Set up an integral for the arc length

Question

I am struggling with a task.

The equation$$(x-y)^2 + 4(x+y)^2 = 1$$represent a conic section and I have to set up an integral for the arc length og the conic section. But I am not sure how to set it up. Im not sure but I think the arc length formula is

$$S = \int \sqrt{(1 + (f’(x))^2}  \,\mathrm{d}x,$$

and this is where I am confused. Am I thinking right? If I am, then $f'(x) = 20x+6$ and all I have to do is to put $f'(x)$ in the formula.

Answer 1

The locus of $(x,y)\in\mathbb{R}^2$ such that $$(x-y)^2+4(x+y)^2 = 1 $$ is isometric to the locus of $(x,y)\in\mathbb{R}^2$ such that $$ 2x^2+8y^2 = 1 $$ i.e. to an ellipse with semi-axis lenghts $\frac{1}{\sqrt{2}}$ and $\frac{1}{2\sqrt{2}}$.
To compute its perimeter it is more practical to employ the formula for the length of parametric curves (rather than the formula for the length of a graph, since an ellipse is not globally a graph) and symmetry: $$ L = 4\int_{0}^{\pi/2}\sqrt{\frac{1}{2}\sin^2\theta+\frac{1}{8}\cos^2\theta}\,d\theta=2\sqrt{2}\int_{0}^{\pi/2}\sqrt{1-\frac{3}{4}\cos^2\theta}\,d\theta =\sqrt{2}\int_{0}^{1}\sqrt{\frac{4-3t^2}{1-t^2}}\,dt$$ This is a complete elliptic integral of the second kind. The numerical evaluation of such objects is an interesting topic, I suggest you to have a look at the dedicated section in my notes (elliptic integrals and the AGM). A practical approximation, which is also pretty accurate for ellipses with a not too large eccentricity, is the following one (due to Muir):

The perimeter of an ellipse with semi-axis lengths $a,b$ is approximately $$ 2\pi\left(\frac{a^{3/2}+b^{3/2}}{2}\right)^{2/3}.$$

In your case Muir's approximation returns a $\color{green}{3.42}475\ldots$ while the actual perimeter is $\color{green}{3.425383717962}\ldots$. The error is less than 2 parts in 10 thousands.