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

- anonymous

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

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- schrodinger

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- anonymous

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

- idku

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

- anonymous

yep thats right

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## More answers

- 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

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

- idku

oh, my trig sub...

- idku

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

- anonymous

trig sub...? never heard of it.

- 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

wolfram is giving me something i dont even understand .

- idku

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

- idku

no simple function answer

- idku

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

- 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

##### 1 Attachment

- anonymous

this is my given question

- idku

this is much better

- idku

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

- idku

You can use a stepsize on that one I think

- anonymous

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

- idku

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

- idku

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

- idku

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

- anonymous

no sorry :(

- 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

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

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

- idku

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

- idku

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

- idku

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

- 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

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

- 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

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

- 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

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

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

- anonymous

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

- 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

because that is the next point

- idku

|dw:1440033265402:dw|

- idku

when you do a u=substitution
du=?

- idku

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

- idku

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

- idku

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

- idku

anyway, i guess back to stepsize?

- idku

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

- idku

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

- anonymous

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

- 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

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

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

- anonymous

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

- idku

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

- idku

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

- anonymous

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

- anonymous

woops

- anonymous

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

- anonymous

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

- idku

yhes, this is correct

- idku

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

- anonymous

woops i mean x2 and y2 at the beginning

- anonymous

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

- idku

yes....

- idku

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

- idku

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

- anonymous

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

- idku

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

- idku

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

- idku

in red the order of operations erro

- idku

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

- anonymous

woops yes i missed that

- idku

yes, and f'(1)=?

- anonymous

1.25992...

- idku

just take 1.26

- anonymous

so (2 , 4.26)

- idku

\[(x_2,y_2)=(2,4.26)\]Correct

- idku

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

- idku

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

- idku

i get (3, 6.935)

- anonymous

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

- anonymous

okay yep got the same thing

- anonymous

so now what?

- idku

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

- idku

4, not 5

- anonymous

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

- 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

2.67526

- 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

oh, my bad

- idku

yeah, 4.26 is the y.

- idku

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

- 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

(3, 5.9699)
y(3)=5.9699

- anonymous

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

- idku

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

- 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

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

- idku

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

- 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

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

##### 1 Attachment

- idku

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

- jim_thompson5910

no not stupid since Euler's method does work here

- idku

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

- anonymous

lol well atleast i learned something new :D

- idku

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

- idku

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

- anonymous

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

- idku

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

- idku

good luck

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