So here is the exam question:
Find an equation for all the points $(x,y)$ for which the function $f(x,y)=x^2+y^2−3x−y$ has a maximum rate of change in a direction parallel to the vector $u = (1, 1)$
And here is my attempt to solve it:
we take the partial derivatives: $$ \dfrac {d}{\delta x} = 2x-3$$ $$ \dfrac {d}{\delta y} = 2y-1 $$
now we find the unit vector $\dfrac {\left|\vec{v} \right|} {\left|v \right|}$ $= \dfrac {1} {\sqrt 2}$
now use the formula that says: $$D_uf(a,b) = f_u′ (a,b) = u·\nabla f(a,b) = u·(f_x(a,b),f_y(a,b))$$
Hence the equation is $$\dfrac {2x+2y-4}{\sqrt 2} = \sqrt 2 \{x+y-2\}$$ which is wrong according to my professor. Any help will be highly appreciated.
EDIT I sent a PM to my professor and he replied with the following:
Your equation/answer can be rewritten as $0x + 0y = 0$ which is trivially true for ALL points $(x,y)$ in $R^2$ . So, no - this is not the answer.
The problem requires that $\nabla f $ is parallel to $(1,1)$, which can be written as grad $f = k(1,1)$. Eliminate k and you should get the linear eq. $y = x - 1$.
Hint: Indeed your equation has same members on RHS and LHS so just says $0=0$. In order to get a gradient in the direction of $(1,1)$ you may indeed seek to solve $(\partial_x f,\partial_y f)=k(1,1)$ but in two dimension there is an easier way: Namely that the gradient is parallel to $(1,1)$ iff it is orthogonal to $(-1,1)$ so take the scalar product and set it equal to zero.
Let $n=(-1,1)$. A vector $v=(v_x,v_y)$ is parallel to $(1,1)$ iff $-v_x+v_y=0$ iff (scalar product) $n \cdot v=0$. So to ensure that the gradient is parallel to $(1,1)$ we calculate: $$ 0= (\partial_x f,\partial_y f)\cdot n = -(2x-3) + (2y-1) \Leftrightarrow y=x-1.$$