Derivative of a vector function with nonfixed bases

 

Derivative of a vector function with nonfixed bases[edit]

The above formulas for the derivative of a vector function rely on the assumption that the basis vectors e₁, e₂, e₃ are constant, that is, fixed in the reference frame in which the derivative of a is being taken, and therefore the e₁, e₂, e₃ each has a derivative of identically zero. This often holds true for problems dealing with vector fields in a fixed coordinate system, or for simple problems in physics. However, many complex problems involve the derivative of a vector function in multiple moving reference frames, which means that the basis vectors will not necessarily be constant. In such a case where the basis vectors e₁, e₂, e₃ are fixed in reference frame E, but not in reference frame N, the more general formula for the ordinary time derivative of a vector in reference frame N is^[1]

$${\displaystyle \frac{{}^{\mathrm{N}}�\mathbf{�}}{��}=\sum _{�=1}^3\frac{�{�}_{�}}{��}{\mathbf{�}}_{�}+\sum _{�=1}^3{�}_{�}\frac{{}^{\mathrm{N}}�{\mathbf{�}}_{�}}{��}}$$

where the superscript N to the left of the derivative operator indicates the reference frame in which the derivative is taken. As shown previously, the first term on the right hand side is equal to the derivative of a in the reference frame where e₁, e₂, e₃ are constant, reference frame E. It also can be shown that the second term on the right hand side is equal to the relative angular velocity of the two reference frames cross multiplied with the vector a itself.^[1] Thus, after substitution, the formula relating the derivative of a vector function in two reference frames is^[1]

$${\displaystyle \frac{{}^{\mathrm{N}}�\mathbf{�}}{��}=\frac{{}^{\mathrm{E}}�\mathbf{�}}{��}+{}^{\mathrm{N}}{�}^{\mathrm{E}}\times \mathbf{�}}$$

where ^Nω^E is the angular velocity of the reference frame E relative to the reference frame N.

One common example where this formula is used is to find the velocity of a space-borne object, such as a rocket, in the inertial reference frame using measurements of the rocket's velocity relative to the ground. The velocity ^Nv^R in inertial reference frame N of a rocket R located at position r^R can be found using the formula

$${\displaystyle \frac{{}^{\mathrm{N}}�}{��}({\mathbf{�}}^{\mathrm{R}})=\frac{{}^{\mathrm{E}}�}{��}({\mathbf{�}}^{\mathrm{R}})+{}^{\mathrm{N}}{�}^{\mathrm{E}}\times {\mathbf{�}}^{\mathrm{R}}.}$$

where ^Nω^E is the angular velocity of the Earth relative to the inertial frame N. Since velocity is the derivative of position, ^Nv^R and ^Ev^R are the derivatives of r^R in reference frames N and E, respectively. By substitution,

$${\displaystyle {}^{\mathrm{N}}{\mathbf{�}}^{\mathrm{R}}={}^{\mathrm{E}}{\mathbf{�}}^{\mathrm{R}}+{}^{\mathrm{N}}{�}^{\mathrm{E}}\times {\mathbf{�}}^{\mathrm{R}}}$$

where ^Ev^R is the velocity vector of the rocket as measured from a reference frame E that is fixed to the Earth.

Derivative and vector multiplication[edit]

The derivative of a product of vector functions behaves similarly to the derivative of a product of scalar functions.^[2] Specifically, in the case of scalar multiplication of a vector, if p is a scalar variable function of q,^[1]

$${\displaystyle \frac{\mathrm{\partial }}{\mathrm{\partial }�}(�\mathbf{�})=\frac{\mathrm{\partial }�}{\mathrm{\partial }�}\mathbf{�}+�\frac{\mathrm{\partial }\mathbf{�}}{\mathrm{\partial }�}.}$$

In the case of dot multiplication, for two vectors a and b that are both functions of q,^[1]

$${\displaystyle \frac{\mathrm{\partial }}{\mathrm{\partial }�}(\mathbf{�}\cdot \mathbf{�})=\frac{\mathrm{\partial }\mathbf{�}}{\mathrm{\partial }�}\cdot \mathbf{�}+\mathbf{�}\cdot \frac{\mathrm{\partial }\mathbf{�}}{\mathrm{\partial }�}.}$$

Similarly, the derivative of the cross product of two vector functions is^[1]

$${\displaystyle \frac{\mathrm{\partial }}{\mathrm{\partial }�}(\mathbf{�}\times \mathbf{�})=\frac{\mathrm{\partial }\mathbf{�}}{\mathrm{\partial }�}\times \mathbf{�}+\mathbf{�}\times \frac{\mathrm{\partial }\mathbf{�}}{\mathrm{\partial }�}.}$$

Derivative of an n-dimensional vector function[edit]

A function f of a real number t with values in the space ${\displaystyle {\mathbb{�}}^{�}}$ can be written as ${\displaystyle �(�)=({�}_1(�),{�}_2(�),\dots ,{�}_{�}(�))}$. Its derivative equals

${\displaystyle {�}^{\prime }(�)=({�}_1^{\prime }(�),{�}_2^{\prime }(�),\dots ,{�}_{�}^{\prime }(�))}$.

If f is a function of several variables, say of ${\displaystyle �\in {\mathbb{�}}^{�}}$, then the partial derivatives of the components of f form a ${\displaystyle �\times �}$ matrix called the Jacobian matrix of f.

Infinite-dimensional vector functions[edit]

Main article: Infinite-dimensional-vector function

If the values of a function f lie in an infinite-dimensional vector space X, such as a Hilbert space, then f may be called an infinite-dimensional vector function.

Functions with values in a Hilbert space[edit]

If the argument of f is a real number and X is a Hilbert space, then the derivative of f at a point t can be defined as in the finite-dimensional case:

${\displaystyle {�}^{\prime }(�)=\underset{ℎ\to 0}{lim}\frac{�(�+ℎ)-�(�)}{ℎ}.}$

Most results of the finite-dimensional case also hold in the infinite-dimensional case too, mutatis mutandis. Differentiation can also be defined to functions of several variables (e.g., ${\displaystyle �\in {\mathbb{�}}^{�}}$ or even ${\displaystyle �\in �}$, where Y is an infinite-dimensional vector space).

N.B. If X is a Hilbert space, then one can easily show that any derivative (and any other limit) can be computed componentwise: if

${\displaystyle �=({�}_1,{�}_2,{�}_3,\dots )}$

(i.e., ${\displaystyle �={�}_1{�}_1+{�}_2{�}_2+{�}_3{�}_3+\cdots }$, where ${\displaystyle {�}_1,{�}_2,{�}_3,\dots }$ is an orthonormal basis of the space X ), and ${\displaystyle {�}^{\prime }(�)}$ exists, then

${\displaystyle {�}^{\prime }(�)=({�}_1^{\prime }(�),{�}_2^{\prime }(�),{�}_3^{\prime }(�),\dots )}$.

However, the existence of a componentwise derivative does not guarantee the existence of a derivative, as componentwise convergence in a Hilbert space does not guarantee convergence with respect to the actual topology of the Hilbert space.

Other infinite-dimensional vector spaces[edit]

Most of the above hold for other topological vector spaces X too. However, not as many classical results hold in the Banach space setting, e.g., an absolutely continuous function with values in a suitable Banach space need not have a derivative anywhere. Moreover, in most Banach spaces setting there are no orthonormal bases.