# Folium of Descartes

1. Sep 6, 2007

### a_Vatar

I'm trying to attack a question on folium of Descartes from J.Stewart Calculus book.
Here are the ones I've been trying so far and got stuck:

btw parametric equations are:
x=$$\frac{3t}{1+t^{3}}$$

y=$$\frac{3t^{2}}{1+t^{3}}$$

(a) Show that if (a, b) lies on the curve, then so does (b, a); that is, the curve is symmetric
with respect to the line y=x...
(c) Show that the line y=-x - 1 is a slant asymptote.
(e) Show that a Cartesian equation of this curve is x$$^{3}$$ + y$$^{3}$$ = 3xy.

A:
I suppose that if (a, b) is on the curve, then we should find a parameter t such that it
solves the following system:
$$\frac{3t}{1+t^{3}}$$ = b
$$\frac{3t^{2}}{1+t^{3}}$$ = a

Adding these up then expanding (1 + t$$^{3}$$) and canceling
I end up with a quadratic equation in this form:
t1,2 = $$\frac{-(3+a+b)+/-\sqrt{(3 + a + b)^{2} - 4(a+b)^{2}}}{2(a+b)}$$

I checked this up for a value of t=3 giving me some values for a and b, one of the roots is 1/3 which gives a point (b, a); But what is the meaning of the second root? I tried to constraint the discriminator but it does not make much sense to impose limitations on value a+b, cause they should lie on the curve, but that does not mean that for every point(a,b) the equation will have a solution.
Am I doing something wrong?

C:
Completely lost here; I tried looking at the limits of both parametric equations as t goes to -$$\infty$$ and +$$\infty$$, but they approach 0 from either left or right, which can be right...Thou I'd expect both limits to be $$\infty$$ as a point approaches the asymptote, what values of t should I look at then; or would it be right look at the ration of both limits? e.g how faster y increases compared to x...

E:
Similar story; From what I've seen before, the way to solve is to eliminate the parameter from one of the equations... I've tried different algebraic manipulations with no success, probably there's another approach or some "trick" :)

I'm trying to crack this task for 2 nights already and it start to become frustrating that I can't really solve most of the questions in it/\

2. Sep 6, 2007

### Emmanuel114

Hints: For the first, $$(a,b)$$ corresponds to a point $$t_1$$.
$$(b,a)$$ corresponds to a point $$t_2$$.

For the second, consider how $$x\rightarrow-\infty$$ and $$y\rightarrow\infty$$.

3. Sep 6, 2007

### ansrivas

a) Let (x0, y0) be a point on the curve. Then there exists t0 such that
$$x_0=\frac{3t_0}{1+t_0^{3}}$$
$$y_0=\frac{3t_0^{2}}{1+t_0^{3}}$$

Now what's the point for t=1/t0 .. that is
$$x_1=\frac{3/t_0}{1+1/t_0^{3}}=y_0$$
$$y_1=\frac{3/t_0^2}{1+1/t_0^{3}}=x_0$$

E) Solve for t by dividing the two equations and then plug back t into either of the two equations.

4. Sep 7, 2007

### a_Vatar

Emmanuel114, I got the expression for t2 ( $$\frac{-(3+a+b)+/-\sqrt{(3 + a + b)^{2} - 4(a+b)^{2}}}{2(a+b)}$$) and one of the roots appear to be 1/t0 as ansrivas
mentions.

ansrivas, how did you guess expression for t of a point (y0, x0)? It is not evident from when you just look at the equations.

E) Solve for t by dividing the two equations and then plug back t into either of the two equations. - I'll try that, thanks! Did you know the answer before, or did you see straight away? That always stunned me how in some derivations and proofs some tricks are applied and viola - it all works out!

Last edited: Sep 7, 2007
5. Sep 7, 2007

### cks

y=-x-1,

we can see that as the line approaches this asymptote, the gradient of the line will be the same as the gradient of this asymptote.

Looking for dy/dx of the folium of Descartes, and looking for the limit of dy/dx where y->
-x-1, then dy/dx =-1,

this verifies.

6. Sep 10, 2007

### a_Vatar

Thanks for help guys! Finally got the trick of dividing one equation the other :)
This yields an expression for t = 1/t0 for A)