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 January 4th, 2011, 07:22 AM #1 Newbie   Joined: Dec 2010 Posts: 11 Thanks: 0 Length of a curve I have a task to calculate the length of the curve. Does anyone know outcome? I big idiot(( http://www.sdilej.eu/pics/ee9c6c683321b ... 9b4eea.jpg Could anyone help me please?
 January 4th, 2011, 08:33 AM #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: Length of a curve You are asked to find the arc length on $y=\frac{1}{3}(x-3)\sqrt{x}$ from x = 0 to x = 3. I would write $y=\frac{1}{3}$$x^{\small{\frac{3}{2}}}-3x^{\small{\frac{1}{2}}}$$$ thus $y'(x)=\frac{1}{3}$$\frac{3}{2}x^{\small{\frac{ 1}{2}}}-\frac{3}{2}x^{\small{-\frac{1}{2}}}$$=\frac{1}{2}x^{\small{\frac{1}{2}}}-\frac{1}{2}x^{\small{-\frac{1}{2}}}=\frac{x-1}{2x^{\small{\frac{1}{2}}}}$ We know the integrand will have as the radicand: $1+$$y'(x)$$^2=1+$$\frac{x-1}{2x^{\small{\frac{1}{2}}}}$$^2=\frac{4x+(x-1)^2}{4x}=\frac{4x+x^2-2x+1}{4x}=$$\frac{x+1}{2x^{\small{\frac{1}{2}}}}$$ ^2$ Since this expression is non-negative on the given interval, the arc length s may be written:  Because the integrand is unbounded at x = 0, we have an improper integral and should use:  
 January 7th, 2011, 08:22 AM #3 Newbie   Joined: Dec 2010 Posts: 11 Thanks: 0 Re: Length of a curve thanks )

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