# [SOLVED] Finding Arc length

• Sep 24th 2008, 04:16 PM
kl.twilleger
[SOLVED] Finding Arc length
find arc length of y=1-x^2 -1<x<1

i used the arc length formula, found the derivative of the function and squared it and plugged it all in and now i'm stuck at this point....

integrating between 1 and -1 the sq.rt of 1+ (x^2/1-x^2) dx

HELP!!
• Sep 24th 2008, 04:37 PM
Jhevon
Quote:

Originally Posted by kl.twilleger
find arc length of y=1-x^2 -1<x<1

i used the arc length formula, found the derivative of the function and squared it and plugged it all in and now i'm stuck at this point....

integrating between 1 and -1 the sq.rt of 1+ (x^2/1-x^2) dx

HELP!!

isn't y' = -2x?

how did you get that integral? it should be $\displaystyle \int_{-1}^1 \sqrt{1 + (-2x)^2 }~dx = \int_{-1}^1 \sqrt{1 + 4x^2}~dx$
• Sep 24th 2008, 04:46 PM
kl.twilleger
y= sq.rt. of 1-x^2. So, using the chain rule, i got 1/2(1-x^2)^-(1/2)*(-2x).
chugging through that i got -x/(sq.rt. 1-x^2). according to the arc length formula i square the derivative and I got x^2/(1-x^2).

Did I follow through right or did I make a mistake?

and how did you make all of your cool signs? I just started and I don't know where they are or how to use them.
• Sep 24th 2008, 04:52 PM
Jhevon
Quote:

Originally Posted by kl.twilleger
y= sq.rt. of 1-x^2. So, using the chain rule, i got 1/2(1-x^2)^-(1/2)*(-2x).
chugging through that i got -x/(sq.rt. 1-x^2). according to the arc length formula i square the derivative and I got x^2/(1-x^2).

Did I follow through right or did I make a mistake?

that's right. (you neglected to mention the square root before!)

hint: try combining the fractions

Quote:

and how did you make all of your cool signs? I just started and I don't know where they are or how to use them.
see here
• Sep 24th 2008, 05:01 PM
kl.twilleger
ok, i tried that and i got \frac{1-x^2}{1-x^2} + \frac{x^2}{1-x^2}

which gave me \sqrt{\frac{1}{1-x^2}}

i'm not sure what i can do algebraically from here

and i can't get that stuff to work for some reason
• Sep 24th 2008, 05:15 PM
Jhevon
Quote:

Originally Posted by kl.twilleger
ok, i tried that and i got $\displaystyle \frac{1-x^2}{1-x^2} + \frac{x^2}{1-x^2}$

which gave me $\displaystyle \sqrt{\frac{1}{1-x^2}}$

i'm not sure what i can do algebraically from here

$\displaystyle \int \frac 1{\sqrt{1 - x^2}}~dx$ is actually an integral you should know by heart. i bet it is on the formula page of your text. besides that, knowing the derivatives of your inverse trig functions should help

Quote:

and i can't get that stuff to work for some reason
you're supposed to type them between  tags. click the $\displaystyle \sum$ buttom to insert them to avoid typing the tags out
• Sep 24th 2008, 05:27 PM
kl.twilleger
Ok, so I looked that identity online and found that it's equal to arcsin(x). So when I plugged it in the got 3.1415... Is that right???
• Sep 24th 2008, 05:51 PM
Jhevon
Quote:

Originally Posted by kl.twilleger
Ok, so I looked that identity online and found that it's equal to arcsin(x). So when I plugged it in the got 3.1415... Is that right???

leave it as arcisn(1)

so in the end it is arcsin(1) - arcsin(-1) = 2 arcsin(1)