Undergrad Directional Derivative: Why Must Vector Be Unit Vector?

[Mr Davis 97]

I know that $D_{\vec{v}} f = \nabla f \cdot \vec{v}$ is the directional derivative. My question is why must the vector $\vec{v}$ be a unit vector? I am sure there is an obvious answer, but my book doesn't really explain it.

 

[fresh_42]

I know that $D_{\vec{v}} f = \nabla f \cdot \vec{v}$ is the directional derivative. My question is why must the vector $\vec{v}$ be a unit vector? I am sure there is an obvious answer, but my book doesn't really explain it.

It doesn't have to be a unit vector. As long as you don't measure things or express them in coordinates, it can be of any length. If you want to measure different things by the same ruler, then you can and should divide it by its length, but the definition doesn't require it. Most authors, however, use the unit vector in their definition.

 

[Mr Davis 97]

It doesn't have to be a unit vector. As long as you don't measure things or express them in coordinates, it can be of any length. If you want to measure different things by the same ruler, then you can and should divide it by its length, but the definition doesn't require it. Most authors, however, use the unit vector in their definition.

Can you elaborate on what would go wrong if I did the multiplication with a non-unit vector? What about $\vec{v}$ being a unit vector allows me to measure things correctly?

 

[fresh_42]

If you have a function $f : \mathbb{R}^n \rightarrow \mathbb{R}$ then $\nabla_v f(x) = \nabla f(x) \cdot v$ which gives you a number that depends on the length of $v$. So if you want to compare the behavior of the directional derivative at different points $x$ and $y$ in a certain direction $v$, then $v$ should always be the same (of any length). And if you want to compare the behavior at the same point $x$ in different directions $v$ and $w$, then you should also have the length of them in mind. Depending on what you want to calculate, they don't necessarily have to be of the same length, e.g. the velocity of driving through a curve. But if you only want to know the rate of change, then this is a division by their length.

It's as always: it depends on what you want to do. Personally I don't see any advantage in the restriction to unit vectors, the limit
$$
\nabla_v f(x) = \lim_{h \to 0} \frac{f(x+hv)-f(x)}{h}
$$
doesn't need it.

One can write the entire thing as $\nabla_v f(x) = f(x+v) - f(x) - r(v)$ where the remainder $r(v)$ means: vanishes faster than linear when approaching zero. So instead of saying "faster than linear" one can write $r(\frac{v}{||v||}) \rightarrow 0$ where the "faster than linear" aspect is divided beforehand.

 

[jedishrfu]

The unit vector approach fits well with physics problems which are always focused on measurement of things and since we all know physics is the most important of sciences that pretty much sums it up.

 

[member 587159]

The unit vector approach fits well with physics problems which are always focused on measurement of things and since we all know physics is the most important of sciences that pretty much sums it up.

That's dangerous thing to say in the math section :P

 

[LCKurtz]

If you choose $\vec v$ to be in the $x$ or $y$ direction, and you want the respective directional derivatives to be $f_x$ and $f_y$, you need $\vec v$ to be a unit vector. To scale it otherwise in a calculus class seems to me to be very unhelpful.

 

[nuuskur]

It is necessary. Furthermore, said formula is valid when f is differentiable at the point.

 

[JonnyG]

No, it doesn't need to be a unit vector. As you know, given $f: \mathbb{R}^m \rightarrow \mathbb{R}^n$, then we define its directional derivative at the point $a \in \mathbb{R}^m$ in the direction $v \in \mathbb{R}^m$ by $D_v f = \frac{d}{dt} \vert_{t = 0} f(a + tv) = \lim\limits_{t \rightarrow 0} \frac{f(a+tv) - f(a)}{t}$, provided this limit exists. As you can see, we are differentiating $f$ along the curve $a + tv$. If, for example, you decided you wanted to take the directional derivative in the direction of $2v$ then you would have $D_{2v}f = \frac{d}{dt}\vert_{t=0} f(a + 2tv)$, so we are differentiating the function along the curve $a + 2tv$. Now what's the difference between the curve $a + tv$ and the curve $a + 2tv$? The image of both curves are the same, but their velocities are different. The first curve has a velocity of $v$ while the second curve has a velocity of $2v$ (just differentiate both curves with respect to $t$). So when we take the directional derivative of $f$ along the second curve, you can think of the function $f$ traveling twice as fast as it would along the first curve.

In fact, the directional derivative operator is linear, so you immediately have $D_{2v}f = 2D_v f$, which shows that scaling $v$ just scales the directional derivative. So if $f$ travels along the curve $a + 2tv$, then it is traveling twice as fast as it would along the curve $a + tv$, which is why we have $D_{2v} f = 2D_v f$.