# 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 $\displaystyle f(x,y)=x^2+y^2-2x+4y$ is $\displaystyle \vec i + \vec j$

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

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

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

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

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.