## Babynini one year ago Vectors.

1. Babynini

2. Babynini

@iambatman :)

3. Babynini

ganeshie it's kind of the same thing we just did except for without angles given.

4. ganeshie8

draw both the velocity vectors break them into component form and simply add them up

5. Babynini

|dw:1433310949412:dw|

6. Babynini

wait, I forgot the component form.

7. ganeshie8

|dw:1433311043791:dw|

8. Babynini

so 19

9. ganeshie8

Nope, you need to add the respective components

10. ganeshie8

x components : -5+0 = -5 y components : 0+24 = 24 so the resultant velocity vector with reference to water is -5i+24j

11. Babynini

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

12. ganeshie8

simply take magnitude of the resultant velocity vector for speed

13. ganeshie8

speed = $$\|-5i+24j\|=\sqrt{(-5)^2+24^2}=?$$

14. Babynini

$\sqrt{601}$ =24.52

15. ganeshie8

looks good

16. Babynini

fabulous. part next? :)

17. ganeshie8

work the angle for any vector $$xi+yj$$ the angle with the positive x axis is given by $$\tan^{-1}(y/x)$$

18. Babynini

so since the five is negative..do i carry the negative with it?

19. Babynini

78.23

20. ganeshie8

try 11.77 for angle

21. Babynini

yeah that worked :P

22. Babynini

how did you get there?

23. ganeshie8

you got angle = $$-78.23$$ right, it is negative

24. ganeshie8

in which quadrant does the angle $$-78.23$$ lies in ?

25. Babynini

4

26. ganeshie8

yes, but the resultant vector lies in 2nd quadrant : |dw:1433312298520:dw|

27. ganeshie8

so you need to add 180 to the angle given by your calculator : $-78.23+180=?$

28. Babynini

101.77

29. ganeshie8

Yes thats the angle from positive x axis : |dw:1433312428785:dw|

30. ganeshie8

but the question is asking for the angle from N : |dw:1433312468055:dw|

31. Babynini

oh oh i see.

32. Babynini

so that - 90 which comes out to 11.77

33. ganeshie8

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

so if it's negative then always yes add 180 yeah? :P

35. ganeshie8

not quite

36. ganeshie8

tell me this in which quadrant does the point (-1, 1) lie in ?

37. Babynini

2

38. ganeshie8

so would you trust me if i say the angle is -45 with positive x axis ?

39. ganeshie8

try working the angles using arctan(y/x) formula for below points : (1, 1) (-1, 1) (1, -1) (-1, -1)

40. Babynini

45 -45 -45 45

41. ganeshie8

do they make sense ?

42. ganeshie8

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

yeah.

44. Babynini

these are good questions to think about :o

45. ganeshie8

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

$$\tan^{-1}(y/x)$$ formula gives you correct angle when the point is in I or IV quadrants

47. Babynini

for III and II add 180?

48. ganeshie8

49. ganeshie8

In our case, the point (-5, 25) is in II quadrant so we added 180 degrees to the answer given by calculator

50. ganeshie8

this is one of the things that trips all students when they begin to learn inverse trig functions

51. Babynini

yeah man, i'm kind of trippin now haha but I think I get it.

52. ganeshie8

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

not now, when ever you have time and feel like... it explains arctan function very nicely

54. Babynini

Thanks :)