Based on some of his other threads, he is not using normalised basis vectors but ##\vec e_i = \partial_i \vec x## (which I usually would denote ##\vec E_i## or similar to underline that they are not necessarily unit vectors - however, there are some places in the literature where ##\vec e_i## is used and instead ##\hat i## or similar is used for normalised basis vectors). With my preferred notation, where ##\vec e_i## are normalised and ##\vec E_i = \partial_i \vec x##, you would have ##\vec E_r = \vec e_r## but ##\vec E_\theta = r \vec e_\theta## so indeed you would have ##\partial_\theta \vec E_r = \vec E_\theta/r##.vela said:You shouldn't get a factor of 1/r. Why do you think there is one?
The derivative of a basis vector is zero. This is because basis vectors are constant and do not change with respect to any variable.
The derivative of a basis vector is important because it helps us understand the rate of change of a vector in a particular direction. This is useful in many areas of science and engineering, such as motion and fluid dynamics.
The derivative of a basis vector is calculated using the rules of vector calculus. It involves taking the derivative of each component of the vector and combining them using the chain rule.
No, the derivative of a basis vector cannot be non-zero. As mentioned before, basis vectors are constant and do not change with respect to any variable, so their derivative is always zero.
The derivative of a basis vector is the basis vector for the derivative of a vector function. In other words, the derivative of a vector function is a vector of the derivatives of each component of the function, and the basis vector for each component is the derivative of the corresponding basis vector.