anonymous
  • anonymous
Hey everyone, need help finding the arc length of the curve y^2=x^3 from (0,0) to (1/4,1/8).
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
I know the formula and I think y'=(3x^2)/(2y)
anonymous
  • anonymous
let me try this, see if it works. \[y=x^\frac{3}{2}\] yes?
anonymous
  • anonymous
ok right, do we only take the +x^3/2?

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anonymous
  • anonymous
\[y'=\frac{3}{2}x^\frac{1}{2}\]
anonymous
  • anonymous
y = x^(3/2) y' = 3/2*x^1/2 (y')^2 = 9/4*x integrate sqrt(1+9/4*x) from 1/8 to 1/4 use u-sub if you have to
anonymous
  • anonymous
Rim would it not be from 0 to 1/4?
anonymous
  • anonymous
oh sorry yes 0 to 1/4
anonymous
  • anonymous
we are going from (0,0) to (1/8,1/4) so it is ok yes? maybe i am wrong, but i will graph it. in any case we have to integrate \[\int_0^{\frac{1}{8}} \sqrt{1-\frac{9}{4}x} dx\]
anonymous
  • anonymous
oops you are right it is 1/4
anonymous
  • anonymous
anti derivative is \[-\frac{(4-9x)^\frac{3}{2}}{27}\]
anonymous
  • anonymous
and integral is \[\frac{8}{27}-\frac{7\sqrt{7}}{216}\]
anonymous
  • anonymous
did you guys get .2824?
anonymous
  • anonymous
let me get a calculator i will check
anonymous
  • anonymous
no in fact i got .21055...
anonymous
  • anonymous
ok, but isnt it also sqrt(1+(9/4)x)?
anonymous
  • anonymous
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anonymous
  • anonymous
yes it is .2824
anonymous
  • anonymous
well i guess i messed up hold on
anonymous
  • anonymous
you are right and i am wrong (0ften the case)
anonymous
  • anonymous
take advantage of Wolfram alpha it is a very useful tool
anonymous
  • anonymous
i put - where i should have had +. sorry
anonymous
  • anonymous
no problem thanks for the help guys
anonymous
  • anonymous
no problem
anonymous
  • anonymous
Just why do we pick positive sqrt all the time?

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