Let's say that we have the implicit equation x^3 - 3xy + y^3 = 0 (this is a folium of Descartes).
How could we go about sketching this function (not using any plotting software!)?
For instance, I could differentiate wrt to x to obtain:
dy/dx = (y - x^2) / (y^2 - x)
with this I could set x = y = 0 (which satisfies the original equation) to obtain dy/dx = 0/0 but I'm not sure how useful that is?
Further, I can see that dy/dx = 0 whenever y = x^2 but of course only when x and y also satisfy the original equation.
I guess in general I'm asking what techniques can be used to get a feel for the shape of this plot, either by solving for certain points, solving for critical points or looking at behavior at large values - the things one would use for an explicit function.
Thanks for any suggestions on how one might start to get a feel for the shape of this graph.
Addendum: I could of course use the parameterization of this curve: x = 3t / (1+t^3) and y = 3t^2 / (1+t^3). I could then substitute in values of t to generate the shape. Maybe that's the best way to go. I was I guess curious about how one would do it straight from the original form of f(x,y) = 0)
How could we go about sketching this function (not using any plotting software!)?
For instance, I could differentiate wrt to x to obtain:
dy/dx = (y - x^2) / (y^2 - x)
with this I could set x = y = 0 (which satisfies the original equation) to obtain dy/dx = 0/0 but I'm not sure how useful that is?
Further, I can see that dy/dx = 0 whenever y = x^2 but of course only when x and y also satisfy the original equation.
I guess in general I'm asking what techniques can be used to get a feel for the shape of this plot, either by solving for certain points, solving for critical points or looking at behavior at large values - the things one would use for an explicit function.
Thanks for any suggestions on how one might start to get a feel for the shape of this graph.
Addendum: I could of course use the parameterization of this curve: x = 3t / (1+t^3) and y = 3t^2 / (1+t^3). I could then substitute in values of t to generate the shape. Maybe that's the best way to go. I was I guess curious about how one would do it straight from the original form of f(x,y) = 0)
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