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Math Help - Length of the Curve

  1. #1
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    Length of the Curve

    Find the Length of The Curve
    y = ln(1 - x2)

    0 < x < (1/7)

    I got the derivative: (-2x)/(1-x^2)

    Then i put it in the formula : sqr*(1+((-2x)/(1-x^2))^2)

    My problem seems to be simplifying the terms to cancel the sqr* before integrating.
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  2. #2
    Super Member Showcase_22's Avatar
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    y=ln(1-x^2) \Rightarrow \ \frac{dy}{dx}=\frac{-2x}{1-x^2}

    \int \sqrt{1+\left( \frac{-2x}{1-x^2} \right)^2}dx

    I'll leave you to put in the limits.

    I'm in a bit of a rush, but have you tried integrating it using \frac{-2x}{1-x^2}=sinh(u)?
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  3. #3
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    0 < x < (1/7)
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  4. #4
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    i dont really see how you use a hyperbolic function to make it possible to integrate
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  5. #5
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    Hello, NickytheNine!

    It seems you need to brush up on your Algebra . . .


    Find the length of the curve: . y\: = \:\ln(1 - x^2),\quad 0 \leq x \leq \tfrac{1}{7}

    I got the derivative: . \frac{dy}{dx} \:=\:\frac{-2x}{1-x^2}

    Then i put it in the formula: . \sqrt{1+\left(\frac{-2x}{1-x^2}\right)^2}

    My problem seems to be simplifying the terms . . .

    Under the radical, we have: . 1 + \frac{4x^2}{(1-x^2)^2} \;=\;\frac{(1-x^2)^2 + 4x^2}{(1-x^2)^2}

    . . . . . = \;\frac{1 - 2x^2 + x^4 + 4x^2}{(1-x^2)^2} \;=\;\frac{1 + 2x^2 + x^4}{(1-x^2)^2} \;=\;\frac{(1+x^2)^2}{(1-x^2)^2}


    Then: . \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \;=\;\sqrt{\frac{(1+x^2)^2}{(1-x^2)^2}} \;=\;\frac{1+x^2}{1-x^2}


    And we must integrate and evaluate: . \int^{\frac{1}{7}}_0 \frac{1+x^2}{1-x^2}\,dx


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    I recommend long division: . \int^{\frac{1}{7}}_0\left(-1 + \frac{2}{1-x^2}\right)\,dx

    Got it?

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  6. #6
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    thank you very much, i just got to work on my algebra i guess lol
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