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

1. anonymous

$\int\limits \sqrt[3]{1 + t^2}$ this is what i cant find

2. idku

$\large \int\limits_{ }^{ }\sqrt[3]{1+t^2}dt$

3. anonymous

yep thats right

4. idku

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

yep, didn't get anywhere. be my guest though im pretty bad at u substitutions :P

6. idku

oh, my trig sub...

7. idku

$$\large \displaystyle\int\limits_{ }^{ }\sqrt[3]{1+t^2}dt$$ $$\large \displaystyle t=\tan\theta$$

8. anonymous

trig sub...? never heard of it.

9. idku

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

wolfram is giving me something i dont even understand .

11. idku
12. idku

13. idku

are you sure you aren't given limits of integration, and this is not a riemman sum type of question?

14. idku

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

16. anonymous

this is my given question

17. idku

this is much better

18. idku

can I use y instead of x (as y(t) ?)

19. idku

You can use a stepsize on that one I think

20. anonymous

interesting i figured i just had to find s(t) and just change the constant and plug in t = 3

21. idku

well, that is the task as far as the steps go, but s(t)=?

22. idku

that is the question where we don't arrive at a simple conclusion just like this, right?

23. idku

have you heard of a step-size, Euler's method or anything like this?

24. anonymous

no sorry :(

25. idku

$$\large \displaystyle f'(t)=\sqrt[3]{1+t^2}$$ 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

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

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

28. idku

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

29. idku

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

30. idku

are you getting it at least a little? (sorry I am bad at explaining)

31. idku

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

lol sort of , i'm not quite sure what the h represents

33. idku

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

we don't know where this step size of 1 would land, and usiong this formula we can find that out.

35. idku

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

$$\large \displaystyle \left(x_n,y_n\right)=\left(x_{n-1}+h,~y_{n-1}~+h\cdot \color{red}{f'(x_{n-1},y_{n-1})}~\right)$$ $$\large \displaystyle \left(x_1,y_1\right)=\left(x_{0}+1,~2~+1\cdot \color{red}{f'(0,2)}~\right)$$ (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

oh, you know x_0 is 0 , i forgot to put that in

38. anonymous

okay wait one sec , why are we even finding (x1,y1)?

39. idku

$$\large \displaystyle \left(x_1,y_1\right)=\left(0+1,~2~+1\cdot \color{red}{\sqrt[3]{1+0^2}}~\right)$$ $$\large \displaystyle \left(x_1,y_1\right)=\left(1,~3\right)$$

40. idku

because that is the next point

41. idku

|dw:1440033265402:dw|

42. idku

when you do a u=substitution du=?

43. idku

du=2t dt so it ends up 1/(2 √(u-1)) du = dt

44. idku

$\int\limits_{ }^{ }\frac{\sqrt[3]{u}}{\sqrt{u-1}}du$

45. idku

if it had 2t next to the cube root, then yes, u sub would be the only and the perfect

46. idku

anyway, i guess back to stepsize?

47. idku

$$\large \displaystyle \left(x_1,y_1\right)=\left(1,~3\right)$$ so here is the result of 1 step

48. idku

see how I am doing the steps? (See how I plug them into the formula?)

49. anonymous

yes , okay so say we do this for a whole bunch of points then what?

50. idku

$$\large \displaystyle \left(x_n,y_n\right)=\left(x_{n-1}+h,~y_{n-1}~+h\cdot \color{red}{f'(x_{n-1},y_{n-1})}~\right)$$ 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

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

i guess you just need to do the dirty work (but do it correctly0

53. anonymous

just out of curiosity in what math class did you learn this method?

54. idku

I learned this method in calculus II, although it also belongs to DE.

55. idku

ok, can you proceed from (1,3) with h=1?

56. anonymous

not sure how to find f('n1) i think i got the rest though.

57. anonymous

woops

58. anonymous

damn , just had the entire thing written out and it just got deleted DX

59. anonymous

$(x _{1},y _{1}) = (x _{1} + 1 , y _{1} + 1 * (f \prime (x _{1})))$

60. idku

yhes, this is correct

61. idku

now, you know x_1 is 1 and y_1 is 3 also, h=1

62. anonymous

woops i mean x2 and y2 at the beginning

63. anonymous

just not sure how to find the f'(x1)

64. idku

yes....

65. idku

you know x_1 is 1 and y_1 is 3 so put that in....

66. idku

$(x_2,y_2)=(1+1,3+1\cdot f'(1))=(2,3+\sqrt[3]{1+(1)^2}~)=(?,?)$

67. anonymous

$(x _{1},y _{1}) = ( 2 , 4 * (f \prime (x _{1})))$

68. idku

(if you want we can do a polynomial approximation for cube root of 2, but it would be too much xD)

69. idku

you mean$(x_2,y_2)=(2,3+\color{red}{1\cdot}f'(1))$

70. idku

in red the order of operations erro

71. idku

and f'(1) is becase $$x_1$$=1 in our case

72. anonymous

woops yes i missed that

73. idku

yes, and f'(1)=?

74. anonymous

1.25992...

75. idku

just take 1.26

76. anonymous

so (2 , 4.26)

77. idku

$(x_2,y_2)=(2,4.26)$Correct

78. idku

Now, next step h=1, x_2 =2 y_2=4.26 go for it...

79. idku

(this is the last step, for this approximation with step size of h=1)

80. idku

i get (3, 6.935)

81. anonymous

one sec only got (3, 5.26 + ? ) so far

82. anonymous

okay yep got the same thing

83. anonymous

so now what?

84. idku

(3, $$\color{red}{4}$$.26 + 1•f'(4.26))

85. idku

4, not 5

86. anonymous

wow thats strange, then how did i get the same answer ?

87. idku

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

2.67526

89. idku

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

91. idku

yeah, 4.26 is the y.

92. idku

(2+1 , (4.26+1\cdot (1+(2)^2)^(1/3)))

93. idku
94. idku

(3, 5.9699) y(3)=5.9699

95. anonymous

interesting we must have done something wrong , none of the answer choices give that result

96. idku

i know it doesn't correspond to options, and you know why?

97. idku

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.

98. anonymous

okay so would it be C because it is the closest smaller answer ?

99. idku

it could be smaller or larger than the actual answer, you never know

100. idku

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)

101. jim_thompson5910

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

102. idku

I am the stupidest person in the world then I guess.... G-wis

103. jim_thompson5910

no not stupid since Euler's method does work here

104. idku

I am overloading it when it is just that simple, just that integral.... om(f)g excuse my lang

105. anonymous

lol well atleast i learned something new :D

106. idku

yes, Euler works, but... but it's too long.....

107. idku

it is just like driving from GA to North Dakota and then to Texas, instead of straight GA-Texas.

108. anonymous

oh well thanks for all the help guys !! :D:D:D:D:D:D:D

109. idku

you welcome, lol, i should literally "fire" myself.

110. idku

good luck