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Time Derivative of Unit Vectors

  1. Jan 23, 2016 #1
    Quick question (a little rusty on this): Why don't unit vectors in Cartesian Coordinates not change with time? For example, suppose [tex] \mathbf{r} (t) = x(t) \mathbf{x} + y(t) \mathbf{y} + z(t) \mathbf{z}[/tex] How exactly do we know that the unit vectors don't change with time?

    Or in other words, what is the argument that justifies this expression: [tex] \frac{d }{dt}\mathbf{x} = 0[/tex]
    Last edited: Jan 23, 2016
  2. jcsd
  3. Jan 23, 2016 #2
    They can change with time. [itex] \mathbb{r(t)} = \sin{t}\mathbb{x} + \cos{t}\mathbb{y} [/itex] is a unit vector that changes with time. If none of the components depend on time, the derivative will be 0. Otherwise the derivative will be another vector.
  4. Jan 23, 2016 #3
    I'm talking about the unit vectors in Cartesian coordinates themselves [tex] \mathbf{e}_1 = \mathbf{x}, \mathbf{e}_2 = \mathbf{y}, \mathbf{e}_3 = \mathbf{z}[/tex]
  5. Jan 23, 2016 #4
    ##\mathbb{x}## for example can be defined as the vector from the origin to the point ## (1,0,0) ##. Since the two points are not changing with time, the vector wont change with time either.
  6. Jan 23, 2016 #5
    okay, that's true, now I remember. Thanks.
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