So I just had this question asked on an exam, but I'm almost sure that it was ill-posed or impossible to solve:
Given the function $$\displaylines{ f:{{R}^2} \to {R} \cr (x,y) \mapsto {x^2} + xy + {y^2} \cr} $$
I was asked the directional derivative at the point (9, 25000) in the direction of $\vec v = (a,b,c)$ and $a,b,c \ne 0$ (I don't remember the exact values of $a, b$ and $c$ because I had to turn in the exam but I'm pretty sure it doesn't matter)
So, what I would normally do is find the gradient vector of $f$ given by $\nabla f$ and evaluate it at the given point $(9,25000)$. Then, calculating the dot product of the directional unit vector $\vec w = \alpha (a,b,c)$ such that $\left\| {\alpha (a,b,c)} \right\| = 1$.
Okay, then $\nabla {f_{(9,25000)}} \cdot \vec w$ should yield the answer.
However, $\nabla {f_{(9,25000)}}$ belongs to $ {R}^2$ while $\vec w$ belongs to ${R}^3$. Therefore, the question doesn't make sense as the direction is given by a vector in a higher-order space than the function's domain. Am I missing something?
The only thing the I can think is if you consider the surface $$F(x,y,f(x,y))=f(x,y)-x^2-xy-y^2=0,$$ and may be the questiong was asking for the derivative of $F$ at the point $$P=(9,2500,f(9,2500)),$$ which it is zero because $F$ is constant.
With this approach at least one can calculate the derivative in the direction of a vector belonging to $\mathbb{R}^3$, but I am aware that this is a little made up.