anonymous
  • anonymous
Find the arc length of y=x^(2/3) for 0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
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amistre64
  • amistre64
\[\int_{}^{}ds\implies \int_{}^{}\sqrt{(x')^2+(y')^2}~dt\]
amistre64
  • amistre64
of instead of dt, we can do with dx, and let x'=1

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amistre64
  • amistre64
what are you getting for your setup?
anonymous
  • anonymous
im getting this \[\int\limits_{0}^{1} \sqrt{(1)^2 + (x^\frac{ 2 }{ 3 })^2} dx\]
anonymous
  • anonymous
oh wait
anonymous
  • anonymous
|dw:1442357936251:dw|
amistre64
  • amistre64
thats better yeah
anonymous
  • anonymous
and then
anonymous
  • anonymous
|dw:1442358056890:dw|
anonymous
  • anonymous
then you find the integral?
amistre64
  • amistre64
then you find the integral ... trying to see if there is a nice form to it tho
amistre64
  • amistre64
the wolfs integral suggests its a fairly simple u-sub
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amistre64
  • amistre64
or not ...
anonymous
  • anonymous
how do you get to that? im trying to integrate
amistre64
  • amistre64
http://www.wolframalpha.com/input/?i=integrate+sqrt%28+1+%2B+4%2F9+x^%28-2%2F3%29+%29
anonymous
  • anonymous
|dw:1442358829336:dw|
amistre64
  • amistre64
i just ask the wolf to integrate it ...
anonymous
  • anonymous
would that be the integral?
anonymous
  • anonymous
im sure its that just trying to figure out how to get to that point
amistre64
  • amistre64
how are you sure that is it? i have to wonder since the wolfs assessment is different
amistre64
  • amistre64
one 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
  • amistre64
t=0 to 1 satisfies the substitution x' = 3t^2, y' =2t sqrt(9t^4 + 4t^2) t sqrt(9t^2 + 4) dt
anonymous
  • anonymous
|dw:1442359448247:dw|
anonymous
  • anonymous
would u sub work here?
amistre64
  • amistre64
i 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
  • amistre64
t 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
  • amistre64
u = 4 to 13 when t=0 to 1
amistre64
  • amistre64
1/27 ((13)^(3/2)-(4^(3/2)), is looking sweet on my end
amistre64
  • amistre64
i 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
  • anonymous
|dw:1442360116489:dw|
anonymous
  • anonymous
is that what you mean?
amistre64
  • amistre64
\[y = x^{2/3}~:~x=(0,1)\] \[x^{1/3}=t~:~x=t^3~:~t=(0,1)\] \[y=t^2\] \[x'=3t^2~:~y'=2t\] \[\int_{0}^{1}\sqrt{(x')^2+(y')^2}~dt\]
amistre64
  • amistre64
that simplifies to: \[\int_{0}^{1}t\sqrt{9t^2+4}~dt\] yes
anonymous
  • anonymous
aaah i see what you did there
anonymous
  • anonymous
|dw:1442360353998:dw|
anonymous
  • anonymous
|dw:1442360476915:dw|
anonymous
  • anonymous
and then you insert u? and then plug in the 1? and that gets you the answer
amistre64
  • amistre64
by substitution u = 9t^2 + 4, as t goes from 0 to 1, u goes from 9(0)+4 to 9(1)+4
amistre64
  • amistre64
so the limits for du stuff is 4 to 13
amistre64
  • amistre64
youcould replace u witn 9t^2 + 4 if you want, but then the 13 and 4 get calculated in there inways
anonymous
  • anonymous
umm im a bit lost but i'll try to see where it is that im confused
amistre64
  • amistre64
k, 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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