Find all $x,y$ for which $\nabla f$ forms an angle of $45°$ with the vector $(1,1)$ Let $f=x^2+y^2 \implies \nabla f=(2x,2y)$.
Find all $x,y$ for which $\nabla f$ forms an angle of $45°$ with the vector $(1,1)$.
So I thought of taking the dot product $2x+2y=\nabla f \cdot v=||\nabla f|\cdot||v|| \cos(45°)=4\sqrt{x^2+y^2}$.
But the I couldn't solve $x+y=2\sqrt{x^2+y^2}$, I know, from inspection that $x=-y$ is a solution, and I think there are no other solutions, but I don't know how to prove it.
Could you guys help me out?
E: Oh god, I see where my mistake was... I miscalculated $\cos(45°)$ as $\sqrt 2$.

Since you are using cartesian co-ordinates, $(1,1)$ forms an angle of $45°$ with $x$ and $y$ co-ordinates.
So $\nabla f$ will form a $45°$ angle with $(1,1)$ if it lies on the positive x or y axis exclusively.
So $\nabla f = c \vec i$
or $\nabla f = c\vec j$
$c > 0$ is any scalar.
so $(c, 0)$ and $(0, c)$.
