1. Problem: Show that the flow with velocity \(\mathbf{v} = x\mathbf{i} + x\mathbf{j} + y\mathbf{k}\) is incompressible. Step 1: Recall that a flow is incompressible if the divergence of velocity \(\nabla \cdot \mathbf{v} = 0\). Step 2: Compute divergence: $$\nabla \cdot \mathbf{v} = \frac{\partial}{\partial x}(x) + \frac{\partial}{\partial y}(x) + \frac{\partial}{\partial z}(y) = 1 + 0 + 0 = 1$$ Step 3: Since divergence is 1, not zero, the flow is not incompressible. 2. Problem: Consider the flow with velocity \(\mathbf{v} = 2x\mathbf{i} + 2x\mathbf{j} + 3y\mathbf{k}\). Is it compressible? Step 1: Compute divergence: $$\nabla \cdot \mathbf{v} = \frac{\partial}{\partial x}(2x) + \frac{\partial}{\partial y}(2x) + \frac{\partial}{\partial z}(3y) = 2 + 0 + 0 = 2$$ Step 2: Since divergence is 2 \(\neq 0\), the flow is compressible. 3. Problem: Determine if the flows are irrotational and incompressible. (i) \(\mathbf{v} = \{0, 3z^2, 0\}\) Step 1: Compute divergence: $$\nabla \cdot \mathbf{v} = \frac{\partial}{\partial x}(0) + \frac{\partial}{\partial y}(3z^2) + \frac{\partial}{\partial z}(0) = 0 + 0 + 0 = 0$$ Step 2: Compute curl: $$\nabla \times \mathbf{v} = \left( \frac{\partial 0}{\partial y} - \frac{\partial 3z^2}{\partial z}, \frac{\partial 0}{\partial z} - \frac{\partial 0}{\partial x}, \frac{\partial 3z^2}{\partial x} - \frac{\partial 0}{\partial y} \right) = (-6z, 0, 0)$$ Step 3: Curl is not zero, so flow is not irrotational. Divergence is zero, so flow is incompressible. (ii) \(\mathbf{v} = \{x, y, -2\}\) Step 1: Compute divergence: $$\nabla \cdot \mathbf{v} = \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} + \frac{\partial (-2)}{\partial z} = 1 + 1 + 0 = 2$$ Step 2: Compute curl: $$\nabla \times \mathbf{v} = \left( \frac{\partial (-2)}{\partial y} - \frac{\partial y}{\partial z}, \frac{\partial x}{\partial z} - \frac{\partial (-2)}{\partial x}, \frac{\partial y}{\partial x} - \frac{\partial x}{\partial y} \right) = (0 - 0, 0 - 0, 1 - 1) = (0, 0, 0)$$ Step 3: Curl is zero, so flow is irrotational. Divergence is 2 \(\neq 0\), so flow is compressible. 4. Problem: Find equation of tangent plane and normal line to surface \(x^2 y + 2 x z^2 = 8\) at point \(P(1,0,2)\). Step 1: Define \(F(x,y,z) = x^2 y + 2 x z^2 - 8 = 0\). Step 2: Compute gradient \(\nabla F = \left( \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z} \right)\): $$\frac{\partial F}{\partial x} = 2 x y + 2 z^2$$ $$\frac{\partial F}{\partial y} = x^2$$ $$\frac{\partial F}{\partial z} = 4 x z$$ Step 3: Evaluate at \(P(1,0,2)\): $$\frac{\partial F}{\partial x} = 2 \cdot 1 \cdot 0 + 2 \cdot 2^2 = 8$$ $$\frac{\partial F}{\partial y} = 1^2 = 1$$ $$\frac{\partial F}{\partial z} = 4 \cdot 1 \cdot 2 = 8$$ Step 4: Equation of tangent plane: $$8(x - 1) + 1(y - 0) + 8(z - 2) = 0$$ Simplify: $$8x - 8 + y + 8z - 16 = 0 \Rightarrow 8x + y + 8z = 24$$ Step 5: Parametric equations of normal line: $$x = 1 + 8t, \quad y = 0 + t, \quad z = 2 + 8t$$ 5. Problem: Find equation of tangent plane and normal line to surface \(2 x z^2 - 3 x y - 4 x = 7\) at point \(P(1,-1,-2)\). Step 1: Define \(G(x,y,z) = 2 x z^2 - 3 x y - 4 x - 7 = 0\). Step 2: Compute gradient \(\nabla G = \left( \frac{\partial G}{\partial x}, \frac{\partial G}{\partial y}, \frac{\partial G}{\partial z} \right)\): $$\frac{\partial G}{\partial x} = 2 z^2 - 3 y - 4$$ $$\frac{\partial G}{\partial y} = -3 x$$ $$\frac{\partial G}{\partial z} = 4 x z$$ Step 3: Evaluate at \(P(1,-1,-2)\): $$\frac{\partial G}{\partial x} = 2 \cdot (-2)^2 - 3 \cdot (-1) - 4 = 8 + 3 - 4 = 7$$ $$\frac{\partial G}{\partial y} = -3 \cdot 1 = -3$$ $$\frac{\partial G}{\partial z} = 4 \cdot 1 \cdot (-2) = -8$$ Step 4: Equation of tangent plane: $$7(x - 1) - 3(y + 1) - 8(z + 2) = 0$$ Simplify: $$7x - 7 - 3y - 3 - 8z - 16 = 0 \Rightarrow 7x - 3y - 8z = 26$$ Step 5: Parametric equations of normal line: $$x = 1 + 7t, \quad y = -1 - 3t, \quad z = -2 - 8t$$ 6. Problem: Find maxima, minima, and saddle points of functions. (i) \(f(x,y) = x^2 + xy + 3x + 2y + 5\) Step 1: Compute partial derivatives: $$f_x = 2x + y + 3, \quad f_y = x + 2$$ Step 2: Set derivatives to zero for critical points: $$2x + y + 3 = 0$$ $$x + 2 = 0 \Rightarrow x = -2$$ Step 3: Substitute \(x = -2\) into first equation: $$2(-2) + y + 3 = 0 \Rightarrow -4 + y + 3 = 0 \Rightarrow y = 1$$ Step 4: Compute second derivatives: $$f_{xx} = 2, \quad f_{yy} = 0, \quad f_{xy} = 1$$ Step 5: Compute Hessian determinant: $$D = f_{xx} f_{yy} - (f_{xy})^2 = 2 \cdot 0 - 1^2 = -1 < 0$$ Step 6: Since \(D < 0\), critical point \((-2,1)\) is a saddle point. (ii) \(f(x,y) = x^2 - 4xy + y^2 + 6y + 2\) Step 1: Compute partial derivatives: $$f_x = 2x - 4y, \quad f_y = -4x + 2y + 6$$ Step 2: Set derivatives to zero: $$2x - 4y = 0 \Rightarrow x = 2y$$ $$-4x + 2y + 6 = 0$$ Step 3: Substitute \(x=2y\) into second equation: $$-4(2y) + 2y + 6 = 0 \Rightarrow -8y + 2y + 6 = 0 \Rightarrow -6y = -6 \Rightarrow y = 1$$ Step 4: Then \(x = 2(1) = 2\). Step 5: Compute second derivatives: $$f_{xx} = 2, \quad f_{yy} = 2, \quad f_{xy} = -4$$ Step 6: Hessian determinant: $$D = 2 \cdot 2 - (-4)^2 = 4 - 16 = -12 < 0$$ Step 7: Since \(D < 0\), critical point \((2,1)\) is a saddle point. (iii) \(f(x,y) = x^3 - y^3 - 2xy + 6\) Step 1: Compute partial derivatives: $$f_x = 3x^2 - 2y, \quad f_y = -3y^2 - 2x$$ Step 2: Set derivatives to zero: $$3x^2 - 2y = 0 \Rightarrow y = \frac{3}{2} x^2$$ $$-3y^2 - 2x = 0$$ Step 3: Substitute \(y\) into second equation: $$-3 \left( \frac{3}{2} x^2 \right)^2 - 2x = 0 \Rightarrow -3 \cdot \frac{9}{4} x^4 - 2x = 0 \Rightarrow -\frac{27}{4} x^4 - 2x = 0$$ Step 4: Factor out \(x\): $$x \left(-\frac{27}{4} x^3 - 2 \right) = 0$$ Step 5: Solutions: \(x=0\) or \(-\frac{27}{4} x^3 - 2 = 0 \Rightarrow x^3 = -\frac{8}{27} \Rightarrow x = -\frac{2}{3}\) Step 6: Corresponding \(y\): - For \(x=0\), \(y=0\) - For \(x=-\frac{2}{3}\), \(y = \frac{3}{2} \left(-\frac{2}{3}\right)^2 = \frac{3}{2} \cdot \frac{4}{9} = \frac{2}{3}\) Step 7: Compute second derivatives: $$f_{xx} = 6x, \quad f_{yy} = -6y, \quad f_{xy} = -2$$ Step 8: Evaluate Hessian determinant \(D = f_{xx} f_{yy} - (f_{xy})^2\) at critical points: - At \((0,0)\): $$D = 6 \cdot 0 \times (-6 \cdot 0) - (-2)^2 = 0 - 4 = -4 < 0$$ saddle point - At \left(-\frac{2}{3}, \frac{2}{3}\right): $$D = (6 \cdot -\frac{2}{3})(-6 \cdot \frac{2}{3}) - 4 = (-4)(-4) - 4 = 16 - 4 = 12 > 0$$ $$f_{xx} = 6 \cdot -\frac{2}{3} = -4 < 0$$ so local maximum. (iv) \(f(x,y) = x^4 + y^4 - 4xy\) Step 1: Compute partial derivatives: $$f_x = 4x^3 - 4y, \quad f_y = 4y^3 - 4x$$ Step 2: Set derivatives to zero: $$4x^3 - 4y = 0 \Rightarrow y = x^3$$ $$4y^3 - 4x = 0 \Rightarrow y^3 = x$$ Step 3: Substitute \(y = x^3\) into second: $$ (x^3)^3 = x \Rightarrow x^9 = x \Rightarrow x^9 - x = 0 \Rightarrow x(x^8 - 1) = 0$$ Step 4: Solutions: $$x=0, \quad x=\pm 1$$ Step 5: Corresponding \(y\): - \(x=0 \Rightarrow y=0\) - \(x=1 \Rightarrow y=1\) - \(x=-1 \Rightarrow y=-1\) Step 6: Compute second derivatives: $$f_{xx} = 12x^2, \quad f_{yy} = 12y^2, \quad f_{xy} = -4$$ Step 7: Hessian determinant: $$D = f_{xx} f_{yy} - (f_{xy})^2 = 12x^2 \cdot 12y^2 - 16 = 144 x^2 y^2 - 16$$ Step 8: Evaluate at critical points: - At (0,0): $$D = 0 - 16 = -16 < 0$$ saddle point - At (1,1): $$D = 144 - 16 = 128 > 0, f_{xx} = 12 > 0$$ local minimum - At (-1,-1): $$D = 144 - 16 = 128 > 0, f_{xx} = 12 > 0$$ local minimum (v) \(f(x,y) = \sqrt{56x^2 - 8y^2 - 16x - 31} + 1 - 8x\) Step 1: Define \(g(x,y) = 56x^2 - 8y^2 - 16x - 31\). Step 2: Domain requires \(g(x,y) \geq 0\). Step 3: To find critical points, compute partial derivatives of \(f\): $$f_x = \frac{1}{2 \sqrt{g}} (112x - 16) - 8, \quad f_y = \frac{1}{2 \sqrt{g}} (-16y)$$ Step 4: Set \(f_x = 0\) and \(f_y = 0\): $$\frac{112x - 16}{2 \sqrt{g}} - 8 = 0 \Rightarrow \frac{112x - 16}{2 \sqrt{g}} = 8$$ $$\frac{-16y}{2 \sqrt{g}} = 0 \Rightarrow y = 0$$ Step 5: From first equation: $$112x - 16 = 16 \sqrt{g}$$ Step 6: Square both sides: $$(112x - 16)^2 = 256 g = 256 (56x^2 - 8y^2 - 16x - 31)$$ Step 7: Substitute \(y=0\) and expand: $$(112x - 16)^2 = 256 (56x^2 - 16x - 31)$$ Step 8: Left side: $$12544 x^2 - 3584 x + 256$$ Right side: $$14336 x^2 - 4096 x - 7936$$ Step 9: Equate and simplify: $$12544 x^2 - 3584 x + 256 = 14336 x^2 - 4096 x - 7936$$ $$0 = 1792 x^2 - 512 x - 8192$$ Step 10: Divide by 64: $$0 = 28 x^2 - 8 x - 128$$ Step 11: Solve quadratic: $$x = \frac{8 \pm \sqrt{64 + 4 \cdot 28 \cdot 128}}{56} = \frac{8 \pm \sqrt{64 + 14336}}{56} = \frac{8 \pm \sqrt{14400}}{56} = \frac{8 \pm 120}{56}$$ Step 12: Solutions: $$x = \frac{128}{56} = \frac{32}{14} = \frac{16}{7} \approx 2.2857, \quad x = \frac{-112}{56} = -2$$ Step 13: Check domain for \(g(x,0)\): - For \(x=2.2857\), $$g = 56 (2.2857)^2 - 16 (2.2857) - 31 > 0$$ - For \(x=-2\), $$g = 56 (4) + 32 - 31 = 224 + 1 = 225 > 0$$ Step 14: Thus critical points at \((16/7, 0)\) and \((-2, 0)\). Step 15: Further classification requires second derivatives and Hessian, which is complex; these points are candidates for extrema. Final answers: - Q1: Flow is compressible (divergence = 1). - Q2: Flow is compressible (divergence = 2). - Q3: (i) incompressible, not irrotational; (ii) irrotational, compressible. - Q4: Tangent plane \(8x + y + 8z = 24\), normal line \(x=1+8t, y=t, z=2+8t\). - Q5: Tangent plane \(7x - 3y - 8z = 26\), normal line \(x=1+7t, y=-1-3t, z=-2-8t\). - Q6: (i) Saddle at (-2,1) (ii) Saddle at (2,1) (iii) Saddle at (0,0), local max at (-2/3, 2/3) (iv) Saddle at (0,0), minima at (1,1) and (-1,-1) (v) Critical points at \((16/7,0)\) and \((-2,0)\) with domain restrictions.