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!!
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.
that's right. (you neglected to mention the square root before!)
hint: try combining the fractions
see hereand 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.
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
you're supposed to type them between [tex][/tex] tags. click the buttom to insert them to avoid typing the tags outand i can't get that stuff to work for some reason