| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 829.1 | Maybe working on the wrong problem? | AKQJ10::YARBROUGH | Why is computing so labor intensive? | Tue Feb 23 1988 12:24 | 14 |
| According to Bronshtein and Semendyayev's *Handbook of Mathematics*,
the Folium of Descartes has the equation
x^3 + y^3 = 3axy,
is asymptotic to a line through (-a,0) and (0,-a) and has a loop which
passes through (0, 0).
It may be easier to compute the area from the parametric form of the
equation,
x = 3at /(1+t**3),
y = 3at**2/(1+t**3)
for -inf<t<-1 and -1<t<inf. The loop corresponds to 0<t<inf.
|
| 829.2 | | CLT::GILBERT | Builder | Tue Feb 23 1988 13:31 | 30 |
| >It may be easier to compute the area from the parametric form of the
>equation,
> x = 3at /(1+t**3),
> y = 3at**2/(1+t**3)
> for -inf<t<-1 and -1<t<inf. The loop corresponds to 0<t<inf.
Sure. Note that 0<t<1 and the line x=y bound half the loop. Thus the
area of the loop is:
1
Area = 2 Integral (x-y) dx
t=0
Now, dx/dt = 3a (1 - 2t^3) / (1+t^3)^2
and y - x = 3at(1-t)/(1+t^3)
So,
1
Area = 18 a^2 Integral t(1-t)(1 - 2t^3) / (1+t^3)^3 dt
t=0
And that integral equals:
- 4 t^3 + 3 t^2 - 1
-------------------
6 (1+t^3)^2
So the area is:
Area = 18 a^2 ( -2/24 - -1/6 ) = 3 a^2 / 2
|
| 829.3 | Thanks, correct solution | COMICS::DEMORGAN | Richard De Morgan, UK CSC/CS | Wed Feb 24 1988 03:46 | 3 |
| Thanks .1, .2. I had mis-remembered the formula. The anser in .2
is correct, but I have seen a 4 line proof which I can't remember
- I think it either used polar or (s, phi) coordinates.
|
| 829.4 | | PSW::WINALSKI | Paul S. Winalski | Mon Mar 07 1988 22:55 | 4 |
| Stan the TECO macro Rabinowitz tells me that the area between the folium
and the asymptote is also 3a**2/2.
--PSW
|