Understanding directional derivative

Question

From my notes:
Given a vector $\vec{v}=(v_1,v_2)$ on $\mathbb{R}^2$, the directional derivative $\nabla_\vec{v}f$ of the function $f=f(x,y)$ in the direction of $f$ is defined by: $$\nabla_\vec{v}f=\vec{v}\cdot\nabla f=v_1f_x+v_2f_y$$ where $|\cdot|$ denotes the inner product. This direction gives the change of $f$ in the direction of $\vec{v}$.

Now, I don't fully understand the last sentence. I don't know what "this direction" is referring to, and am getting confused. Is it referring to the directional derivative - in which case, the sentence says the following?

The directional derivative gives the change of function f in the direction of $\vec{v}$?

Answer 1

Suppose you are standing on a hillside.

In one direction is "up the hill", in the opposite direction is "down the hill" In another direction is across the hill.

The direction derivative then indicates the slope of the hill for a given direction.

Answer 2

Yes, that is right. The directional derivative $\nabla_v f(x)$ gives the rate of change of $f$ at $x$ in the $v$ direction. Imagine drawing a line through the point $x$ in the $v$ direction; the 2D function $f$ restricts to a one-dimensional function on the line. The directional derivative then measures the ordinary, 1D derivative of the restricted function along that line. In fact, when I teach I usually define the directional derivative in this way:

$$\nabla_v f(x) = \left(\frac{d}{dt} f(x + tv)\right)\Bigg\vert_{t\to 0}.$$

Answer 3

Consider a line through the point $(x_0,y_0)$ in the direction $(a,b)$ this line is given parametrically by

$$x=x_0+at$$ $$y=y_0+bt.$$ So along this line we the the (one variable !) function $$f(x_0+at,y_0+bt)$$

Taking the derivative and using the chain rule we have

$$\frac{d}{dt} f(x_0+at,y_0+bt)=af_x+bf_y$$