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Chapter 9: Vector Calculus

Section 9.2: Vector Objects

Example 9.2.10

Express in polar coordinates the Cartesian vector field $F = f (x, y) i + g (x, y) j$ .

Solution

Mathematical Solution

•	Figure 9.2.10(a) provides a schematic of the polar coordinate curves pulled back onto the Cartesian plane.

•	At one point, namely $(r, θ) = (3 / 2, π / 3)$ , the unit basis vectors for polar coordinates are drawn.

•	The green vector is $e_{r}$ , while the black one is $e_{θ}$ .

•	Clearly, as this pair of mutually orthogonal unit vectors is drawn at different points on the polar coordinate curves, they will change orientation, that is, exhibit position dependence.

use plots, Student:-VectorCalculus in
module()
local p1,p2,p3,vr,vt;
p1:=coordplot(polar,[0..3,0..2*Pi],labeling=true,scaling=constrained);vr:=RootedVector(root=Vector(<1.5,Pi/3>,polar),<cos(Pi/3),sin(Pi/3)>);
vt:=RootedVector(root=Vector(<1.5,Pi/3>,polar),<-sin(Pi/3),cos(Pi/3)>);
p2:=PlotVector([vr,vt],width=.1,color=[green,black]);
p3:=display(p1,p2,scaling=constrained);
print(p3);
end module:
end use:

Figure 9.2.10(a) Polar coordinates

The basis vectors suggested by Figure 9.2.10(a) are the normalized versions of $\frac{\partial R}{\partial r}$ and $\frac{\partial R}{\partial θ}$ , where R is the position vector to the point $(x, y) = (r \cos (θ), r \sin (θ))$ . Consequently,

$e_{r} = \frac{\partial R}{\partial r}$ = $[\begin{array}{c} \cos (θ) \\ \sin (θ) \end{array}]$ and $e_{θ} = \frac{1}{r} \frac{\partial R}{\partial θ}$ = $\frac{1}{r} [\begin{array}{c} - r \sin (θ) \\ r \cos (θ) \end{array}] = [\begin{array}{c} - \sin (θ) \\ \cos (θ) \end{array}]$

Writing $e_{r} = \cos (θ) i + \sin (θ) j$ and $e_{θ} = - \sin (θ) i + \cos (θ) j$ , and solving for i and j in terms of $e_{r}$ and $e_{θ}$ gives

$i = \cos (θ) e_{r} - \sin (θ) e_{θ}$ and $j = \sin (θ) e_{r} + \cos (θ) e_{θ}$

In polar coordinates, the Cartesian vector field $F = f (x, y) i + g (x, y) j$ is given by

$f (r \cos (θ), r \sin (θ)) [\begin{array}{c} \cos (θ) \\ - \sin (θ) \end{array}] + g (r \cos (θ), r \sin (θ)) [\begin{array}{c} \sin (θ) \\ \cos (θ) \end{array}]$

which simplifies to

$[\begin{array}{c} f (r \cos (θ), r \sin (θ)) \cos (θ) + g (r \cos (θ), r \sin (θ)) \sin (θ) \\ - f (r \cos (θ), r \sin (θ)) \sin (θ) + g (r \cos (θ), r \sin (θ)) \cos (θ) \end{array}]$

Consequently, when changing coordinates in a vector or vector field, it is not enough to change coordinates in the components. The change in the basis vectors must also be taken into account.

Maple Solution - Interactive

Initialize

•	Tools≻Load Package: Student Vector Calculus

Loading Student:-VectorCalculus

•	Tools≻Tasks≻Browse: Calculus - Vector≻ Vector Algebra and Settings≻ Display Format for Vectors

•	Press the Access Settings button and select "Display as Column Vector"

Display Format for Vectors

Define the position vector R

•	Context Panel: Assign to a Name≻R

$⟨r \cos (θ), r \sin (θ)⟩$ $\overset{assign to a name}{\to}$ $R$

Obtain a representation of $e_{r}$

•	Write the name R. Context Panel: Evaluate and Display Inline

•	Context Panel: Student Vector Calculus≻Differentiate≻With Respect To≻ $r$

•	Context Panel: Assign to a Name≻Er

$R$ = $\overset{differentiate}{\to}$ $\overset{assign to a name}{\to}$ $Er$

Obtain a representation of $e_{θ}$

•	Write the name R. Context Panel: Evaluate and Display Inline

•	Context Panel: Student Vector Calculus≻Differentiate≻With Respect To≻ $θ$

•	Context Panel: Student Vector Calculus≻Normalize≻Euclidean

•	Context Panel: Simplify≻Assuming Positive

•	Context Panel: Assign to a Name≻Et

$R$ = $\overset{differentiate}{\to}$ $\overset{Euclidean-normalize}{\to}$ $\overset{assuming positive}{\to}$ $\overset{assign to a name}{\to}$ $Et$

Solve for i and j in terms of $e_{r}$ and $e_{θ}$

•	Write a sequence of two equations relating i and j to $e_{r}$ and $e_{θ}$ ; press the Enter key.

•	Context Panel: Solve≻Solve for Variables≻ $\{i, j\}$

•	Context Panel: Simplify≻Trig

•	Context Panel: Assign to a Name≻ $S$

$Er \cdot ⟨i, j⟩ = e_{r}, Et \cdot ⟨i, j⟩ = e_{θ}$

$\cos (θ) i + \sin (θ) j = e_{r}, - \sin (θ) i + \cos (θ) j = e_{θ}$

$\overset{solve (specified)}{\to}$

$\{i = - \frac{\sin (θ) e_{θ} - \cos (θ) e_{r}}{{\sin (θ)}^{2} + {\cos (θ)}^{2}}, j = \frac{\sin (θ) e_{r} + e_{θ} \cos (θ)}{{\sin (θ)}^{2} + {\cos (θ)}^{2}}\}$

$\overset{simplify trig}{=}$

$\{i = - \sin (θ) e_{θ} + \cos (θ) e_{r}, j = \sin (θ) e_{r} + e_{θ} \cos (θ)\}$

$\overset{assign to a name}{\to}$

$S$

Introduce polar coordinates in the components

•	Expression palette: Evaluation template Press the Enter key.

•	Context Panel: Assign to a Name≻temp

$\binom{f (x, y) i + g (x, y) j}{} | \binom{}{x = r \cos (θ), y = r \sin (θ)}$ = $f (r \cos (θ), r \sin (θ)) i + g (r \cos (θ), r \sin (θ)) j$ $\overset{assign to a name}{\to}$ $temp$

Change the basis vectors

•	Expression palette: Evaluation template Press the Enter key.

•	Context Panel: Collect≻Name≻e[r]

•	Context Panel: Collect≻Name≻e[theta]

$\binom{temp}{} | \binom{}{S}$

$f (r \cos (θ), r \sin (θ)) (- \sin (θ) e_{θ} + \cos (θ) e_{r}) + g (r \cos (θ), r \sin (θ)) (\sin (θ) e_{r} + e_{θ} \cos (θ))$

$\overset{collect w.r.t. e[r]}{=}$

$(f (r \cos (θ), r \sin (θ)) \cos (θ) + g (r \cos (θ), r \sin (θ)) \sin (θ)) e_{r} - f (r \cos (θ), r \sin (θ)) \sin (θ) e_{θ} + g (r \cos (θ), r \sin (θ)) e_{θ} \cos (θ)$

$\overset{collect w.r.t. e[theta]}{=}$

$(- f (r \cos (θ), r \sin (θ)) \sin (θ) + g (r \cos (θ), r \sin (θ)) \cos (θ)) e_{θ} + (f (r \cos (θ), r \sin (θ)) \cos (θ) + g (r \cos (θ), r \sin (θ)) \sin (θ)) e_{r}$

Maple Solution - Coded

Initialize

•	Install the Student VectorCalculus package and execute the BasisFormat command.

$with (Student :- VectorCalculus) &colon; BasisFormat (false) &colon;$

Define the vector field F

•	Apply the VectorField command.

$F ≔ VectorField (⟨f (x, y), g (x, y)⟩) &colon;$

Change to polar coordinates by applying the MapToBasis command

$MapToBasis (F, polar [r, θ])$

The column vector returned by the MapToBasis command gives the components of the vector field in polar coordinates. The basis vectors would be Maple's "barred" vectors, the "moving" basis vectors that are a function of position.

A solution from first principles is more enlightening.

Define R, the position vector in polar coordinates

•	Use the PositionVector command with the optional name of a coordinate system.

$R ≔ PositionVector ([r, θ], polar)$

By differentiation and normalization, obtain unit tangent vectors along the polar coordinate curves

$Er ≔ diff (R, r)$

$Et ≔ Normalize (diff (R, θ)) assuming r > 0$

Express these unit vectors in terms of i and j, equate to the names er and et, and solve for i and j

$q_{1} ≔ DotProduct (Er, ⟨i, j⟩) = er$

$\cos (θ) i + \sin (θ) j = er$

$q_{2} ≔ DotProduct (Et, ⟨i, j⟩) = et$

$- \sin (θ) i + \cos (θ) j = et$

•	Apply the solve and simplify commands.

$S ≔ simplify (solve (\{q_{1}, q_{2}\}, \{i, j\}))$

$\{i = \cos (θ) er - \sin (θ) et, j = et \cos (θ) + \sin (θ) er\}$

In the field $F = x y i + x / y j$ , replace i and j with their equivalents in terms of er and et and $x$ and $y$ with their equivalents in polar coordinates

$temp ≔ collect (eval (⟨f (x, y), g (x, y)⟩ \cdot ⟨i, j⟩, S), [er, et])$

$(f (x, y) \cos (θ) + g (x, y) \sin (θ)) er + (\cos (θ) g (x, y) - \sin (θ) f (x, y)) et$

$eval (temp, [x = r \cos (θ), y = r \sin (θ)])$

$(f (r \cos (θ), r \sin (θ)) \cos (θ) + g (r \cos (θ), r \sin (θ)) \sin (θ)) er + (- f (r \cos (θ), r \sin (θ)) \sin (θ) + g (r \cos (θ), r \sin (θ)) \cos (θ)) et$

Compare

$MapToBasis (F, polar [r, θ])$

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