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Math Help - Find at which the direction of fastest change of F is <1,1>. Gradient problem.

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    Find at which the direction of fastest change of F is <1,1>. Gradient problem.

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

    So I think the question is asking me to find the points in \mathbb{R}^2 such that \vec\nabla f(x,y)=<1,1>

    \vec\nabla f(x,y)=<f_x,f_y>; f_x = 2x+2, f_y = 2y+4
    \vec\nabla f(x,y)=<2x+2,2y+4>

    The text is saying all points on y =x+1 work. i'm only getting the one point (\frac{3}{2},\frac{5}{2}).

    While (\frac{3}{2},\frac{5}{2}) is on y =x+1 something like (2,3) is on y =x+1 but the gradient at that point is <1,5> which isn't equal to <1,1>?

    Would someone care to explain to me what I'm doing wrong?
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    Re: Find at which the direction of fastest change of F is <1,1>. Gradient problem.

    OK I think I've found something that fixes my problem with this question.

    I wrote the problem down wrong! This kind of makes sense to me now.
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