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anonymous

  • one year ago

can't find the integral of this function v(t) = (1+t^2)^(1/3) anyone know how?

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  1. anonymous
    • one year ago
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    \[\int\limits \sqrt[3]{1 + t^2}\] this is what i cant find

  2. idku
    • one year ago
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    \[\large \int\limits_{ }^{ }\sqrt[3]{1+t^2}dt\]

  3. anonymous
    • one year ago
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    yep thats right

  4. idku
    • one year ago
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    well, seems as though it doesn't have closed form based on my first impression. Lets set u=1+t² and see wat we get, did you try that?

  5. anonymous
    • one year ago
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    yep, didn't get anywhere. be my guest though im pretty bad at u substitutions :P

  6. idku
    • one year ago
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    oh, my trig sub...

  7. idku
    • one year ago
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    \(\large \displaystyle\int\limits_{ }^{ }\sqrt[3]{1+t^2}dt\) \(\large \displaystyle t=\tan\theta \)

  8. anonymous
    • one year ago
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    trig sub...? never heard of it.

  9. idku
    • one year ago
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    lets see what then, u=tan(theta) du=sec^2theta \(\large \displaystyle\int\limits_{ }^{ }\sec^2\theta\sqrt[3]{1+\tan^2\theta}~d\theta \) \(\large \displaystyle\int\limits_{ }^{ }\sec^{2+2/3}\theta\) that is bullsh... as well

  10. anonymous
    • one year ago
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    wolfram is giving me something i dont even understand .

  11. idku
    • one year ago
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    acc. to wolfram http://www.wolframalpha.com/input/?i=integral+of+%281%2Bx%5E2%29%5E%281%2F3%29

  12. idku
    • one year ago
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    no simple function answer

  13. idku
    • one year ago
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    are you sure you aren't given limits of integration, and this is not a riemman sum type of question?

  14. idku
    • one year ago
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    or maybe it is sort of \(\large \displaystyle f(x)=\int\limits_{2 }^{ \cos(x)}\sqrt[3]{1+t^2}dt\) and you need f'(x) ?

  15. anonymous
    • one year ago
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  16. anonymous
    • one year ago
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    this is my given question

  17. idku
    • one year ago
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    this is much better

  18. idku
    • one year ago
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    can I use y instead of x (as y(t) ?)

  19. idku
    • one year ago
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    You can use a stepsize on that one I think

  20. anonymous
    • one year ago
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    interesting i figured i just had to find s(t) and just change the constant and plug in t = 3

  21. idku
    • one year ago
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    well, that is the task as far as the steps go, but s(t)=?

  22. idku
    • one year ago
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    that is the question where we don't arrive at a simple conclusion just like this, right?

  23. idku
    • one year ago
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    have you heard of a step-size, Euler's method or anything like this?

  24. anonymous
    • one year ago
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    no sorry :(

  25. idku
    • one year ago
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    \(\large \displaystyle f'(t)=\sqrt[3]{1+t^2}\) y(0)=2, find y(3)=? lets use a step-size of h=1 for now (but with h=1/2 or even less you can get a better approximation. Each step of size h is as follows: \(\large \displaystyle \left(x_n,y_n\right)=\left(x_{n-1}+h,~y_{n-1}~+h\cdot \color{red}{f'(x_{n-1},y_{n-1})}~\right)\)

  26. idku
    • one year ago
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    maybe we won't understand why it is like this right now, but you can see the formula, without completely going like wt(f)....

  27. idku
    • one year ago
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    \(\large \displaystyle \left(t_n,y_n\right)=\left(t_{n-1}+h,~y_{n-1}~+h\cdot \color{red}{f'(t_{n-1},y_{n-1})}~\right)\) excuse me it should be t.

  28. idku
    • one year ago
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    \(\large \displaystyle \left(x_n,y_n\right)=\left(x_{n-1}+h,~y_{n-1}~+h\cdot \color{red}{f'(x_{n-1},y_{n-1})}~\right)\) you are given: \(\large \displaystyle f'(t)=\sqrt[3]{1+t^2}\) and a first point of (t=0, y=2)

  29. idku
    • one year ago
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    \(\large \displaystyle \left(x_1,y_1\right)=\left(x_{0}+1,~2~+1\cdot \color{red}{f'(0,2)}~\right)\) this would be the first step starting from (0,2), and h=1

  30. idku
    • one year ago
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    are you getting it at least a little? (sorry I am bad at explaining)

  31. idku
    • one year ago
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    ok, I will give you the entire process wth h=1, and then you will tell me how much you get out of it.

  32. anonymous
    • one year ago
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    lol sort of , i'm not quite sure what the h represents

  33. idku
    • one year ago
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    you are starting from (0,2) right? that is the first given point |dw:1440032865183:dw| so it is the size of the step, in other words, we are carefully drawing the approximation for the graph of y(t) based on y'(t) taking it with little stpes of size h (of size 1 in this case)

  34. idku
    • one year ago
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    we don't know where this step size of 1 would land, and usiong this formula we can find that out.

  35. idku
    • one year ago
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    and from there, when we do the step and find the next point, we will take another step... ultimately, we wil get to the y(3)

  36. idku
    • one year ago
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    \(\large \displaystyle \left(x_n,y_n\right)=\left(x_{n-1}+h,~y_{n-1}~+h\cdot \color{red}{f'(x_{n-1},y_{n-1})}~\right)\) \(\large \displaystyle \left(x_1,y_1\right)=\left(x_{0}+1,~2~+1\cdot \color{red}{f'(0,2)}~\right)\) (there just isn't a y in this case, so would just plug in 0 for t) I haven't ever done Euler's for explicitly defined functions. \(\large \displaystyle \left(x_1,y_1\right)=\left(x_{0}+1,~2~+1\cdot \color{red}{\sqrt[3]{1+0^2}}~\right)\)

  37. idku
    • one year ago
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    oh, you know x_0 is 0 , i forgot to put that in

  38. anonymous
    • one year ago
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    okay wait one sec , why are we even finding (x1,y1)?

  39. idku
    • one year ago
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    \(\large \displaystyle \left(x_1,y_1\right)=\left(0+1,~2~+1\cdot \color{red}{\sqrt[3]{1+0^2}}~\right)\) \(\large \displaystyle \left(x_1,y_1\right)=\left(1,~3\right)\)

  40. idku
    • one year ago
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    because that is the next point

  41. idku
    • one year ago
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    |dw:1440033265402:dw|

  42. idku
    • one year ago
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    when you do a u=substitution du=?

  43. idku
    • one year ago
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    du=2t dt so it ends up 1/(2 √(u-1)) du = dt

  44. idku
    • one year ago
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    \[\int\limits_{ }^{ }\frac{\sqrt[3]{u}}{\sqrt{u-1}}du\]

  45. idku
    • one year ago
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    if it had 2t next to the cube root, then yes, u sub would be the only and the perfect

  46. idku
    • one year ago
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    anyway, i guess back to stepsize?

  47. idku
    • one year ago
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    \(\large \displaystyle \left(x_1,y_1\right)=\left(1,~3\right)\) so here is the result of 1 step

  48. idku
    • one year ago
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    see how I am doing the steps? (See how I plug them into the formula?)

  49. anonymous
    • one year ago
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    yes , okay so say we do this for a whole bunch of points then what?

  50. idku
    • one year ago
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    \(\large \displaystyle \left(x_n,y_n\right)=\left(x_{n-1}+h,~y_{n-1}~+h\cdot \color{red}{f'(x_{n-1},y_{n-1})}~\right) \) can you use this formula to get from (1,3) with h=1 to the next point? Note: in this case: \(\large \displaystyle \left(x_n,y_n\right)=\left(x_{n-1}+h,~y_{n-1}~+h\cdot \color{red}{f'(x_{n-1})}~\right) \) (since there is no y - it is explicit function)

  51. idku
    • one year ago
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    yes, we have to do it a bunch of time. On mathematica, there is a function that can do it instantly for many steps ahead right away, but I totally forgot how

  52. idku
    • one year ago
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    i guess you just need to do the dirty work (but do it correctly0

  53. anonymous
    • one year ago
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    just out of curiosity in what math class did you learn this method?

  54. idku
    • one year ago
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    I learned this method in calculus II, although it also belongs to DE.

  55. idku
    • one year ago
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    ok, can you proceed from (1,3) with h=1?

  56. anonymous
    • one year ago
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    not sure how to find f('n1) i think i got the rest though.

  57. anonymous
    • one year ago
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    woops

  58. anonymous
    • one year ago
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    damn , just had the entire thing written out and it just got deleted DX

  59. anonymous
    • one year ago
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    \[(x _{1},y _{1}) = (x _{1} + 1 , y _{1} + 1 * (f \prime (x _{1})))\]

  60. idku
    • one year ago
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    yhes, this is correct

  61. idku
    • one year ago
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    now, you know x_1 is 1 and y_1 is 3 also, h=1

  62. anonymous
    • one year ago
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    woops i mean x2 and y2 at the beginning

  63. anonymous
    • one year ago
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    just not sure how to find the f'(x1)

  64. idku
    • one year ago
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    yes....

  65. idku
    • one year ago
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    you know x_1 is 1 and y_1 is 3 so put that in....

  66. idku
    • one year ago
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    \[(x_2,y_2)=(1+1,3+1\cdot f'(1))=(2,3+\sqrt[3]{1+(1)^2}~)=(?,?)\]

  67. anonymous
    • one year ago
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    \[(x _{1},y _{1}) = ( 2 , 4 * (f \prime (x _{1}))) \]

  68. idku
    • one year ago
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    (if you want we can do a polynomial approximation for cube root of 2, but it would be too much xD)

  69. idku
    • one year ago
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    you mean\[(x_2,y_2)=(2,3+\color{red}{1\cdot}f'(1))\]

  70. idku
    • one year ago
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    in red the order of operations erro

  71. idku
    • one year ago
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    and f'(1) is becase \(x_1\)=1 in our case

  72. anonymous
    • one year ago
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    woops yes i missed that

  73. idku
    • one year ago
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    yes, and f'(1)=?

  74. anonymous
    • one year ago
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    1.25992...

  75. idku
    • one year ago
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    just take 1.26

  76. anonymous
    • one year ago
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    so (2 , 4.26)

  77. idku
    • one year ago
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    \[(x_2,y_2)=(2,4.26)\]Correct

  78. idku
    • one year ago
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    Now, next step h=1, x_2 =2 y_2=4.26 go for it...

  79. idku
    • one year ago
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    (this is the last step, for this approximation with step size of h=1)

  80. idku
    • one year ago
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    i get (3, 6.935)

  81. anonymous
    • one year ago
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    one sec only got (3, 5.26 + ? ) so far

  82. anonymous
    • one year ago
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    okay yep got the same thing

  83. anonymous
    • one year ago
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    so now what?

  84. idku
    • one year ago
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    (3, \(\color{red}{4}\).26 + 1•f'(4.26))

  85. idku
    • one year ago
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    4, not 5

  86. anonymous
    • one year ago
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    wow thats strange, then how did i get the same answer ?

  87. idku
    • one year ago
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    so, you do the same thing, you plug in the \(x_{n-1}\) (in this case \(x_2\)which is equivalent to 4.26, into the f'(t))

  88. anonymous
    • one year ago
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    2.67526

  89. idku
    • one year ago
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    this is what i entered for \(x_3,y_3\) http://www.wolframalpha.com/input/?i=%282%2B1+%2C+%284.26%2B1%5Ccdot+%281%2B%284.26%29%5E2%29%5E%281%2F3%29%29%29

  90. idku
    • one year ago
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    oh, my bad

  91. idku
    • one year ago
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    yeah, 4.26 is the y.

  92. idku
    • one year ago
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    (2+1 , (4.26+1\cdot (1+(2)^2)^(1/3)))

  93. idku
    • one year ago
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    (3, 5.9699) y(3)=5.9699

  94. anonymous
    • one year ago
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    interesting we must have done something wrong , none of the answer choices give that result

  95. idku
    • one year ago
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    i know it doesn't correspond to options, and you know why?

  96. idku
    • one year ago
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    We are going from y(0) to y(3) with a step size of h=1 the step size is too big! Take a smaller step-size to get a better approximation.

  97. anonymous
    • one year ago
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    okay so would it be C because it is the closest smaller answer ?

  98. idku
    • one year ago
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    it could be smaller or larger than the actual answer, you never know

  99. idku
    • one year ago
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    I knew it wouldn't suffice from the beginning, i did with h=1, so that it will be easy to understand. now, you just need to redo the dirty work, starting from \(\left(x_0=0,~y_0=2 \right)\) but, you need to pick h=½ (or something like this, but h=1 is just too big)

  100. jim_thompson5910
    • one year ago
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    I'm not sure why you're using Euler's method when a numeric integration is much quicker (see attached for how to type it into the TI calculator). You'll find that \[\Large \int_{0}^{3}\sqrt[3]{1+t^2}dt \approx 4.511532459 \approx 4.512\] which is the net change in position from t = 0 to t = 3. Add that to the initial position to figure out where this object ends up at t = 3 seconds. http://tibasicdev.wikidot.com/fnint

  101. idku
    • one year ago
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    I am the stupidest person in the world then I guess.... G-wis

  102. jim_thompson5910
    • one year ago
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    no not stupid since Euler's method does work here

  103. idku
    • one year ago
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    I am overloading it when it is just that simple, just that integral.... om(f)g excuse my lang

  104. anonymous
    • one year ago
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    lol well atleast i learned something new :D

  105. idku
    • one year ago
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    yes, Euler works, but... but it's too long.....

  106. idku
    • one year ago
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    it is just like driving from GA to North Dakota and then to Texas, instead of straight GA-Texas.

  107. anonymous
    • one year ago
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    oh well thanks for all the help guys !! :D:D:D:D:D:D:D

  108. idku
    • one year ago
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    you welcome, lol, i should literally "fire" myself.

  109. idku
    • one year ago
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    good luck

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