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The folium of descartes is defined by $x^3+y^3-3xy=0$ (cartesian), $r=\frac{3\sec \theta \tan \theta}{1+\tan^3 \theta}$ (polar) or $x=\frac{3at}{1+t^3}, y=\frac{3at^2}{1+t^3}$ (parametric).

It looks like this and has the slant asymptote $y=-x-1$. I'm trying to show that the area between the asymptote and the infinite branches of the curve is the same as the area of the loop. I already know the area of the loop, it is $1.5$ units.

I managed to show that the area between the asymptote and the branches is also $1.5$ units, but the method I used was last-resort inelegant** (I, or more specifically wolfram-alpha, solved the cartesian equation for $y$ and then the resultant integral).

Are there different ways to show that the asymptote-branches area is 1.5 units? Showing that its area is the same as the area of the loop is fine as well, but probably harder/needing a special trick.

Thank you!

** The problem source says about this problem: "Use a CAS to evaluate the resultant integral" so I didn't feel too guilty about making WA do it. The problem source is James Stewart Calculus, in the polar and parametric equations chapter.

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Introduce new coordinates $\xi$, $\eta$ such that the asymptote becomes the $\xi$-axis: $$\xi={1\over\sqrt{2}}(x-y),\quad \eta={1\over\sqrt{2}}(x+y+1)\ .$$ (This amounts to a $45^\circ$ rotation and a shift in the $\eta$ direction.) The parametric representation of the folium then is $$\xi(t)={1\over\sqrt{2}}{3(t-t^2)\over1+t^3},\quad\eta={1\over\sqrt{2}}\left({3(t+t^2)\over1+t^3}+1\right)\ .\tag{1}$$ The branch $\gamma$ of the folium going from $\bigl(0,{1\over\sqrt{2}}\bigr)$ to $(+\infty,0)$ can now be viewed as a graph over the $\xi$-axis, albeit not parametrized by $\xi$. Instead it is parametrized by $(1)$ with the $t$-interval $-\infty<t<-1$. In order to find the area between $\gamma$ and the $\xi$-axis we have to compute $$\int_\gamma \eta\>d\xi=\int_{-\infty}^{-1}\eta(t)\,\xi'(t)\>dt=\int_{-\infty}^{-1}{3(1-2t-2t^3+t^4)\over2(1-t+t^2)^3}\>dt={3\over4}\ ,$$ which is half the desired area. (I have used Mathematica to obtain the final value.)

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  • $\begingroup$ What maths is this? $\endgroup$ – Lug Gian Feb 1 '18 at 23:20
  • $\begingroup$ I saw something similar, where the curve is rotated by $\theta=5\pi/4$ and $a\mapsto \sqrt{2}a$ to get the parametric equations $x=3a\frac{t^2-1}{3t^2+1},~y=xt$. $\endgroup$ – Cye Waldman Feb 2 '18 at 20:20
  • $\begingroup$ --great answer . $\endgroup$ – Vicrobot Nov 19 '18 at 19:07

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