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Babynini

  • one year ago

Vectors.

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  1. Babynini
    • one year ago
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  2. Babynini
    • one year ago
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    @iambatman :)

  3. Babynini
    • one year ago
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    ganeshie it's kind of the same thing we just did except for without angles given.

  4. ganeshie8
    • one year ago
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    draw both the velocity vectors break them into component form and simply add them up

  5. Babynini
    • one year ago
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    |dw:1433310949412:dw|

  6. Babynini
    • one year ago
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    wait, I forgot the component form.

  7. ganeshie8
    • one year ago
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    |dw:1433311043791:dw|

  8. Babynini
    • one year ago
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    so 19

  9. ganeshie8
    • one year ago
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    Nope, you need to add the respective components

  10. ganeshie8
    • one year ago
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    x components : -5+0 = -5 y components : 0+24 = 24 so the resultant velocity vector with reference to water is -5i+24j

  11. Babynini
    • one year ago
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    haha right. sorry. So now how do we change that to decimal numbers? (note how the first question wants to be answered)

  12. ganeshie8
    • one year ago
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    simply take magnitude of the resultant velocity vector for speed

  13. ganeshie8
    • one year ago
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    speed = \(\|-5i+24j\|=\sqrt{(-5)^2+24^2}=?\)

  14. Babynini
    • one year ago
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    \[\sqrt{601}\] =24.52

  15. ganeshie8
    • one year ago
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    looks good

  16. Babynini
    • one year ago
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    fabulous. part next? :)

  17. ganeshie8
    • one year ago
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    work the angle for any vector \(xi+yj\) the angle with the positive x axis is given by \(\tan^{-1}(y/x)\)

  18. Babynini
    • one year ago
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    so since the five is negative..do i carry the negative with it?

  19. Babynini
    • one year ago
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    78.23

  20. ganeshie8
    • one year ago
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    try 11.77 for angle

  21. Babynini
    • one year ago
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    yeah that worked :P

  22. Babynini
    • one year ago
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    how did you get there?

  23. ganeshie8
    • one year ago
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    you got angle = \(-78.23\) right, it is negative

  24. ganeshie8
    • one year ago
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    in which quadrant does the angle \(-78.23\) lies in ?

  25. Babynini
    • one year ago
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    4

  26. ganeshie8
    • one year ago
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    yes, but the resultant vector lies in 2nd quadrant : |dw:1433312298520:dw|

  27. ganeshie8
    • one year ago
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    so you need to add 180 to the angle given by your calculator : \[-78.23+180=?\]

  28. Babynini
    • one year ago
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    101.77

  29. ganeshie8
    • one year ago
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    Yes thats the angle from positive x axis : |dw:1433312428785:dw|

  30. ganeshie8
    • one year ago
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    but the question is asking for the angle from N : |dw:1433312468055:dw|

  31. Babynini
    • one year ago
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    oh oh i see.

  32. Babynini
    • one year ago
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    so that - 90 which comes out to 11.77

  33. ganeshie8
    • one year ago
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    thats it! with \(\tan^{-1}\) function you always need to check if 180 degrees needs to be added based on the quadrant of the actual point

  34. Babynini
    • one year ago
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    so if it's negative then always yes add 180 yeah? :P

  35. ganeshie8
    • one year ago
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    not quite

  36. ganeshie8
    • one year ago
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    tell me this in which quadrant does the point (-1, 1) lie in ?

  37. Babynini
    • one year ago
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    2

  38. ganeshie8
    • one year ago
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    so would you trust me if i say the angle is -45 with positive x axis ?

  39. ganeshie8
    • one year ago
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    try working the angles using arctan(y/x) formula for below points : (1, 1) (-1, 1) (1, -1) (-1, -1)

  40. Babynini
    • one year ago
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    45 -45 -45 45

  41. ganeshie8
    • one year ago
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    do they make sense ?

  42. ganeshie8
    • one year ago
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    look at the last point and last angle : (-1, -1) 45 how can a point in III rd quadrant make an angle 45 ?

  43. Babynini
    • one year ago
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    yeah.

  44. Babynini
    • one year ago
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    these are good questions to think about :o

  45. ganeshie8
    • one year ago
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    it doesn't make sense, so you need to add 180 to 45 : 45+180 = 225 so the angle made by vector (-1, -1) is 225 and not 45 ok

  46. ganeshie8
    • one year ago
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    \(\tan^{-1}(y/x)\) formula gives you correct angle when the point is in I or IV quadrants

  47. Babynini
    • one year ago
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    for III and II add 180?

  48. ganeshie8
    • one year ago
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    When th point lies in II or III quadrants, you need to add 180 degrees to the answer given by your calculator

  49. ganeshie8
    • one year ago
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    In our case, the point (-5, 25) is in II quadrant so we added 180 degrees to the answer given by calculator

  50. ganeshie8
    • one year ago
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    this is one of the things that trips all students when they begin to learn inverse trig functions

  51. Babynini
    • one year ago
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    yeah man, i'm kind of trippin now haha but I think I get it.

  52. ganeshie8
    • one year ago
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    watch this video if u have time https://www.khanacademy.org/math/trigonometry/unit-circle-trig-func/inverse_trig_functions/v/inverse-trig-functions-arctan

  53. ganeshie8
    • one year ago
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    not now, when ever you have time and feel like... it explains arctan function very nicely

  54. Babynini
    • one year ago
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    Thanks :)

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