Parametrizing the folium of Descartes

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In summary, the student was unable to solve a problem in Calc III that involved parametrizing the folium of Descartes. They found help online and were able to solve the problem using x(t) = 3at/(t3 + 1) and y(t) = 3at2/(t3 + 1).
  • #1
Bohrok
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After recently studying Calc III on my own, I came across this problem with parametrizing the folium of Descartes.

480px-Folium_Of_Descartes.svg.png


Homework Statement



Show that for t ≠ -1, 0, the line y = tx intersects the folium at the origin and at one other point P. Express the coordinates of P in terms of t to obtain a parametrization of the folium.

Homework Equations



The folium of Descartes is the curve with the equation x3 + y3 = 3axy, where a ≠ 0 is a constant.

The Attempt at a Solution



I couldn't really get anywhere with this problem. I understand why t ≠ -1, 0, and y = tx → t = y/x.
Using the formula for the slope between two points with the origin (0, 0) and point P(x1, y1) on the folium went back to t = y/x

I know that x(t) = 3at/(t3 + 1) and y(t) = 3at2/(t3 + 1) from the Internet but I couldn't find a good explanation of the steps in deriving the parametric equations. They look quite similar to the previous problem in the book which would probably help with this problem, but I couldn't do anything useful with that problem either:
Show that the line of slope t through (-1, 0) intersects the unit circle in the point with coordinates x = (1-t2)/(t2 + 1), y = 2t/(t2 + 1)
Conclude that these equations parametrize the unit circle with the point (-1, 0) excluded. Show further that t = y/(x + 1)
 
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  • #2
Hi Bohrok! :smile:
Bohrok said:
The folium of Descartes is the curve with the equation x3 + y3 = 3axy, where a ≠ 0 is a constant.

I know that x(t) = 3at/(t3 + 1) and y(t) = 3at2/(t3 + 1) from the Internet but I couldn't find a good explanation of the steps in deriving the parametric equations.

Just put t = y/x into x3 + y3 = 3axy :wink:
 
  • #3
Thanks so much tiny-tim! So annoying when I can't see the right path and start down a wrong one that doesn't work out... :redface:
 

1. What is the folium of Descartes?

The folium of Descartes is a mathematical curve named after French philosopher and mathematician René Descartes. It is also known as the "Descartes' spiral" or "folium of Descartes' spiral". It is a plane curve defined by the equation x^3 + y^3 = 3axy.

2. How is the folium of Descartes parametrized?

The folium of Descartes can be parametrized using the following equations: x = t + t^3 and y = t - t^3. This parametrization results in a looping curve that intersects itself at the origin.

3. What is the significance of the folium of Descartes?

The folium of Descartes has been studied extensively by mathematicians for its interesting properties. It is a transcendental curve, meaning it cannot be expressed by a finite number of algebraic operations. It also has a cusp at the origin, where the tangent line is vertical, making it a point of inflection. It has applications in areas such as physics and engineering.

4. How is the folium of Descartes related to the Cartesian coordinate system?

The folium of Descartes is named after René Descartes, who is known as the father of analytic geometry. The Cartesian coordinate system, also known as the Cartesian plane, was invented by Descartes and is used to graph the folium of Descartes. The curve is defined by an algebraic equation, making it a useful tool in the study of analytic geometry.

5. Can the folium of Descartes be generalized to higher dimensions?

Yes, the folium of Descartes can be generalized to higher dimensions by using the equation x^n + y^n = nax^m, where n and m are positive integers. This results in a hypersurface in n-dimensional space. The folium of Descartes itself is a special case of this equation when n = 3 and m = 1.

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