We are looking for the length of a curve, also known as the arc length. Before we get to the formula for arc length, it would help if we re-wrote the equation in y = form.
We are given:
![36 y^{2} =( x^{2} -4)^3](https://tex.z-dn.net/?f=36%20y%5E%7B2%7D%20%3D%28%20x%5E%7B2%7D%20-4%29%5E3%20)
We divide by 36 and take the root of both sides to obtain:
![y = \sqrt{ \frac{( x^{2} -4)^3}{36} }](https://tex.z-dn.net/?f=y%20%3D%20%5Csqrt%7B%20%5Cfrac%7B%28%20x%5E%7B2%7D%20-4%29%5E3%7D%7B36%7D%20%7D%20)
Note that the square root can be written as an exponent of 1/2 and so we can further simplify the above to obtain:
![y = \frac{( x^{2} -4)^{3/2}}{6} }=( \frac{1}{6} )(x^{2} -4)^{3/2}}](https://tex.z-dn.net/?f=y%20%3D%20%20%5Cfrac%7B%28%20x%5E%7B2%7D%20-4%29%5E%7B3%2F2%7D%7D%7B6%7D%20%7D%3D%28%20%5Cfrac%7B1%7D%7B6%7D%20%29%28x%5E%7B2%7D%20-4%29%5E%7B3%2F2%7D%7D)
Let's leave that for the moment and look at the formula for arc length. The formula is
![L= \int\limits^c_d {ds}](https://tex.z-dn.net/?f=L%3D%20%5Cint%5Climits%5Ec_d%20%7Bds%7D)
where ds is defined differently for equations in rectangular form (which is what we have), polar form or parametric form.
Rectangular form is an equation using x and y where one variable is defined in terms of the other. We have y in terms of x. For this, we define ds as follows:
![ds= \sqrt{1+( \frac{dy}{dx})^2 } dx](https://tex.z-dn.net/?f=ds%3D%20%5Csqrt%7B1%2B%28%20%5Cfrac%7Bdy%7D%7Bdx%7D%29%5E2%20%7D%20dx%20)
As a note for a function x in terms of y simply switch each dx in the above to dy and vice versa.
As you can see from the formula we need to find dy/dx and square it. Let's do that now.
We can use the chain rule: bring down the 3/2, keep the parenthesis, raise it to the 3/2 - 1 and then take the derivative of what's inside (here
![x^2-4](https://tex.z-dn.net/?f=x%5E2-4)
). More formally, we can let
![u=x^{2} -4](https://tex.z-dn.net/?f=u%3Dx%5E%7B2%7D%20-4)
and then consider the derivative of
![u^{3/2}du](https://tex.z-dn.net/?f=u%5E%7B3%2F2%7Ddu)
. Either way, we obtain,
![\frac{dy}{dx}=( \frac{1}{6})( x^{2} -4)^{1/2}(2x)=( \frac{x}{2})( x^{2} -4)^{1/2}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bdy%7D%7Bdx%7D%3D%28%20%5Cfrac%7B1%7D%7B6%7D%29%28%20x%5E%7B2%7D%20-4%29%5E%7B1%2F2%7D%282x%29%3D%28%20%5Cfrac%7Bx%7D%7B2%7D%29%28%20x%5E%7B2%7D%20-4%29%5E%7B1%2F2%7D%20%20%20)
Looking at the formula for ds you see that dy/dx is squared so let's square the dy/dx we just found.
![( \frac{dy}{dx}^2)=( \frac{x^2}{4})( x^{2} -4)= \frac{x^4-4 x^{2} }{4}](https://tex.z-dn.net/?f=%28%20%5Cfrac%7Bdy%7D%7Bdx%7D%5E2%29%3D%28%20%5Cfrac%7Bx%5E2%7D%7B4%7D%29%28%20x%5E%7B2%7D%20-4%29%3D%20%5Cfrac%7Bx%5E4-4%20x%5E%7B2%7D%20%7D%7B4%7D%20%20%20)
This means that in our case:
![ds= \sqrt{1+\frac{x^4-4 x^{2} }{4}} dx](https://tex.z-dn.net/?f=ds%3D%20%5Csqrt%7B1%2B%5Cfrac%7Bx%5E4-4%20x%5E%7B2%7D%20%7D%7B4%7D%7D%20dx)
![ds= \sqrt{\frac{4}{4}+\frac{x^4-4 x^{2} }{4}} dx](https://tex.z-dn.net/?f=ds%3D%20%5Csqrt%7B%5Cfrac%7B4%7D%7B4%7D%2B%5Cfrac%7Bx%5E4-4%20x%5E%7B2%7D%20%7D%7B4%7D%7D%20dx)
![ds= \sqrt{\frac{x^4-4 x^{2}+4 }{4}} dx](https://tex.z-dn.net/?f=ds%3D%20%5Csqrt%7B%5Cfrac%7Bx%5E4-4%20x%5E%7B2%7D%2B4%20%7D%7B4%7D%7D%20dx)
![ds= \sqrt{\frac{( x^{2} -2)^2 }{4}} dx](https://tex.z-dn.net/?f=ds%3D%20%5Csqrt%7B%5Cfrac%7B%28%20x%5E%7B2%7D%20-2%29%5E2%20%7D%7B4%7D%7D%20dx)
![ds= \frac{x^2-2}{2}dx =( \frac{1}{2} x^{2} -1)dx](https://tex.z-dn.net/?f=ds%3D%20%20%5Cfrac%7Bx%5E2-2%7D%7B2%7Ddx%20%3D%28%20%5Cfrac%7B1%7D%7B2%7D%20x%5E%7B2%7D%20-1%29dx%20)
Recall, the formula for arc length:
![L= \int\limits^c_d {ds}](https://tex.z-dn.net/?f=L%3D%20%5Cint%5Climits%5Ec_d%20%7Bds%7D)
Here, the limits of integration are given by 5 and 9 from the initial problem (the values of x over which we are computing the length of the curve). Putting it all together we have:
![L= \int\limits^9_5 { \frac{1}{2} x^{2} -1 } \, dx = (\frac{1}{2}) ( \frac{x^3}{3}) -x](https://tex.z-dn.net/?f=L%3D%20%5Cint%5Climits%5E9_5%20%7B%20%5Cfrac%7B1%7D%7B2%7D%20x%5E%7B2%7D%20-1%20%7D%20%5C%2C%20dx%20%3D%20%28%5Cfrac%7B1%7D%7B2%7D%29%20%28%20%5Cfrac%7Bx%5E3%7D%7B3%7D%29%20-x)
evaluated from 9 to 5 (I cannot seem to get the notation here but usually it is a straight line with the 9 up top and the 5 on the bottom -- just like the integral with the 9 and 5 but a straight line instead). This means we plug 9 into the expression and from that subtract what we get when we plug 5 into the expression.
That is,