anonymous
  • anonymous
can't find the integral of this function v(t) = (1+t^2)^(1/3) anyone know how?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\int\limits \sqrt[3]{1 + t^2}\] this is what i cant find
idku
  • idku
\[\large \int\limits_{ }^{ }\sqrt[3]{1+t^2}dt\]
anonymous
  • anonymous
yep thats right

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idku
  • 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?
anonymous
  • anonymous
yep, didn't get anywhere. be my guest though im pretty bad at u substitutions :P
idku
  • idku
oh, my trig sub...
idku
  • idku
\(\large \displaystyle\int\limits_{ }^{ }\sqrt[3]{1+t^2}dt\) \(\large \displaystyle t=\tan\theta \)
anonymous
  • anonymous
trig sub...? never heard of it.
idku
  • 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
anonymous
  • anonymous
wolfram is giving me something i dont even understand .
idku
  • idku
acc. to wolfram http://www.wolframalpha.com/input/?i=integral+of+%281%2Bx%5E2%29%5E%281%2F3%29
idku
  • idku
no simple function answer
idku
  • idku
are you sure you aren't given limits of integration, and this is not a riemman sum type of question?
idku
  • 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) ?
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
this is my given question
idku
  • idku
this is much better
idku
  • idku
can I use y instead of x (as y(t) ?)
idku
  • idku
You can use a stepsize on that one I think
anonymous
  • anonymous
interesting i figured i just had to find s(t) and just change the constant and plug in t = 3
idku
  • idku
well, that is the task as far as the steps go, but s(t)=?
idku
  • idku
that is the question where we don't arrive at a simple conclusion just like this, right?
idku
  • idku
have you heard of a step-size, Euler's method or anything like this?
anonymous
  • anonymous
no sorry :(
idku
  • 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: \(\large \displaystyle \left(x_n,y_n\right)=\left(x_{n-1}+h,~y_{n-1}~+h\cdot \color{red}{f'(x_{n-1},y_{n-1})}~\right)\)
idku
  • 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)....
idku
  • 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.
idku
  • 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)
idku
  • 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
idku
  • idku
are you getting it at least a little? (sorry I am bad at explaining)
idku
  • 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.
anonymous
  • anonymous
lol sort of , i'm not quite sure what the h represents
idku
  • 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)
idku
  • idku
we don't know where this step size of 1 would land, and usiong this formula we can find that out.
idku
  • 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)
idku
  • 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. \(\large \displaystyle \left(x_1,y_1\right)=\left(x_{0}+1,~2~+1\cdot \color{red}{\sqrt[3]{1+0^2}}~\right)\)
idku
  • idku
oh, you know x_0 is 0 , i forgot to put that in
anonymous
  • anonymous
okay wait one sec , why are we even finding (x1,y1)?
idku
  • idku
\(\large \displaystyle \left(x_1,y_1\right)=\left(0+1,~2~+1\cdot \color{red}{\sqrt[3]{1+0^2}}~\right)\) \(\large \displaystyle \left(x_1,y_1\right)=\left(1,~3\right)\)
idku
  • idku
because that is the next point
idku
  • idku
|dw:1440033265402:dw|
idku
  • idku
when you do a u=substitution du=?
idku
  • idku
du=2t dt so it ends up 1/(2 √(u-1)) du = dt
idku
  • idku
\[\int\limits_{ }^{ }\frac{\sqrt[3]{u}}{\sqrt{u-1}}du\]
idku
  • idku
if it had 2t next to the cube root, then yes, u sub would be the only and the perfect
idku
  • idku
anyway, i guess back to stepsize?
idku
  • idku
\(\large \displaystyle \left(x_1,y_1\right)=\left(1,~3\right)\) so here is the result of 1 step
idku
  • idku
see how I am doing the steps? (See how I plug them into the formula?)
anonymous
  • anonymous
yes , okay so say we do this for a whole bunch of points then what?
idku
  • 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)
idku
  • 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
idku
  • idku
i guess you just need to do the dirty work (but do it correctly0
anonymous
  • anonymous
just out of curiosity in what math class did you learn this method?
idku
  • idku
I learned this method in calculus II, although it also belongs to DE.
idku
  • idku
ok, can you proceed from (1,3) with h=1?
anonymous
  • anonymous
not sure how to find f('n1) i think i got the rest though.
anonymous
  • anonymous
woops
anonymous
  • anonymous
damn , just had the entire thing written out and it just got deleted DX
anonymous
  • anonymous
\[(x _{1},y _{1}) = (x _{1} + 1 , y _{1} + 1 * (f \prime (x _{1})))\]
idku
  • idku
yhes, this is correct
idku
  • idku
now, you know x_1 is 1 and y_1 is 3 also, h=1
anonymous
  • anonymous
woops i mean x2 and y2 at the beginning
anonymous
  • anonymous
just not sure how to find the f'(x1)
idku
  • idku
yes....
idku
  • idku
you know x_1 is 1 and y_1 is 3 so put that in....
idku
  • idku
\[(x_2,y_2)=(1+1,3+1\cdot f'(1))=(2,3+\sqrt[3]{1+(1)^2}~)=(?,?)\]
anonymous
  • anonymous
\[(x _{1},y _{1}) = ( 2 , 4 * (f \prime (x _{1}))) \]
idku
  • idku
(if you want we can do a polynomial approximation for cube root of 2, but it would be too much xD)
idku
  • idku
you mean\[(x_2,y_2)=(2,3+\color{red}{1\cdot}f'(1))\]
idku
  • idku
in red the order of operations erro
idku
  • idku
and f'(1) is becase \(x_1\)=1 in our case
anonymous
  • anonymous
woops yes i missed that
idku
  • idku
yes, and f'(1)=?
anonymous
  • anonymous
1.25992...
idku
  • idku
just take 1.26
anonymous
  • anonymous
so (2 , 4.26)
idku
  • idku
\[(x_2,y_2)=(2,4.26)\]Correct
idku
  • idku
Now, next step h=1, x_2 =2 y_2=4.26 go for it...
idku
  • idku
(this is the last step, for this approximation with step size of h=1)
idku
  • idku
i get (3, 6.935)
anonymous
  • anonymous
one sec only got (3, 5.26 + ? ) so far
anonymous
  • anonymous
okay yep got the same thing
anonymous
  • anonymous
so now what?
idku
  • idku
(3, \(\color{red}{4}\).26 + 1•f'(4.26))
idku
  • idku
4, not 5
anonymous
  • anonymous
wow thats strange, then how did i get the same answer ?
idku
  • 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))
anonymous
  • anonymous
2.67526
idku
  • 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
idku
  • idku
oh, my bad
idku
  • idku
yeah, 4.26 is the y.
idku
  • idku
(2+1 , (4.26+1\cdot (1+(2)^2)^(1/3)))
idku
  • idku
http://www.wolframalpha.com/input/?i=%282%2B1+%2C+%284.26%2B1%5Ccdot+%281%2B%282%29%5E2%29%5E%281%2F3%29%29%29
idku
  • idku
(3, 5.9699) y(3)=5.9699
anonymous
  • anonymous
interesting we must have done something wrong , none of the answer choices give that result
idku
  • idku
i know it doesn't correspond to options, and you know why?
idku
  • 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.
anonymous
  • anonymous
okay so would it be C because it is the closest smaller answer ?
idku
  • idku
it could be smaller or larger than the actual answer, you never know
idku
  • 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)
jim_thompson5910
  • 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
idku
  • idku
I am the stupidest person in the world then I guess.... G-wis
jim_thompson5910
  • jim_thompson5910
no not stupid since Euler's method does work here
idku
  • idku
I am overloading it when it is just that simple, just that integral.... om(f)g excuse my lang
anonymous
  • anonymous
lol well atleast i learned something new :D
idku
  • idku
yes, Euler works, but... but it's too long.....
idku
  • idku
it is just like driving from GA to North Dakota and then to Texas, instead of straight GA-Texas.
anonymous
  • anonymous
oh well thanks for all the help guys !! :D:D:D:D:D:D:D
idku
  • idku
you welcome, lol, i should literally "fire" myself.
idku
  • idku
good luck

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