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Find all points at which the direction of fastest change of rate of the function $f(x,y) = x^2+y^2-2x-4y$ is $i+j$.

So $\nabla f(x,y) = \langle2x-2, 2y-4 \rangle$ and $u = \dfrac{i}{\sqrt{2}} + \dfrac{j}{\sqrt{2}}$. How can I further solve this? The fastest change of rate must be in the direction of the gradient itself, but how can I use that info?

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You just have to find all points (x,y) $\in$ $\mathbb{R}^2$ such that $\nabla{f}$(x,y) has direction $i+j$.
$\nabla{f}(x,y)=(2x-2,2y-4)$
so all points such that $\nabla{f}(x,y)=(2x-2,2y-4)$ = (t , t) , $t\in \mathbb{R}\setminus$ {0} have direction of fastest change along $i+j$ .
So we get x = (2+t)/2 and y = (4+t)/2 , $t\in \mathbb{R}\setminus$ {0}. So the points satisfying such conditions are all the points on line y = 1 + x.

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