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exo1/exo1.tex
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\subsection{}
On sait que le vecteur tangante unitaire est donné par la formule \\
$\vec{T}(t) = \frac{\vec{r}'(t)}{\|\vec{r}'(t)\|}$\\
Calculons $\vec{r}'(t)$ soit: $\vec{r}'(t) = ((t\sin{t})', (t\cos{t})', (t)')$\\
\begin{math}
(t\sin{t})' = \sin{t} + t\cos{t}\\
(t\cos{t})' = \cos{t} - t\sin{t}\\
(t)' = 1\\
\vec{r}'(t) = (\sin{t} + t\cos{t}, \cos{t} - t\sin{t}, 1)\\
\end{math}
Déterminons à présent $\|\vec{r}'(t)\|$ soit:\\
\begin{math}
\|\vec{r}'(t)\| = \sqrt{(\sin{t}+t\cos{t})^2 + (\cos{t}-t\sin{t})^2 + 1^2}\\
\Leftrightarrow \|\vec{r}'(t)\|^2 = \sin^2{t}+\cos^2{t} + 2t\sin{t}\cos{t}-2t\sin{t}\cos{t} + t^2\cos^2{t}+t^2\sin^2{t}+1\\
\Leftrightarrow 1 + 0 + t^2\cos^2{t}+t^2\sin^2{t}+1\\
\Leftrightarrow 1 + 0 + t^2 + 1\\
\Leftrightarrow t^2 + 2
\end{math}
On a donc:\\
\begin{math}
\vec{T}(t) = \frac{(\sin{t}+t\cos{t}, \cos{t}-t\sin{t}, 1)}{\sqrt{t^2+2}}\\
\boxed{\Leftrightarrow \frac{1}{\sqrt{t^2+2}}(\sin{t}+t\cos{t}, \cos{t}-t\sin{t}, 1)}
\end{math}
\subsection{}
On a
\[
\vec{r'}(t_0) \text{ -> vectur directeur}
\]
\[\vec{r}(t_0) \text{ -> points de passage}\]
Soit\\
\begin{math}
\vec{r}(t_0) = (t_0\sin{t_0}, t_0\cos{t_0}, t_0) = (\frac{\pi}{2}, 0, \frac{\pi}{2})\\
\Leftrightarrow (\frac{\pi}{2}\sin{\frac{\pi}{2}}, 0, \frac{\pi}{2})\\
\Leftrightarrow (\frac{\pi}{2}, 0, \frac{\pi}{2}) \Rightarrow t_0 = \frac{\pi}{2}\\
\end{math}
On remplace dans \(\vec{R}'(t)\) soit\\
\begin{math}
\vec{r}'(t_0) = (\sin{\frac{\pi}{2}} + \frac{\pi}{2}\cos{\frac{\pi}{2}}, \cos{\frac{\pi}{2}}-\frac{\pi}{2}\sin{\frac{\pi}{2}}, 1)\\
\Leftrightarrow (1 + 0, -\frac{\pi}{2}, 1)
\end{math}
On a donc
\[
\left\{
\begin{array}{l}
x = \frac{\pi}{2} + r \\
y = 0 - \frac{\pi}{2} r \\
z = \frac{\pi}{2} + r
\end{array}
\right.
\]
\subsection{}
On cherche \(\vec{r}(t)\) tel que \(\vec{r}'(t) = r(t)\) soit:\\
\begin{math}
\vec{r}(t) = (t\sin{t_0}, t\cos{t_0}, t)\\
\int{\vec{r}'(t)} = (\int{t\sin{t}}, \int{t\cos{t}}, \int{t})
\end{math}
\begin{itemize}
\item \(\int{t\sin{t}dt} = -t\cos{t} - \sin{t} + C_1\)
\item \(\int{t\cos{t}dt} = t\cos{t}-\cos{t} + C_2\)
\item \(\int{tdt} = \frac{t^2}{2} + C_3\)
\end{itemize}
\begin{math}
\int{\vec{r}(t) = (-t\cos{t}-\sin{t}, t\cos{t}-\cos{t}, \frac{t-2}{2}) + C \text{ (C = $C_1 + C_2 + C_3$)}}
\end{math}

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\subsection{}
On a:
\begin{itemize}
\item \(x = e^t \cos t\)
\item \(y = e^t\)
\item \(z = e^t \sin t\)
\end{itemize}
Calculons les dérivées soit:
\begin{itemize}
\item \(\frac{d}{dt} (e^t \cos t) = e^t \cos t - e^t \sin t = e^t (\cos t - \sin t)\)
\item \(\frac{d}{dt} (e^t) = e^t\)
\item \(\frac{d}{dt} (e^t \sin t) = e^t \sin t + e^t \cos t = e^t (\sin t + \cos t)\)
\end{itemize}
On a alors: \[
\vec{r'}(t) = \left( e^t (\cos t - \sin t), e^t, e^t (\sin t + \cos t) \right)
\]
Calculons la norme de \(\vec{r'}(t)\) soit:
\[\|\mathbf{r'}(t)\| = \sqrt{(e^t (\cos t - \sin t))^2 + (e^t)^2 + (e^t (\sin t + \cos t))^2}\]
\begin{itemize}
\item \((e^t (\cos t - \sin t))^2 = e^{2t} (\cos t - \sin t)^2 = e^{2t} (\cos^2 t - 2 \sin t \cos t + \sin^2 t) = e^{2t} (1 - 2 \sin t \cos t)\)
\item \(e^{2t}\)
\item \((e^t (\sin t + \cos t))^2 = e^{2t} (\sin^2 t + 2 \sin t \cos t + \cos^2 t) = e^{2t} (1 + 2 \sin t \cos t)\)
\end{itemize}
On a donc\\
\begin{math}
\|\mathbf{r'}(t)\| = \sqrt{e^{2t} (1 - 2 \sin t \cos t) + e^{2t} + e^{2t} (1 + 2 \sin t \cos t)}\\
\|\mathbf{r'}(t)\| = \sqrt{e^{2t} (1 - 2 \sin t \cos t + 1 + 1 + 2 \sin t \cos t)}\\
\|\mathbf{r'}(t)\| = \sqrt{3 e^{2t}} = \sqrt{3} e^t
\end{math}
L'abscice curviligne est donné par la formule \[s(t) = \int_{t_0}^{t} \sqrt{3} \, e^u \, du\] On a donc:\\
\begin{math}
s(t) = \sqrt{3} \int_{t_0}^{t} e^u \, du \\
\Leftrightarrow \sqrt{3} \left[ e^u \right]_{t_0}^{t} \\
\Leftrightarrow \sqrt{3} \left( e^t - e^{t_0} \right)\\
\end{math}
Fixons \(t_0 = 0 \Rightarrow s(t) = \sqrt{3}(e^t-1)\)
\subsection{}
Sur l'intervale \(0 \leq t\leq 2\pi\) on a
\[\sqrt{3}(e^{2\pi}-e^0) = \sqrt{3}(e^{2\pi}) - 1\]

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\subsection{}
\subsubsection*{Calculons $\vec{T}(t) = \frac{\vec{r'}(t)}{||\vec{r'}(t)||}$}
Nous commençons par calculer la dérivée du vecteur position $\vec{r}(t)$ soit \[\vec{r'}(t) = ((3t^2)', (6t)', (t^3)')\]
\begin{itemize}
\item \((3t^2)' = 6t\)
\item \((6t)' = 6\)
\item \((t^3)' = 3t^2\)
\end{itemize}
\[\boxed{\vec{r'}(t) = (6t, 6, 3t^2)}\]
Nous allons maintenant calculer la norme de la dérivée du vecteur positition $\vec{r}(t)$ soit\\
\begin{math}
||\vec{r'}(t)|| = \sqrt{(6t)^2+6^2+(3t^2)^2}\\
\Leftrightarrow \sqrt{36t^2 + 36 + 9t^4}\\
\Leftrightarrow \sqrt{9t^4 + 36t^2 + 36}
\end{math}
Nous avons donc:\\
\begin{math}
\vec{T}(t) = \frac{(6t, 6, 3t^2)}{\sqrt{9t^4 + 36t^2 + 36}}\\
\Leftrightarrow \frac{(6t, 6, 3t^2)}{\sqrt{(3t^2+6)^2}}\\
\Leftrightarrow \frac{(6t, 6, 3t^2)}{3t^2+6}\\
\end{math}
Donc, \[\boxed{\vec{T}(t) =\frac{(6t, 6, 3t^2)}{3t^2+6}}\]
\subsubsection*{Calculons $\vec{N}(t) = \frac{\vec{T'}(t)}{||\vec{T'}(t)||}$}
On pose :
\[
\vec{T}(t) = \left( \frac{6t}{3t^2+6},\, \frac{6}{3t^2+6},\, \frac{3t^2}{3t^2+6} \right)
\]
On dérive chaque composante avec la règle du quotient :
\[
T_1'(t) = \frac{d}{dt}\left( \frac{6t}{3t^2+6} \right) = \frac{6(3t^2+6) - 6t(6t)}{(3t^2+6)^2} = \frac{-18t^2 + 36}{(3t^2+6)^2}
\]
\[
T_2'(t) = \frac{d}{dt}\left( \frac{6}{3t^2+6} \right) = \frac{-36t}{(3t^2+6)^2}
\]
\[
T_3'(t) = \frac{d}{dt}\left( \frac{3t^2}{3t^2+6} \right) = \frac{36t}{(3t^2+6)^2}
\]
Ainsi :
\[
\vec{T}'(t) = \left( \frac{-18t^2 + 36}{(3t^2 + 6)^2},\; \frac{-36t}{(3t^2 + 6)^2},\; \frac{36t}{(3t^2 + 6)^2} \right)
\]
\[
\|\vec{T}'(t)\| = \sqrt{ \left( \frac{-18t^2 + 36}{(3t^2+6)^2} \right)^2 + \left( \frac{-36t}{(3t^2+6)^2} \right)^2 + \left( \frac{36t}{(3t^2+6)^2} \right)^2 }
\]
\[
= \frac{ \sqrt{(-18t^2 + 36)^2 + 2(36t)^2} }{(3t^2 + 6)^2}
\]
\[
= \frac{ \sqrt{324t^4 - 1296t^2 + 1296 + 2592t^2} }{(3t^2 + 6)^2}
= \frac{ \sqrt{324t^4 + 1296t^2 + 1296} }{(3t^2 + 6)^2}
\]
\[
\vec{N}(t) = \frac{\vec{T}'(t)}{\|\vec{T}'(t)\|} = \left(
\frac{-18t^2 + 36}{\sqrt{324t^4 + 1296t^2 + 1296}},\;
\frac{-36t}{\sqrt{324t^4 + 1296t^2 + 1296}},\;
\frac{36t}{\sqrt{324t^4 + 1296t^2 + 1296}}
\right)
\]
\subsubsection*{Calculons $\vec{B}(t)$}
Nous avons les vecteurs suivants :
\[
\vec{T}(t) =\frac{(6t, 6, 3t^2)}{3t^2+6}
\]
\[
\vec{N}(t) = \frac{\vec{T}'(t)}{\|\vec{T}'(t)\|} = \left(
\frac{-18t^2 + 36}{\sqrt{324t^4 + 1296t^2 + 1296}},\;
\frac{-36t}{\sqrt{324t^4 + 1296t^2 + 1296}},\;
\frac{36t}{\sqrt{324t^4 + 1296t^2 + 1296}}
\right)
\]
On calcule les composantes du produit vectoriel :
\[
\vec{B}(t) =
\begin{vmatrix}
\vec{i} & \vec{j} & \vec{k} \\
\frac{6t}{3t^2 + 6} & \frac{6}{3t^2 + 6} & \frac{3t^2}{3t^2 + 6} \\
\frac{-18t^2 + 36}{\sqrt{324t^4 + 1296t^2 + 1296}} & \frac{-36t}{\sqrt{324t^4 + 1296t^2 + 1296}} & \frac{36t}{\sqrt{324t^4 + 1296t^2 + 1296}}
\end{vmatrix}
\]
\textbf{Composante en \( \vec{i} \)} :
\[
\vec{i} \cdot \left( \frac{6}{3t^2 + 6} \cdot \frac{36t}{\sqrt{324t^4 + 1296t^2 + 1296}} - \frac{3t^2}{3t^2 + 6} \cdot \frac{-36t}{\sqrt{324t^4 + 1296t^2 + 1296}} \right)
\]
\[
= \vec{i} \cdot \frac{1}{3t^2 + 6} \cdot \frac{36t(6 + t^2)}{\sqrt{324t^4 + 1296t^2 + 1296}}
= \vec{i} \cdot \frac{36t(6 + t^2)}{(3t^2 + 6)\sqrt{324t^4 + 1296t^2 + 1296}}
\]
\textbf{Composante en \( \vec{j} \)} :
\[
- \vec{j} \cdot \left( \frac{6t}{3t^2 + 6} \cdot \frac{36t}{\sqrt{324t^4 + 1296t^2 + 1296}} - \frac{3t^2}{3t^2 + 6} \cdot \frac{-18t^2 + 36}{\sqrt{324t^4 + 1296t^2 + 1296}} \right)
\]
\[
= -\vec{j} \cdot \left( \frac{1}{3t^2 + 6} \cdot \frac{36t^2 + (3t^2)(-18t^2 + 36)}{\sqrt{324t^4 + 1296t^2 + 1296}} \right)
\]
Calculons :
\[
(3t^2)(-18t^2 + 36) = -54t^4 + 108t^2
\quad \text{et} \quad
36t^2 + (-54t^4 + 108t^2) = -54t^4 + 144t^2
\]
Donc la composante en \( \vec{j} \) vaut :
\[
- \vec{j} \cdot \frac{-54t^4 + 144t^2}{(3t^2 + 6)\sqrt{324t^4 + 1296t^2 + 1296}}
\]
\textbf{Composante en \( \vec{k} \)} :
\[
\vec{k} \cdot \left( \frac{6t}{3t^2 + 6} \cdot \frac{-36t}{\sqrt{324t^4 + 1296t^2 + 1296}} - \frac{6}{3t^2 + 6} \cdot \frac{-18t^2 + 36}{\sqrt{324t^4 + 1296t^2 + 1296}} \right)
\]
\[
= \vec{k} \cdot \frac{1}{3t^2 + 6} \cdot \frac{-36t^2 + 108 - 6t^2}{\sqrt{324t^4 + 1296t^2 + 1296}}
= \vec{k} \cdot \frac{-36t^2 + 108 - 6t^2}{(3t^2 + 6)\sqrt{324t^4 + 1296t^2 + 1296}}
= \vec{k} \cdot \frac{-42t^2 + 108}{(3t^2 + 6)\sqrt{324t^4 + 1296t^2 + 1296}}
\]
\subsubsection*{Résultat final}
\begin{math}
\vec{B}(t) =(
\frac{36t(t^2 + 6)}{(3t^2 + 6)\sqrt{324t^4 + 1296t^2 + 1296}}\\
\frac{54t^4 - 144t^2}{(3t^2 + 6)\sqrt{324t^4 + 1296t^2 + 1296}}\\
\frac{-42t^2 + 108}{(3t^2 + 6)\sqrt{324t^4 + 1296t^2 + 1296}})
\end{math}
\subsection{}
Pour calculer la courbure, nous allons utiliser la formule \(k(t) = \frac{||\vec{T'}(t)||}{||\vec{r'}(t)}||\)
\begin{itemize}
\item \(||\vec{r'}(t)|| = 3t^2 + 6\)
\item \(\frac{4}{t^2 + 2}\)
\end{itemize}
\begin{math}
K(t) = \frac{\frac{4}{t^2 + 2}}{3t^2+6}\\
\Leftrightarrow \frac{4}{t^2 + 2} \frac{1}{3t^2+6}\\
\Leftrightarrow \frac{4(1)}{(t^2 + 2)(3t^2+6)}\\
\Leftrightarrow \frac{4}{3(t^2+2)^2}\\
\end{math}
\subsection{}
\subsubsection*{a)}
Soit la courbe définie par la fonction vectorielle :
\[
\vec{r}(t) = \left(3t^2,\ 6t,\ t^3\right)
\]
Le point donné est \( P = \left(3,\ 6,\ 1\right) \) et correspond à \( t = 1 \).
On commence par calculer le vecteur tangent au point \( t = 1 \) :
\[
\vec{r}'(t) = \left(6t,\ 6,\ 3t^2\right) \Rightarrow \vec{r}'(1) = (6,\ 6,\ 3)
\]
On prend \(\vec{n} = \vec{T}(1) = (6,6,3)\) comme vecteur normal.
L'équation du plan passant par \(P = (3,6,1)\) et de vecteur normal \(\vec{n} = (a,b,c)\) s'écrit :
\[
a(x - x_0) + b(y - y_0) + c(z - z_0) = 0
\]
On remplace :
\[
6(x - 3) + 6(y - 6) + 3(z - 1) = 0
\Rightarrow 6x + 6y + 3z = 57
\]
Donc, l'équation du \textbf{plan normal} est :
\[
\boxed{6x + 6y + 3z = 57}
\]
\subsubsection*{b)}
Le plan osculateur est défini par les vecteurs \(\vec{T}(t)\) et \(\vec{N}(t)\). On a :
\[
\vec{r}'(1) = (6,\ 6,\ 3), \quad
\vec{r}''(1) = \left(6,\ 0,\ 6\right)
\]
On calcule le produit vectoriel :
\[
\vec{B}(1) = \vec{r}'(1) \wedge \vec{r}''(1) =
\begin{vmatrix}
\vec{i} & \vec{j} & \vec{k} \\
6 & 6 & 3 \\
6 & 0 & 6 \\
\end{vmatrix}
= (36,\ -18,\ -36)
\]
On a donc un vecteur normal \(\vec{n} = \vec{B}(1) = (36,\ -18,\ -36)\) et le point \(P = (3,\ 6,\ 1)\).
Équation du plan :
\[
36(x - 3) - 18(y - 6) - 36(z - 1) = 0
\Rightarrow 36x - 18y - 36z = -36
\Rightarrow \boxed{2x - y - 2z = -2}
\]

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Soit la position d’un point donnée par :
\[
\vec{r}(t) = (3t,\ t^3,\ 3t^2)
\]
On commence par calculer les dérivées :
\[
\vec{r}'(t) = (3,\ 3t^2,\ 6t), \quad
\vec{r}''(t) = (0,\ 6t,\ 6)
\]
\subsubsection*{Composante tangentielle \( a_T(t) \)}
\[
a_T(t) = \frac{\vec{r}'(t) \cdot \vec{r}''(t)}{\|\vec{r}'(t)\|}
\]
Calcul du produit scalaire :
\[
\vec{r}'(t) \cdot \vec{r}''(t) = 3 \cdot 0 + 3t^2 \cdot 6t + 6t \cdot 6 = 18t^3 + 36t
\]
Norme de \( \vec{r}'(t) \) :
\[
\|\vec{r}'(t)\| = \sqrt{3^2 + (3t^2)^2 + (6t)^2} = \sqrt{9 + 9t^4 + 36t^2}
\]
Donc :
\[
\boxed{a_T(t) = \frac{18t^3 + 36t}{\sqrt{9 + 9t^4 + 36t^2}}}
\]
\subsection*{2. Composante normale \( a_N(t) \)}
\[
a_N(t) = \frac{\|\vec{r}'(t) \wedge \vec{r}''(t)\|}{\|\vec{r}'(t)\|}
\]
Produit vectoriel :
\[
\vec{r}'(t) \wedge \vec{r}''(t) =
\begin{vmatrix}
\vec{i} & \vec{j} & \vec{k} \\
3 & 3t^2 & 6t \\
0 & 6t & 6
\end{vmatrix}
= (-27t^2,\ -18,\ 18t)
\]
Norme du produit vectoriel :
\[
\|\vec{r}'(t) \wedge \vec{r}''(t)\| = \sqrt{(-27t^2)^2 + (-18)^2 + (18t)^2}
= \sqrt{729t^4 + 324 + 324t^2}
\]
Donc :
\[
\boxed{a_N(t) = \frac{\sqrt{729t^4 + 324 + 324t^2}}{\sqrt{9 + 9t^4 + 36t^2}}}
\]

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\documentclass[12pt]{article}
\usepackage{graphicx} % Required for inserting images
\usepackage{graphicx} % Required for inserting images
\usepackage[T1]{fontenc}
\usepackage{fancyhdr}
\usepackage{float}
\usepackage{fancyvrb}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{systeme,mathtools}
\usepackage{hyperref}
\hypersetup{
colorlinks,
citecolor=black,
filecolor=black,
linkcolor=black,
urlcolor=black
}
\renewcommand{\thesubsection}{\thesection.\alph{subsection}}
\title{Devoir n°4 - 8MAP107}
\author{Louis Gallet \and{Damien Placé} \and {Sacha Sitbon}}
\date{Mars 2025}
\begin{document}
\maketitle
% Liste des étudiants
\begin{description}
\item[Louis Gallet] GALL08010500
\item[Damien Placé] PLAD25070500
\item[Sacha Sitbon] SITS18100500 \\
\end{description}
% Espace pour l'évaluation
\begin{description}
\item[Note :] \hrulefill\ /100\% \hspace{8cm}
\item[Commentaire :] \par % Démarre un nouveau paragraphe
\end{description}
\newpage
\tableofcontents
\newpage
\section{Exercice 1}
\input{exo1/exo1.tex}
\newpage
\section{Exercice 2}
\input{exo2/exo2.tex}
\newpage
\section{Exercice 3}
\input{exo3/exo3.tex}
\newpage
\section{Exercice 4}
\input{exo4/exo4.tex}
\newpage
\end{document}