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

1. ## 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 $\vec\nabla f(x,y)=<1,1>$

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

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?

2. ## 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.