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 February 16th, 2011, 10:07 PM #1 Member   Joined: Jan 2011 Posts: 30 Thanks: 0 The length of a curve with logaritm (Natural) Hi guys the problem is to get the length of the curve y= ln(x^2-1) 0<= x <= 1/2 As far as I am concerned we can use a formula L = \$(from a to b) SQRT [1+ f'(x)] dx I don't know how to get the derivative of the function, could someone help me complete this task
 February 16th, 2011, 10:37 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: The length of a curve with logaritm (Natural) In your formula, you need to square f'(x) in the radicand of the integrand. $\frac{d}{dx}$$\ln\(x^2-1$$\)=\frac{2x}{x^2-1}$ While the integral will return a real value, is it meaningful? The curve is not defined on [-1,1], although W|A shows it has a real and imaginary part on this interval. It looks like the same result would be returned by using $y=\ln$$1-x^2$$$ which is defined on (-1,1). Also confusing to me is that W|A shows the same graph for both functions.

 Tags curve, length, logaritm, natural

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