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