1. **Problem statement:** Calculate the second partial derivatives $\frac{\partial^2 f}{\partial x^2}$, $\frac{\partial^2 f}{\partial \theta^2}$, and $\frac{\partial^2 f}{\partial r \partial \theta}$ for the function $$f(x,y) = \frac{x^2 + y^2 - x}{x^2 + y^2}$$ where $x = r \cos \theta$ and $y = r \sin \theta$. 2. **Rewrite the function:** Note that $x^2 + y^2 = r^2$, so $$f(x,y) = \frac{r^2 - x}{r^2} = 1 - \frac{x}{r^2}$$ Substitute $x = r \cos \theta$: $$f(r,\theta) = 1 - \frac{r \cos \theta}{r^2} = 1 - \frac{\cos \theta}{r}$$ 3. **Calculate $\frac{\partial f}{\partial x}$ and $\frac{\partial^2 f}{\partial x^2}$:** Rewrite $f$ in terms of $x$ and $y$: $$f = \frac{x^2 + y^2 - x}{x^2 + y^2}$$ Let $g = x^2 + y^2$, then $$f = \frac{g - x}{g} = 1 - \frac{x}{g}$$ Calculate first derivative: $$\frac{\partial f}{\partial x} = - \frac{\partial}{\partial x} \left( \frac{x}{g} \right) = - \frac{g \cdot 1 - x \cdot 2x}{g^2} = - \frac{g - 2x^2}{g^2}$$ Calculate second derivative: $$\frac{\partial^2 f}{\partial x^2} = - \frac{\partial}{\partial x} \left( \frac{g - 2x^2}{g^2} \right)$$ Calculate numerator derivative: $$\frac{\partial}{\partial x} (g - 2x^2) = 2x + 2y \frac{\partial y}{\partial x} - 4x = 2x - 4x = -2x$$ Since $y$ is independent of $x$, $\frac{\partial y}{\partial x} = 0$. Calculate denominator derivative: $$\frac{\partial}{\partial x} (g^2) = 2g \cdot 2x = 4x g$$ Use quotient rule: $$\frac{\partial^2 f}{\partial x^2} = - \frac{(-2x) g^2 - (g - 2x^2) 4x g}{g^4} = - \frac{-2x g^2 - 4x g (g - 2x^2)}{g^4}$$ Simplify numerator: $$-2x g^2 - 4x g^2 + 8x^3 g = -6x g^2 + 8x^3 g$$ So $$\frac{\partial^2 f}{\partial x^2} = - \frac{-6x g^2 + 8x^3 g}{g^4} = \frac{6x g^2 - 8x^3 g}{g^4} = \frac{2x g (3g - 4x^2)}{g^4} = \frac{2x (3g - 4x^2)}{g^3}$$ 4. **Calculate $\frac{\partial^2 f}{\partial \theta^2}$:** Recall $$f(r,\theta) = 1 - \frac{\cos \theta}{r}$$ First derivative w.r.t. $\theta$: $$\frac{\partial f}{\partial \theta} = - \frac{1}{r} (-\sin \theta) = \frac{\sin \theta}{r}$$ Second derivative: $$\frac{\partial^2 f}{\partial \theta^2} = \frac{1}{r} \cos \theta$$ 5. **Calculate $\frac{\partial^2 f}{\partial r \partial \theta}$:** First derivative w.r.t. $\theta$ is $$\frac{\partial f}{\partial \theta} = \frac{\sin \theta}{r}$$ Now differentiate w.r.t. $r$: $$\frac{\partial^2 f}{\partial r \partial \theta} = \frac{\partial}{\partial r} \left( \frac{\sin \theta}{r} \right) = - \frac{\sin \theta}{r^2}$$ --- 6. **Problem statement:** Write the equations of the tangent plane and normal line to the surface $$S: x^2 + y^2 = 2z$$ at the point $P(1, -1, 1)$. 7. **Find gradient vector $\nabla F$:** Rewrite surface as $$F(x,y,z) = x^2 + y^2 - 2z = 0$$ Calculate partial derivatives: $$F_x = 2x, \quad F_y = 2y, \quad F_z = -2$$ At $P(1,-1,1)$: $$F_x = 2(1) = 2, \quad F_y = 2(-1) = -2, \quad F_z = -2$$ Gradient vector: $$\nabla F = (2, -2, -2)$$ 8. **Equation of tangent plane:** Using point-normal form: $$2(x - 1) - 2(y + 1) - 2(z - 1) = 0$$ Simplify: $$2x - 2 - 2y - 2 - 2z + 2 = 0 \Rightarrow 2x - 2y - 2z - 2 = 0$$ Divide by 2: $$x - y - z - 1 = 0$$ 9. **Equation of normal line:** Parametric form using gradient vector as direction: $$x = 1 + 2t, \quad y = -1 - 2t, \quad z = 1 - 2t$$ **Final answers:** $$\frac{\partial^2 f}{\partial x^2} = \frac{2x (3(x^2 + y^2) - 4x^2)}{(x^2 + y^2)^3}$$ $$\frac{\partial^2 f}{\partial \theta^2} = \frac{\cos \theta}{r}$$ $$\frac{\partial^2 f}{\partial r \partial \theta} = - \frac{\sin \theta}{r^2}$$ Tangent plane: $$x - y - z - 1 = 0$$ Normal line: $$x = 1 + 2t, \quad y = -1 - 2t, \quad z = 1 - 2t$$