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thez25

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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  1. vf321
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    By kappa do you mean curvature?

    • one year ago
  2. TuringTest
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    kappa is curvature and what is tau...

    • one year ago
  3. vf321
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    Does he maybe mean T, the unit tangent?

    • one year ago
  4. thez25
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    Yes to both sorry for the confusion

    • one year ago
  5. vf321
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    Really? I never heard of that use of tau before...

    • one year ago
  6. TuringTest
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    me neither, I've only seen big T

    • one year ago
  7. TuringTest
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    anyway, it's very straightforward with the formulas

    • one year ago
  8. vf321
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    Well anyway ill handle the tangent u can do kappa, Turing.

    • one year ago
  9. TuringTest
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    you need tangent for kappa, so...

    • one year ago
  10. thez25
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    its a T in like greek format i believe it is the unit tangent though

    • one year ago
  11. vf321
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    lol ur right i was thinking about 2D formula for curvature

    • one year ago
  12. vf321
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    BTW how do you do a vector overscript in this?

    • one year ago
  13. TuringTest
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    ok you don't need T for kappa, but it helps

    • one year ago
  14. vf321
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    superscript*

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

    • one year ago
  16. vf321
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    \[r(t)=(acost)i+(asint)j+btk\]\[r'(t)=v(t)=D_t(acost)i+D_t(asint)j+D_t(bt)k\]

    • one year ago
  17. vf321
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    yeah howd u get the arrow and hat superscripts?

    • one year ago
  18. TuringTest
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    \vec r\[\vec r\]

    • one year ago
  19. TuringTest
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    right-click any LaTeX you don't know and you can do show as ->tex commands then see the source code

    • one year ago
  20. vf321
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    sweet thx was always too lazy to learn latex lol

    • one year ago
  21. TuringTest
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    me too I just pick it up as I go along on here...

    • one year ago
  22. vf321
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    \[\vec v(t) = <-a \sin(t), a \cos(t), b>\]

    • one year ago
  23. vf321
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    By what i showed u earlier, @OP

    • one year ago
  24. thez25
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    thanks im following you just fine thus far!

    • one year ago
  25. vf321
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    \[\vec T(t) = \frac{\vec v(t)}{\left| \vec v(t) \right|}\]

    • one year ago
  26. vf321
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    I trust you can handle all that fun algebra yourself...

    • one year ago
  27. thez25
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    haha yes!

    • one year ago
  28. vf321
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    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
  29. vf321
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    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
  30. thez25
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    no just for what the largest value of T is Thanks!

    • one year ago
  31. vf321
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    ds lol soz

    • one year ago
  32. vf321
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    \kappa = \left| \frac{d\vec T}{ds}\right|

    • one year ago
  33. TuringTest
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    I usually get curvature with\[\kappa=\frac{\|\vec T'(t)\|}{\|\vec r'(t)\|}\]

    • one year ago
  34. TuringTest
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    keeps arc length out of it

    • one year ago
  35. vf321
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    \[\kappa = \left| \frac{d\vec T}{ds}\right|\]

    • one year ago
  36. vf321
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    I suppose, if you want to avoid chain rule, that TT's method's better.

    • one year ago
  37. thez25
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    Thats a good idea :p

    • one year ago
  38. vf321
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    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
  39. thez25
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    Lol I liked calculus till i got into some of this stuff....

    • one year ago
  40. TuringTest
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    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
  41. TuringTest
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    product of*...

    • one year ago
  42. vf321
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    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
  43. vf321
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    Man its been a while!

    • one year ago
  44. thez25
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    yea thats right!

    • one year ago
  45. TuringTest
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    ah, thanks for bridging the gap :)

    • one year ago
  46. vf321
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    but @thez25 the most important thing here is understanding the formulas. Do you get em?

    • one year ago
  47. thez25
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    yeah im familiar with those its just really the end part thats getting me lol

    • one year ago
  48. vf321
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    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||

    • one year ago
  49. vf321
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    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
  50. thez25
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    I got ya thanks :p

    • one year ago
  51. vf321
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    scratch that last statement tho its only valid in some cases.

    • one year ago
  52. thez25
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    okay got ya

    • one year ago
  53. thez25
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    So how do you maximize T for a given value of a?

    • one year ago
  54. vf321
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    what do you mean 'maximize'? T is a unit vector.

    • one year ago
  55. thez25
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    The question asks what is the largest value t can have for a given value of a?

    • one year ago
  56. vf321
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    Strange. |T| = 1 always...

    • one year ago
  57. vf321
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    if it was velocity, as in \[\vec v(t)\] Then I'd say look at the physics!

    • one year ago
  58. TuringTest
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    well a and b are susceptible to change, so perhaps we need the partial wrt b ??

    • one year ago
  59. vf321
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    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
  60. vf321
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    for 2D circles. the k-component just adds a constant, orthogonal component to motion.

    • one year ago
  61. thez25
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    I think since were dealing with a helix T could actually be torsion not unit tangent sorry!

    • one year ago
  62. vf321
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    haha yeah! i forgot. It probably was torsion!

    • one year ago
  63. thez25
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    Torsion is b/(a^2+b^2) hmm

    • one year ago
  64. vf321
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    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
  65. thez25
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    Lets just do it your way i can follow those formulas better

    • one year ago
  66. vf321
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    \[\vec v(t)=<−a sin(t),a cos(t),b>\] \[\vec B(t) = \vec T(t) \times \vec N(t)\] \[\vec N(t) = \frac{d\vec T(t)}{ds}\]

    • one year ago
  67. vf321
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    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
  68. thez25
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    so we still have to find T lol

    • one year ago
  69. vf321
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    Hopefully you already know the unit binormal formula and the unit normal formula

    • one year ago
  70. thez25
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    The T in those formulas is the unit tangent right not torsion?

    • one year ago
  71. thez25
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    Yeah ive seen all of those before!

    • one year ago
  72. vf321
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    Yes. and BTW my formula for N should be divided by magnitude, forgot that.

    • one year ago
  73. thez25
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    right right!

    • one year ago
  74. vf321
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    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
  75. thez25
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    just to make sure its the magnitude of V right?

    • one year ago
  76. vf321
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    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
  77. thez25
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    ohh that makes it easier to just cancel rather than figure out T right away good work!

    • one year ago
  78. vf321
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    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
  79. vf321
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    Sorry if this is hard to understand. If you don't get it, I can go back...

    • one year ago
  80. thez25
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    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
  81. vf321
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    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
  82. thez25
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    I got that earlier when i was finding them :p

    • one year ago
  83. thez25
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    is V^2 in the previous equation equal to 1 somehow?

    • one year ago
  84. vf321
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    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
  85. thez25
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    sorry didnt see that you explained that at the end of that paragraph

    • one year ago
  86. vf321
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    wait i messed my cross sry.. it shouldnt have no z value.

    • one year ago
  87. thez25
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    isnt z the 0 there? so no value right

    • one year ago
  88. vf321
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    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
  89. thez25
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    okay thanks!

    • one year ago
  90. vf321
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    Now that we have B, we can find the torsion!

    • one year ago
  91. vf321
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    \[\tau = \left|\left|\frac{d\vec B(t)/dt}{\vec v(t)}\right|\right|\] 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
  92. vf321
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    I think that you can handle the torsion from here, right?

    • one year ago
  93. thez25
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    its alright I got what you meant by it!

    • one year ago
  94. vf321
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    dB/dt = {b Cos[t], b Sin[t], 0}/V ||v(t)|| = V V = sqrt(a^2+b^2)

    • one year ago
  95. vf321
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    k i gotta go now im sure u can handle it from there.

    • one year ago
  96. thez25
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    well if you get a chance later could u explain how to maximize it lol thanks for all your help though!

    • one year ago
  97. TuringTest
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    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
  98. TuringTest
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    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
  99. thez25
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    sorry i left but they want it with respect to any given value of a...

    • one year ago
  100. TuringTest
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    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
  101. thez25
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    alright thanks so that would give the largest value torsion can have for a given value of a then correct?

    • one year ago
  102. TuringTest
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    since tau only depends on a and b, yes

    • one year ago
  103. TuringTest
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    according to the earlier work, which I'm not going to independently verify

    • one year ago
  104. thez25
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    okay thank you for your help!

    • one year ago
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