The main objects upon which this package operates are the free vector, the PositionVector, the RootedVector, the VectorField, curves, surfaces, and scalar fields.  Lists cannot be used in place of vectors, and scalar fields are ordinary Maple functions and expressions.  Information about vector objects - coordinate systems and names of coordinates - are stored as attributes of the vector objects.  The About, attributes, and GetCoordinates commands can be used to recover this information.

The free vector is is created in one of two ways.  The angle-bracket notation creates a free vector in the extant coordinate system.  At this point in this worksheet, that coordinate system is Cartesian.

The explicit use of the Vector command permits attaching coordinates to the free vector.

In each case, the free vector is really a point.  In Cartesian coordinates, the free vector ${\mathbf{v}}_1$ is a vector from the origin to the point $\left(a\,b\,c\right)$, and the identification of the point with the free vector is permissible because the unit basis vectors are everywhere parallel.

By analogy, the free vector ${\mathbf{v}}_2$ is a point in a rectangular frame with axes labeled $r$ (horizontal) and $\mathrm{\θ}$ (vertical).  In this frame, the point with polar coordinates $\left(r\,\mathrm{\θ}\right)\=\left(a\,b\right)$ is connected to the origin with an arrow, and this arrow is ${\mathbf{v}}_2$.

Information about ${\mathbf{v}}_1$ and ${\mathbf{v}}_2$ is obtained as follows.

The VectorField command is used to create a vector field.

The unit basis vectors ${\stackrel{\&conjugate0;}{\mathbf{e}}}_x$ and ${\stackrel{\&conjugate0;}{\mathbf{e}}}_y$ are used in place of $\mathbf{i}$ and $\mathbf{j}$, respectively.  The bar denotes that the domain of these basis vectors is the underlying Cartesian space, and that the vectors themselves are position dependent.  Of course, in Cartesian coordinates, these basis vectors are independent of position, but the bar is still used to denote that these are the "moving" basis vectors that are defined at each point in the space supporting the vector field, and that are given in the coordinate system in which the vector field is given.

Alternatively, the vector field

defines a field of vectors whose domain is each point (except for the origin) in the Cartesian plane on which the coordinate curves of the polar coordinate system have been "pulled back."  This is the familiar view of polar coordinates superimposed on the Cartesian plane - concentric circles as lines of constant $r$, and radial rays as lines of constant $\mathrm{\θ}$.  In this view of the change of coordinates to polar coordinates, the vectors ${\stackrel{\&conjugate0;}{\mathbf{e}}}_r$ and ${\stackrel{\&conjugate0;}{\mathbf{e}}}_{\mathrm{\θ}}$ are position dependent.  These "moving" basis vectors change direction from point to point, and are not to be confused with the basis vectors appearing in

If the evalVF command is used to evaluate the vector field G at a point in the Cartesian plane, the result is a RootedVector.  For example, evaluating $\mathbf{G}$ at the Cartesian point whose polar-coordinate description is $\left(r\,\mathrm{\θ}\right)\=\left(a\,b\right)$ is effected by

because the free vector ${\mathbf{v}}_2$ is the Maple representation of the point whose polar coordinates are $\left(a\,b\right)$.  Indeed, we have

The root point is the point of "attachment" for the single vector ${\mathbf{G}}_1$, which is actually a "bound" vector, bound to the root point.

A rooted vector can be defined directly with the RootedVector command.

To verify that this is a vector in polar coordinates, rooted at the polar point $\left(r\,\mathrm{\θ}\right)\=\left(3\,\frac{\mathrm{\π}}{5}\right)\,$ use

Information about the rooted vector is actually stored in a module that could be created with the VectorSpace command.

The vector space is the set of all vectors of appropriate dimension that have their root point at $\left(r\,\mathrm{\θ}\right)\=\left(3\,\frac{\mathrm{\π}}{5}\right)\.$  The basis for this vector space are the vectors $\left\{{\stackrel{\&conjugate0;}{\mathbf{e}}}_r\,{\stackrel{\&conjugate0;}{\mathbf{e}}}_{\mathrm{\θ}}\right\}\,$the moving basis vectors defined at the root point.  Individual vectors in this space can be created as module exports via the syntax

That ${\mathbf{v}}_4$ is the vector $\mathrm{\α} {\stackrel{\&conjugate0;}{\mathbf{e}}}_r\+\mathrm{\β} {\stackrel{\&conjugate0;}{\mathbf{e}}}_{\mathrm{\θ}}$ rooted at the point $\left(r\,\mathrm{\θ}\right)\=\left(3\,\frac{\mathrm{\π}}{5}\right)$ can be see with

Creating rooted vectors via the VectorSpace command works well when many such vectors will have the same root point.  The RootedVector command is most convenient for single rooted vectors.

Finally, there is the PositionVector command that creates the radius-vector form of curves and surfaces.  The vectors

are position vectors, vectors from the origin of a Cartesian frame to the relevant point in that frame. Points can be parameterized by up to two parameters;  a one-parameter family of points represents a curve, and a two-parameter family, a surface.  Indeed, we have

The PositionVector construct applies only in Cartesian coordinates.  The Position Vector

describes the origin-centered circle of radius 2 drawn in the Cartesian plane, as we see from

Consequently, in the Cartesian plane the PositionVector, the free vector, and the vector rooted at the origin are essentially the same arrow, but because these are represented in the VectorCalculus package by distinct data structures, they are not necessarily recognized by Maple as being the same.  Thus, the package provides the ConvertVector command for converting one form of the Cartesian vector to another.

A final word about the visual display of vectors and vector fields in the VectorCalculus package.  The default display for free vectors and vector fields is explicitly in terms of basis vectors.  This display can be changed to a column-vector format with the BasisFormat command.  The setting

causes the displays to be

respectively, for a free vector and a vector field.  The PositionVector is always displayed as a column vector.

The MapToBasis command imposes a change of coordinates on a vector or vector field.  For example, to express the Cartesian vector field

in polar coordinates, use

This is equivalent to expressing the unit basis vectors $\mathbf{i}$ and $\mathbf{j}$ in terms of the vectors ${\stackrel{\&conjugate0;}{\mathbf{e}}}_r$ and ${\stackrel{\&conjugate0;}{\mathbf{e}}}_{\mathrm{\θ}}\,$then making the substitutions $x\=r \cos \left(\mathrm{\θ}\right)\,y\=r \sin \left(\mathrm{\θ}\right)$.  This calculation can be carried out from first principles if the following position vector is first defined.

The basis vectors ${\stackrel{\&conjugate0;}{\mathbf{e}}}_r$ and ${\stackrel{\&conjugate0;}{\mathbf{e}}}_{\mathrm{\θ}}$ are obtained from R via

Writing these vectors in terms of $\mathbf{i}$ and $\mathbf{j}$ can be done with

Solving for $\mathbf{i}$ and $\mathbf{j}$ in terms of vectors ${\stackrel{\&conjugate0;}{\mathbf{e}}}_r$ and ${\stackrel{\&conjugate0;}{\mathbf{e}}}_{\mathrm{\θ}}\,$we get

Representing the vector field $\mathbf{F}$ in terms of $\mathbf{i}$ and $\mathbf{j}$, then replacing $\mathbf{i}$ and $\mathbf{j}$ with their equivalents in terms of the vectors ${\stackrel{\&conjugate0;}{\mathbf{e}}}_r$ and ${\stackrel{\&conjugate0;}{\mathbf{e}}}_{\mathrm{\θ}}\,$we get

Making the substitutions $x\=r \cos \left(\mathrm{\θ}\right)\, y\=r \sin \left(\mathrm{\θ}\right)$, we get

A final rearrangement of terms, and an identification of $\mathrm{er}$ and $\mathrm{et}$ with vectors ${\stackrel{\&conjugate0;}{\mathbf{e}}}_r$ and ${\stackrel{\&conjugate0;}{\mathbf{e}}}_{\mathrm{\θ}}\,$respectively, gives

which compares favorably with

The MapToBasis command can also be used to transform a point in one coordinate system to its representation in another.  Setting

will make the result of

easier to understand.  The Cartesian point $\left(x\,y\right)\=\left(1\,2\right)$, represented as a free vector, has been transformed to $\left(r\,\mathrm{\θ}\right)\=\left(\sqrt{5}\,\arctan \left(2\right)\right)\,$also given as a free vector.  This is not a vector transformation.  The MapToBasis command applies the contravariant transformation law to a rooted vector or a vector field.  Applied to a free vector, it transforms just the components.  The basis vectors are not so transformed because the free vector is not a vector; it is really just a representation of a point in curvilinear coordinates.

Three of the top level operators, `+` ( addition ), `*` ( scalar multiplication ), and `.` ( dot product ), have been overloaded to operate on the Vector/attribute combinations created by the package.  Within the package, the diff command is also modified.  It will act directly on vectors and vector fields without having to be mapped onto the target structures.

Also, a new operator, `&x` ( cross product ), is exported by the package.  Where possible, these operators have been extended to understand the standard vector differential operator, Del ( or Nabla ), which is also a package export.

Basic Arithmetic

The top level operators $\&plus;$ addition $-$ subtraction  * scalar multiplication have been overloaded to operate on the Vector/attribute combinations created by the package.  To illustrate this, first define the following vectors with a common root point. >  $\mathrm{w1}:=\mathrm{RootedVector}\left(\mathrm{root}\&equals;\left[1\&comma;2\right]\&comma;\left[2\&comma;3\right]\&comma;\mathrm{polar}\left[r\&comma;\mathrm{\&theta;}\right]\right)\&semi;$$\mathrm{About}\left(\mathrm{w1}\right)$ >  $\mathrm{w2}:=\mathrm{RootedVector}\left(\mathrm{root}\&equals;\left[1\&comma;2\right]\&comma;\left[5\&comma;-4\right]\&comma;\mathrm{polar}\left[r\&comma;\mathrm{\&theta;}\right]\right)\&semi;$$\mathrm{About}\left(\mathrm{w2}\right)$ The basic operations of addition, subtraction, and scalar multiplication are then obtained as follows. >  $\mathrm{w3}:=\mathrm{w1}\&plus;\mathrm{w2}\&semi;$$\mathrm{About}\left(\mathrm{w3}\right)$ >  $\mathrm{w4}:=\mathrm{w1}-\mathrm{w2}\&semi;$$\mathrm{About}\left(\mathrm{w4}\right)$ >  $\mathrm{w5}:=\mathrm{\&alpha;}\mathrm{w1}\&semi;$$\mathrm{About}\left(\mathrm{w5}\right)$ The addition ${\mathbf{w}}_3\&equals;{\mathbf{w}}_1\&plus;{\mathbf{w}}_2$ can be corroborated by performing the sum in Cartesian coordinates via   >  $\mathrm{w6}:=\mathrm{MapToBasis}\left(\mathrm{w1}\&comma;\mathrm{cartesian}\right)\&colon;$$\mathrm{w7}≔\mathrm{MapToBasis}\left(\mathrm{w2}\&comma;\mathrm{cartesian}\right)\&colon;$$\mathrm{w8}≔\mathrm{w6}\&plus;\mathrm{w7}\&colon;$$\mathrm{simplify}\left(\mathrm{MapToBasis}\left(\mathrm{w8}\&comma;\mathrm{polar}\right)\right)$ The lengths of the vectors $w_1$ and $w_5\&equals;\mathrm{\&alpha;} w_1$ are respectively >  $\mathrm{Norm}\left(\mathrm{w1}\right)$ $\sqrt{13}$ (3.1.1)   and >  $\mathrm{assuming}\left(\left[\mathrm{Norm}\left(\mathrm{w5}\right)\right]\&comma;\left[0<\mathrm{\&alpha;}\right]\right)$ $\sqrt{13}\mathrm{\&alpha;}$ (3.1.2) In particular, this calculation highlights the possible misconception that the radial component of a vector expressed in polar coordinates determines the length of that vector.  Graphs of the vectors ${\mathbf{w}}_1$ and ${\mathbf{w}}_6$ (the Cartesian equivalent of ${\mathbf{w}}_1$) appear in Figure 1, with ${\mathbf{w}}_1$ in black, and ${\mathbf{w}}_6$ in red. >  $\mathrm{PlotVector}\left(\mathrm{w1}\&comma;\mathrm{scaling}\&equals;\mathrm{constrained}\&comma;\mathrm{view}\&equals;\left[-4..0\&comma;0..2\right]\&comma;\mathrm{labels}\&equals;\left[x\&comma;y\right]\&comma;\mathrm{axes}\&equals;\mathrm{frame}\right)$ >  $\mathrm{PlotVector}\left(\mathrm{w6}\&comma;\mathrm{scaling}\&equals;\mathrm{constrained}\&comma;\mathrm{color}\&equals;\mathrm{red}\&comma;\mathrm{view}\&equals;\left[-4..0\&comma;0..2\right]\&comma;\mathrm{labels}\&equals;\left[x\&comma;y\right]\&comma;\mathrm{axes}\&equals;\mathrm{frame}\right)$ Figure 1   The vectors $w_1$ in black and $w_6$ (the Cartesian equivalent of $w_1$) in red Since the length of ${\mathbf{w}}_6$ is   >  $\mathrm{Norm}\left(\mathrm{w6}\right)$ $\sqrt{13}$ (3.1.3) it is clear that the length of ${\mathbf{w}}_1$ is a function of both components in polar coordinates, not just the radial component.

Differentiation of a Vector Field

Consider the vector field   >  $\mathrm{unassign}\left(\'v\'\right)\;$$F:=\mathrm{VectorField}\left(⟨1\,1⟩\,\mathrm{parabolic}\left[u\,v\right]\right)$ defined in the parabolic coordinate system for which the transformation equations are   >  $X:=\frac{u^2-v^2}{2}\;$$Y≔u v$ $X:=\frac{1}{2}{u}^2-\frac{1}{2}{v}^2$ $Y:=uv$ (3.2.1) Parabolic coordinates are not as familiar as polar coordinates, so fewer background assumptions will interfere with the following calculations.   The derivative of $\mathbf{F}$ with respect to $u$ is given by   >  $\mathrm{ans}≔\mathrm{simplify}\left(\mathrm{diff}\left(F\,u\right)\right)$ This result can be checked by computing from first principles.  To begin, obtain expressions for the unit basis vectors ${\stackrel{\&conjugate0;}{\mathbf{e}}}_u$ and ${\stackrel{\&conjugate0;}{\mathbf{e}}}_v$.  For this, start with the position vector $\mathbf{R}\=x\left(u\,v\right) \mathbf{i}\+y\left(u\,v\right) \mathbf{j}$, defined in Maple via   >  $R:=\mathrm{PositionVector}\left(\left[X\,Y\right]\right)$ The unit basis vectors are obtained from the position vector by the following calculations. >  $\mathrm{temp1}:=\mathrm{Normalize}\left(\mathrm{diff}\left(R\,u\right)\right)\;$$\mathrm{eu}:=\mathrm{assuming}\left(\left[\mathrm{simplify}\left(\mathrm{temp1}\right)\right]\,\left[\mathrm{real}\right]\right)$ >  $\mathrm{temp2}:=\mathrm{Normalize}\left(\mathrm{diff}\left(R\,v\right)\right)\;$$\mathrm{ev}:=\mathrm{assuming}\left(\left[\mathrm{simplify}\left(\mathrm{temp2}\right)\right]\,\left[\mathrm{real}\right]\right)$ (Differentiation of the PositionVector creates a RootedVector, an object captured in a module data structure.  If the normalized form of the vector $\frac{\partial \mathbf{R}}{\partial u}$ were simplified with an assumption in the step that creates the module, the root point in the module would be stored with the assumption.  Since this later leads to difficulties, the creation of the module is done first, and the simplification with the assumption is done in a second and separate step.)   Noting that ${\stackrel{\&conjugate0;}{\mathbf{e}}}_u$ and ${\stackrel{\&conjugate0;}{\mathbf{e}}}_v$ are both linear combinations of $\mathbf{i}$ and $\mathbf{j}$, differentiation of $\mathbf{F}\={\stackrel{\&conjugate0;}{\mathbf{e}}}_u\+{\stackrel{\&conjugate0;}{\mathbf{e}}}_v$ with respect to $u$ can take place componentwise.  In fact, the derivative of $\mathbf{F}$ componentwise is   >  $A≔\mathrm{collect}\left(\mathrm{diff}\left(\mathrm{eu}\+\mathrm{ev}\,u\right)\,\left[i\,j\right]\,\mathrm{simplify}\right)$ To express $\mathbf{A}\=\frac{\partial \mathbf{F}}{\partial u}$ in terms of  ${\stackrel{\&conjugate0;}{\mathbf{e}}}_u$ and ${\stackrel{\&conjugate0;}{\mathbf{e}}}_v\,$solve the equations   >  $\mathrm{eqn1}:=\mathrm{Eu}\=\mathrm{eu}\.⟨i\,j⟩$$\mathrm{eqn2}:=\mathrm{Ev}\=\mathrm{ev}\.⟨i\,j⟩$ $\mathrm{eqn1}:=\mathrm{Eu}\=\frac{ui}{\sqrt{v^2\+u^2}}\+\frac{vj}{\sqrt{v^2\+u^2}}$ $\mathrm{eqn2}:=\mathrm{Ev}\=-\frac{vi}{\sqrt{v^2\+u^2}}\+\frac{uj}{\sqrt{v^2\+u^2}}$ (3.2.2) for $\mathbf{i}$ and $\mathbf{j}$ in terms of  ${\stackrel{\&conjugate0;}{\mathbf{e}}}_u$ and ${\stackrel{\&conjugate0;}{\mathbf{e}}}_v\.$  Doing this gives   >  $B:=\mathrm{solve}\left(\left\{\mathrm{eqn1}\,\mathrm{eqn2}\right\}\,\left\{i\,j\right\}\right)$ $B:=\left\{i\=\frac{\mathrm{Eu}u-v\mathrm{Ev}}{\sqrt{v^2\+u^2}}\,j\=\frac{\mathrm{Ev}u\+v\mathrm{Eu}}{\sqrt{v^2\+u^2}}\right\}$ (3.2.3) Make these replacements in $\mathbf{A}$ to obtain $\frac{\partial \mathbf{F}}{\partial u}$ as   >  $\mathrm{collect}\left(\mathrm{eval}\left(A\.⟨i\,j⟩\,B\right)\,\left[\mathrm{Eu}\,\mathrm{Ev}\right]\,\mathrm{simplify}\right)$ $\frac{v\mathrm{Eu}}{v^2\+u^2}-\frac{v\mathrm{Ev}}{v^2\+u^2}$ (3.2.4) which compares favorably with   >  $\mathrm{ans}$

Tangent Vector along a Curve

Let   >  $R:=\mathrm{PositionVector}\left(\left[a\left(s\right)\,b\left(s\right)\,c\left(s\right)\right]\right)$ define a space curve $\mathbf{R}\left(s\right)$ in Euclidean space.  The "secant vector"   $\frac{\mathbf{R}\left(s\+h\right)-\mathbf{R}\left(s\right)}{h}$  is given by   >  $\mathrm{SV}:=\frac{\genfrac{}{}{0ex}{}{R}{\phantom{s\=s\+h}}|\genfrac{}{}{0ex}{}{\phantom{R}}{s\=s\+h}-R}{h}$ The vector   >  $\mathrm{TV}:=\underset{h\→0}{lim}\mathrm{SV}$ tangent to the curve $\mathbf{R}$, can also be written as   >  $\mathrm{convert}\left(\mathrm{TV}\,\mathrm{diff}\right)$ Of course, this tangent vector could also be obtained with the built-in TangentVector command, as shown by   >  $\mathrm{TangentVector}\left(R\right)$

Dot Products

The dot product of free vectors is defined for Euclidean space only.  For example, we have   >  $⟨a\,b⟩\.⟨c\,d⟩$ $ac\+bd$ (3.4.1) or even   >  $\mathrm{DotProduct}\left(⟨a\,b⟩\,⟨c\,d⟩\right)$ $ac\+bd$ (3.4.2) However, in curvilinear coordinates, the dot product between free vectors is not defined because the free vector represents a point, not a vector, in non-Cartesian coordinates.  Maple gives the following error for such a calculation.   >  $\mathrm{DotProduct}\left(\mathrm{Vector}\left(⟨a\,b⟩\,\mathrm{coordinates}\=\mathrm{polar}\right)\,\mathrm{Vector}\left(⟨c\,d⟩\,\mathrm{coordinates}\=\mathrm{polar}\right)\right)$ Error, (in VectorCalculus:-DotProduct) cannot take the dot product of free Vectors in non-cartesian coordinates The divergence of the vector field   >  $F$ is given by   >  $\mathrm{Divergence}\left(F\right)\;$ $\frac{\frac{u}{\sqrt{v^2\+u^2}}\+\frac{v}{\sqrt{v^2\+u^2}}}{v^2\+u^2}$ (3.4.3) even though $\mathbf{F}$ was defined in parabolic coordinates!  Alternatively, the divergence of $\mathbf{F}$ can be obtained with   >  $\mathrm{Del}\.F$ $\frac{\frac{u}{\sqrt{v^2\+u^2}}\+\frac{v}{\sqrt{v^2\+u^2}}}{v^2\+u^2}$ (3.4.4) Recalling that $\mathbf{F}$ is given in terms of the unit vectors $\mathbf{i}$ and $\mathbf{j}$ by the sum   >  $\mathrm{Fij}:=\mathrm{simplify}\left(\mathrm{eu}\+\mathrm{ev}\right)$ the divergence can also be computed in Cartesian coordinates by use of the chain rule.  The computations are slightly simplified if the components of $\mathbf{F}$ are named   >  $f:={\mathrm{Fij}}_1$$g:={\mathrm{Fij}}_2$ $f:=\frac{u-v}{\sqrt{v^2\+u^2}}$ $g:=\frac{v\+u}{\sqrt{v^2\+u^2}}$ (3.4.5) The divergence of $F$, computed in Cartesian coordinates, is given by   $\nabla ·\mathbf{F}\=\left(\frac{\partial f}{\partial u} \frac{\partial u}{\partial x}\+\frac{\partial f}{\partial v} \frac{\partial v}{\partial x}\right)\+\left(\frac{\partial g}{\partial u} \frac{\partial u}{\partial y}\+\frac{\partial g}{\partial v} \frac{\partial v}{\partial y}\right)$    where the derivatives of $u$ and $v$ are   >  $\mathrm{dx}:=\mathrm{implicitdiff}\left(\left\{x\=X\,y\=Y\right\}\,\left\{u\left(x\,y\right)\,v\left(x\,y\right)\right\}\,\left\{u\,v\right\}\,x\right)$$\mathrm{dy}:=\mathrm{implicitdiff}\left(\left\{x\=X\,y\=Y\right\}\,\left\{u\left(x\,y\right)\,v\left(x\,y\right)\right\}\,\left\{u\,v\right\}\,y\right)$ $\mathrm{dx}:=\left\{{\mathrm{D}}_1\left(u\right)\=\frac{u}{v^2\+u^2}\,{\mathrm{D}}_1\left(v\right)\=-\frac{v}{v^2\+u^2}\right\}$ $\mathrm{dy}:=\left\{{\mathrm{D}}_2\left(u\right)\=\frac{v}{v^2\+u^2}\,{\mathrm{D}}_2\left(v\right)\=\frac{u}{v^2\+u^2}\right\}$ (3.4.6) Consequently, the chain rule gives   >  $\mathrm{DF}:=\left(\mathrm{diff}\left(f\,u\right)\right)\mathrm{D}\left[1\right]\left(u\right)\+\left(\mathrm{diff}\left(f\,v\right)\right)\mathrm{D}\left[1\right]\left(v\right)\+\left(\mathrm{diff}\left(g\,u\right)\right)\mathrm{D}\left[2\right]\left(u\right)\+\left(\mathrm{diff}\left(g\,v\right)\right)\mathrm{D}\left[2\right]\left(v\right)$ $\mathrm{DF}:=\left(\frac{1}{\sqrt{v^2\+u^2}}-\frac{\left(u-v\right)u}{{\left(v^2\+u^2\right)}^{3\/2}}\right){\mathrm{D}}_1\left(u\right)\+\left(-\frac{1}{\sqrt{v^2\+u^2}}-\frac{v\left(u-v\right)}{{\left(v^2\+u^2\right)}^{3\/2}}\right){\mathrm{D}}_1\left(v\right)\+\left(\frac{1}{\sqrt{v^2\+u^2}}-\frac{\left(v\+u\right)u}{{\left(v^2\+u^2\right)}^{3\/2}}\right){\mathrm{D}}_2\left(u\right)\+\left(\frac{1}{\sqrt{v^2\+u^2}}-\frac{v\left(v\+u\right)}{{\left(v^2\+u^2\right)}^{3\/2}}\right){\mathrm{D}}_2\left(v\right)$ (3.4.7) which, upon substitution for the derivatives of $u$ and $v$, becomes   >  $\mathrm{simplify}\left(\mathrm{eval}\left(\mathrm{DF}\,\mathrm{dx}\∪\mathrm{dy}\right)\right)$ $\frac{v\+u}{{\left(v^2\+u^2\right)}^{3\/2}}$ (3.4.8) clearly the same as   >  $\mathrm{simplify}\left(\mathrm{Del}\.F\right)$ $\frac{v\+u}{{\left(v^2\+u^2\right)}^{3\/2}}$ (3.4.9) Other operations with the $\nabla \,$the nabla, and the dot product appear in Table 1. ${\nabla }^2\=\nabla ·\nabla \,$the Laplacian operator >  $\mathrm{SetCoordinates}\left(\mathrm{cartesian}\left[x\,y\,z\right]\right)\:$$\mathrm{Del}\.\mathrm{Del}$ $\mathrm{VectorCalculus:-Laplacian}$ (3.4.10) ${\nabla }^{2}h\left(x\,y\,z\right)$ >  $\mathrm{Del}\.\mathrm{Del}\left(h\left(x\,y\,z\right)\right)$ $\frac{{\partial }^2}{\partial x^2}h\left(x\,y\,z\right)\+\frac{{\partial }^2}{\partial y^2}h\left(x\,y\,z\right)\+\frac{{\partial }^2}{\partial z^2}h\left(x\,y\,z\right)$ (3.4.11) The directional derivative operator  $L\mathbf{\=}\mathbf{v}·\nabla$ >  $\mathrm{unassign}\left(\'v\'\right)$$\:$ >  $$\begin{gathered} L≔\mathrm{VectorField}\left(⟨u\,v\,w⟩\right) \. \mathrm{Del}\: \\ L\left(h\left(x\,y\,z\right)\right) \end{gathered}$$ $u\left(\frac{\partial }{\partial x}h\left(x\,y\,z\right)\right)\+v\left(\frac{\partial }{\partial y}h\left(x\,y\,z\right)\right)\+w\left(\frac{\partial }{\partial z}h\left(x\,y\,z\right)\right)$ (3.4.12) Table 1   Additional operations with ∇, the nabla operator, and the dot product

Cross-Product

In Cartesian coordinates, the cross-product of the free vectors   >  $$\begin{gathered} \mathrm{BasisFormat}\left(\mathrm{false}\right)\: \\ \mathrm{unassign}\left(\'f\'\right)\; \\ v≔⟨a\,b\,c⟩\: \\ w≔⟨d\,e\,f⟩\: \\ v\,w \end{gathered}$$ can be computed by any of the following three approaches. >  $v\&xw$ >  $\mathrm{CrossProduct}\left(v\,w\right)$ >  $\mathbf{v}\times \mathbf{w}$ In non-Cartesian coordinates, the cross-product between free vectors is not defined because the free vector is merely a notational device for a point.  However, the cross-product is defined for rooted vectors such as   >  $\mathrm{unassign}\left(\'v\'\right)$$$\begin{gathered} \; \\ A≔\mathrm{RootedVector}\left(\mathrm{point}\=\left[a\,b\,c\right]\,\left[u_1\,u_2\,u_3\right]\,\mathrm{spherical}\left[\mathrm{\ρ}\,\mathrm{\φ}\,\mathrm{\θ}\right]\right)\: \end{gathered}$$$B:=\mathrm{RootedVector}\left(\mathrm{point}\=\left[a\,b\,c\right]\,\left[v_1\,v_2\,v_3\right]\,\mathrm{spherical}\left[\mathrm{\ρ}\,\mathrm{\φ}\,\mathrm{\θ}\right]\right)\:$$A\,B$ Surprisingly, the formalism for the cross-product in such non-Cartesian coordinates as the spherical system is the same as it is in the Cartesian system.   >  $A\&xB$ The curl of the vector field   >  $\mathrm{unassign}\left(\'g\'\right)\;$$F≔\mathrm{VectorField}\left(⟨f\left(x\,y\,z\right)\,g\left(x\,y\,z\right)\,h\left(x\,y\,z\right)⟩\,\mathrm{cartesian}\left[x\,y\,z\right]\right)$ can be computed by any of the following three approaches. >  $\mathrm{Curl}\left(F\right)$ >  $\mathrm{Del}\&xF$ >  $\nabla \times \mathbf{F}$ Other cross-products in which the $\nabla \,$or nabla, appear can be found in Table 2.   Define the operator $\mathbf{F}\times \nabla$ and apply it to the scalar $f\left(x\,y\,z\right)\.$ >  $$\begin{gathered} \mathrm{unassign}\left(\'v\'\,\'w\'\right)\: \\ L≔\mathrm{VectorField}\left(⟨u\,v\,w⟩\right) \&x \mathrm{Del}\: \\ L\left(f\left(x\,y\,z\right)\right) \end{gathered}$$ Define the operator $\nabla \times \nabla$ and apply it to the scalar $f\left(x\,y\,z\right)\.$ >  $L≔\mathrm{Del} \&x \mathrm{Del}\:$$L\left(f\left(x\,y\,z\right)\right)$ Table 2   Cross-products in which the $\nabla \,$or nabla, appear In the each case in Table 2, the cross-product is with $\nabla f\,$the gradient of the scalar $f$.