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s3a

  • 4 years ago

UNIT NORMAL VECTOR Question: For this question ( http://f.imgtmp.com/BIMPQ.jpg), given that I have the unit tangent, I tried dT/dt/|dT/dt| and that does not yield the correct answer for the normal vector. For the x components (to be brief): I get 2sin(-pi/3)/sqrt(13) (for unit tangent - correct answer) and -4cos(-pi/3)/sqrt(32) (for unit normal - wrong answer). but my book and this video ( http://www.youtube.com/watch?v=vsKVV5_Hchs) at 1:20 say N = (dT/dt)/|dT/dt| (each vector is a unit vector here). Any help would be greatly appreciated! Thanks in advance!

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  1. anonymous
    • 4 years ago
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    by definition T'(t)/|T'(t)| is the normal vector of r(t)=x(t)i+y(t)j+k(t)k and N(t) = T'(t)/|T'(t)| so to compute normal vector first compute Tangent vector and calculate its unit vector then evaluate the tangent and normal vector at pi/6 NOTE than tangent vector should be in vector form when taking its unit vector to calculate normal vector

  2. s3a
    • 4 years ago
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    that's exactly what i did or think i did. so I guess i should show you more specific work for you to find a mistake?

  3. anonymous
    • 4 years ago
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    \[\text{r'(t)=-2sin(2t)i-2cos(2t)j+3k}\] \[\text{T'(t)}=\frac{\text{(-2sin(2t)i-2cos(2t)j+3k)}}{\text{sqrt((-2sin(2t))^2+(-2cos(2t))^2+3^2)}}\] N(t) =(-(4 cos(2 t))/sqrt(13)+(sin(2 t))/sqrt(13))/sqrt(16/13 cos^2(2 t)+1/13 sin^2(2 t)) \[\href{ http://www.wolframalpha.com/input/?i=%28-%284+cos%282+t%29%29%2Fsqrt%2813%29%2B%28sin%282+t%29%29%2Fsqrt%2813%29%29%2Fsqrt%2816%2F13+cos%5E2%282+t%29%2B1%2F13+sin%5E2%282+t%29%29&lk=1&a=ClashPrefs_*Math- }{wolfram result}\] just add i and j and evaluate at pi/6

  4. s3a
    • 4 years ago
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    inside the sin and cos there is a negative not that it matters for the cos part.

  5. s3a
    • 4 years ago
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    (-(4 cos(-pi/3))/sqrt(13)) for the x component is wrong.

  6. amistre64
    • 4 years ago
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    given vector equation r r' = tangent vector r'' = normal vector

  7. amistre64
    • 4 years ago
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    or simply cross r and r'

  8. amistre64
    • 4 years ago
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    .... thats a stray thought that might be erroneous :)

  9. s3a
    • 4 years ago
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    N = B x T

  10. s3a
    • 4 years ago
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    which i think is different from what you said

  11. amistre64
    • 4 years ago
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    just a bit, I was thinking 2d; not 3d

  12. amistre64
    • 4 years ago
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    cant see the img you posted to be sure

  13. s3a
    • 4 years ago
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    o that's crucial

  14. s3a
    • 4 years ago
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    wait

  15. s3a
    • 4 years ago
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    remove the comma at the end of the link and it should work

  16. amistre64
    • 4 years ago
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    better lol, ), was messing it up

  17. amistre64
    • 4 years ago
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    \[k=\frac{|r'xr"|}{|r'|^3}\] if memory serves, but id have to dbl chk that

  18. s3a
    • 4 years ago
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    i checked myself to test :)

  19. s3a
    • 4 years ago
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    ya that's correct. k = |v x a|/v^3

  20. s3a
    • 4 years ago
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    but why isn't my way working?

  21. s3a
    • 4 years ago
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    my book and the video confirm my formula.

  22. amistre64
    • 4 years ago
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    r = < cos(-2t), sin(-2t), 3t > r' = < 2sin(-2t), -2cos(2t), 3 > r'' = < -4cos(-2t), 4sin(2t), 0 >

  23. amistre64
    • 4 years ago
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    r' = tangent vector r'' = normal vector

  24. s3a
    • 4 years ago
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    is r'' really the normal vector or is it just the acceleration vector which just happens to be parallel and is only the same when it's a unit vector?

  25. amistre64
    • 4 years ago
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    we can unit them up afterwards I believe

  26. amistre64
    • 4 years ago
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    http://tutorial.math.lamar.edu/Classes/CalcIII/TangentNormalVectors.aspx

  27. amistre64
    • 4 years ago
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    r' = < 2sin(-2t), -2cos(2t), 3 > |r'| = \(\sqrt{4sin^2(-2t)+4cos^2(-2t)+9}=\sqrt{13}\) right?

  28. s3a
    • 4 years ago
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    sry it froze. 2 secs.

  29. s3a
    • 4 years ago
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    yes.

  30. amistre64
    • 4 years ago
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    r'/|r'| = unit tangent then in my view

  31. s3a
    • 4 years ago
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    but 2sin(-pi/3)/sqrt(13) is wrong for the x component.

  32. amistre64
    • 4 years ago
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    -pi/3 ?

  33. s3a
    • 4 years ago
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    -2 * pi/6 = -pi/3

  34. amistre64
    • 4 years ago
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    compute at t = pi/6

  35. s3a
    • 4 years ago
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    i did. -2*t = -2 * pi/6 = -pi/3

  36. amistre64
    • 4 years ago
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    does sin(-2t) = -sin(2t)?

  37. s3a
    • 4 years ago
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    ya. also cos(x) = cos(-x)

  38. amistre64
    • 4 years ago
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    r = < cos(-2t), sin(-2t), 3t > r = < cos(2t), -sin(2t), 3t > r' = < -2sin(2t), -2cos(2t), 3 > would amout to the same then, dunno if that improves the results tho

  39. amistre64
    • 4 years ago
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    given r; that has to be the tangent to it ..... at any rate :)

  40. amistre64
    • 4 years ago
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    unit still give us a /sqrt(13) tho

  41. amistre64
    • 4 years ago
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    x = -2sin(pi/3)/sqrt(13) = -2*1/2*1/sqrt(13) = sqrt(13)/13

  42. amistre64
    • 4 years ago
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    sqrt(3)/2 can never really do that in my head right lol -sqrt(3/13) would be tangent x

  43. s3a
    • 4 years ago
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    wait. i'm looking for N not T.

  44. s3a
    • 4 years ago
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    o i found my mistake. -4cos(pi/3)/sqrt(16) is the answer for the x component.

  45. s3a
    • 4 years ago
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    my mistake was basically i factored the denominator wrong and got 2*16=32 when i should have gotten 16*1

  46. s3a
    • 4 years ago
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    thanks for the help.

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