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Curvature of r(t) = (3 sin t) i + (3 cos t) j + 4t k

  1. Oct 30, 2016 #1
    1. The problem statement, all variables and given/known data

    find the curvature of the vector valued function r(t) = 3sint i + 3cost j +4t k
    2. Relevant equations


    3. The attempt at a solution
    For the unit tangent vector , i got T(t) = (3cost i -3sint j +4k) / sqrt (9 ((sint)^2 ) + 9 ((cost)^2 ) + 4^2 ) =
    (3cost i -3sint j +4k) / 5

    For the principal unit vector , N(t) , i got -3sint i -3cost j ) / sqrt (9 ((sint)^2 ) + 9 ((cost)^2 )
    = -sint i -cost j

    We can conclude that |r'(t) | = 5 , |T'(t) | = 3 , so , curvature = |T'(t) | / |r'(t) | = 3/5 , but the ans is 3/25 , which part of my working is wrong ?
     
    Last edited by a moderator: Oct 30, 2016
  2. jcsd
  3. Oct 30, 2016 #2

    PeroK

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    You need ##|\hat{T}'|## not ##|T'|##.
     
    Last edited by a moderator: Oct 30, 2016
  4. Oct 30, 2016 #3
    what's the difference between ##|\hat{T}'|## and ##|T'|## ???
     
  5. Oct 30, 2016 #4

    PeroK

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    The first is the time derivative of the unit tangent vector. The second is the time derivative of the tangent vector.
     
  6. Oct 30, 2016 #5
    Unit tangent vector and tangent vector are not the same ?
     
  7. Oct 30, 2016 #6

    PeroK

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    No, why would they be? The tangent vector can have any length.
     
  8. Oct 30, 2016 #7
    so , how to get time derivative of the tangent vector. ?
     
  9. Oct 30, 2016 #8

    PeroK

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    It's the time derivative of the unit tangent vector you want. You just calculate the unit tangent vector, then differentiate it.
     
  10. Oct 30, 2016 #9
    do you mean it's |T'(t) | ? if so , then my ans of |T'(t) | is 3 , since k = |T'(t) | / |r'(t) | , so i gt k = 3/5 , but not 3/25
     
  11. Oct 30, 2016 #10

    PeroK

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    That's just what you did the first time! You need to differentiate ##\hat{T}##.
     
  12. Oct 30, 2016 #11
    removed
     
  13. Oct 30, 2016 #12
    do you mean differentiate |T| to get |T'(t) | ?

    If so , then my |T'(t) | = 3
     
  14. Oct 30, 2016 #13

    PeroK

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    What happened to the ##1/5##?
     
  15. Oct 30, 2016 #14
    1/5 ???
    i dont have 1/5 in my working
     
  16. Oct 30, 2016 #15

    PeroK

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    Yes you do!

     
  17. Oct 30, 2016 #16
    OK , but I didn't get 3 /25 . . . .
     
  18. Oct 30, 2016 #17

    PeroK

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    That's because you dropped the 1/5!!!!!!
     
  19. Oct 31, 2016 #18
    ok , i have tried in another way , but i didnt get the ans though ...
    here's my ans , i gt k = 0.1357
    instead of 4/25
     

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  20. Oct 31, 2016 #19
    for the previous question , i gt the question already , thanks!
     
  21. Nov 1, 2016 #20
    bump , what's wrong with my working ?
     

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