Need help with setup for arc length of x^(2/3) + y^(2/3) = 4

H

hank

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Sep 13, 2006
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Find the arc length in the second quadrant of the curve

x^(2/3) + y^(2/3) = 4 from x = -8 to x = -1.

Ok what I'm thinking i should do is:

y^(2/3) = 4 - x^(2/3)

But here's where I get stuck.

I'm thinking I should ^(1/6) both sides, but that gets really ugly. It seems to me there should be an easier way to set this up.
 
This thing is an astroid, I believe. There's symmetry, so you can change your limits of integration to positive values, 1 and 8.

Solve for y, then use the arc length formula integral.

\(\displaystyle \L\\y=(4-x^{\frac{2}{3}})^{\frac{3}{2}}\)

The derivative is:

\(\displaystyle \L\\\frac{-\sqrt{4-x^{\frac{2}{3}}}}{x^{\frac{1}{3}}}\)

Square it and get:

\(\displaystyle \L\\\frac{4-x^{\frac{2}{3}}}{x^{\frac{2}{3}}}\)

\(\displaystyle 1+\frac{4-x^{\frac{2}{3}}}{x^{\frac{2}{3}}}=\frac{4}{x^{\frac{2}{3}}}\)

\(\displaystyle \L\\\sqrt{\frac{4}{x^{\frac{2}{3}}}}=\frac{2}{x^{\frac{1}{3}}}\)

Now, integrate:

\(\displaystyle \L\\\int_{1}^{8}\frac{2}{x^{\frac{1}{3}}}dx=3x^{\frac{2}{3}}\)

\(\displaystyle \L\\3(8)^{\frac{2}{3}}-3(1)^{\frac{2}{3}}=12-3=9\)
 
arc length =[1+[dy/dx]^2]^1/2 dx integrated from -8 to -1

x^2/3 + y^2/3 =4
let us determine dy /dx
2/3 x^(-1/3)dx +2/3 y^(-1/3)dy=0
dy/y^1/3 =- dx/x^1/3
dy/dx = - [y/x]^1/3

arc length = [1+ [-[y/x]^1/3]^2 ]^1/2 dx integrated from -8 to -1
arc length = [1+y^2/3/x^2/3]^1/2 dx int. from -8 to -1

but y^2/3 = 4-x^2/3
arc length = [1 + [4-x^2/3]/x^2/3 ] ^1/2 dx int from -8 to -1
arc length = [x^2/3 +4-x^2/3]/x^2/3 ]^1/2 dx int fro -8 to -1
arc length = [4/x^2/3]^1/2 dx int. from -8 to -1
arc length = 2/x^1/3 dx int. from -8 to -1
arc length = 2 int [-8,-1]x^-1/3 dx
arc length =2 x^2/3 [3/2] evaluated at -1,-8
arc length =2[1] - 2[4]
arc length = -6 answer

I assume I lost a minus sign, but the technique is right
Arthur
 
Re: Need help with setup...

Hello, Hank!

Find the arc length in the second quadrant of the curve:
. . \(\displaystyle x^{\frac{2}{3}}\,+\,y^{\frac{2}{3}} \:=\:4\) from \(\displaystyle x\,=\,-8\) to \(\displaystyle x\,=\,-1\)
Click to expand...

\(\displaystyle \text{Solve for }y:\;\;y^{\frac{2}{3}}\:=\:4\,-\,x^{\frac{2}{3}}\;\;\Rightarrow\;\;y\;=\;\left(4\,-\,x^{\frac{2}{3}}\right)^{\frac{3}{2}}\)

Hence: \(\displaystyle \L\:\frac{dy}{dx} \:=\:\frac{3}{2}\left(4\,-\,x^{\frac{2}{3}}\right)^{\frac{1}{2}}\left(-\frac{2}{3}x^{-\frac{1}{3}}\right) \;=\;-\frac{\sqrt{4\,-\,x^{\frac{2}{3}}}}{x^{\frac{1}{3}}}\)

Then: \(\displaystyle \L\:1\,+\,\left(\frac{dy}{dx}\right)^2 \:=\:1\,+\,\frac{4\,-\,x^{\frac{2}{3}}}{x^{\frac{2}{3}}} \;=\;\frac{4}{x^{\frac{2}{3}}}\)

Hence: \(\displaystyle \L\:ds\;=\;\sqrt{\frac{4}{x^{\frac{2}{3}}}} \;=\;\frac{2}{x^{\frac{1}{3}}} \;=\;2x^{-\frac{1}{3}}\)

Can you finish it now?

 
I found my error
from arc length = [1+[4-x^2/3]/x^2/3]^1/2 dx int. from -8 to -1
arc length = [4/x^2/3]^1/2 dx int from -8 to -1
arc length = +/-2 x^(-1/3) dx int from -8 to -1
arc length = +/-2 x^2/3 / 2/3 evaluated at -1,-8
arc length = +/-3 x^2/3 evaluated at -1,-8
arc length = +/- 3 [1-4]
arc length= +/- 9
arc length =9 answer, -9 is a extraneous answer, arc length must be positive

Arthur
 
Thanks tons, everyone.

What screwed me up was simply solving for y.

Shoulda been 3/2 instead of 1/6. Thx.
 
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