Help with parametric equation for cartesian x^3 + y^3 = 3xy

bedigursimran

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Sep 12, 2017
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The folium of Descartes has the graph shown below in Figure 1, the Cartesian equation is x^3 + y^3 = 3xy

Use the substitution y/x = t to find the parametric equation of the folium.



folium.jpg

The problem that I have with it is understanding how to go about doing it. Like can I substitute in y = tx (rearranged y/x=t) in the equation of folium?
Which gives.

x^3 + (tx)^3 = 3tx^2
=> x^2 + (t^3)(x^2) = 3tx

Is that right? If so, how do I turn it into a parametric equation?

Thanks guys!! I am really confused guys. :D
 
The folium of Descartes has the graph shown below in Figure 1, the Cartesian equation is x^3 + y^3 = 3xy

Use the substitution y/x = t to find the parametric equation of the folium.



View attachment 8486

The problem that I have with it is understanding how to go about doing it. Like can I substitute in y = tx (rearranged y/x=t) in the equation of folium?
Which gives.

x^3 + (tx)^3 = 3tx^2
=> x^2 + (t^3)(x^2) = 3tx

Is that right? If so, how do I turn it into a parametric equation?

Thanks guys!! I am really confused guys. :D

x^3 + (tx)^3 = 3tx^2

x^3 * (1+ t^3) = 3t * x^2 x = ? → y = ?
 
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