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anonymous

  • 5 years ago

A particle moves on the x-axis so that its velocity at any time t is greater than or equal to zero is given by v(t)=12t²-36t+15. At t=1, the particle is at the origin. a) Fine the position x(t) of the particle at any time t is greater than or equal to zero. b) Find all values of t for which the particle is at rest. c) Find the maximum velocity of the particle for 0<t<2. (<-- both of those are "less than or equal to") d) Find the total distance traveled by the particle from t=0 to t=2.

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  1. anonymous
    • 5 years ago
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    Velocity is the derivative of the function. Find the derivative first.

  2. anonymous
    • 5 years ago
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    well for part a), I took the integral of v(t) to find position

  3. amistre64
    • 5 years ago
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    good, integral and an initial condition...good

  4. amistre64
    • 5 years ago
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    b is the values for t from teh quadratic formula

  5. anonymous
    • 5 years ago
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    how do you determine position when t is greater than or equal to 0?

  6. amistre64
    • 5 years ago
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    the position is determined by the integral of the function; using at t=1; p=0 as the anchor

  7. amistre64
    • 5 years ago
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    F(t) = 4t^3-18t^2+15t +C 0 = 4-18 +15 +C; solve for C :)

  8. anonymous
    • 5 years ago
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    ok, that makes sense

  9. anonymous
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    so C=-1

  10. anonymous
    • 5 years ago
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    giving us the answer as p(t) = 4t^3-18t^2+15t-1?

  11. amistre64
    • 5 years ago
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    p(t) = 4t^3-18t^2+15t -1 ; yes unless there is some other interpretation for at t=1 the particle is at the origin. To me that means the particle is at (t,0) wehn t=1

  12. amistre64
    • 5 years ago
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    http://www.wolframalpha.com/input/?i=f%28t%29+%3D+4t%5E3-18t%5E2%2B15t+-1

  13. amistre64
    • 5 years ago
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    the derivative tells tha the particle is at rest on the hump and in the valley of that graph right?

  14. anonymous
    • 5 years ago
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    I don't understand that part

  15. amistre64
    • 5 years ago
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    well, i typed it wrong to begin with :) when velocity = 0 it is at rest correct?

  16. anonymous
    • 5 years ago
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    wait, so if you take the max. and min. of the position graph, the derivative (or velocity) will be 0, correct?

  17. amistre64
    • 5 years ago
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    if you take the derivative=0...which is velocity = 0; it is at rest; if only for a brief moment :)

  18. amistre64
    • 5 years ago
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    the max min of the position graph just tells you "where" it is.. not how fast it is moving...

  19. anonymous
    • 5 years ago
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    can I find when it is at rest by plugging v(t) = 0?

  20. amistre64
    • 5 years ago
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    http://www.wolframalpha.com/input/?i=v%28t%29%3D12t%C2%B2-36t%2B15

  21. amistre64
    • 5 years ago
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    yes

  22. anonymous
    • 5 years ago
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    ok, according to that graph, it would be at t=0.5 and t=2.5.

  23. amistre64
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    yes

  24. amistre64
    • 5 years ago
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    which corresponds to the hump and valley of the position :)

  25. amistre64
    • 5 years ago
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    since the derivative of the position tells the slope at any given point; then yes, the hump and valley of the position are the zeros of the derivative....

  26. anonymous
    • 5 years ago
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    That makes sense. Now to find the max velocity between 0<t<2 would be where the velocity graph is highest? and the max velocity would be the highest y value?

  27. amistre64
    • 5 years ago
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    yes

  28. amistre64
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    http://www.wolframalpha.com/input/?i=12t%5E2-36t%2B15++from+0+to+2

  29. amistre64
    • 5 years ago
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    it appears to be at zero..

  30. anonymous
    • 5 years ago
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    how would you find this algebraically? like if it is non calculator

  31. amistre64
    • 5 years ago
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    determine the local min/max of the velocity in the interval by the v(t); and use the 2nd derivative to tell you if its a (-)MAX or a (+)MIN

  32. amistre64
    • 5 years ago
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    the 2nd derivative tells you concavity points...

  33. amistre64
    • 5 years ago
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    we know that 1/2 and 5/2 are critical points; so which one is between 0 and 2?

  34. anonymous
    • 5 years ago
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    1/2 is in the bounds

  35. amistre64
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    12t^2-36t+15 12(1/2)^2-36(1/2)+15 12/4 -36/2 + 15 = ?? for starters :)

  36. anonymous
    • 5 years ago
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    okay we get 0 :D

  37. amistre64
    • 5 years ago
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    the 2nd derivative is: 24t -36 right?; so test for "signs" in this one.. 24(1/2) - 36 12 - 36 = (-).... this appears to be a MAX when in doubt just derive x^2 twice since its bowl shaped; you get a (+) for a MIN

  38. amistre64
    • 5 years ago
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    im remembering this stuff as i go lol

  39. amistre64
    • 5 years ago
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    check p(t) at 0 and at 2 as well.... :)

  40. anonymous
    • 5 years ago
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    haha I know what you mean :) and hmm why does the (-) give you a MAX?

  41. amistre64
    • 5 years ago
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    its just the nature of the game there.... recall the shape of the x^2 graph? looks like a big "U" whichi means the MIN point is at the bottom of it.... D(x^2) = 2x D(2x) = 2...which is (+); therefore, the 2nd derivative is (+) we get a MIN

  42. amistre64
    • 5 years ago
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    the maximum velocity of the particle is going to be when v(t) is the biggest..

  43. anonymous
    • 5 years ago
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    okay. and to answer c) the max velocity would be v(t)=15?

  44. amistre64
    • 5 years ago
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    regardless of position; so the 2nd derivative is actually the 1st derivative of v(t) :)

  45. amistre64
    • 5 years ago
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    the MAX velocity is 15 yes :)

  46. anonymous
    • 5 years ago
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    okay :D and somebody in the questions scroll down is saying they need you for a minute ;)

  47. anonymous
    • 5 years ago
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    if you don't mind, I'd appreciate if you also walked me through part d).

  48. amistre64
    • 5 years ago
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    .....lol :) can you hold? or you think you got this?

  49. anonymous
    • 5 years ago
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    but go ahead and see what they need :D

  50. amistre64
    • 5 years ago
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    the total distance traveled is going to be an integral of sqrt(1-x^2) dx ...but that might be the position function...

  51. anonymous
    • 5 years ago
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    could you explain how you got that?

  52. amistre64
    • 5 years ago
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    how i got that? hmmm.... back on my head :) the way you find the length of a curve is by integrating the position really really close like: when you get real close; the curve becomes straight and you see that the length is just s = sqrt(1+x^2) which is the pythag thrm; s^2 = x^2+y^2

  53. amistre64
    • 5 years ago
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    this pythag is primed for either parameteric equations: sqrt(f'(x)^2 + g'(x)^2); or regular curves as well, as in: sqrt(1+[f'(x)]^2)

  54. anonymous
    • 5 years ago
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    I'm confused at how you "see that the length is just s=sqrt(1+x^2).." etc

  55. amistre64
    • 5 years ago
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    the length of a curve is the sum of all its straight parts..... do you agree?

  56. anonymous
    • 5 years ago
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    ok that makes sense

  57. amistre64
    • 5 years ago
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    calculus allows us to look so close to a curve, that it straightens out right?

  58. amistre64
    • 5 years ago
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    all these "straight" parts are just a bunch of pythag identites added togehter

  59. anonymous
    • 5 years ago
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    what are pythag identities?

  60. amistre64
    • 5 years ago
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    x^2 + y^2 = z^2 is the pythagorean theorum right? use any letters you want, its the same thing :)

  61. anonymous
    • 5 years ago
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    okay, I'm associating that with a right triangle though.. ><

  62. amistre64
    • 5 years ago
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    the length of a curve; lets call it "s" is up close; s^2 = x^2 + y^2 to get up real close, we take a derivative... ds^2 = dx^2 + dy^2 ds = sqrt(dx^2 + dy^2)

  63. amistre64
    • 5 years ago
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    you should associate it with a right triangle; because the "s" part is the hypotenuse

  64. amistre64
    • 5 years ago
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    and the legs are the derivatives of x and y...squared

  65. anonymous
    • 5 years ago
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    do you mean that each of the little "straight" segments in a graph are like the hypotenuses of right triangles?? haha I'm a bit confused at this part

  66. amistre64
    • 5 years ago
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    let me draw a picture :)

  67. anonymous
    • 5 years ago
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    awesome

  68. amistre64
    • 5 years ago
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  69. amistre64
    • 5 years ago
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    calculus get us so close to this that x become dx and y becoems dy and s becomes ds

  70. anonymous
    • 5 years ago
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    wow, so I'm not crazy for imagining that ^^'

  71. amistre64
    • 5 years ago
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    f(x) = x; f'(x) = dx g(y) = y; g'(y) = dy h(s) = s; h'(s) = ds

  72. amistre64
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    ds^2 = dx^2 + dy^2 [h'(s)]^2 = [f'(x)]^2 + [g'(y)]^2

  73. amistre64
    • 5 years ago
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    ds = sqrt( [f'(x)]^2 + [g'(y)]^2)

  74. amistre64
    • 5 years ago
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    if the length of one piece is "sqrt( [f'(x)]^2 + [g'(y)]^2)" then the total length is the sum of all the little pieces of ds.... the integral of: \[\int\limits_{a}^{b}\sqrt{ [f'(x)]^2 + [g'(y)]^2}ds\]

  75. amistre64
    • 5 years ago
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    since we only have a g(y) = p(t) then x=x giveing us D(x) = 1 the length of the curve is then... \[\int\limits_{a}^{b} \sqrt{1+[p'(t)]^2} dt\]

  76. amistre64
    • 5 years ago
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    so whats our length from 0 to 2? it amounts to ln(sec(t)+tan(t)) from 0 to 2

  77. anonymous
    • 5 years ago
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    uhh would you plug in [ln(sec(2)+tan(2))]-[ln(sec(0)+tan(0))]?

  78. amistre64
    • 5 years ago
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    yes

  79. amistre64
    • 5 years ago
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    we can say u = v(t) = tan^2 sqrt(1+tan^2) = sec [S] sec(u) = ln(sec(u)+tan(u)) but we would have to modify the interval if we leave it like this :)

  80. amistre64
    • 5 years ago
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    resubstitute in for sec(u) and tan(u)

  81. amistre64
    • 5 years ago
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    p'(t) = tan(u) is what i mean to get across lol.... p'(t) = v(t) right?

  82. anonymous
    • 5 years ago
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    ehh you lost me here. ><

  83. amistre64
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  84. anonymous
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    so we're working with the hypotenuse of that?

  85. amistre64
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    ln[sec(u)+tan(u)] has to be turned back into the original values for p'(t) :) sec(u) = sqrt[(12t^2 -36t +15)^2 +1] and tan(u) = (12t^2 -36t +15) to amount to..... ln [ sqrt((12t^2 -36t +15)^2 +1) + (12t^2 -36t +15) ] \[\ln[(\sqrt{(12t^2 -36t+15)^2 +1}) + (12t^2 -36t +15)]\]

  86. amistre64
    • 5 years ago
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    no, that triangle is how we substitute and re integrate our values for p'(t)

  87. amistre64
    • 5 years ago
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    F(0) = ln[ sqrt(226) +15 ] F(2) = ln[ sqrt(82) - 9 ]

  88. amistre64
    • 5 years ago
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    length from 0 to 2 = F(2) - F(0)

  89. amistre64
    • 5 years ago
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    -2.89 - 3.40 = |-6.29| = 6.29

  90. anonymous
    • 5 years ago
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    wow, and that's the final answer?

  91. amistre64
    • 5 years ago
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    i believe that is the answer; as long as my fingers dint trip on the calculator ;)

  92. anonymous
    • 5 years ago
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    ok, thank you so much! :)

  93. amistre64
    • 5 years ago
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    is there a way to see if its right?

  94. anonymous
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    when my teacher grades it, I'll find out

  95. amistre64
    • 5 years ago
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    ha!! .... tell your teacher to hurry up ;)

  96. anonymous
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    lol :)

  97. amistre64
    • 5 years ago
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    the other method is to re evaluate the interval to match the substitution... tan(u) = 12t^2 -36t +15 u = tan^-1(12t^2 -36t +15) when t = 0 what does u =? when t = 2 what does u =??

  98. amistre64
    • 5 years ago
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    t = 0; u = 86.1859 t = 2; u = -83.6598 soooo...... lets see if thats good ;) ln[sec(-83.6598) + tan(-83.6598)] - ln[sec(86.1859)+tan(86.1859)]

  99. amistre64
    • 5 years ago
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    2.079 - 2.77 ...... i never get that right lol; which is why i always revert it back to the original :)

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