## 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

1. vf321 Group Title

By kappa do you mean curvature?

2. TuringTest Group Title

kappa is curvature and what is tau...

3. vf321 Group Title

Does he maybe mean T, the unit tangent?

4. thez25 Group Title

Yes to both sorry for the confusion

5. vf321 Group Title

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

6. TuringTest Group Title

me neither, I've only seen big T

7. TuringTest Group Title

anyway, it's very straightforward with the formulas

8. vf321 Group Title

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

9. TuringTest Group Title

you need tangent for kappa, so...

10. thez25 Group Title

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

11. vf321 Group Title

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

12. vf321 Group Title

BTW how do you do a vector overscript in this?

13. TuringTest Group Title

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

14. vf321 Group Title

superscript*

15. TuringTest Group Title

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

16. vf321 Group Title

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

17. vf321 Group Title

yeah howd u get the arrow and hat superscripts?

18. TuringTest Group Title

\vec r$\vec r$

19. TuringTest Group Title

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

20. vf321 Group Title

sweet thx was always too lazy to learn latex lol

21. TuringTest Group Title

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

22. vf321 Group Title

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

23. vf321 Group Title

By what i showed u earlier, @OP

24. thez25 Group Title

thanks im following you just fine thus far!

25. vf321 Group Title

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

26. vf321 Group Title

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

27. thez25 Group Title

haha yes!

28. vf321 Group Title

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

29. vf321 Group Title

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?

30. thez25 Group Title

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

31. vf321 Group Title

ds lol soz

32. vf321 Group Title

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

33. TuringTest Group Title

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

34. TuringTest Group Title

keeps arc length out of it

35. vf321 Group Title

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

36. vf321 Group Title

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

37. thez25 Group Title

Thats a good idea :p

38. vf321 Group Title

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...

39. thez25 Group Title

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

40. TuringTest Group Title

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

41. TuringTest Group Title

product of*...

42. vf321 Group Title

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

43. vf321 Group Title

Man its been a while!

44. thez25 Group Title

yea thats right!

45. TuringTest Group Title

ah, thanks for bridging the gap :)

46. vf321 Group Title

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

47. thez25 Group Title

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

48. vf321 Group Title

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||

49. vf321 Group Title

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

50. thez25 Group Title

I got ya thanks :p

51. vf321 Group Title

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

52. thez25 Group Title

okay got ya

53. thez25 Group Title

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

54. vf321 Group Title

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

55. thez25 Group Title

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

56. vf321 Group Title

Strange. |T| = 1 always...

57. vf321 Group Title

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

58. TuringTest Group Title

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

59. vf321 Group Title

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

60. vf321 Group Title

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

61. thez25 Group Title

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

62. vf321 Group Title

haha yeah! i forgot. It probably was torsion!

63. thez25 Group Title

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

64. vf321 Group Title

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

65. thez25 Group Title

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

66. vf321 Group Title

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

67. vf321 Group Title

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

68. thez25 Group Title

so we still have to find T lol

69. vf321 Group Title

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

70. thez25 Group Title

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

71. thez25 Group Title

Yeah ive seen all of those before!

72. vf321 Group Title

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

73. thez25 Group Title

right right!

74. vf321 Group Title

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

75. thez25 Group Title

just to make sure its the magnitude of V right?

76. vf321 Group Title

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

77. thez25 Group Title

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

78. vf321 Group Title

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>/

79. vf321 Group Title

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

80. thez25 Group Title

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!

81. vf321 Group Title

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}}$

82. thez25 Group Title

I got that earlier when i was finding them :p

83. thez25 Group Title

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

84. vf321 Group Title

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

85. thez25 Group Title

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

86. vf321 Group Title

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

87. thez25 Group Title

isnt z the 0 there? so no value right

88. vf321 Group Title

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

89. thez25 Group Title

okay thanks!

90. vf321 Group Title

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

91. vf321 Group Title

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

92. vf321 Group Title

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

93. thez25 Group Title

its alright I got what you meant by it!

94. vf321 Group Title

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

95. vf321 Group Title

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

96. thez25 Group Title

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

97. TuringTest Group Title

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

98. TuringTest Group Title

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

99. thez25 Group Title

sorry i left but they want it with respect to any given value of a...

100. TuringTest Group Title

so the only thing to take the derivative with respect to is b. so set$\frac{\partial\tau}{\partial b}=0$solve for b

101. thez25 Group Title

alright thanks so that would give the largest value torsion can have for a given value of a then correct?

102. TuringTest Group Title

since tau only depends on a and b, yes

103. TuringTest Group Title

according to the earlier work, which I'm not going to independently verify

104. thez25 Group Title

okay thank you for your help!