I have an exercise in my textbook saying the following:
Let $f : R^2 → R$ be defined by $z(x, y) = x^3 − 2x^2y + xy^2 + 1$. Find the directional derivative at $(1, 2)$ along the direction towards $(4, 6)$.
While I understand the process and the evaluation process, I don't understand my teachers approach to finding the unit vector.
According to the solution it is this:
We have $x = (1, 2), y = (3, 4)$ because the direction is given by the difference of the two points.
Yet, using the normal approach, I would get
$u=\frac{(4,6)}{\sqrt{4^2+6^2}} = (\frac{2}{\sqrt{13}},\frac{3}{\sqrt{13}})$
Meaning his solution is $5$, while mine is $\frac{4\sqrt{13}}{13}$
Who is right and if he is, why is my approach wrong and how does his' work?
I think you are wrong. $(4,6)$ is not the direction, it’s a point and you have to find the direction by subtracting your two points.
We want to find the directional derivative in the direction $<3,4>$ since it’s a direction you have to take to reach that point. (This was found by subtracting our two points to find the vector that points in the correct direction).
The unit vector is $<3/5,4/5>$. The gradient is $(f_x,f_y)$ and is equal to $(-1,2)$ at your point.
Take the dot product of your gradient vector (normal to your level curve) with the unit normal vector to find the answer.