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Thread: Arc length question

  1. #1
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    Arc length question

    A curve has equation $\displaystyle 9ay^2= x(3a-x)^2$

    where a is a positive constant

    a) Sketch a graph of the curve and showing stationary points and intersection with the axes

    b) show that the arc length of C in the first quadrant form the point $\displaystyle x = 4a$ up to $\displaystyle x = 9a$ is given by

    $\displaystyle \frac{1}{2} \int_{4a}^{9a}{\left[ \left( \frac{a}{x} \right )^{\frac{1}{2}} + \left( \frac{x}{a} \right )^{\frac{1}{2}} \right] dx}$
    I have no problems with drawing the graph. I got that function is symmetrical about the x axis, roots at $\displaystyle ( 0 , 0 )$ and $\displaystyle (3a , 0 )$ and a turning point $\displaystyle \left( a , \pm \frac{2}{3} a \right)$ you have a closed loop between 0 to 3a and then y just increases with x increasing.

    My only problem is the arc length part. I could do with a pointer.

    Thanks Bobak
    Last edited by bobak; Mar 20th 2008 at 03:59 AM. Reason: fixed a few typos
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  2. #2
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    Quote Originally Posted by bobak View Post
    I have no problems with drawing the graph. I got that function is symmetrical about the x axis, roots at $\displaystyle ( 0 , 0 )$ and $\displaystyle (3a , 0 )$ and a turning point $\displaystyle \left( a , \pm \frac{2}{3} a \right)$ you have a closed loop between 0 to 3a and then y just increases with x increasing.

    My only problem is the arc length part. I could do with a pointer.

    Thanks Bobak
    $\displaystyle L = \int_{x_1}^{x_2} \sqrt{1 + \left( \frac{dy}{dx}\right)^2 } \, dx$.

    In the positive quadrant the rule can be written as $\displaystyle y = \frac{\sqrt{x^3 + 6ax^2 + 9a^2x}}{3 \sqrt{a}}$.

    After (quite) a bit of algebra, you have:


    $\displaystyle \frac{dy}{dx} = \frac{1}{6\sqrt{a}} \, \frac{3x^2 - 12ax + 9a^2}{\sqrt{x^3 - 6ax^2 + 9a^2x}}$


    $\displaystyle \Rightarrow \left( \frac{dy}{dx} \right)^2 = \frac{(x-a)^2}{4 a x} \Rightarrow 1 + \left( \frac{dy}{dx}\right)^2 = \frac{(x+a)^2}{4ax}$.


    Then:


    $\displaystyle \sqrt{1 + \left( \frac{dy}{dx}\right)^2 } = \frac{1}{2} \, \frac{(x + a)}{\sqrt{ax}} = \frac{1}{2} \, \left( \frac{x}{\sqrt{ax}} + \frac{a}{\sqrt{ax}} \right) = ......$
    Last edited by mr fantastic; Mar 20th 2008 at 04:36 AM.
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