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Folium of Descartes

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Algebraic curve of the form x³ + y³ – 3axy = 0

The folium of Descartes (green) with asymptote (blue), when ${\textstyle a=1}$

The folium of Descartes (green) with asymptote (blue), when ${\textstyle a=1}$

Ingeometry, thefolium of Descartes (from Latin folium 'leaf'; named forRené Descartes) is analgebraic curve defined by theimplicit equation ${\begin{aligned}x^{3}+y^{3}-3axy&=0,\\3(x^{2}dx+y^{2}dy)-3a(xdy+ydx)&=0\\(x^{2}-ay)dx&=(ax-y^{2})dy,\\{\frac {dy}{dx}}&={\frac {x^{2}-ay}{ax-y^{2}}},\qquad {\frac {dx}{dy}}={\frac {ax-y^{2}}{x^{2}-ay}}.\end{aligned}}$

History

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René Descartes
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The curve was first proposed and studied byRené Descartes in 1638.^[1] Its claim to fame lies in an incident in the development ofcalculus. Descartes challengedPierre de Fermat to find the tangent line to the curve at an arbitrary point, since Fermat had recently discovered a method for finding tangent lines. Fermat solved the problem easily, something Descartes was unable to do.^[2] Since the invention of calculus, the slope of the tangent line can be found easily usingimplicit differentiation.^[3] MayorJohan(nes) Hudde's second letter on maxima and minima (1658) mentions his calculation of the maximum width of the closed loop, part ofMathematical Exercitions, 5 books (1656/57 Leyden) p. 498, byFrans van Schooten Jnr.

Graphing the curve

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Folium of Descartes in polar coordinatesr-φ: ${\textstyle r(\phi )/a,\,a=1}$

Folium of Descartes in polar coordinatesr-φ: ${\textstyle r(\phi )/a,\,a=1}$

The folium of Descartes can be expressed inpolar coordinates as $r={\frac {3a\sin \theta \cos \theta }{\sin ^{3}\theta +\cos ^{3}\theta }},$ which is plotted on the left. This is equivalent to^[4]

$r={\frac {3a\sec \theta \tan \theta }{1+\tan ^{3}\theta }}.$

Another technique is to write $y=px$ and solve for $x {\displaystyle x}$ and $y {\displaystyle y}$ in terms of $p {\displaystyle p}$ . This yields therational parametric equations:^[5]

x={{3ap} \over {1+p^{3}}},\,y={{3ap^{2}} \over {1+p^{3}}}.

We can see that the parameter is related to the position on the curve as follows:

$p<-1$ corresponds to $x>0$ , $y<0$ : the right, lower, "wing".
$-1<p<0$ corresponds to $x<0$ , $y>0$ : the left, upper "wing".
$p>0$ corresponds to $x>0$ , $y>0$ : the loop of the curve.

Another way of plotting the function can be derived from symmetry over $y=x$ . The symmetry can be seen directly from its equation (x and y can be interchanged). By applying rotation of 45° clockwise for example, one can plot the function symmetric over rotated x axis.

This operation is equivalent to a substitution: $x={{u+v} \over {\sqrt {2}}},\,y={{u-v} \over {\sqrt {2}}}$ and yields $v=\pm u{\sqrt {\frac {3a{\sqrt {2}}-2u}{6u+3a{\sqrt {2}}}}}\,,\,u<3a/{\sqrt {2}}.$ Plotting in the Cartesian system of $(u,v)$ gives thefolium rotated by 45° and therefore symmetric by $u {\displaystyle u}$ -axis.

Properties

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It forms a loop in the first quadrant with adouble point at the origin and hasasymptote $x+y=-a\,.$ It is symmetrical about the line $y=x$ . As such, the curve and this line intersect at the origin and at the point ${\textstyle (3a/2,3a/2).}$

Implicit differentiation gives the formula for the slope of the tangent line to this curve to be^[3]

${\frac {dy}{dx}}={\frac {ay-x^{2}}{y^{2}-ax}}\,,$ with poles ${\textstyle x=y^{2}/a}$ and value 0 or ±∞ at origin (0,0).

Using either one of the polar representations above, the area of the interior of the loop is found to be ${\textstyle 1{\frac {1}{2}}a\cdot a.}$ Moreover, the area between the "wings" of the curve and its slanted asymptote is also ${\textstyle 3a^{2}/2.}$ ^[1]

Relationship to the trisectrix ofMaclaurin

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Trisectrix of Maclaurin: top angle is⁠2/3⁠ φ. Length⁠1+1/2⁠a on hor. axis,O to turning point.

The folium of Descartes is related to thetrisectrix of Maclaurin byaffine transformation. To see this, start with the equation $x^{3}+y^{3}=3a\cdot xy\,,$ and change variables to find the equation in a coordinate system rotated 45 degrees. This amounts to setting

$x={{X+Y} \over {\sqrt {2}}},y={{X-Y} \over {\sqrt {2}}}.$ In the $X, Y {\displaystyle X,Y}$ plane the equation is $2X(X^{2}+3Y^{2})=3{\sqrt {2}}a(X^{2}-Y^{2}).$

If we stretch the curve in the $Y {\displaystyle Y}$ direction by a factor of ${\sqrt {3}}$ this becomes $2X(X^{2}+Y^{2})=a{\sqrt {2}}(3X^{2}-Y^{2}),$ which is the equation of the trisectrix of Maclaurin.

Notes

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^^a ^b"Folium of Descartes".Encyclopedia of Mathematics. June 5, 2020. RetrievedJanuary 30, 2021.
^Simmons, p. 101
^^a ^bStewart, James (2012). "S. 3.5: Implicit Differentiation".Calculus: Early Transcendentals (7th ed.). United States of America: c-Engage Learning. pp. 209–11.ISBN 978-0-538-49790-9.
^Stewart, James (2012). "Ch. 10: Parametric Equations and Polar Coordinates".Calculus: Early Transcendentals (7th ed.). USA: C-engage Learning. p. 687.ISBN 978-0-538-49790-9.
^Wildberger, N.J. (4 August 2013)."DiffGeom3: Parametrized curves and algebraic curves".www.youtube.com.University of New South Wales.Archived from the original on 2021-12-21. Retrieved5 September 2013.

References

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J. Dennis Lawrence:A catalog of special plane curves. 1972, Dover Publications.ISBN 0-486-60288-5 pp. 106–108
George F. Simmons:Calculus Gems: Brief Lives and Memorable Mathematics. 1992, New York: McGraw-Hill.ISBN 0-07-057566-5 xiv, 355; new edition 2007, The Mathematical Association of America (MAA)