# why normalize and the definition of directional derivative

I'm not understanding how to solve this problem.

I think the problem lies in the fact that I don't understand why normalize and the definition of directional derivative... It's in Portuguese. The question would be: Determine the directional derivative of f in P in direction of vector u: • What do you not understand about calculating a directional derivative? – Pockets May 25 '14 at 23:39
• Normalization is cosmetic. Some authors to not require normalization. Your lecturer apparently does, that's just how he defines it. – Git Gud May 25 '14 at 23:40
• @GitGud: I think there's more to it than that. normalization makes the units easier to interpret: it's the rate of change of the function per unit distance in the specified direction. you also can't compare the rate of change in different directions unless you use vectors of the same length. – symplectomorphic May 25 '14 at 23:45
• @symple: If you want to know how much you change by moving a unit distance, you ask for the directional derivative of a unit vector. If you ask for the directional derivative of a vector that has length 2, you want to know how much you change by moving 2 units. – user14972 May 26 '14 at 0:20
• @Hurkyl: I'm aware... maybe I shouldn't have said "easier" to interpret, but more convenient. – symplectomorphic May 26 '14 at 0:21

The directional derivative of a scalar function $f:\mathbb{R}^n\to\mathbb{R}$ at a point $P$ in the direction of a vector $\mathbf{u}$ is usually (but not always) defined as $$\nabla f_P\cdot\frac{\mathbf{u}}{\|\mathbf{u}\|}$$ where $\nabla f_P$ is the gradient of $f$ evaluated at $P$.

The normalization in this definition is motivated for at least two reasons:

1. Normalizing allows you to interpret the directional derivative as the rate of change of the function per unit distance in the direction of $\mathbf{u}$.

2. You can't meaningfully compare the rates of change of the function in different directions unless you use vectors of the same length. So why not use vectors of unit length? That correctly captures the fact that a directional derivative should depend only on the direction of the vector, not on its magnitude.

EDIT

There is a nice footnote on the issue in Hubbard's book:

To compare derivatives in different directions one must first normalize the vectors to have the same length. If $\mathbf{v_1}$ is a kilometer long, $\mathbf{v_2}$ is a centimeter long, and "time 1" is a minute, then $\mathbf{x}$ is traveling 60 kilometers an hour in the direction $\mathbf{v_1}$ compared to 60 centimeters an hour in direction $\mathbf{v_2}$; knowing how much $\mathbf{f(x)}$ varies when $\mathbf{x}$ travels one minute in these two directions does not tell you in which direction $\mathbf{f(x)}$ is varying fastest at time 0 (i.e., as $h\to0$). For this reason, some authors allow only vectors $\mathbf{v}$ of length 1 to be used to define directional derivatives. We feel this restriction is undesirable, as it loses the essential linear character of the directional derivative as a function of $\mathbf{v}$.

• But some of us prefer the directional derivative to be linear in $\mathbf u$, so don't act like this is universal. – Ted Shifrin May 26 '14 at 0:03
• @Ted: I thought the comments on the question made it clear that it wasn't universal, but I've revised my answer. – symplectomorphic May 26 '14 at 0:05
• Another definition of directional derivative that I think is instructive is: the directional derivative of $f$ in the direction of $\mathbf{u}$ at the point $\mathbf{x}$ is $\frac{d}{dt}\mid_{t=0} f(\mathbf{x} + t \mathbf{u})$. If $\mathbf{u}$ is a unit vector, this is equivalent to the definition above. – Phillip Andreae May 26 '14 at 0:14
• @user49048: just to be clear, even if $\mathbf{u}$ isn't a unit vector, your definition still agrees with taking the dot product of $\nabla f$ and $\mathbf{u}$. – symplectomorphic May 26 '14 at 0:16