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

Question

Find all points at which the direction of fastest change of the function   f(x,y) = x^2+y^2-2x-4y  is  <1,1>

anonymous · Answer

I know I should get the gradient vector of f... That is:
$$\nabla f(x,y)=<2x-2,2y-4>$$
Now what?

anonymous · Answer

$$\nabla f(x,y)=(2x-2,2y-4)=\lambda(\hat{i}+\hat{k}),\lambda>0$$
So
$$2x-2=2y-4$$
Hence
$$y=x+1$$
Thus, all points lie on this line.

anonymous · Answer

Typo which $$(\hat{i}+\hat{k})$$ should be $$(\hat{i}+\hat{j})$$

anonymous · Answer

@agostino, you're correct.
Thanks for the help, guys!

anonymous · Answer

@agostino, could you perhaps explain the first line of your answer to me?

anonymous · Answer

The direction is literally given as (1,1) hence i+j

anonymous · Answer

Okay, and what does the part with the lamdas mean?

anonymous · Answer

We just know the direction but not the magnitude, hence multiplying by an arbitrary real number.

anonymous · Answer

Oh is see...
"something" times (i + j) gives the values we want.  This "something" is > 0, and since "someting" is multiplied to both i and j, we can set the two equal to each other?

anonymous · Answer

Exactly.

anonymous · Answer

Thanks!