1. **Find $\frac{\partial u}{\partial x}$ and $\frac{\partial u}{\partial y}$ for $u = x^2 - y^2$, and evaluate at $(-2,-2)$.** - The function is $u = x^2 - y^2$. - Partial derivatives treat other variables as constants. - Formula: $\frac{\partial u}{\partial x} = \lim_{h \to 0} \frac{u(x+h,y) - u(x,y)}{h}$. Step 1: Differentiate w.r.t. $x$: $$\frac{\partial u}{\partial x} = 2x$$ Step 2: Differentiate w.r.t. $y$: $$\frac{\partial u}{\partial y} = -2y$$ Step 3: Evaluate at $(-2,-2)$: $$\frac{\partial u}{\partial x}(-2,-2) = 2(-2) = -4$$ $$\frac{\partial u}{\partial y}(-2,-2) = -2(-2) = 4$$ 2. **Discuss continuity of $f(x,y) = \begin{cases} \frac{x^2 - y^2}{x^2 + y^2}, & (x,y) \neq (0,0) \\ 0, & (x,y) = (0,0) \end{cases}$ at $(0,0)$.** - To check continuity at $(0,0)$, check if $\lim_{(x,y) \to (0,0)} f(x,y) = f(0,0) = 0$. Step 1: Use polar coordinates $x = r\cos\theta$, $y = r\sin\theta$: $$f(r,\theta) = \frac{r^2 \cos^2\theta - r^2 \sin^2\theta}{r^2} = \cos^2\theta - \sin^2\theta = \cos(2\theta)$$ Step 2: Limit as $r \to 0$ depends on $\theta$: $$\lim_{r \to 0} f(r,\theta) = \cos(2\theta)$$ Since this depends on $\theta$, the limit does not exist. Step 3: Therefore, $f$ is **not continuous** at $(0,0)$. 3. **Show $f(x,y) = \frac{2x^2 y}{x^4 + y^2}$ has no limit as $(x,y) \to (0,0)$.** Step 1: Approach along $y = 0$: $$f(x,0) = \frac{2x^2 \cdot 0}{x^4 + 0} = 0$$ Step 2: Approach along $y = x^2$: $$f(x,x^2) = \frac{2x^2 \cdot x^2}{x^4 + x^4} = \frac{2x^4}{2x^4} = 1$$ Step 3: Limits differ (0 vs 1), so limit does not exist. 4. **If $u = x^2 y + y^2 z + z^2 x$, find $\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z}$.** Step 1: Compute each partial derivative: $$\frac{\partial u}{\partial x} = 2xy + z^2$$ $$\frac{\partial u}{\partial y} = x^2 + 2yz$$ $$\frac{\partial u}{\partial z} = y^2 + 2zx$$ Step 2: Sum them: $$2xy + z^2 + x^2 + 2yz + y^2 + 2zx$$ 5. **Check continuity of $f(x,y) = \begin{cases} \frac{x^2 y^2}{4x^2 + 5 y^2}, & (x,y) \neq (0,0) \\ 0, & (x,y) = (0,0) \end{cases}$ at $(0,0)$.** Step 1: Use polar coordinates $x = r\cos\theta$, $y = r\sin\theta$: $$f(r,\theta) = \frac{r^4 \cos^2\theta \sin^2\theta}{4 r^2 \cos^2\theta + 5 r^2 \sin^2\theta} = \frac{r^4 \cos^2\theta \sin^2\theta}{r^2 (4 \cos^2\theta + 5 \sin^2\theta)} = r^2 \frac{\cos^2\theta \sin^2\theta}{4 \cos^2\theta + 5 \sin^2\theta}$$ Step 2: As $r \to 0$, $f(r,\theta) \to 0$ for all $\theta$. Step 3: Since $f(0,0) = 0$, $f$ is **continuous** at $(0,0)$. 6. **Find $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ at $(1,2)$ for $f(x,y) = x^2 + 2xy + 3y^2 - 1$.** Step 1: Compute partial derivatives: $$\frac{\partial f}{\partial x} = 2x + 2y$$ $$\frac{\partial f}{\partial y} = 2x + 6y$$ Step 2: Evaluate at $(1,2)$: $$\frac{\partial f}{\partial x}(1,2) = 2(1) + 2(2) = 2 + 4 = 6$$ $$\frac{\partial f}{\partial y}(1,2) = 2(1) + 6(2) = 2 + 12 = 14$$ 7. **Find $\frac{\partial u}{\partial x}$ and $\frac{\partial u}{\partial y}$ at $(-2,-2)$ for $f(x,y) = x^2 - y^2$.** (Same as problem 1) $$\frac{\partial u}{\partial x} = 2x, \quad \frac{\partial u}{\partial y} = -2y$$ At $(-2,-2)$: $$\frac{\partial u}{\partial x} = 2(-2) = -4$$ $$\frac{\partial u}{\partial y} = -2(-2) = 4$$ 8. **Find equations of tangent plane and normal line to surface $\frac{x^2}{4} + y^2 + \frac{z^2}{9} = 3$ at $(-2,1,-3)$.** Step 1: Define function: $$F(x,y,z) = \frac{x^2}{4} + y^2 + \frac{z^2}{9} - 3 = 0$$ Step 2: Compute gradient $\nabla F$: $$\frac{\partial F}{\partial x} = \frac{x}{2}, \quad \frac{\partial F}{\partial y} = 2y, \quad \frac{\partial F}{\partial z} = \frac{2z}{9}$$ Step 3: Evaluate gradient at $(-2,1,-3)$: $$\nabla F(-2,1,-3) = \left( \frac{-2}{2}, 2(1), \frac{2(-3)}{9} \right) = (-1, 2, -\frac{2}{3})$$ Step 4: Equation of tangent plane: $$-1(x + 2) + 2(y - 1) - \frac{2}{3}(z + 3) = 0$$ Simplify: $$-x - 2 + 2y - 2 - \frac{2}{3}z - 2 = 0$$ $$-x + 2y - \frac{2}{3}z - 6 = 0$$ Or: $$-x + 2y - \frac{2}{3}z = 6$$ Step 5: Parametric equations of normal line: $$x = -2 - t, \quad y = 1 + 2t, \quad z = -3 - \frac{2}{3} t$$ --- Final answers summarized: 1. $\frac{\partial u}{\partial x} = 2x$, $\frac{\partial u}{\partial y} = -2y$, at $(-2,-2)$: $-4$, $4$. 2. $f$ not continuous at $(0,0)$. 3. Limit does not exist for $f(x,y) = \frac{2x^2 y}{x^4 + y^2}$. 4. Sum of partials: $2xy + z^2 + x^2 + 2yz + y^2 + 2zx$. 5. $f$ continuous at $(0,0)$. 6. $\frac{\partial f}{\partial x}(1,2) = 6$, $\frac{\partial f}{\partial y}(1,2) = 14$. 7. Same as 1. 8. Tangent plane: $-x + 2y - \frac{2}{3}z = 6$, normal line parametric: $x = -2 - t$, $y = 1 + 2t$, $z = -3 - \frac{2}{3} t$.