Another arc length

**tttcomrader** · October 25th 2007, 10:48 AM

Find the length of the arc formed by

$y= \frac{1}{8}(-4x^{2}+2ln(x))$ from x=2 to x=6.

**Krizalid** · October 25th 2007, 10:50 AM

Arc lenght formula states

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

Apply it.

**tttcomrader** · October 25th 2007, 11:03 AM

I did, but then I have a crazy integral to deal with.

**Soroban** · October 25th 2007, 02:14 PM

Hello, tttcomrader!

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

Find the length of the arc formed by

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

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

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

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

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

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

I assume you can finish it . . .

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