respect to t find\[\frac{d\tau}{dt}=0\]

Answer 98

but\[\tau=\frac b{\sqrt{a^2+b^2}}\]and you said you are "given a value of a" so perhaps they want\[\frac{\partial\tau}{\partial b}\]?