1. Problem: Sketch the level curves of the function $$f(x,y) = \sqrt{9 - x^2 - y^2}$$ for $$k = 0,1,2,3$$. Step 1: Set $$f(x,y) = k$$, so $$\sqrt{9 - x^2 - y^2} = k$$. Step 2: Square both sides: $$9 - x^2 - y^2 = k^2$$. Step 3: Rearrange: $$x^2 + y^2 = 9 - k^2$$. Step 4: For different values of $$k$$: - When $$k=0$$, $$x^2 + y^2 = 9$$ (circle radius 3). - When $$k=1$$, $$x^2 + y^2 = 8$$ (circle radius $$\sqrt{8}$$). - When $$k=2$$, $$x^2 + y^2 = 5$$ (circle radius $$\sqrt{5}$$). - When $$k=3$$, $$x^2 + y^2 = 0$$ (point at origin). These are concentric circles with decreasing radius. 2. Problem: Examine the continuity of $$f(x,y) = \begin{cases}\frac{x^3 + y^3}{x^2 + y^2}, &(x,y) \neq (0,0) \\ 0, &(x,y) = (0,0)\end{cases}$$ at $$(0,0)$$ and check if $$f_x(0,0)$$ and $$f_y(0,0)$$ exist. Step 1: Check continuity at $$(0,0)$$ by checking $$\lim_{(x,y) \to (0,0)} f(x,y)$$. Step 2: Use polar coordinates $$x = r\cos\theta$$, $$y = r\sin\theta$$: $$f(r,\theta) = \frac{r^3(\cos^3\theta + \sin^3\theta)}{r^2} = r(\cos^3\theta + \sin^3\theta)$$. Step 3: As $$r \to 0$$, $$f(r,\theta) \to 0$$ regardless of $$\theta$$, so the limit is 0. Step 4: Since $$f(0,0) = 0$$, $$f$$ is continuous at $$(0,0)$$. Step 5: Find partial derivatives at $$(0,0)$$: - $$f_x(0,0) = \lim_{h \to 0} \frac{f(h,0) - f(0,0)}{h} = \lim_{h \to 0} \frac{\frac{h^3}{h^2} - 0}{h} = \lim_{h \to 0} \frac{h - 0}{h} = 1$$. - $$f_y(0,0) = \lim_{k \to 0} \frac{f(0,k) - f(0,0)}{k} = \lim_{k \to 0} \frac{\frac{k^3}{k^2} - 0}{k} = \lim_{k \to 0} \frac{k - 0}{k} = 1$$. Hence both exist and equal 1. 3. Problem: Find $$\lim_{(x,y) \to (0,1)} \tan^{-1} \left[ \frac{x^2 +1}{(x^2 + (y-1)^2)^2} \right]$$. Step 1: Let $$h = y - 1$$, so the limit becomes $$\lim_{(x,h) \to (0,0)} \tan^{-1} \left[ \frac{x^2 + 1}{(x^2 + h^2)^2} \right]$$. Step 2: Notice denominator tends to zero, numerator tends to 1: $$\frac{x^2 + 1}{(x^2 + h^2)^2} \to \infty$$ as $$(x,h) \to (0,0)$$. Step 3: Therefore, inside arctan goes to $$+\infty$$, so limit equals $$\frac{\pi}{2}$$. 4. Problem: Sketch domain of (a) $$g(x,y) = \ln(1 - x^2 - y^2)$$ Step 1: Domain requires $$1 - x^2 - y^2 > 0$$. Step 2: Rewrite: $$x^2 + y^2 < 1$$, the interior of circle radius 1. Boundary is included (solid line) because $$\ln$$ is undefined at zero. Actually, the boundary $$1 - x^2 - y^2 = 0$$ is excluded as $$\ln(0)$$ undefined. (b) $$w(x,y) = \sqrt{x^2 + y^2 - 4}$$ Step 1: Domain requires $$x^2 + y^2 -4 \geq 0$$ or $$x^2 + y^2 \geq 4$$. Step 2: This is the exterior and boundary of circle radius 2 centered at origin. Step 3: Boundary included (solid line), outside domain is dashed. Summary: - For $$g(x,y)$$ domain is inside circle radius 1 without boundary. - For $$w(x,y)$$ domain is outside circle radius 2 including boundary.