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