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- Thread starter Mr Davis 97
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fresh_42

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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.

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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?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.

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fresh_42

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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.

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.

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jedishrfu

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

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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 travelling twice as fast as it would along the curve ##a + tv##, which is why we have ##D_{2v} f = 2D_v f##.

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