diff --git a/exo1/exo1.tex b/exo1/exo1.tex new file mode 100644 index 0000000..8cd4360 --- /dev/null +++ b/exo1/exo1.tex @@ -0,0 +1,70 @@ +\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} \ No newline at end of file diff --git a/exo2/exo2.tex b/exo2/exo2.tex new file mode 100644 index 0000000..ec4d9da --- /dev/null +++ b/exo2/exo2.tex @@ -0,0 +1,40 @@ +\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 +\[\sqrt{3}(e^{2\pi}-e^0) = \sqrt{3}(e^{2\pi}) - 1\] \ No newline at end of file diff --git a/exo3/exo3.tex b/exo3/exo3.tex new file mode 100644 index 0000000..263a59f --- /dev/null +++ b/exo3/exo3.tex @@ -0,0 +1,224 @@ +\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 : +\[ +36(x - 3) - 18(y - 6) - 36(z - 1) = 0 +\Rightarrow 36x - 18y - 36z = -36 +\Rightarrow \boxed{2x - y - 2z = -2} +\] \ No newline at end of file diff --git a/exo4/exo4.tex b/exo4/exo4.tex new file mode 100644 index 0000000..bad8494 --- /dev/null +++ b/exo4/exo4.tex @@ -0,0 +1,59 @@ +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 : +\[ +\boxed{a_N(t) = \frac{\sqrt{729t^4 + 324 + 324t^2}}{\sqrt{9 + 9t^4 + 36t^2}}} +\] \ No newline at end of file diff --git a/main.pdf b/main.pdf new file mode 100644 index 0000000..e7116c4 --- /dev/null +++ b/main.pdf @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:7ab423c797d30e72d10809a7da25657183cf69bfd346fdf4528453b39bd05835 +size 235033 diff --git a/main.tex b/main.tex new file mode 100644 index 0000000..599d448 --- /dev/null +++ b/main.tex @@ -0,0 +1,66 @@ +\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}