Maximum Abscissa for Loop of Folium of Descartes

Theorem

Consider the folium of Descartes defined in parametric form as:

$\begin {cases} x = \dfrac {3 a t} {1 + t^3} \\ y = \dfrac {3 a t^2} {1 + t^3} \end {cases}$

The point on the loop at which the $x$ value is at a maximum occurs when $t = \sqrt [3] {\dfrac 1 2}$, corresponding to the point $P$ defined as:

$P = \tuple {2^{2/3} a, 2^{1/3} a}$

Proof

We calculate the derivative of $x$ with respect to $t$:

\(\ds \dfrac {\d x} {\d t}\)

\(=\)

\(\ds \map {\dfrac \d {\d t} } {\dfrac {3 a t} {1 + t^3} }\)

\(\ds \)

\(=\)

\(\ds \dfrac {\paren {1 + t^3} \times 3 a - 3 a t \paren {3 t^2} } {\paren {1 + t^3}^2}\)

Quotient Rule for Derivatives

\(\ds \)

\(=\)

\(\ds \dfrac {3 a - 6 a t^3} {\paren {1 + t^3}^2}\)

simplifying

Thus $x$ is stationary when:

\(\ds 3 a - 6 a t^3\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds t\)

\(=\)

\(\ds \paren {\dfrac 1 2}^{1/3}\)

From Behaviour of Parametric Equations for Folium of Descartes according to Parameter, it is clear from the geometry that $x$ is a local maximum for this value of $t$.

Then we have:

\(\ds x\)

\(=\)

\(\ds \dfrac {3 a \times \paren {1/2}^{1/3} } {1 + \paren {\paren {1/2}^{1/3} }^3}\)

\(\ds \)

\(=\)

\(\ds \dfrac {3 a \times \paren {1/2}^{1/3} } {1 + 1/2}\)

\(\ds \)

\(=\)

\(\ds 2 a \times \paren {\dfrac 1 2}^{1/3}\)

\(\ds \)

\(=\)

\(\ds \paren {\dfrac {2^3} 2}^{1/3} a\)

\(\ds \)

\(=\)

\(\ds 2^{2/3} a\)

and:

\(\ds y\)

\(=\)

\(\ds \dfrac {3 a \times \paren {\paren {1/2}^{1/3} }^2} {1 + \paren {\paren {1/2}^{1/3} }^3}\)

\(\ds \)

\(=\)

\(\ds \dfrac {3 a \times \paren {1/2}^{2/3} } {1 + 1/2}\)

\(\ds \)

\(=\)

\(\ds 2 a \times \paren {\dfrac 1 2}^{2/3}\)

\(\ds \)

\(=\)

\(\ds \paren {\dfrac {2^3} {2^2} }^{1/3} a\)

\(\ds \)

\(=\)

\(\ds 2^{1/3} a\)

Hence the result.

$\blacksquare$