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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?
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits \sqrt[3]{1 + t^2}\] this is what i cant find

idku
 one year ago
Best ResponseYou've already chosen the best response.3\[\large \int\limits_{ }^{ }\sqrt[3]{1+t^2}dt\]

idku
 one year ago
Best ResponseYou've already chosen the best response.3well, 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
 one year ago
Best ResponseYou've already chosen the best response.0yep, didn't get anywhere. be my guest though im pretty bad at u substitutions :P

idku
 one year ago
Best ResponseYou've already chosen the best response.3\(\large \displaystyle\int\limits_{ }^{ }\sqrt[3]{1+t^2}dt\) \(\large \displaystyle t=\tan\theta \)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0trig sub...? never heard of it.

idku
 one year ago
Best ResponseYou've already chosen the best response.3lets 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
 one year ago
Best ResponseYou've already chosen the best response.0wolfram is giving me something i dont even understand .

idku
 one year ago
Best ResponseYou've already chosen the best response.3acc. to wolfram http://www.wolframalpha.com/input/?i=integral+of+%281%2Bx%5E2%29%5E%281%2F3%29

idku
 one year ago
Best ResponseYou've already chosen the best response.3no simple function answer

idku
 one year ago
Best ResponseYou've already chosen the best response.3are you sure you aren't given limits of integration, and this is not a riemman sum type of question?

idku
 one year ago
Best ResponseYou've already chosen the best response.3or 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
 one year ago
Best ResponseYou've already chosen the best response.0this is my given question

idku
 one year ago
Best ResponseYou've already chosen the best response.3can I use y instead of x (as y(t) ?)

idku
 one year ago
Best ResponseYou've already chosen the best response.3You can use a stepsize on that one I think

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0interesting i figured i just had to find s(t) and just change the constant and plug in t = 3

idku
 one year ago
Best ResponseYou've already chosen the best response.3well, that is the task as far as the steps go, but s(t)=?

idku
 one year ago
Best ResponseYou've already chosen the best response.3that is the question where we don't arrive at a simple conclusion just like this, right?

idku
 one year ago
Best ResponseYou've already chosen the best response.3have you heard of a stepsize, Euler's method or anything like this?

idku
 one year ago
Best ResponseYou've already chosen the best response.3\(\large \displaystyle f'(t)=\sqrt[3]{1+t^2}\) y(0)=2, find y(3)=? lets use a stepsize 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: \(\large \displaystyle \left(x_n,y_n\right)=\left(x_{n1}+h,~y_{n1}~+h\cdot \color{red}{f'(x_{n1},y_{n1})}~\right)\)

idku
 one year ago
Best ResponseYou've already chosen the best response.3maybe we won't understand why it is like this right now, but you can see the formula, without completely going like wt(f)....

idku
 one year ago
Best ResponseYou've already chosen the best response.3\(\large \displaystyle \left(t_n,y_n\right)=\left(t_{n1}+h,~y_{n1}~+h\cdot \color{red}{f'(t_{n1},y_{n1})}~\right)\) excuse me it should be t.

idku
 one year ago
Best ResponseYou've already chosen the best response.3\(\large \displaystyle \left(x_n,y_n\right)=\left(x_{n1}+h,~y_{n1}~+h\cdot \color{red}{f'(x_{n1},y_{n1})}~\right)\) you are given: \(\large \displaystyle f'(t)=\sqrt[3]{1+t^2}\) and a first point of (t=0, y=2)

idku
 one year ago
Best ResponseYou've already chosen the best response.3\(\large \displaystyle \left(x_1,y_1\right)=\left(x_{0}+1,~2~+1\cdot \color{red}{f'(0,2)}~\right)\) this would be the first step starting from (0,2), and h=1

idku
 one year ago
Best ResponseYou've already chosen the best response.3are you getting it at least a little? (sorry I am bad at explaining)

idku
 one year ago
Best ResponseYou've already chosen the best response.3ok, I will give you the entire process wth h=1, and then you will tell me how much you get out of it.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lol sort of , i'm not quite sure what the h represents

idku
 one year ago
Best ResponseYou've already chosen the best response.3you 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
 one year ago
Best ResponseYou've already chosen the best response.3we don't know where this step size of 1 would land, and usiong this formula we can find that out.

idku
 one year ago
Best ResponseYou've already chosen the best response.3and 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
 one year ago
Best ResponseYou've already chosen the best response.3\(\large \displaystyle \left(x_n,y_n\right)=\left(x_{n1}+h,~y_{n1}~+h\cdot \color{red}{f'(x_{n1},y_{n1})}~\right)\) \(\large \displaystyle \left(x_1,y_1\right)=\left(x_{0}+1,~2~+1\cdot \color{red}{f'(0,2)}~\right)\) (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
 one year ago
Best ResponseYou've already chosen the best response.3oh, you know x_0 is 0 , i forgot to put that in

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay wait one sec , why are we even finding (x1,y1)?

idku
 one year ago
Best ResponseYou've already chosen the best response.3\(\large \displaystyle \left(x_1,y_1\right)=\left(0+1,~2~+1\cdot \color{red}{\sqrt[3]{1+0^2}}~\right)\) \(\large \displaystyle \left(x_1,y_1\right)=\left(1,~3\right)\)

idku
 one year ago
Best ResponseYou've already chosen the best response.3because that is the next point

idku
 one year ago
Best ResponseYou've already chosen the best response.3when you do a u=substitution du=?

idku
 one year ago
Best ResponseYou've already chosen the best response.3du=2t dt so it ends up 1/(2 √(u1)) du = dt

idku
 one year ago
Best ResponseYou've already chosen the best response.3\[\int\limits_{ }^{ }\frac{\sqrt[3]{u}}{\sqrt{u1}}du\]

idku
 one year ago
Best ResponseYou've already chosen the best response.3if it had 2t next to the cube root, then yes, u sub would be the only and the perfect

idku
 one year ago
Best ResponseYou've already chosen the best response.3anyway, i guess back to stepsize?

idku
 one year ago
Best ResponseYou've already chosen the best response.3\(\large \displaystyle \left(x_1,y_1\right)=\left(1,~3\right)\) so here is the result of 1 step

idku
 one year ago
Best ResponseYou've already chosen the best response.3see how I am doing the steps? (See how I plug them into the formula?)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes , okay so say we do this for a whole bunch of points then what?

idku
 one year ago
Best ResponseYou've already chosen the best response.3\(\large \displaystyle \left(x_n,y_n\right)=\left(x_{n1}+h,~y_{n1}~+h\cdot \color{red}{f'(x_{n1},y_{n1})}~\right) \) can you use this formula to get from (1,3) with h=1 to the next point? Note: in this case: \(\large \displaystyle \left(x_n,y_n\right)=\left(x_{n1}+h,~y_{n1}~+h\cdot \color{red}{f'(x_{n1})}~\right) \) (since there is no y  it is explicit function)

idku
 one year ago
Best ResponseYou've already chosen the best response.3yes, 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
 one year ago
Best ResponseYou've already chosen the best response.3i guess you just need to do the dirty work (but do it correctly0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0just out of curiosity in what math class did you learn this method?

idku
 one year ago
Best ResponseYou've already chosen the best response.3I learned this method in calculus II, although it also belongs to DE.

idku
 one year ago
Best ResponseYou've already chosen the best response.3ok, can you proceed from (1,3) with h=1?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0not sure how to find f('n1) i think i got the rest though.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0damn , just had the entire thing written out and it just got deleted DX

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[(x _{1},y _{1}) = (x _{1} + 1 , y _{1} + 1 * (f \prime (x _{1})))\]

idku
 one year ago
Best ResponseYou've already chosen the best response.3now, you know x_1 is 1 and y_1 is 3 also, h=1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0woops i mean x2 and y2 at the beginning

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0just not sure how to find the f'(x1)

idku
 one year ago
Best ResponseYou've already chosen the best response.3you know x_1 is 1 and y_1 is 3 so put that in....

idku
 one year ago
Best ResponseYou've already chosen the best response.3\[(x_2,y_2)=(1+1,3+1\cdot f'(1))=(2,3+\sqrt[3]{1+(1)^2}~)=(?,?)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[(x _{1},y _{1}) = ( 2 , 4 * (f \prime (x _{1}))) \]

idku
 one year ago
Best ResponseYou've already chosen the best response.3(if you want we can do a polynomial approximation for cube root of 2, but it would be too much xD)

idku
 one year ago
Best ResponseYou've already chosen the best response.3you mean\[(x_2,y_2)=(2,3+\color{red}{1\cdot}f'(1))\]

idku
 one year ago
Best ResponseYou've already chosen the best response.3in red the order of operations erro

idku
 one year ago
Best ResponseYou've already chosen the best response.3and f'(1) is becase \(x_1\)=1 in our case

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0woops yes i missed that

idku
 one year ago
Best ResponseYou've already chosen the best response.3\[(x_2,y_2)=(2,4.26)\]Correct

idku
 one year ago
Best ResponseYou've already chosen the best response.3Now, next step h=1, x_2 =2 y_2=4.26 go for it...

idku
 one year ago
Best ResponseYou've already chosen the best response.3(this is the last step, for this approximation with step size of h=1)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0one sec only got (3, 5.26 + ? ) so far

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay yep got the same thing

idku
 one year ago
Best ResponseYou've already chosen the best response.3(3, \(\color{red}{4}\).26 + 1•f'(4.26))

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wow thats strange, then how did i get the same answer ?

idku
 one year ago
Best ResponseYou've already chosen the best response.3so, you do the same thing, you plug in the \(x_{n1}\) (in this case \(x_2\)which is equivalent to 4.26, into the f'(t))

idku
 one year ago
Best ResponseYou've already chosen the best response.3this 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
 one year ago
Best ResponseYou've already chosen the best response.3(2+1 , (4.26+1\cdot (1+(2)^2)^(1/3)))

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0interesting we must have done something wrong , none of the answer choices give that result

idku
 one year ago
Best ResponseYou've already chosen the best response.3i know it doesn't correspond to options, and you know why?

idku
 one year ago
Best ResponseYou've already chosen the best response.3We are going from y(0) to y(3) with a step size of h=1 the step size is too big! Take a smaller stepsize to get a better approximation.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay so would it be C because it is the closest smaller answer ?

idku
 one year ago
Best ResponseYou've already chosen the best response.3it could be smaller or larger than the actual answer, you never know

idku
 one year ago
Best ResponseYou've already chosen the best response.3I 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
 one year ago
Best ResponseYou've already chosen the best response.2I'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
 one year ago
Best ResponseYou've already chosen the best response.3I am the stupidest person in the world then I guess.... Gwis

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2no not stupid since Euler's method does work here

idku
 one year ago
Best ResponseYou've already chosen the best response.3I am overloading it when it is just that simple, just that integral.... om(f)g excuse my lang

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lol well atleast i learned something new :D

idku
 one year ago
Best ResponseYou've already chosen the best response.3yes, Euler works, but... but it's too long.....

idku
 one year ago
Best ResponseYou've already chosen the best response.3it is just like driving from GA to North Dakota and then to Texas, instead of straight GATexas.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh well thanks for all the help guys !! :D:D:D:D:D:D:D

idku
 one year ago
Best ResponseYou've already chosen the best response.3you welcome, lol, i should literally "fire" myself.
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