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nickymarden

  • 2 years ago

does anyone want to teach me Multivariable Calculus? :P

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  1. amistre64
    • 2 years ago
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    you know what you did to x in calc 1 and 2? now do it to y and z

  2. nickymarden
    • 2 years ago
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    im good with matrices and all that. it's when it comes a part of vectors that i'm not understanding really well

  3. amistre64
    • 2 years ago
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    |dw:1334382554391:dw|

  4. amistre64
    • 2 years ago
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    vector just point in a direction

  5. nickymarden
    • 2 years ago
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    wait, ill show u

  6. amistre64
    • 2 years ago
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    good, casue i aint none to goog at mindreading :)

  7. amistre64
    • 2 years ago
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    ... or typing lol

  8. nickymarden
    • 2 years ago
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    haha

  9. ald0024
    • 2 years ago
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    I'm in the same position! Some teacher failed me along the way in the vector department!

  10. nickymarden
    • 2 years ago
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    given vectors a=2xi+2xj+xk, b=xi-2xj+2xk and c=2xi-xj-2xk show that {a,b,c} is a negative orthogonal base for x<0

  11. amistre64
    • 2 years ago
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    an orthogonal base is such that each vector is orthogonal to each other

  12. amistre64
    • 2 years ago
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    since x is a common factor of everyting; we can ignore it

  13. nickymarden
    • 2 years ago
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    for wich value(s) of x, will {a,b,c} be an orthornomal base?

  14. amistre64
    • 2 years ago
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    a.b =2xi+2xj+xk, xi-2xj+2xk ----------- 2 - 4 + 2 = 0 b.c=2xi-xj -2xk xi-2xj+2xk ----------- 2 + 2 - 4 = 0 a.c =2xi+2xj+xk, xi-2xj+2xk ----------- 2 - 4 + 2 = 0 so it is orthonormal

  15. nickymarden
    • 2 years ago
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    i did that, but when it comes to the second part i get lost.

  16. amistre64
    • 2 years ago
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    and since x is just a common "scalar" id assume x can be anything; but maybe zero since people hate zeros

  17. nickymarden
    • 2 years ago
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    i got this:

  18. amistre64
    • 2 years ago
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    " Moreover, they are all required to have length one" http://mathworld.wolfram.com/OrthonormalBasis.html

  19. nickymarden
    • 2 years ago
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    \[||a||=\sqrt{4x^2+4x^2+x^2}=\sqrt{9x^2}=3x\]

  20. amistre64
    • 2 years ago
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    looks like x needs to be the 1/|...| yeah, that

  21. nickymarden
    • 2 years ago
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    so i did it for all 3 vectors and they all were 3x

  22. amistre64
    • 2 years ago
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    x is the scalar that makes them go unit i believe; which means the x = 1 over magnitude

  23. nickymarden
    • 2 years ago
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    so since ||a||=||b||=||c||=3x=1 ---> x=1/3

  24. amistre64
    • 2 years ago
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    x is not part of the length; its factored out and eqauted to the reciprocal of the magnitude

  25. amistre64
    • 2 years ago
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    each vector has the same basic components so they are going to have the same lengths <2,2,1> ; sqrt(4+4+1) = sqrt(9) = 3 x = 1/3 yep

  26. nickymarden
    • 2 years ago
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    there's a third part :X

  27. amistre64
    • 2 years ago
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    and the negative is just -1/3 right?

  28. nickymarden
    • 2 years ago
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    Find the coordinates to v in the orthornomal base obtained, in wich v in the canonical base has coordinates (1,-2,-3)

  29. amistre64
    • 2 years ago
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    now your just making up words :P

  30. nickymarden
    • 2 years ago
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    im sorry, its in portuguese, i'm translating :P

  31. nickymarden
    • 2 years ago
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    or trying anyway

  32. amistre64
    • 2 years ago
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    standard basis perhaps?

  33. nickymarden
    • 2 years ago
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    yeees

  34. amistre64
    • 2 years ago
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    row reduce your vector basis next to this new vector from the standard bases

  35. amistre64
    • 2 years ago
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    how should we define our vectors in the basis? what should our x y and z parts correlate to?

  36. amistre64
    • 2 years ago
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    or does it matter?

  37. nickymarden
    • 2 years ago
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    i did it like this

  38. amistre64
    • 2 years ago
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    it matters; switching columns of a matrix alters things; switching rows doent

  39. nickymarden
    • 2 years ago
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    for the first vector and i did the same to the other 2

  40. amistre64
    • 2 years ago
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    v1=2,2,1 v2=1,-2,2 v3=2,-1,-2 \[\frac{1}{3}\begin{vmatrix}v_1&v_2&v_3\\2&1&2\\2&-2&-1\\1&2&-2\end{vmatrix}\ \begin{vmatrix}c_1\\c_2\\c_3\end{vmatrix}=\begin{vmatrix}1\\-2\\-3\end{vmatrix}\]

  41. amistre64
    • 2 years ago
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    your vectors need to be put in columns, not rows

  42. amistre64
    • 2 years ago
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    multiply both sides by 3 to get rid of the scalar

  43. nickymarden
    • 2 years ago
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    ooh yeah

  44. amistre64
    • 2 years ago
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    \[RREF\begin{vmatrix}2&1&2&|&3\\2&-2&-1&|&-6\\1&2&-2&|&-9\end{vmatrix}\]

  45. amistre64
    • 2 years ago
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    the left goes to identity and the right goes to the coord vector in the basis

  46. nickymarden
    • 2 years ago
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    got it.

  47. nickymarden
    • 2 years ago
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    guess im not as bad as i thought? at least i got the first 2 parts correct :p and thanks for explaining the rest :))

  48. amistre64
    • 2 years ago
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    youre welcome :)

  49. Rogue
    • 2 years ago
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    I'd like to learn multi-variate as well to get a head start on next year, but I'm not a fan of matrices and vectors :(

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