feat: ✨ Finish DM
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exo1/exo1.tex
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exo1/exo1.tex
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\subsection{}
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On sait que le vecteur tangante unitaire est donné par la formule \\
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$\vec{T}(t) = \frac{\vec{r}'(t)}{\|\vec{r}'(t)\|}$\\
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Calculons $\vec{r}'(t)$ soit: $\vec{r}'(t) = ((t\sin{t})', (t\cos{t})', (t)')$\\
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\begin{math}
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(t\sin{t})' = \sin{t} + t\cos{t}\\
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(t\cos{t})' = \cos{t} - t\sin{t}\\
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(t)' = 1\\
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\vec{r}'(t) = (\sin{t} + t\cos{t}, \cos{t} - t\sin{t}, 1)\\
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\end{math}
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Déterminons à présent $\|\vec{r}'(t)\|$ soit:\\
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\begin{math}
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\|\vec{r}'(t)\| = \sqrt{(\sin{t}+t\cos{t})^2 + (\cos{t}-t\sin{t})^2 + 1^2}\\
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\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\\
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\Leftrightarrow 1 + 0 + t^2\cos^2{t}+t^2\sin^2{t}+1\\
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\Leftrightarrow 1 + 0 + t^2 + 1\\
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\Leftrightarrow t^2 + 2
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\end{math}
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On a donc:\\
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\begin{math}
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\vec{T}(t) = \frac{(\sin{t}+t\cos{t}, \cos{t}-t\sin{t}, 1)}{\sqrt{t^2+2}}\\
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\boxed{\Leftrightarrow \frac{1}{\sqrt{t^2+2}}(\sin{t}+t\cos{t}, \cos{t}-t\sin{t}, 1)}
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\end{math}
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\subsection{}
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On a
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\[
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\vec{r'}(t_0) \text{ -> vectur directeur}
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\]
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\[\vec{r}(t_0) \text{ -> points de passage}\]
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Soit\\
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\begin{math}
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\vec{r}(t_0) = (t_0\sin{t_0}, t_0\cos{t_0}, t_0) = (\frac{\pi}{2}, 0, \frac{\pi}{2})\\
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\Leftrightarrow (\frac{\pi}{2}\sin{\frac{\pi}{2}}, 0, \frac{\pi}{2})\\
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\Leftrightarrow (\frac{\pi}{2}, 0, \frac{\pi}{2}) \Rightarrow t_0 = \frac{\pi}{2}\\
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\end{math}
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On remplace dans \(\vec{R}'(t)\) soit\\
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\begin{math}
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\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)\\
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\Leftrightarrow (1 + 0, -\frac{\pi}{2}, 1)
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\end{math}
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On a donc
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\[
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\left\{
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\begin{array}{l}
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x = \frac{\pi}{2} + r \\
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y = 0 - \frac{\pi}{2} r \\
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z = \frac{\pi}{2} + r
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\end{array}
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\right.
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\]
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\subsection{}
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On cherche \(\vec{r}(t)\) tel que \(\vec{r}'(t) = r(t)\) soit:\\
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\begin{math}
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\vec{r}(t) = (t\sin{t_0}, t\cos{t_0}, t)\\
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\int{\vec{r}'(t)} = (\int{t\sin{t}}, \int{t\cos{t}}, \int{t})
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\end{math}
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\begin{itemize}
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\item \(\int{t\sin{t}dt} = -t\cos{t} - \sin{t} + C_1\)
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\item \(\int{t\cos{t}dt} = t\cos{t}-\cos{t} + C_2\)
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\item \(\int{tdt} = \frac{t^2}{2} + C_3\)
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\end{itemize}
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\begin{math}
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\int{\vec{r}(t) = (-t\cos{t}-\sin{t}, t\cos{t}-\cos{t}, \frac{t-2}{2}) + C \text{ (C = $C_1 + C_2 + C_3$)}}
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\end{math}
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exo2/exo2.tex
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exo2/exo2.tex
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\subsection{}
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On a:
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\begin{itemize}
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\item \(x = e^t \cos t\)
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\item \(y = e^t\)
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\item \(z = e^t \sin t\)
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\end{itemize}
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Calculons les dérivées soit:
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\begin{itemize}
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\item \(\frac{d}{dt} (e^t \cos t) = e^t \cos t - e^t \sin t = e^t (\cos t - \sin t)\)
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\item \(\frac{d}{dt} (e^t) = e^t\)
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\item \(\frac{d}{dt} (e^t \sin t) = e^t \sin t + e^t \cos t = e^t (\sin t + \cos t)\)
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\end{itemize}
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On a alors: \[
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\vec{r'}(t) = \left( e^t (\cos t - \sin t), e^t, e^t (\sin t + \cos t) \right)
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\]
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Calculons la norme de \(\vec{r'}(t)\) soit:
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\[\|\mathbf{r'}(t)\| = \sqrt{(e^t (\cos t - \sin t))^2 + (e^t)^2 + (e^t (\sin t + \cos t))^2}\]
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\begin{itemize}
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\item \((e^t (\cos t - \sin t))^2 = e^{2t} (\cos t - \sin t)^2 = e^{2t} (\cos^2 t - 2 \sin t \cos t + \sin^2 t) = e^{2t} (1 - 2 \sin t \cos t)\)
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\item \(e^{2t}\)
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\item \((e^t (\sin t + \cos t))^2 = e^{2t} (\sin^2 t + 2 \sin t \cos t + \cos^2 t) = e^{2t} (1 + 2 \sin t \cos t)\)
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\end{itemize}
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On a donc\\
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\begin{math}
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\|\mathbf{r'}(t)\| = \sqrt{e^{2t} (1 - 2 \sin t \cos t) + e^{2t} + e^{2t} (1 + 2 \sin t \cos t)}\\
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\|\mathbf{r'}(t)\| = \sqrt{e^{2t} (1 - 2 \sin t \cos t + 1 + 1 + 2 \sin t \cos t)}\\
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\|\mathbf{r'}(t)\| = \sqrt{3 e^{2t}} = \sqrt{3} e^t
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\end{math}
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L'abscice curviligne est donné par la formule \[s(t) = \int_{t_0}^{t} \sqrt{3} \, e^u \, du\] On a donc:\\
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\begin{math}
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s(t) = \sqrt{3} \int_{t_0}^{t} e^u \, du \\
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\Leftrightarrow \sqrt{3} \left[ e^u \right]_{t_0}^{t} \\
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\Leftrightarrow \sqrt{3} \left( e^t - e^{t_0} \right)\\
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\end{math}
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Fixons \(t_0 = 0 \Rightarrow s(t) = \sqrt{3}(e^t-1)\)
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\subsection{}
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Sur l'intervale \(0 \leq t\leq 2\pi\) on a
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\[\sqrt{3}(e^{2\pi}-e^0) = \sqrt{3}(e^{2\pi}) - 1\]
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exo3/exo3.tex
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\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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\[
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36(x - 3) - 18(y - 6) - 36(z - 1) = 0
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\Rightarrow 36x - 18y - 36z = -36
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\Rightarrow \boxed{2x - y - 2z = -2}
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\]
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exo4/exo4.tex
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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 :
|
||||
\[
|
||||
\boxed{a_N(t) = \frac{\sqrt{729t^4 + 324 + 324t^2}}{\sqrt{9 + 9t^4 + 36t^2}}}
|
||||
\]
|
||||
66
main.tex
Normal file
66
main.tex
Normal file
@ -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}
|
||||
Loading…
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Reference in New Issue
Block a user