Let φ(x, y) = arctan (y/x) . (a) Find the vector field F := ∇φ, and simplify its components as much as possible. (b) What is the domain of definition of F? (c) Find all the points (a, b) (belonging to the domain of definition of F) at which the vector F(a, b) is parallel to the line y = x. (Note: For a real number u, arctan u is also denoted by tan−1 u. Use the notation that you prefer.)

Answer:a) [tex]\large F(x,y)=(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2})[/tex]b) [tex]\large \mathbb{R}^2-\{(0,0)\}[/tex]c) the points of the form (x, -x) for x≠0 Step-by-step explanation:a) If φ(x, y) = arctan (y/x), the vector field F = ∇φ would be [tex]\large F(x,y)=(\frac{\partial \phi}{\partial x},\frac{\partial \phi}{\partial y})[/tex] On one hand we have, [tex]\large \frac{\partial \phi}{\partial x}=\frac{\partial arctan(y/x)}{\partial x}=\frac{-y/x^2}{1+(y/x)^2}=-\frac{y/x^2}{1+y^2/x^2}=\\\\=-\frac{y/x^2}{(x^2+y^2)/x^2}=-\frac{y}{x^2+y^2}[/tex] On the other hand, [tex]\large \frac{\partial \phi}{\partial y}=\frac{\partial arctan(y/x)}{\partial y}=\frac{1/x}{1+(y/x)^2}=\frac{1/x}{1+y^2/x^2}=\\\\=\frac{1/x}{(x^2+y^2)/x^2}=\frac{x}{x^2+y^2}[/tex] So [tex]\large F(x,y)=(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2})[/tex] b) The domain of definition of F is  [tex]\large \mathbb{R}^2-\{(0,0)\}[/tex] i.e., all the plane X-Y except the (0,0) c) Here we want to find all the points such that [tex]\large (-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2})=(k,k)[/tex] where k is a real number other than 0. But this means [tex]\large -\frac{y}{x^2+y^2}=\frac{x}{x^2+y^2}\Rightarrow y=-x[/tex] So, all the points in the line y = -x except (0,0) are parallel to the vector field F, that is, the points (x, -x) with x≠ 0