1. **Exercice 5**: Soient \(\overrightarrow{F} = 5\overrightarrow{i} + 2\overrightarrow{j} + 2\overrightarrow{k}\) et \(\overrightarrow{L} = 3\overrightarrow{i} - 4\overrightarrow{j} - 6\overrightarrow{k}\). 1. Calcul du travail \(W = \overrightarrow{F} \cdot \overrightarrow{L} = 5 \times 3 + 2 \times (-4) + 2 \times (-6) = 15 - 8 - 12 = -5\). 2. Projection de \(\overrightarrow{F}\) sur \(\overrightarrow{L}\) : \[\text{proj}_{\overrightarrow{L}} \overrightarrow{F} = \frac{\overrightarrow{F} \cdot \overrightarrow{L}}{\|\overrightarrow{L}\|^2} \overrightarrow{L}\] Calcul de \(\|\overrightarrow{L}\| = \sqrt{3^2 + (-4)^2 + (-6)^2} = \sqrt{9 + 16 + 36} = \sqrt{61}\). Donc \[ \mathrm{proj}_{\overrightarrow{L}} \overrightarrow{F} = \frac{-5}{61} (3 \overrightarrow{i} - 4 \overrightarrow{j} - 6 \overrightarrow{k}) = \left( -\frac{15}{61} \right) \overrightarrow{i} + \frac{20}{61} \overrightarrow{j} + \frac{30}{61} \overrightarrow{k} \]. 3. Travail nul signifie \(\overrightarrow{F} \cdot \overrightarrow{D} = 0\), pour \(\overrightarrow{D} = \alpha \overrightarrow{i} + \beta \overrightarrow{j} + \gamma \overrightarrow{k}\). Donc \[ 5\alpha + 2\beta + 2\gamma = 0 \] C'est la relation entre \(\alpha, \beta, \gamma\). --- 2. **Exercice 6**: Soit \(F(x,y,z) = 3x^2 - 7xy^2 + 2yz + 5xyz^3\). 1. Évaluation en \(M(2,0,-1)\): \[ F(2,0,-1) = 3(2)^2 - 7(2)(0)^2 + 2(0)(-1) + 5(2)(0)(-1)^3 = 3\times4 + 0 + 0 + 0 = 12 \] 2. Gradient de \(F\) : \[ \nabla F = \left(\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z}\right) \] Calculons chaque: - \(\frac{\partial F}{\partial x} = 6x - 7y^2 + 5yz^3\) - \(\frac{\partial F}{\partial y} = -14xy + 2z + 5xz^3\) - \(\frac{\partial F}{\partial z} = 2y + 15xyz^2\) 3. Divergence de \(\overrightarrow{V} = \nabla F\): Les composantes de \(\overrightarrow{V}\) sont les dérivées partielles ci-dessus. Calcul de divergence: \[ \mathrm{Div}\, \overrightarrow{V} = \frac{\partial^2 F}{\partial x^2} + \frac{\partial^2 F}{\partial y^2} + \frac{\partial^2 F}{\partial z^2} \] Nous calculons: - \(\frac{\partial^2 F}{\partial x^2} = 6\) - \(\frac{\partial^2 F}{\partial y^2} = -14x\) - \(\frac{\partial^2 F}{\partial z^2} = 30xyz\) Donc \[ \mathrm{Div} \overrightarrow{V} = 6 - 14x + 30xyz \] 4. Rotationnel de \(\overrightarrow{V}\) (noté \(\overrightarrow{rot}(\overrightarrow{V})\)) On calcule: \[ \overrightarrow{rot}(\overrightarrow{V}) = \left( \frac{\partial V_z}{\partial y} - \frac{\partial V_y}{\partial z}, \frac{\partial V_x}{\partial z} - \frac{\partial V_z}{\partial x}, \frac{\partial V_y}{\partial x} - \frac{\partial V_x}{\partial y} \right) \] Avec \(V_x = \frac{\partial F}{\partial x}\), \(V_y = \frac{\partial F}{\partial y}\), \(V_z = \frac{\partial F}{\partial z}\). Calculs: - \(\frac{\partial V_z}{\partial y} = \frac{\partial}{\partial y}(2y + 15xyz^2) = 2 + 15xz^2\) - \(\frac{\partial V_y}{\partial z} = \frac{\partial}{\partial z}(-14xy + 2z + 5xz^3) = 2 + 15xz^2\) - Donc première composante : \(2 + 15xz^2 - (2 + 15xz^2) = 0\) - \(\frac{\partial V_x}{\partial z} = \frac{\partial}{\partial z}(6x - 7y^2 + 5yz^3) = 15yz^2\) - \(\frac{\partial V_z}{\partial x} = \frac{\partial}{\partial x}(2y + 15xyz^2) = 15yz^2\) - Deuxième composante : \(15yz^2 - 15yz^2 = 0\) - \(\frac{\partial V_y}{\partial x} = \frac{\partial}{\partial x}(-14xy + 2z + 5xz^3) = -14y + 5z^3\) - \(\frac{\partial V_x}{\partial y} = \frac{\partial}{\partial y}(6x - 7y^2 + 5yz^3) = -14y + 5z^3\) - Troisième composante : \((-14y + 5z^3) - (-14y + 5z^3) = 0\) Donc \[ \overrightarrow{rot}(\overrightarrow{V}) = \overrightarrow{0} \] 5. Valeurs en \(M(2,0,-1)\) : - \(\overrightarrow{V}(2,0,-1) = (6\times 2 -7\times0 +5\times0\times (-1)^3, -14\times 2\times 0 + 2\times (-1) + 5 \times 2 \times (-1)^3, 2 \times 0 + 15 \times 2 \times 0 \times (-1)^2) = (12, -2 - 10, 0) = (12, -12, 0)\) - \(\mathrm{Div} \overrightarrow{V}(2,0,-1) = 6 - 14 \times 2 + 30 \times 2 \times 0 \times (-1) = 6 - 28 + 0 = -22\) - \(\overrightarrow{rot}(\overrightarrow{V})(2,0,-1) = \overrightarrow{0}\). **Réponses finales :** - Travail \(W = -5\). - Projection de \(\overrightarrow{F}\) sur \(\overrightarrow{L}\) est \(\left( -\frac{15}{61} , \frac{20}{61} , \frac{30}{61} \right)\). - Relation pour travail nul: \(5\alpha + 2\beta + 2\gamma = 0\). - \(F(2,0,-1) = 12\). - \(\nabla F = (6x - 7y^2 + 5 y z^3, -14 x y + 2 z + 5 x z^3, 2 y + 15 x y z^2)\). - \(\mathrm{Div} \overrightarrow{V} = 6 - 14 x + 30 x y z\). - \(\overrightarrow{rot}(\overrightarrow{V}) = \overrightarrow{0}\). - En \(M\), \[ \overrightarrow{V} = (12, -12, 0), \quad \mathrm{Div} \overrightarrow{V} = -22, \quad \overrightarrow{rot}(\overrightarrow{V}) = \overrightarrow{0} \]