# Implicit differentiation and the Folium of Descartes [duplicate]

Given the function (Folium of Descartes) $x^3 + y^3 = 3xy$, how would I find the equations for tangent and normal lines at the point $\left( \frac{3}{2},\frac{3}{2} \right)$?

I know that I must use the Dt function, but how do I calculate the tangent and normal lines?

Also, how would I graph these lines using the ContourPlot function?

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– user9660
Commented Mar 12, 2015 at 21:45
• Please, consider updating your question to include what you have tried and where you are getting stuck. That way, people on this site will know exactly what help you need.
– user9660
Commented Mar 12, 2015 at 21:45

  Folium = Plot[y /. Solve[x^3 + y^3 == 3 x y, y], {x, -3, 3},
Epilog -> {PointSize[Large], Red, Point[{3/2, 3/2}]}]


Once you determine which of the three solution segments goes through $\{3/2, 3/2\}$ (it is the first) you find the derivative with respect to $x$. You can either take the derivative with respect to $x$ for the first solution, or merely cut and paste it as follows and substitute the $x$ value $3/2$:

Re@N@Simplify@
D[(2^(1/3) x)/(-x^3 + Sqrt[-4 x^3 + x^6])^(1/3)
+ (-x^3 + Sqrt[-4 x^3 + x^6])^(1/3)/2^(1/3), x] /. x -> 3/2


(* -1 *)

So the tangent line is $y = -x + b$, where $b$ ensures the line goes through $\{3/2, 3/2\}$. A simple Solve shows that $b = 3$.

Thus the tangent line can be plotted as:

mylinePlot = Plot[-x + 3, {x, -3, 3}, PlotStyle -> Red];

Show[Folium, myLine]


The perpendicular has slope $-1$ divided by the slope of the tangent, and thus the slope is $+1$ and its intercept must ensure the perpendicular line go through $\{3/2, 3/2\}$. As before, a simple Solve reveals $b = 0$:

myPerp = Plot[x, {x, -3, 3}, PlotStyle -> Green];

Show[Folium, myLine, myPerp, AspectRatio->1]