70 lines
2.2 KiB
TeX
70 lines
2.2 KiB
TeX
\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} |