Time derivatives of unit vectors are always zero since unit vectors’ lengths are constant, regardless of coordinate system. True False
The correct answer and explanation is:
Correct Answer: False
Explanation (Approx. 300 words):
The statement “Time derivatives of unit vectors are always zero since unit vectors’ lengths are constant, regardless of coordinate system” is false because it confuses the constancy of a vector’s magnitude with the constancy of its direction.
While a unit vector indeed has a constant magnitude of 1, its direction can change over time, especially in non-Cartesian (curvilinear) coordinate systems, such as polar, cylindrical, or spherical coordinates. The time derivative of a vector accounts for changes in both magnitude and direction.
In Cartesian coordinates, the standard unit vectors i^,j^,k^\hat{i}, \hat{j}, \hat{k} (pointing in the x, y, and z directions) are constant in both magnitude and direction. Thus, their time derivatives are indeed zero: di^dt=dj^dt=dk^dt=0\frac{d\hat{i}}{dt} = \frac{d\hat{j}}{dt} = \frac{d\hat{k}}{dt} = 0
However, in polar coordinates, the unit vectors r^\hat{r} and θ^\hat{\theta} change direction as the point moves. Their magnitudes remain 1, but the directions vary with the angular position θ(t)\theta(t). The time derivatives of these unit vectors are not zero: dr^dt=θ˙θ^,dθ^dt=−θ˙r^\frac{d\hat{r}}{dt} = \dot{\theta} \hat{\theta}, \quad \frac{d\hat{\theta}}{dt} = -\dot{\theta} \hat{r}
This shows that the directional change leads to a non-zero time derivative of the unit vector, even though its length stays constant.
Hence, it is incorrect to assume that the time derivative of a unit vector is always zero just because its magnitude doesn’t change. The key is whether the vector’s direction changes with time, which it often does in curvilinear coordinate systems.
Conclusion: The time derivative of a unit vector is not always zero. It depends on the coordinate system and whether the vector’s direction changes over time.