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anonymous
 one year ago
Find the arc length of y=x^(2/3) for 0<x<1
anonymous
 one year ago
Find the arc length of y=x^(2/3) for 0<x<1

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amistre64
 one year ago
Best ResponseYou've already chosen the best response.0\[\int_{}^{}ds\implies \int_{}^{}\sqrt{(x')^2+(y')^2}~dt\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0of instead of dt, we can do with dx, and let x'=1

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0what are you getting for your setup?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0im getting this \[\int\limits_{0}^{1} \sqrt{(1)^2 + (x^\frac{ 2 }{ 3 })^2} dx\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442357936251:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442358056890:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0then you find the integral?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0then you find the integral ... trying to see if there is a nice form to it tho

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0the wolfs integral suggests its a fairly simple usub

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how do you get to that? im trying to integrate

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=integrate+sqrt%28+1+%2B+4%2F9+x^%282%2F3%29+%29

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442358829336:dw

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0i just ask the wolf to integrate it ...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0would that be the integral?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0im sure its that just trying to figure out how to get to that point

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0how are you sure that is it? i have to wonder since the wolfs assessment is different

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0one thing i am thing of is trying to work out a parametric like x = t^3, y=t^2 this way, t = x^(1/3), letting y = x^(2/3) by eliminating the parameter

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0t=0 to 1 satisfies the substitution x' = 3t^2, y' =2t sqrt(9t^4 + 4t^2) t sqrt(9t^2 + 4) dt

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442359448247:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0would u sub work here?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0i already tried working it like that ... i was getting nowhere. but i believe if we split it into parameters it works out better for us

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0t sqrt(9t^2 + 4) dt let u = 9t^2+4 du = 18t dt t sqrt(u) du/18t u^(1/2) du/18 2/3(18) u^(3/2) 1/3(9) u^(3/2) 1/27 u^(3/2)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0u = 4 to 13 when t=0 to 1

amistre64
 one year ago
Best ResponseYou've already chosen the best response.01/27 ((13)^(3/2)(4^(3/2)), is looking sweet on my end

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0i cant make it work if we keep y as a function of x as its defined. so i wondered if parameters would make life easier, and it does

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442360116489:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is that what you mean?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0\[y = x^{2/3}~:~x=(0,1)\] \[x^{1/3}=t~:~x=t^3~:~t=(0,1)\] \[y=t^2\] \[x'=3t^2~:~y'=2t\] \[\int_{0}^{1}\sqrt{(x')^2+(y')^2}~dt\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0that simplifies to: \[\int_{0}^{1}t\sqrt{9t^2+4}~dt\] yes

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0aaah i see what you did there

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442360353998:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442360476915:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and then you insert u? and then plug in the 1? and that gets you the answer

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0by substitution u = 9t^2 + 4, as t goes from 0 to 1, u goes from 9(0)+4 to 9(1)+4

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0so the limits for du stuff is 4 to 13

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0youcould replace u witn 9t^2 + 4 if you want, but then the 13 and 4 get calculated in there inways

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0umm im a bit lost but i'll try to see where it is that im confused

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0k, and dont worry about being lost, my approach was prolly not 'standard' it is just what worked out in my head.
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