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anonymous

  • one year ago

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  1. anonymous
    • one year ago
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    |dw:1433512332090:dw|

  2. anonymous
    • one year ago
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    @Loser66

  3. Loser66
    • one year ago
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    R can be understood as direction \(\mathbb R\) is one direction |dw:1433512991259:dw|

  4. Loser66
    • one year ago
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    \(\mathbb R^2\) is 2 dimensions

  5. Loser66
    • one year ago
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    |dw:1433513046896:dw|

  6. Loser66
    • one year ago
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    \(\mathbb R^3\) is 3 dimensions |dw:1433513079018:dw|

  7. anonymous
    • one year ago
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    that's a really stupid notation when u could just say we're working in 3 dimensions!!!

  8. Loser66
    • one year ago
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    \(\vec r(t)=< f(t), g(t) >\) is parametric vector

  9. anonymous
    • one year ago
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    what's the meaning of those less than greater than symbols?

  10. Loser66
    • one year ago
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    That is just the notation for YOU to realize that is a parametric vector form

  11. anonymous
    • one year ago
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    I know \[\vec r(f(t),g(t))=f(t)i+g(t)j\]

  12. Loser66
    • one year ago
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    For example: you have distance , velocity and time. You can graph a distance w.r.t time by the graph with x-axis represents|dw:1433513283150:dw| time and y-axis represents distance

  13. Loser66
    • one year ago
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    same as velocity and time |dw:1433513309349:dw|

  14. Loser66
    • one year ago
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    What if I want you to graph distanc/ velocity and both w.r.t time?

  15. Loser66
    • one year ago
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    Use parametric equations to jot them out.

  16. anonymous
    • one year ago
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    I simply know that \[\vec r(x,y,z)=x i+yj+zk\] now be it x or x(t) either way we will have \[\vec r(x(t),y(t),z(t))=x(t)i+y(t)j+z(t)k\] I just don't understand those curly braces

  17. Loser66
    • one year ago
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    Think of this, an accident happens. A car felt down a cliff. |dw:1433513446289:dw|

  18. Loser66
    • one year ago
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    if it is just an accident, the car's position should be as shown. But it is not, it is here |dw:1433513523649:dw|

  19. Loser66
    • one year ago
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    Parametric equations answer those question. If you understand why we have to know them, it inspires you a lllllllllllot

  20. anonymous
    • one year ago
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    tell me one thing, \[\vec r=<3,4,5>\] is this the same as \[3i+4j+5k\]

  21. Loser66
    • one year ago
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    yup

  22. Loser66
    • one year ago
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    Have you ever take linear algebra??

  23. anonymous
    • one year ago
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    I've studied vectors but I've never seen this notation in my school

  24. Loser66
    • one year ago
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    Do you accept 1 dimension?

  25. anonymous
    • one year ago
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    what do u mean ?

  26. Loser66
    • one year ago
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    Do you understand number line? it is 1 dimension |dw:1433513798994:dw|

  27. anonymous
    • one year ago
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    of course I do

  28. Loser66
    • one year ago
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    Do you accept 2 dimensions? |dw:1433513841395:dw|

  29. anonymous
    • one year ago
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    yep...

  30. Loser66
    • one year ago
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    How about 3 dimensions?

  31. anonymous
    • one year ago
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    yes, but how is that relevant?I simply wanted to know what those notations were for, I've never seen them before.

  32. Loser66
    • one year ago
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    1D---yes 2D----yes 3D----yes why not 4D, 5D.......nD??? Because you don't know them, you reject them? so that you said "that's a really stupid notation when u could just say we're working in 3 dimensions!!!"

  33. anonymous
    • one year ago
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    Is it also correct to say that if \[\vec r(t)=<f(t),g(t),h(t)>\]\[\vec r(t)=f(t)i+g(t)j+h(t)k\]

  34. Loser66
    • one year ago
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    yes

  35. anonymous
    • one year ago
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    Alright I understand this notation now

  36. anonymous
    • one year ago
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    thx

  37. Loser66
    • one year ago
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    the t in \(r(t),f(t), g(t)\) shows all function are w.r.t t

  38. anonymous
    • one year ago
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    I'm not rejection higher dimensions but if you're working in 4 dimensions u can simply say in 4 dimensional space

  39. Loser66
    • one year ago
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    and the comma or i, j, k show they are linearly independent variable.

  40. Loser66
    • one year ago
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    But they are all relate to each other. In \(\mathbb R^3\) when x =0, it can't become \(\mathbb R^2\)

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