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Math Help - Arc length question

  1. #1
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    Arc length question

    A curve has equation 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 x = 4a up to x = 9a is given by

    \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 ( 0 , 0 ) and (3a , 0 ) and a turning point \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; March 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 ( 0 , 0 ) and (3a , 0 ) and a turning point \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
    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 y = \frac{\sqrt{x^3 + 6ax^2 + 9a^2x}}{3 \sqrt{a}}.

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


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


    \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:


    \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; March 20th 2008 at 04:36 AM.
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