# Another arc length

• Oct 25th 2007, 10:48 AM
Another arc length
Find the length of the arc formed by

$\displaystyle y= \frac{1}{8}(-4x^{2}+2ln(x))$ from x=2 to x=6.
• Oct 25th 2007, 10:50 AM
Krizalid
Arc lenght formula states

$\displaystyle \int_a^b\sqrt{1+(y')^2}\,dx$

Apply it.
• Oct 25th 2007, 11:03 AM
I did, but then I have a crazy integral to deal with.
• Oct 25th 2007, 02:14 PM
Soroban

Watch your algebra . . . The integral comes out rather neat . . .

Quote:

Find the length of the arc formed by

$\displaystyle y\:= \:\frac{1}{8}\left[-4x^{2}+2\ln(x))\right]$ .from $\displaystyle x=2$ to $\displaystyle x=6.$

$\displaystyle y' \:=\:\frac{1}{8}\left(-8x + \frac{2}{x}\right) \;=\;\frac{1}{4x} - x$

$\displaystyle (y')^2 \;=\;\left[\frac{1}{4x} - x\right]^2 \;=\;\frac{1}{16x^2} - \frac{1}{2} + x^2$

$\displaystyle 1 + (y')^2\;=\;1 + \left(\frac{1}{16x^2} - \frac{1}{2} + x^2\right) \;=\;\frac{1}{16x^2} + \frac{1}{2} + x^2 \;=\;\left(\frac{1}{4x} + x\right)^2$

. . Hence: .$\displaystyle \sqrt{1 + (y')^2} \;=\;\sqrt{\left(\frac{1}{4x} + x\right)^2} \;=\;\frac{1}{4x} + x$

Therefore: .$\displaystyle L \;=\;\int^6_2\left(\frac{1}{4x} + x\right)\,dx$

I assume you can finish it . . .