# Math Help - Arc length question

1. ## 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

2. Originally Posted by bobak
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:

$\sqrt{1 + \left( \frac{dy}{dx}\right)^2 } = \frac{1}{2} \, \frac{(x + a)}{\sqrt{ax}} = \frac{1}{2} \, \left( \frac{x}{\sqrt{ax}} + \frac{a}{\sqrt{ax}} \right) = ......$