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Math Help - Stuck: Arc Length

  1. #1
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    Stuck: Arc Length

    Hello. I need help with a problem.

    The problem is: Show that the circumference of the unit circle is equal to

    2 (integral, 1, -1) dx/sqrt(1-x^2) (an improper integral). Evaluate, thus verifying that the circumference is 2pi.

    This is how far I've gotten:

    x^2 + y^2 = 1 --> y = sqrt(1-x^2)

    y = sqrt(1-x^2)
    y'= -x/sqrt(1-x^2)

    (y')^2 + 1 = (-x/sqrt(1-x^2))^2 + 1

    = (x^2)/(1-x^2) + 1

    Arclength = s = (integral, 1, -1) sqrt(1 + (x^2)/(1-x^2)) dx

    = (integral, 1, -1) sqrt(1/(1-x^2)) dx

    This is where I am stuck. Thanks!
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  2. #2
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    Quote Originally Posted by wyhwang7 View Post
    Hello. I need help with a problem.

    The problem is: Show that the circumference of the unit circle is equal to

    2 (integral, 1, -1) dx/sqrt(1-x^2) (an improper integral). Evaluate, thus verifying that the circumference is 2pi.

    This is how far I've gotten:

    x^2 + y^2 = 1 --> y = sqrt(1-x^2)

    y = sqrt(1-x^2)
    y'= -x/sqrt(1-x^2)

    (y')^2 + 1 = (-x/sqrt(1-x^2))^2 + 1

    = (x^2)/(1-x^2) + 1

    Arclength = s = (integral, 1, -1) sqrt(1 + (x^2)/(1-x^2)) dx

    = (integral, 1, -1) sqrt(1/(1-x^2)) dx

    This is where I am stuck. Thanks!
    \sqrt{\frac{1}{1 - x^2}} = \frac{1}{\sqrt{1 - x^2}} and I hope you now recognise the integrand as a standard form.
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