# Understanding directional derivative

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 $\vec{v}$?

• It should not say "this direction", that doesn't make sense. Your edit is ok, except I would say "rate of change" or "instantaneous rate of change" rather than just "change". – littleO Oct 9 '17 at 21:10
• $v$ should be a unit vector, or if it is not the directional derivative is $\frac {v}{\|v\|}\cdot \nabla f$ – Doug M Oct 9 '17 at 21:13
• @DougM unfortunately I've observed that there is no universal convention on whether the direction of the directional derivative must be a unit vector or not... – user7530 Oct 9 '17 at 21:14
• One benefit of allowing $v$ to be a non-unit vector is that then the function $v \mapsto D_v f(x)$ is a linear transformation. – littleO Oct 9 '17 at 21:17
• This image may help. – Andrew Tawfeek Oct 9 '17 at 21:18

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.

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:

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

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

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