Exercise 15.1.4

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

Prove the arc length formula stated in (15.6)

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof}
 Here the equation of the ellipse $E$ is
 $$x^2 + \frac{y^2}{b^2} = 1,$$
 with eccentricity $k = \sqrt{1- b^2}$.

We compute the arc length $l$ of (E) between $x = u, y=v\ (-1<u<v<1)$ on the upper part of the curve. Then
$$l =  \int_u^v \sqrt{1+\left( \frac{\D y}{\D x}ight)^2}\  \D x,$$
where $y = f(x) = b \sqrt{1-x^2}.$ Then $f'(x) =  \frac{\D y}{\D x }= -\frac{2x}{\sqrt{1-x^2}}$, thus
\begin{align*}
l &=  \int_u^v \sqrt{1+\left(\frac{bx}{\sqrt{1-x^2}}ight)^2}\  \D x\
&=\int_u^v \sqrt{\frac{1 - x^2 + b^2 x^2}{1-x^2}}\  \D x\
&=\int_u^v \sqrt{\frac{1 - k^2 x^2}{1-x^2}}\  \D x\
\end{align*}
We have proved
$$l =  \int_u^v \sqrt{1+\left( \frac{\D y}{\D x}ight)^2}\  \D x = \int_u^v \sqrt{\frac{1 - k^2 x^2}{1-x^2}}\  \D x = \int_u^v \frac{\sqrt{(1-x^2)(1 - k^2 x^2)}}{1-x^2}\  \D x.$$
The arc length of the ellipse is given by an elliptic integral.
\end{proof}

Exercise 15.1.4

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Exercise 15.1.4

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