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Zela101

  • one year ago

Vector Valued Functions: Describe the graph of the equation

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  1. Zela101
    • one year ago
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    |dw:1443300346520:dw|

  2. Zela101
    • one year ago
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    @dan815

  3. Zela101
    • one year ago
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    |dw:1443300596235:dw|

  4. dan815
    • one year ago
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    hello

  5. dan815
    • one year ago
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    you can use t as an angle and think about it liek a polar graph

  6. Zela101
    • one year ago
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    |dw:1443301007077:dw|

  7. dan815
    • one year ago
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    or you can use t as a 3rd dimension and graph it as a 3d plot whatever u like

  8. Zela101
    • one year ago
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    Was this a correct approach, dan?

  9. dan815
    • one year ago
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    both are good

  10. Zela101
    • one year ago
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    |dw:1443301167273:dw|

  11. dan815
    • one year ago
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    yes thats right

  12. Zela101
    • one year ago
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    Alright, thanks! I was making sure of i'm getting the concept before going ahead and begin with the other problems.

  13. dan815
    • one year ago
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    if u want to think in polar also one more thing to note is that

  14. dan815
    • one year ago
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    you are starting with the tangent vectors along this circle then u have the perpendiculars along the circle

  15. dan815
    • one year ago
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    <x,-y> is perpendicular to <y,x>

  16. dan815
    • one year ago
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    u just have a different sclaring factor on these vectors

  17. Zela101
    • one year ago
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    I see. Does it also apply in 3d graphs?

  18. dan815
    • one year ago
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    hmm

  19. dan815
    • one year ago
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    it would be a slightly different interprataion if t wasnt seen as the angle

  20. dan815
    • one year ago
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    but r'(t) should still be perpendicular to r(t)

  21. Zela101
    • one year ago
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    |dw:1443301651054:dw|

  22. dan815
    • one year ago
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    sint r(t) dot r'(t) = 0 no matter how u plot this

  23. dan815
    • one year ago
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    since*

  24. dan815
    • one year ago
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    not all r(t) dot r'|(t) = 0 ofcourse though

  25. dan815
    • one year ago
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    |dw:1443301819148:dw|

  26. dan815
    • one year ago
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    so this would be a plane with that normal direction vector

  27. dan815
    • one year ago
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    its a line or a curve i mean not a plane

  28. Zela101
    • one year ago
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    |dw:1443302036857:dw|

  29. Zela101
    • one year ago
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    Thanks Dan!

  30. dan815
    • one year ago
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    close its z=1+3(x/2)

  31. dan815
    • one year ago
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    but ya

  32. dan815
    • one year ago
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    its just that line along the plane y=-3

  33. Zela101
    • one year ago
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    |dw:1443302189787:dw|

  34. Zela101
    • one year ago
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    Lol that was a silly mistake I made

  35. dan815
    • one year ago
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    |dw:1443302201369:dw|

  36. dan815
    • one year ago
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    |dw:1443302270988:dw|

  37. dan815
    • one year ago
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    like that on that y=-3 plane we have this line existing

  38. Zela101
    • one year ago
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    And this line would be z=1+(3/2)x

  39. dan815
    • one year ago
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    yes

  40. dan815
    • one year ago
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    it looks flipped because x is positive when u go left on the planed noe

  41. Zela101
    • one year ago
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    Thanks again :)

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