Maximum Abscissa for Loop of Folium of Descartes
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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.
$\blacksquare$