# At what points is the direction of fastest change of the function f(x,y)=x^2+

At what points is the direction of fastest change of the function $$\displaystyle{f{{\left({x},{y}\right)}}}={x}^{{2}}+{y}^{{2}}-{4}{x}-{8}{y}$$ is i+j.

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avortarF
The needed direction is determined by $$\displaystyle\nabla{f{{\left({x},{y}\right)}}}$$
$$\displaystyle\nabla{f{{\left({x},{y}\right)}}}={<}\frac{{\partial{f}}}{{\partial{x}}},\frac{{\partial{f}}}{\partial}{y}{)}{>}$$
$$\displaystyle\frac{{\partial{f}}}{{\partial{x}}}=\frac{{\partial}}{{\partial{x}}}{\left({x}^{{2}}+{y}^{{2}}-{4}{x}-{8}{y}\right)}$$
=2x-4
$$\displaystyle\frac{{\partial{f}}}{{\partial{y}}}=\frac{{\partial}}{{\partial{y}}}{\left({x}^{{2}}+{y}^{{2}}-{4}{x}-{8}{y}\right)}$$
=2y-8
Therefore,
$$\displaystyle\nabla{f{{\left({x},{y}\right)}}}={<}\frac{{\partial{f}}}{{\partial{x}}},\frac{{\partial{f}}}{\partial}{y}{)}{>}$$
$$\displaystyle={<}{2}{x}-{4},{2}{y}-{8}{>}$$
gradf(x,y) has to be parallel to i+j
The direction vector of i+j is $$\displaystyle{<}{1},{1}{>}$$
Thus,
$$\displaystyle{<}{2}{x}-{4},{2}{y}-{8}{>}$$ has to be parallel to <1, 1>
Since in $$\displaystyle{<}{1},{1}{>}$$ the coordinates are equal, 2x-4=2y-8
2x-4=2y-8
2(x-2)=2(y-4)
x-2=y-4
y=x+2