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anonymous

  • 5 years ago

P(t) = 1000+360sin((π/6)t)+180sin((π/3)t) Find the derivative and then solve for when P(t)=0

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  1. amistre64
    • 5 years ago
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    its alot of sums; so deriving is easy

  2. amistre64
    • 5 years ago
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    0 +360pi/6 cos(t.pi/6) +180pi/3cos(t.pi/3)

  3. anonymous
    • 5 years ago
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    ok i got that part right. now the solving for p(t)=0 part is where im stuck

  4. amistre64
    • 5 years ago
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    looks tricky ;)

  5. amistre64
    • 5 years ago
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    360/6 = 180/3 so that all the same

  6. amistre64
    • 5 years ago
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    cos(t.pi/6) + cos(t.pi/3) = 0 when.....

  7. amistre64
    • 5 years ago
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    you agree with this so far?

  8. anonymous
    • 5 years ago
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    yep

  9. amistre64
    • 5 years ago
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    we gotta get them at 180 from each other than right?

  10. amistre64
    • 5 years ago
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    t.60 = t.30 + 180; or, t.30 = t.60 + 180

  11. amistre64
    • 5 years ago
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    t(60-30) = 180 t(30) = 180 t = 60 maybe?

  12. amistre64
    • 5 years ago
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    err...6 lol

  13. amistre64
    • 5 years ago
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    6.60 = 360 6.30 = 180; thats a good one right?

  14. amistre64
    • 5 years ago
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    a(1) + a(-1) = 0 right

  15. amistre64
    • 5 years ago
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    so every multiple of 6 should be good for that one...

  16. amistre64
    • 5 years ago
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    t.30 = t.60 + 180 t(30-60) = 180 t = 180/-30 t = -6; same thing

  17. amistre64
    • 5 years ago
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    so whenever t = a multiple of 6, we should be good

  18. anonymous
    • 5 years ago
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    i kinda lost you at t.60 = t.30 + 180; or, t.30 = t.60 + 180. from cos(t.pi/6) + cos(t.pi/3) = 0 in fact, for that^ i simplified and got 60cos((t x pi/6)+ (t x 2pi/6) = 0

  19. amistre64
    • 5 years ago
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    given that t not equal zero perhaps; or some such restriction

  20. amistre64
    • 5 years ago
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    there are bound to be multples of 6 that make t.60 = t.30; in that case its of no use

  21. amistre64
    • 5 years ago
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    ok; we need angles that are 180 from each other in order for there cosines to be opposites of each other right?

  22. anonymous
    • 5 years ago
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    uh huh

  23. amistre64
    • 5 years ago
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    so we start at 60 and 30 and need to at least find the angles that are 180 from each other.... t.60 = t.30 + 180 means that for every multiple of t, we move by 60 and 30 till we find angles that are 180 from each other right?

  24. amistre64
    • 5 years ago
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    t.60 - t.30 = 180 t(60 - 30) = 180 t = 180/30 t = 6

  25. anonymous
    • 5 years ago
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    ooh ok

  26. amistre64
    • 5 years ago
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    now when the angels are 360 from eah other they are the same...so we exclude doing the same thing but with +360 instead of +180 right?

  27. amistre64
    • 5 years ago
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    360/30 = 12; so every multiple of 12 wont work.... right?

  28. anonymous
    • 5 years ago
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    uh huh

  29. amistre64
    • 5 years ago
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    so odd multiples of 6 work; but even multiples fail; so we restricct it to when t is an odd multiple of 6

  30. amistre64
    • 5 years ago
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    i can also think of one more condition, but havent found a way to describe it yet......... it when they angle to the right matches the angle on the left; like a 60 and a 120 would have opposite cosines

  31. anonymous
    • 5 years ago
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    lol i think im good. thank you.

  32. amistre64
    • 5 years ago
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    :) ok

  33. anonymous
    • 5 years ago
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    another question if you dont mind? lol

  34. amistre64
    • 5 years ago
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    i dunno, im pretty much good for one question and then go stupid lol

  35. amistre64
    • 5 years ago
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    whats the q

  36. anonymous
    • 5 years ago
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    LOL, it's a position, velocity, acceleration question.

  37. anonymous
    • 5 years ago
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    Find the position function X(t), of a particle moving along a straight line, with constant acceleration equal to 2, knowing that the position and velocity vanish at t = 2.

  38. amistre64
    • 5 years ago
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    is it posted already or you wanna hit me with it

  39. anonymous
    • 5 years ago
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    Find the initial position and the time t > 0 at which the particle passes again through the initial position.

  40. anonymous
    • 5 years ago
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    i know i should do anti derivative for the first part in order to find the velocity equation.

  41. amistre64
    • 5 years ago
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    acceleration is the derivative of velocity is the derivative of position

  42. anonymous
    • 5 years ago
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    yep

  43. amistre64
    • 5 years ago
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    with a constant acceleration of 2; that means we got linear velocity of 2x+c right?

  44. anonymous
    • 5 years ago
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    yeah

  45. anonymous
    • 5 years ago
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    would i plug in velocity = 0 since it vanishes or no?

  46. amistre64
    • 5 years ago
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    and a position of x^2 +bx +cthen

  47. amistre64
    • 5 years ago
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    the only way for a linear function to vanish at some point it to make a hole iin the graph right?

  48. anonymous
    • 5 years ago
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    yeah

  49. amistre64
    • 5 years ago
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    which means making the velocity function an extention of a rational function

  50. amistre64
    • 5 years ago
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    2x +c ------ at best d-2

  51. amistre64
    • 5 years ago
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    but with a d-2 up top

  52. anonymous
    • 5 years ago
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    what's d-2?

  53. amistre64
    • 5 years ago
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    something to make a whole at 2 to make the graph vanish

  54. anonymous
    • 5 years ago
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    lol i think you might be over thinking the problem. this was suppose to be an easy math problem my prof gave on the test. haha

  55. amistre64
    • 5 years ago
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    i could be over thinking it tho :)

  56. anonymous
    • 5 years ago
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    lmao would it be wrong to set the velocity - 0?

  57. anonymous
    • 5 years ago
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    velocity = 0*

  58. amistre64
    • 5 years ago
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    cant have a constant acceleration whout velcoty can we?

  59. amistre64
    • 5 years ago
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    my keyboard hates me lol

  60. anonymous
    • 5 years ago
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    haha its okay

  61. anonymous
    • 5 years ago
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    there was a similar question to this, and my sis helped me with that.

  62. anonymous
    • 5 years ago
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    and in that other problem, my sis taught me to set velocity = 0

  63. anonymous
    • 5 years ago
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    then solve for the constant

  64. amistre64
    • 5 years ago
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    cant say that i ever came across the problem before, so it might be plausible :)

  65. anonymous
    • 5 years ago
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    and then from that, do another anti derivative to find the position fuction

  66. anonymous
    • 5 years ago
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    lol alrighty, thanks anyways! enjoy the rest of your day :]

  67. amistre64
    • 5 years ago
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    :) yw

  68. toxicsugar22
    • 5 years ago
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    amristre can u help me

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