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!!

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- Sep 24th 2008, 04:16 PMkl.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 PMJhevon
- Sep 24th 2008, 04:46 PMkl.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 PMJhevon
that's right. (you neglected to mention the square root before!)

hint: try combining the fractions

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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, 05:01 PMkl.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 PMJhevon
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

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and i can't get that stuff to work for some reason

- Sep 24th 2008, 05:27 PMkl.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 PMJhevon