Arc length query

**MarkOfEternity** · October 17th 2009, 11:23 AM

I have been trying to find the archlength of $y^2=4x$ , where 0 ≤ x ≤ 2.

I first solved $y=\sqrt{4x}$ .

Then I solved for y'.
$y'=\frac{4x^\frac{-1}{2}}{2}*4=2(4x)^\frac{-1}{2}$ .

The formula I have to find arclength is $\int_a^b \sqrt{1+(y')^2} dx$ .

So I plugged in and got this:

$\int_0^2 \sqrt{1+(2(4x)^\frac{-1}{2})^2}=\int_0^2 \sqrt{1+(4(4x)^{-1})}=\int_0^2 \sqrt{1+\frac{4}{4x}}=\int_0^2 \sqrt{1+\frac{1}{x}}$ .

I evaluated $\frac{1}{x}=(\frac{1}{\sqrt{x}})^2$ , because my book offers a formula for the form $\sqrt{a^2+u^2}$ .

The formula was

My $a=1$ and my $u=\frac{1}{\sqrt{x}}$ , so I plugged those into the formula.

What I got out was this:

$\int_0^2 \sqrt{1^2+\frac{1}{\sqrt{x}}^2} du=\frac{\frac{1}{\sqrt{x}}}{2}\sqrt{1^2+\frac{1}{ \sqrt{x}}^2}+\frac{1^2}{2}\ln(\frac{1}{\sqrt{x}}+\ sqrt{1^2+\frac{1}{\sqrt{x}}^2})=$
$\frac{1}{2\sqrt{x}}\sqrt{1+\frac{1}{\sqrt{x}}^2}+\ frac{1}{2}\ln(\frac{1}{\sqrt{x}}+\sqrt{1+(\frac{1} {\sqrt{x}})^2})$

This is what I got to. Now, I need to evaluate it from 0 to 2. But when I plug in 2 for the value of x, my answer is undefined. Did I do something wrong? Is there a different way to do this? Please...PLEASE help!

**Scott H** · October 17th 2009, 12:46 PM

Alternatively, you can set the independent variable to $y$ and integrate from $0$ to $2\sqrt{2}$ .

Edit: You'd multiply it by $2$ to get both sides' worth.

**apcalculus** · October 17th 2009, 12:47 PM

http://www.wolframalpha.com/input/?i...1%2B(1/x))^0.5

click on 'Show Steps'

I hope it helps.

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