# Calculating Arc length of a curve.

• Feb 5th 2010, 08:44 PM
jfinkbei
Calculating Arc length of a curve.
The curve: $\displaystyle y=\frac{x^2}{2} - \frac{lnx}{4}, 2\leq x \leq4$

my work so far:
$\displaystyle x(t)=\frac{x^2}{2} - \frac{ln}{4}$

$\displaystyle x'(t)=x - \frac{1}{4x}$

$\displaystyle y(t)=t$

$\displaystyle y'(t)=1$

$\displaystyle \int^4_2 \sqrt{1^2 + (x - \frac{1}{4x})^2}dx$

$\displaystyle \int^4_2 \sqrt{1+\frac{9}{16}x^2}dx$

anyways after that mess I am really not sure how to integrate this with u-substitution. If I made a mistake please let me know or some direction if I am right so far would be very helpful!! Thanks a lot!
• Feb 5th 2010, 09:33 PM
Shananay
Quote:

Originally Posted by jfinkbei
$\displaystyle \int^4_2 \sqrt{1^2 + (x - \frac{1}{4x})^2}dx$

$\displaystyle \int^4_2 \sqrt{1+\frac{9}{16}x^2}dx$

How did you make this step? What is $\displaystyle \left(x-\frac{1}{4x}\right)^2$?

$\displaystyle \left(x-\frac{1}{4x}\right)\left(x-\frac{1}{4x}\right)$

I'm getting something different under the radical.

By the way, the 1 doesn't need to be squared. The formula for arclength has a $\displaystyle 1+[f'(x)]^2$ under the radical always.
• Feb 5th 2010, 09:46 PM
Shananay
for the integrand I'm getting $\displaystyle \sqrt{\frac{16x^4+8x^2+1}{16x^2}}$

Both the top and bottom are perfect squares, so factor and simplify and let $\displaystyle u=x^2$

EDIT: It might not be that simple actually, I haven't done the whole problem. You may need to make an additional substitution. Let me know if you get stuck again and I'll try to work it out.

EDIT2: you shouldn't need to sub at all for this integral. Sorry, I'm all mixed up right now.
• Feb 6th 2010, 02:28 AM
HallsofIvy
$\displaystyle \left(x-\frac{1}{4x}\right)^2= x^2- \frac{1}{2}+ \frac{1}{16x^2}$ so

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

which is just the first formula with "$\displaystyle -\frac{1}{2}$" changed to "$\displaystyle +\frac{1}{2}$". It is
$\displaystyle \left(x+ \frac{1}{4x}\right)^2$ so the square root just gives

$\displaystyle \int x+ \frac{1}{4x} dx$
• Feb 6th 2010, 06:27 AM
Shananay
Quote:

Originally Posted by HallsofIvy
$\displaystyle \left(x-\frac{1}{4x}\right)^2= x^2- \frac{1}{2}+ \frac{1}{16x^2}$ so

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

which is just the first formula with "$\displaystyle -\frac{1}{2}$" changed to "$\displaystyle +\frac{1}{2}$". It is
$\displaystyle \left(x+ \frac{1}{4x}\right)^2$ so the square root just gives

$\displaystyle \int x+ \frac{1}{4x} dx$

Faster and easier than what I was doing. This is the way to go.