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Math Help - Perfect squares for arc lenght

  1. #1
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    Perfect squares for arc lenght

    hi, I'm having problems recognizing what is a perfect square and what is not. Could anyone share so basic stuff i need to know about them , i'm very lost with them and I have no idea what to do with them. For exemple : \frac{5x^4}{6}-\frac{3x^{-4}}{10}

    Can i get a full demonstration of why this is a perfect square and tip to recognize it thanks alot
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  2. #2
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    Hello, Larrioto!

    I'm having problems recognizing what is a perfect square and what is not.

    For example: . \frac{5x^4}{6}-\frac{3x^{-4}}{10}

    Can i get a full demonstration of why this is a perfect square and tip to recognize it?
    First of all, it is not a perfect square . . . Let's start at the very beginning.


    We have a function: . y \;=\;\frac{1}{6}x^5 + \frac{1}{10}x^{\:\!\text{-}3} .
    . . . not a perfect square
    . . and we want its arc length over some interval (a,b).

    We use the formula: . L \;=\;\int^b_a\sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx

    So we form the expession under the radical . . .


    We have: . \frac{dy}{dx} \;=\;\frac{5}{6}x^4 - \frac{3}{10}x^{\:\!\text{-}4} .
    . . . not a perfect square

    Then: . \left(\frac{dy}{dx}\right)^2 \;=\;\left(\frac{5}{6}x^4 - \frac{3}{10}x^{\:\!\text{-}4}\right)^2  \;=\;\frac{25}{36}x^8 - \frac{1}{2} + \frac{9}{100}x^{\:\!\text{-}8}

    \text{And: }\;1 + \left(\frac{dy}{dx}\right)^2 \;=\;1 + \frac{25}{36}x^8 - \frac{1}{2} + \frac{9}{100}x^{\;\!\text{-}8} \;=\;\underbrace{\frac{25}{36}x^8 + \frac{1}{2} + \frac{9}{100}x^{\;\!\text{-}8}}_{{\color{blue}\text{This is a perfect square!}}} . =\;\left(\frac{5}{6}x^4 + \frac{3}{10}x^{\;\!\text{-}4}\right)^2

    Hence: . \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \;=\;\sqrt{\left(\frac{5}{6}x^4 + \frac{3}{10}x^{\;\!\text{-}4}\right)^2} \;=\;\frac{5}{6}x^4 + \frac{3}{10}x^{\;\!\text{-}4}


    Therefore: . L \;=\;\int^b_a\left(\frac{5}{6}x^4 + \frac{3}{10}x^{\;\!\text{-}4}\right)\,dx


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

    So they gave us a rather ugly function,
    . . but it produces a simple integral.

    I've learned to expect this in an Arc Length problem.
    But, as far as I know, there's no way to recognize it in advance.

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