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thez25

  • 2 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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  1. vf321
    • 2 years ago
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    By kappa do you mean curvature?

  2. TuringTest
    • 2 years ago
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    kappa is curvature and what is tau...

  3. vf321
    • 2 years ago
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    Does he maybe mean T, the unit tangent?

  4. thez25
    • 2 years ago
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    Yes to both sorry for the confusion

  5. vf321
    • 2 years ago
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    Really? I never heard of that use of tau before...

  6. TuringTest
    • 2 years ago
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    me neither, I've only seen big T

  7. TuringTest
    • 2 years ago
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    anyway, it's very straightforward with the formulas

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

  9. TuringTest
    • 2 years ago
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    you need tangent for kappa, so...

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

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

  12. vf321
    • 2 years ago
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    BTW how do you do a vector overscript in this?

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

  14. vf321
    • 2 years ago
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    superscript*

  15. TuringTest
    • 2 years ago
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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

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

  17. vf321
    • 2 years ago
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    yeah howd u get the arrow and hat superscripts?

  18. TuringTest
    • 2 years ago
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    \vec r\[\vec r\]

  19. TuringTest
    • 2 years ago
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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

  20. vf321
    • 2 years ago
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    sweet thx was always too lazy to learn latex lol

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

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

  23. vf321
    • 2 years ago
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    By what i showed u earlier, @OP

  24. thez25
    • 2 years ago
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    thanks im following you just fine thus far!

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

  26. vf321
    • 2 years ago
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    I trust you can handle all that fun algebra yourself...

  27. thez25
    • 2 years ago
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    haha yes!

  28. vf321
    • 2 years ago
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    And then curvature is pretty easy too. I forgot that it's torsion that's hard. It's just \[\kappa = \left| \frac{d\vec T}{dt}\right|\]

  29. vf321
    • 2 years ago
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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?

  30. thez25
    • 2 years ago
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    no just for what the largest value of T is Thanks!

  31. vf321
    • 2 years ago
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    ds lol soz

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

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

  34. TuringTest
    • 2 years ago
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    keeps arc length out of it

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

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

  37. thez25
    • 2 years ago
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    Thats a good idea :p

  38. vf321
    • 2 years ago
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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...

  39. thez25
    • 2 years ago
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    Lol I liked calculus till i got into some of this stuff....

  40. TuringTest
    • 2 years ago
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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

  41. TuringTest
    • 2 years ago
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    product of*...

  42. vf321
    • 2 years ago
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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: \[\kappa = \left| \left| \frac{d\vec T}{ds} \right| \right|=\left| \left| \frac{d\vec T/dt}{ds/dt} \right| \right|=\left| \left| \frac{d\vec T/dt}{v(t)} \right| \right|\]

  43. vf321
    • 2 years ago
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    Man its been a while!

  44. thez25
    • 2 years ago
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    yea thats right!

  45. TuringTest
    • 2 years ago
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    ah, thanks for bridging the gap :)

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

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

  48. vf321
    • 2 years ago
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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||

  49. vf321
    • 2 years ago
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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

  50. thez25
    • 2 years ago
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    I got ya thanks :p

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

  52. thez25
    • 2 years ago
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    okay got ya

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

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

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

  56. vf321
    • 2 years ago
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    Strange. |T| = 1 always...

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

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

  59. vf321
    • 2 years ago
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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

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

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

  62. vf321
    • 2 years ago
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    haha yeah! i forgot. It probably was torsion!

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

  64. vf321
    • 2 years ago
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    Well it might be i dunno. Let's solve for it, unless your book told you that's it... \[\tau = \left| \left| \frac{d\vec B(t)}{ds} \right| \right| = \left| \left| \frac{d\vec B(t)/dt}{v(t)} \right| \right|\]

  65. thez25
    • 2 years ago
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    Lets just do it your way i can follow those formulas better

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

  67. vf321
    • 2 years ago
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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

  68. thez25
    • 2 years ago
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    so we still have to find T lol

  69. vf321
    • 2 years ago
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    Hopefully you already know the unit binormal formula and the unit normal formula

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

  71. thez25
    • 2 years ago
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    Yeah ive seen all of those before!

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

  73. thez25
    • 2 years ago
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    right right!

  74. vf321
    • 2 years ago
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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... \[\vec T(t) = \frac{\vec v(t)}{V} = \frac{<−a sin(t),a cos(t),b>}{V} \]

  75. thez25
    • 2 years ago
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    just to make sure its the magnitude of V right?

  76. vf321
    • 2 years ago
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    V IS the magnitude - ||v(t)|| = V \[\vec N(t) = \frac{d\vec T(t)/dt}{ds/dt} =\frac{ <-a cos(t), -a sin(t), 0>/V+\vec v(t) *V'/V^2}{\vec v(t)}\]

  77. thez25
    • 2 years ago
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    ohh that makes it easier to just cancel rather than figure out T right away good work!

  78. vf321
    • 2 years ago
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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>/

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

  80. thez25
    • 2 years ago
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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!

  81. vf321
    • 2 years ago
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    Okay. Either way, some way or another, you get: \[\vec N = <-cos(t), -sin(t) ,0>\]\[\vec T = \frac{<−asin(t),acos(t),b>}{\sqrt{a^2+b^2}}\]

  82. thez25
    • 2 years ago
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    I got that earlier when i was finding them :p

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

  84. vf321
    • 2 years ago
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    When we cross them, the work for which I won't bother to show, we get \[\vec B = \frac{<a sin(t) cos(t), -a sin(t)cos(t), 0>}{\sqrt{a^2+b^2}}\]

  85. thez25
    • 2 years ago
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    sorry didnt see that you explained that at the end of that paragraph

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

  87. thez25
    • 2 years ago
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    isnt z the 0 there? so no value right

  88. vf321
    • 2 years ago
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    Yes, but that's wrong. This is the right cross product (I checked with mma) \[\vec B = \frac{<b sin(t),- bcos(t), a>}{\sqrt{a^2+b^2}}]\]

  89. thez25
    • 2 years ago
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    okay thanks!

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

  91. vf321
    • 2 years ago
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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

  92. vf321
    • 2 years ago
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    I think that you can handle the torsion from here, right?

  93. thez25
    • 2 years ago
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    its alright I got what you meant by it!

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

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

  96. thez25
    • 2 years ago
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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!

  97. TuringTest
    • 2 years ago
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    maximize it with respect to what? if you want to maximize it with respect to t find\[\frac{d\tau}{dt}=0\]

  98. TuringTest
    • 2 years ago
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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\[\frac{\partial\tau}{\partial b}\]?

  99. thez25
    • 2 years ago
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    sorry i left but they want it with respect to any given value of a...

  100. TuringTest
    • 2 years ago
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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

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

  102. TuringTest
    • 2 years ago
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    since tau only depends on a and b, yes

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

  104. thez25
    • 2 years ago
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    okay thank you for your help!

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