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Math Help - [SOLVED] Finding Arc length

  1. #1
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    [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!!
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    Quote Originally Posted by kl.twilleger View Post
    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 \int_{-1}^1 \sqrt{1 + (-2x)^2 }~dx = \int_{-1}^1 \sqrt{1 + 4x^2}~dx
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  3. #3
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    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.
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    Quote Originally Posted by kl.twilleger View Post
    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

    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
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    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
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    Quote Originally Posted by kl.twilleger View Post
    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
    \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


    and i can't get that stuff to work for some reason
    you're supposed to type them between [tex][/tex] tags. click the \sum buttom to insert them to avoid typing the tags out
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    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???
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    Quote Originally Posted by kl.twilleger View Post
    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)
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