Derivatives of Vector Functions - Revision history

Khanh at 01:20, 12 November 2021

2021-11-12T01:20:35Z

Lila at 16:57, 19 October 2021

2021-10-19T16:57:23Z

Lila at 16:56, 19 October 2021

2021-10-19T16:56:17Z

Khanh: Created page with "* [https://openstax.org/books/calculus-volume-3/pages/3-2-calculus-of-vector-valued-functions Calculus of Vector-Valued Functions], OpenStax"

2021-09-26T02:17:52Z

Created page with "* [https://openstax.org/books/calculus-volume-3/pages/3-2-calculus-of-vector-valued-functions Calculus of Vector-Valued Functions], OpenStax"

New page

* [https://openstax.org/books/calculus-volume-3/pages/3-2-calculus-of-vector-valued-functions Calculus of Vector-Valued Functions], OpenStax

← Older revision		Revision as of 01:20, 12 November 2021
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	==Resources==		==Resources==
−	* [https://en.wikipedia.org/wiki/Vector-valued_function Vector-valued function], Wikipedia
	* [https://openstax.org/books/calculus-volume-3/pages/3-2-calculus-of-vector-valued-functions Calculus of Vector-Valued Functions], OpenStax		* [https://openstax.org/books/calculus-volume-3/pages/3-2-calculus-of-vector-valued-functions Calculus of Vector-Valued Functions], OpenStax
		+
		+	== Licensing ==
		+	Content obtained and/or adapted from:
		+	* [https://en.wikipedia.org/wiki/Vector-valued_function Vector-valued function, Wikipedia] under a CC BY-SA license

@@ Line 1: / Line 1: @@
 ==Derivative of a three-dimensional vector function==
-{{see also|Gradient}}
 Many vector-valued functions, like scalar-valued functions, can be differentiated by simply differentiating the components in the Cartesian coordinate system.  Thus, if
 :<math>\mathbf{r}(t) = f(t)\mathbf{i} + g(t)\mathbf{j} + h(t)\mathbf{k}</math>
@@ Line 11: / Line 11: @@
 ===Partial derivative===
-The partial derivative of a vector function '''a''' with respect to a scalar variable ''q'' is defined as<ref name="dynon19">{{harvnb|Kane|Levinson|1996|pp=29–37}}</ref>
+The partial derivative of a vector function '''a''' with respect to a scalar variable ''q'' is defined as
 :<math>\frac{\partial\mathbf{a}}{\partial q} = \sum_{i=1}^{n}\frac{\partial a_i}{\partial q}\mathbf{e}_i</math>
@@ Line 18: / Line 18: @@
 ===Ordinary derivative===
-If '''a''' is regarded as a vector function of a single scalar variable, such as time ''t'', then the equation above reduces to the first ordinary time derivative of '''a''' with respect to ''t'',<ref name="dynon19"/>
+If '''a''' is regarded as a vector function of a single scalar variable, such as time ''t'', then the equation above reduces to the first ordinary time derivative of '''a''' with respect to ''t'',
 :<math>\frac{d\mathbf{a}}{dt} = \sum_{i=1}^{n}\frac{da_i}{dt}\mathbf{e}_i.</math>
 ===Total derivative===
-If the vector '''a''' is a function of a number ''n'' of scalar variables ''q''<sub>''r''</sub> (''r'' = 1,...,''n''), and each ''q''<sub>''r''</sub> is only a function of time ''t'', then the ordinary derivative of '''a''' with respect to ''t'' can be expressed, in a form known as the total derivative, as<ref name="dynon19"/>
+If the vector '''a''' is a function of a number ''n'' of scalar variables ''q''<sub>''r''</sub> (''r'' = 1,...,''n''), and each ''q''<sub>''r''</sub> is only a function of time ''t'', then the ordinary derivative of '''a''' with respect to ''t'' can be expressed, in a form known as the total derivative, as
 :<math>\frac{d\mathbf a}{dt} = \sum_{r=1}^{n}\frac{\partial \mathbf a}{\partial q_r} \frac{dq_r}{dt} + \frac{\partial \mathbf a}{\partial t}.</math>
@@ Line 33: / Line 32: @@
 ===Derivative of a vector function with nonfixed bases===
-The above formulas for the derivative of a vector function rely on the assumption that the basis vectors '''e'''<sub>1</sub>,'''e'''<sub>2</sub>,'''e'''<sub>3</sub> are constant, that is, fixed in the reference frame in which the derivative of '''a''' is being taken, and therefore the '''e'''<sub>1</sub>,'''e'''<sub>2</sub>,'''e'''<sub>3</sub> 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 '''e'''<sub>1</sub>,'''e'''<sub>2</sub>,'''e'''<sub>3</sub> 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<ref name="dynon19"/>
+The above formulas for the derivative of a vector function rely on the assumption that the basis vectors '''e'''<sub>1</sub>,'''e'''<sub>2</sub>,'''e'''<sub>3</sub> are constant, that is, fixed in the reference frame in which the derivative of '''a''' is being taken, and therefore the '''e'''<sub>1</sub>,'''e'''<sub>2</sub>,'''e'''<sub>3</sub> 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 '''e'''<sub>1</sub>,'''e'''<sub>2</sub>,'''e'''<sub>3</sub> 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
 :<math>\frac{{}^\mathrm{N}d\mathbf{a}}{dt} = \sum_{i=1}^{3}\frac{da_i}{dt}\mathbf{e}_i + \sum_{i=1}^{3}a_i\frac{{}^\mathrm{N}d\mathbf{e}_i}{dt}</math>
-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 '''e'''<sub>1</sub>,'''e'''<sub>2</sub>,'''e'''<sub>3</sub> 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.<ref name="dynon19"/>  Thus, after substitution, the formula relating the derivative of a vector function in two reference frames is<ref name="dynon19"/>
+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 '''e'''<sub>1</sub>,'''e'''<sub>2</sub>,'''e'''<sub>3</sub> 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. Thus, after substitution, the formula relating the derivative of a vector function in two reference frames is
 :<math>\frac{{}^\mathrm Nd\mathbf a}{dt} =  \frac{{}^\mathrm Ed\mathbf a }{dt} + {}^\mathrm N \mathbf \omega^\mathrm E \times \mathbf a</math>
@@ Line 54: / Line 53: @@
 ===Derivative and vector multiplication===
-The derivative of the products of vector functions behaves similarly to the derivative of the products of scalar functions. Specifically, in the case of scalar multiplication of a vector, if ''p'' is a scalar variable function of ''q'',<ref name="dynon19"/>
+The derivative of the products of vector functions behaves similarly to the derivative of the products of scalar functions. Specifically, in the case of scalar multiplication of a vector, if ''p'' is a scalar variable function of ''q'',
 :<math>\frac{\partial}{\partial q}(p\mathbf a) = \frac{\partial p}{\partial q}\mathbf a + p\frac{\partial \mathbf a}{\partial q}.</math>
-In the case of dot multiplication, for two vectors '''a''' and '''b''' that are both functions of ''q'',<ref name="dynon19"/>
+In the case of dot multiplication, for two vectors '''a''' and '''b''' that are both functions of ''q'',
 :<math>\frac{\partial}{\partial q}(\mathbf a \cdot \mathbf b) = \frac{\partial \mathbf a }{\partial q} \cdot \mathbf b + \mathbf a \cdot \frac{\partial \mathbf b}{\partial q}.</math>
-Similarly, the derivative of the cross product of two vector functions is<ref name="dynon19"/>
+Similarly, the derivative of the cross product of two vector functions is
 :<math>\frac{\partial}{\partial q}(\mathbf a \times \mathbf b) = \frac{\partial \mathbf a }{\partial q} \times \mathbf b + \mathbf a \times \frac{\partial \mathbf b}{\partial q}.</math>

← Older revision		Revision as of 16:56, 19 October 2021
Line 1:		Line 1:
		+	==Derivative of a three-dimensional vector function==
		+	{{see also\|Gradient}}
		+	Many vector-valued functions, like scalar-valued functions, can be differentiated by simply differentiating the components in the Cartesian coordinate system. Thus, if
		+	:<math>\mathbf{r}(t) = f(t)\mathbf{i} + g(t)\mathbf{j} + h(t)\mathbf{k}</math>
		+	is a vector-valued function, then
		+	:<math>\frac{d\mathbf{r}(t)}{dt} = f'(t)\mathbf{i} + g'(t)\mathbf{j} + h'(t)\mathbf{k}.</math>
		+	The vector derivative admits the following physical interpretation: if '''r'''(''t'') represents the position of a particle, then the derivative is the velocity of the particle
		+	:<math>\mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt} .</math>
		+	Likewise, the derivative of the velocity is the acceleration
		+	:<math>\frac{d\mathbf{v}(t)}{d t}=\mathbf{a}(t).</math>
		+
		+	===Partial derivative===
		+	The partial derivative of a vector function '''a''' with respect to a scalar variable ''q'' is defined as<ref name="dynon19">{{harvnb\|Kane\|Levinson\|1996\|pp=29–37}}</ref>
		+
		+	:<math>\frac{\partial\mathbf{a}}{\partial q} = \sum_{i=1}^{n}\frac{\partial a_i}{\partial q}\mathbf{e}_i</math>
		+
		+	where ''a''<sub>''i''</sub> is the ''scalar component'' of '''a''' in the direction of '''e'''<sub>''i''</sub>. It is also called the direction cosine of '''a''' and '''e'''<sub>''i''</sub> or their dot product. The vectors '''e'''<sub>1</sub>,'''e'''<sub>2</sub>,'''e'''<sub>3</sub> form an orthonormal basis fixed in the reference frame in which the derivative is being taken.
		+
		+	===Ordinary derivative===
		+	If '''a''' is regarded as a vector function of a single scalar variable, such as time ''t'', then the equation above reduces to the first ordinary time derivative of '''a''' with respect to ''t'',<ref name="dynon19"/>
		+
		+	:<math>\frac{d\mathbf{a}}{dt} = \sum_{i=1}^{n}\frac{da_i}{dt}\mathbf{e}_i.</math>
		+
		+	===Total derivative===
		+	If the vector '''a''' is a function of a number ''n'' of scalar variables ''q''<sub>''r''</sub> (''r'' = 1,...,''n''), and each ''q''<sub>''r''</sub> is only a function of time ''t'', then the ordinary derivative of '''a''' with respect to ''t'' can be expressed, in a form known as the total derivative, as<ref name="dynon19"/>
		+
		+	:<math>\frac{d\mathbf a}{dt} = \sum_{r=1}^{n}\frac{\partial \mathbf a}{\partial q_r} \frac{dq_r}{dt} + \frac{\partial \mathbf a}{\partial t}.</math>
		+
		+	Some authors prefer to use capital D to indicate the total derivative operator, as in ''D''/''Dt''. The total derivative differs from the partial time derivative in that the total derivative accounts for changes in '''a''' due to the time variance of the variables ''q''<sub>''r''</sub>.
		+
		+	===Reference frames===
		+	Whereas for scalar-valued functions there is only a single possible reference frame, to take the derivative of a vector-valued function requires the choice of a reference frame (at least when a fixed Cartesian coordinate system is not implied as such). Once a reference frame has been chosen, the derivative of a vector-valued function can be computed using techniques similar to those for computing derivatives of scalar-valued functions. A different choice of reference frame will, in general, produce a different derivative function. The derivative functions in different reference frames have a specific kinematical relationship.
		+
		+	===Derivative of a vector function with nonfixed bases===
		+	The above formulas for the derivative of a vector function rely on the assumption that the basis vectors '''e'''<sub>1</sub>,'''e'''<sub>2</sub>,'''e'''<sub>3</sub> are constant, that is, fixed in the reference frame in which the derivative of '''a''' is being taken, and therefore the '''e'''<sub>1</sub>,'''e'''<sub>2</sub>,'''e'''<sub>3</sub> 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 '''e'''<sub>1</sub>,'''e'''<sub>2</sub>,'''e'''<sub>3</sub> 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<ref name="dynon19"/>
		+
		+	:<math>\frac{{}^\mathrm{N}d\mathbf{a}}{dt} = \sum_{i=1}^{3}\frac{da_i}{dt}\mathbf{e}_i + \sum_{i=1}^{3}a_i\frac{{}^\mathrm{N}d\mathbf{e}_i}{dt}</math>
		+
		+	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 '''e'''<sub>1</sub>,'''e'''<sub>2</sub>,'''e'''<sub>3</sub> 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.<ref name="dynon19"/> Thus, after substitution, the formula relating the derivative of a vector function in two reference frames is<ref name="dynon19"/>
		+
		+	:<math>\frac{{}^\mathrm Nd\mathbf a}{dt} = \frac{{}^\mathrm Ed\mathbf a }{dt} + {}^\mathrm N \mathbf \omega^\mathrm E \times \mathbf a</math>
		+
		+	where <sup>N</sup>'''''ω'''''<sup>E</sup> 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 <sup>N</sup>'''v'''<sup>R</sup> in inertial reference frame N of a rocket R located at position '''r'''<sup>R</sup> can be found using the formula
		+
		+	:<math> \frac{{}^\mathrm Nd}{dt}(\mathbf r^\mathrm R) = \frac{{}^\mathrm Ed}{dt}(\mathbf r^\mathrm R) + {}^\mathrm N \mathbf \omega^\mathrm E \times \mathbf r^\mathrm R.</math>
		+
		+	where <sup>N</sup>'''''ω'''''<sup>E</sup> is the angular velocity of the Earth relative to the inertial frame N. Since velocity is the derivative of position, <sup>N</sup>'''v'''<sup>R</sup> and <sup>E</sup>'''v'''<sup>R</sup> are the derivatives of '''r'''<sup>R</sup> in reference frames N and E, respectively. By substitution,
		+
		+	:<math>{}^\mathrm N \mathbf v^\mathrm R = {}^\mathrm E \mathbf v^\mathrm R + {}^\mathrm N \mathbf \omega^\mathrm E \times \mathbf r^\mathrm R</math>
		+
		+	where <sup>E</sup>'''v'''<sup>R</sup> is the velocity vector of the rocket as measured from a reference frame E that is fixed to the Earth.
		+
		+	===Derivative and vector multiplication===
		+	The derivative of the products of vector functions behaves similarly to the derivative of the products of scalar functions. Specifically, in the case of scalar multiplication of a vector, if ''p'' is a scalar variable function of ''q'',<ref name="dynon19"/>
		+
		+	:<math>\frac{\partial}{\partial q}(p\mathbf a) = \frac{\partial p}{\partial q}\mathbf a + p\frac{\partial \mathbf a}{\partial q}.</math>
		+
		+	In the case of dot multiplication, for two vectors '''a''' and '''b''' that are both functions of ''q'',<ref name="dynon19"/>
		+
		+	:<math>\frac{\partial}{\partial q}(\mathbf a \cdot \mathbf b) = \frac{\partial \mathbf a }{\partial q} \cdot \mathbf b + \mathbf a \cdot \frac{\partial \mathbf b}{\partial q}.</math>
		+
		+	Similarly, the derivative of the cross product of two vector functions is<ref name="dynon19"/>
		+
		+	:<math>\frac{\partial}{\partial q}(\mathbf a \times \mathbf b) = \frac{\partial \mathbf a }{\partial q} \times \mathbf b + \mathbf a \times \frac{\partial \mathbf b}{\partial q}.</math>
		+
		+	==Derivative of an ''n''-dimensional vector function==
		+	A function ''f'' of a real number ''t'' with values in the space <math>\R^n</math> can be written as <math>f(t)=(f_1(t),f_2(t),\ldots,f_n(t))</math>. Its derivative equals
		+	:<math>f'(t)=(f_1'(t),f_2'(t),\ldots,f_n'(t))</math>.
		+	If ''f'' is a function of several variables, say of <math>t\in\R^m</math>, then the partial derivatives of the components of ''f'' form a <math>n\times m</math> matrix called the ''Jacobian matrix of f''.
		+
		+	==Resources==
		+	* [https://en.wikipedia.org/wiki/Vector-valued_function Vector-valued function], Wikipedia
	* [https://openstax.org/books/calculus-volume-3/pages/3-2-calculus-of-vector-valued-functions Calculus of Vector-Valued Functions], OpenStax		* [https://openstax.org/books/calculus-volume-3/pages/3-2-calculus-of-vector-valued-functions Calculus of Vector-Valued Functions], OpenStax