## 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.

1. anonymous

Velocity is the derivative of the function. Find the derivative first.

2. anonymous

well for part a), I took the integral of v(t) to find position

3. amistre64

good, integral and an initial condition...good

4. amistre64

b is the values for t from teh quadratic formula

5. anonymous

how do you determine position when t is greater than or equal to 0?

6. amistre64

the position is determined by the integral of the function; using at t=1; p=0 as the anchor

7. amistre64

F(t) = 4t^3-18t^2+15t +C 0 = 4-18 +15 +C; solve for C :)

8. anonymous

ok, that makes sense

9. anonymous

so C=-1

10. anonymous

giving us the answer as p(t) = 4t^3-18t^2+15t-1?

11. amistre64

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
13. amistre64

the derivative tells tha the particle is at rest on the hump and in the valley of that graph right?

14. anonymous

I don't understand that part

15. amistre64

well, i typed it wrong to begin with :) when velocity = 0 it is at rest correct?

16. anonymous

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

17. amistre64

if you take the derivative=0...which is velocity = 0; it is at rest; if only for a brief moment :)

18. amistre64

the max min of the position graph just tells you "where" it is.. not how fast it is moving...

19. anonymous

can I find when it is at rest by plugging v(t) = 0?

20. amistre64
21. amistre64

yes

22. anonymous

ok, according to that graph, it would be at t=0.5 and t=2.5.

23. amistre64

yes

24. amistre64

which corresponds to the hump and valley of the position :)

25. amistre64

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

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

yes

28. amistre64
29. amistre64

it appears to be at zero..

30. anonymous

how would you find this algebraically? like if it is non calculator

31. amistre64

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

the 2nd derivative tells you concavity points...

33. amistre64

we know that 1/2 and 5/2 are critical points; so which one is between 0 and 2?

34. anonymous

1/2 is in the bounds

35. amistre64

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

36. anonymous

okay we get 0 :D

37. amistre64

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

im remembering this stuff as i go lol

39. amistre64

check p(t) at 0 and at 2 as well.... :)

40. anonymous

haha I know what you mean :) and hmm why does the (-) give you a MAX?

41. amistre64

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

the maximum velocity of the particle is going to be when v(t) is the biggest..

43. anonymous

okay. and to answer c) the max velocity would be v(t)=15?

44. amistre64

regardless of position; so the 2nd derivative is actually the 1st derivative of v(t) :)

45. amistre64

the MAX velocity is 15 yes :)

46. anonymous

okay :D and somebody in the questions scroll down is saying they need you for a minute ;)

47. anonymous

if you don't mind, I'd appreciate if you also walked me through part d).

48. amistre64

.....lol :) can you hold? or you think you got this?

49. anonymous

but go ahead and see what they need :D

50. amistre64

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

could you explain how you got that?

52. amistre64

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

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

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

55. amistre64

the length of a curve is the sum of all its straight parts..... do you agree?

56. anonymous

ok that makes sense

57. amistre64

calculus allows us to look so close to a curve, that it straightens out right?

58. amistre64

all these "straight" parts are just a bunch of pythag identites added togehter

59. anonymous

what are pythag identities?

60. amistre64

x^2 + y^2 = z^2 is the pythagorean theorum right? use any letters you want, its the same thing :)

61. anonymous

okay, I'm associating that with a right triangle though.. ><

62. amistre64

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

you should associate it with a right triangle; because the "s" part is the hypotenuse

64. amistre64

and the legs are the derivatives of x and y...squared

65. anonymous

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

let me draw a picture :)

67. anonymous

awesome

68. amistre64

69. amistre64

calculus get us so close to this that x become dx and y becoems dy and s becomes ds

70. anonymous

wow, so I'm not crazy for imagining that ^^'

71. amistre64

f(x) = x; f'(x) = dx g(y) = y; g'(y) = dy h(s) = s; h'(s) = ds

72. amistre64

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

73. amistre64

ds = sqrt( [f'(x)]^2 + [g'(y)]^2)

74. amistre64

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

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

so whats our length from 0 to 2? it amounts to ln(sec(t)+tan(t)) from 0 to 2

77. anonymous

uhh would you plug in [ln(sec(2)+tan(2))]-[ln(sec(0)+tan(0))]?

78. amistre64

yes

79. amistre64

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

resubstitute in for sec(u) and tan(u)

81. amistre64

p'(t) = tan(u) is what i mean to get across lol.... p'(t) = v(t) right?

82. anonymous

ehh you lost me here. ><

83. amistre64

84. anonymous

so we're working with the hypotenuse of that?

85. amistre64

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

no, that triangle is how we substitute and re integrate our values for p'(t)

87. amistre64

F(0) = ln[ sqrt(226) +15 ] F(2) = ln[ sqrt(82) - 9 ]

88. amistre64

length from 0 to 2 = F(2) - F(0)

89. amistre64

-2.89 - 3.40 = |-6.29| = 6.29

90. anonymous

wow, and that's the final answer?

91. amistre64

i believe that is the answer; as long as my fingers dint trip on the calculator ;)

92. anonymous

ok, thank you so much! :)

93. amistre64

is there a way to see if its right?

94. anonymous

when my teacher grades it, I'll find out

95. amistre64

ha!! .... tell your teacher to hurry up ;)

96. anonymous

lol :)

97. amistre64

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

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

2.079 - 2.77 ...... i never get that right lol; which is why i always revert it back to the original :)