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Thread: Lengths of Curves

  1. #1
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    Lengths of Curves

    Hello everyone. Much to my chagrin, I am having an algebraic problem. Here is the question:

    Question:

    Find the length of the curve analytically by anti-differentiation. You will need to simplify the integrand algebraically before finding an anti-derivative.

    $\displaystyle y = \frac{x^3}{3} + x^2 + x + \frac{1}{4x + 4}, 0 \leq x \leq 2 $

    Solution:

    $\displaystyle L = \int_{a}^{b} \sqrt{1 + (\frac{dy}{dx})^2} dx $

    $\displaystyle \frac{dy}{dx} = x^2 + 2x + 1 - \frac{1}{4(x + 1)^2} = (x+1)^2 - \frac{1}{4(x+1)^2} $

    $\displaystyle L = \int_{0}^{2} \sqrt{1 + ( (x+1)^2 - \frac{1}{4(x+1)^2} )^2} $

    I understand everything up to this point. In the answer, this is what it does after the prior step:

    $\displaystyle = \int_{0}^{2} \sqrt{( (x+1)^2 + \frac{1}{4(x+1)^2} )^2} $

    What happened to the 1? What happened to the negative sign? Any help is appreciated. Thanks in advance.
    Last edited by lilaziz1; Apr 3rd 2010 at 05:09 PM.
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  2. #2
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    Sorry I get

    $\displaystyle \frac{dy}{dx}$ = $\displaystyle \frac{x^2}{2}+2x+1-\frac{1}{4{(x+1)}^2}$
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  3. #3
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    Oh. Sorry. I mistyped. :s ment to write x^3/3
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by lilaziz1 View Post
    $\displaystyle L = \int_{0}^{2} \sqrt{1 + ( (x+1)^2 - \frac{1}{4(x+1)^2} )^2} $

    I understand everything up to this point. In the answer, this is what it does after the prior step:

    $\displaystyle = \int_{0}^{2} \sqrt{( (x+1)^2 + \frac{1}{4(x+1)^2} )^2} $

    What happened to the 1? What happened to the negative sign? Any help is appreciated. Thanks in advance.
    $\displaystyle 1 + ( (x+1)^2 - \frac{1}{4(x+1)^2} )^2 = 1+(x+1)^4-2\frac{(x+1)^2}{4(x+1)^2}+\frac{1}{16(x+1)^4} $

    $\displaystyle = 1+(x+1)^4-\frac12+\frac{1}{16(x+1)^4} =(x+1)^4+\frac12+\frac{1}{16(x+1)^4} $

    $\displaystyle = (x+1)^4+2\frac{(x+1)^2}{4(x+1)^2}+\frac{1}{16(x+1) ^4} = \left((x+1)^2+\frac{1}{4(x+1)^2}\right)^2 $
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  5. #5
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    :O thx!
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