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Home : Support : Online Help : Education : Study Guides : MultivariateCalculus : Chapter 9 : Examples : Section 9-3 : Example 9-3-3

Chapter 9: Vector Calculus

Section 9.3: Differential Operators

Example 9.3.3

Derive the expression for the gradient in polar coordinates.

Solution

Mathematical Solution

Starting with the definition of the gradient in Cartesian coordinates, there are two parts to the required derivation. First, the chain rule must be applied to obtain the polar-coordinate equivalents of the Cartesian derivatives with respect to $x$ and $y$ . Second, the Cartesian basis vectors $\{i, j\}$ must be expressed in terms of the polar basis vectors $\{e_{r}, e_{θ}\}$ . Of course, once both these steps have been completed, it is a matter of algebraic simplification to pass from $\nabla F = F_{x} i + F_{y} j$ to the polar-coordinate equivalent $f_{r} e_{r} + \frac{f_{θ}}{r} e_{θ}$ .

To apply the chain rule, use the notation

$f (r, θ) = f (r (x, y), θ (x, y)) = F (x, y)$

where the function $F$ , while related to the function $f$ , is not the same! For example, if $f (r, θ) = r θ$ , then $F (x, y) = \sqrt{x^{2} + y^{2}} \arctan (y / x)$ . For $f$ , the two arguments are simply multiplied. That is not what happens with $F$ . The two arguments $x$ and $y$ are not simply added. Hence, the use of a different name, a usage that is not always present in the literature.

The chain rule then gives $F_{x} = f_{r} r_{x} + f_{θ} θ_{x}$ and $F_{y} = f_{r} r_{y} + f_{θ} θ_{y}$ . Table 9.3.3(a) then provides the derivatives of $r$ and $θ$ with respect to $x$ and $y$ .

$r_{x} = \frac{x}{\sqrt{x^{2} + y^{2}}} = \frac{r \cos (θ)}{r} = \cos (θ)$

$r_{y} = \frac{y}{\sqrt{x^{2} + y^{2}}} = \frac{r \sin (θ)}{r} = \sin (θ)$

$θ_{x} = \frac{- y / x^{2}}{1 + {(y / x)}^{2}} = - \frac{y}{x^{2} + y^{2}} = - \frac{r \sin (θ)}{r^{2}} = - \frac{\sin (θ)}{r}$

$θ_{y} = \frac{1 / x}{1 + {(y / x)}^{2}} = \frac{x}{x^{2} + y^{2}} = \frac{r \cos (θ)}{r^{2}} = \frac{\cos (θ)}{r}$

Table 9.3.3(a) Derivatives of $r$ and $θ$ with respect to $x$ and $y$

From Example 9.2.8, $i = \cos (θ) e_{r} - \sin (θ) e_{θ}, j = \sin (θ) e_{r} + \cos (θ) e_{θ}$ , so

$F_{x} i + F_{y} j$	$= (f_{r} r_{x} + f_{θ} θ_{x}) (\cos (θ) e_{r} - \sin (θ) e_{θ}) + (f_{r} r_{y} + f_{θ} θ_{y}) (\sin (θ) e_{r} + \cos (θ) e_{θ})$
	$= (f_{r} r_{x} + f_{θ} θ_{x}) [\begin{array}{c} \cos (θ) \\ - \sin (θ) \end{array}] + (f_{r} r_{y} + f_{θ} θ_{y}) [\begin{array}{c} \sin (θ) \\ \cos (θ) \end{array}]$
	$= (f_{r} \cos (θ) + f_{θ} (- \frac{\sin (θ)}{r})) [\begin{array}{c} \cos (θ) \\ - \sin (θ) \end{array}] + (f_{r} \sin (θ) + f_{θ} \frac{\cos (θ)}{r}) [\begin{array}{c} \sin (θ) \\ \cos (θ) \end{array}]$
	$= [\begin{array}{c} f_{r} (\cos^{2} (θ) + \sin^{2} (θ)) + f_{θ} (\sin (θ) \cos (θ) - \sin (θ) \cos (θ)) / r \\ f_{r} (\sin (θ) \cos (θ) - \sin (θ) \cos (θ)) + f_{θ} (\sin^{2} (θ) + \cos^{2} (θ)) / r \end{array}]$
	$= [\begin{array}{c} f_{r} \\ f_{θ} / r \end{array}]$

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 polar coordinates

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

$[x = r \cos (θ), y = r \sin (θ)]$ $\overset{assign to a name}{\to}$ $Pc$

Define $\{i, j\}$ in terms of $\{e_{r}, e_{θ}\}$

•	Copy and paste the set $S$ from Example 9.2.8.

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

$\{i = \cos (θ) e_{r} - \sin (θ) e_{θ}, j = \sin (θ) e_{r} + \cos (θ) e_{θ}\}$ $\overset{assign to a name}{\to}$ $S$

Define $F = f (r (x, y), θ (x, y))$

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

$f (\sqrt{x^{2} + y^{2}}, \arctan (y, x))$ $\overset{assign to a name}{\to}$ $F$

Apply the chain rule to obtain $F_{x}$ and $F_{y}$ , convert to polar coordinates, and simplify.

•	Expression palette: Evaluation template Calculus palette: Partial-differentiation operator Press the Enter key.

•	Context Panel: Simplify≻Assuming Positive

•	Context Panel: Simplify≻Assuming Real Range (Complete dialog as shown to the right.)

•	Context Panel: Assign to a Name≻ $Fx$ (or $Fy$ )

$\binom{\frac{\partial}{\partial x} F}{} | \binom{}{Pc}$

$\frac{D_{1} (f) (\sqrt{r^{2} {\cos (θ)}^{2} + r^{2} {\sin (θ)}^{2}}, \arctan (r \sin (θ), r \cos (θ))) r \cos (θ)}{\sqrt{r^{2} {\cos (θ)}^{2} + r^{2} {\sin (θ)}^{2}}} - \frac{D_{2} (f) (\sqrt{r^{2} {\cos (θ)}^{2} + r^{2} {\sin (θ)}^{2}}, \arctan (r \sin (θ), r \cos (θ))) \sin (θ)}{r {\cos (θ)}^{2} (1 + \frac{{\sin (θ)}^{2}}{{\cos (θ)}^{2}})}$

$\overset{assuming positive}{\to}$

$\frac{D_{1} (f) (r, \arctan (\sin (θ), \cos (θ))) r \cos (θ) - D_{2} (f) (r, \arctan (\sin (θ), \cos (θ))) \sin (θ)}{r}$

$\overset{assuming real range}{\to}$

$\frac{D_{1} (f) (r, θ) r \cos (θ) - D_{2} (f) (r, θ) \sin (θ)}{r}$

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

$Fx$

$\binom{\frac{\partial}{\partial y} F}{} | \binom{}{Pc}$

$\frac{D_{1} (f) (\sqrt{r^{2} {\cos (θ)}^{2} + r^{2} {\sin (θ)}^{2}}, \arctan (r \sin (θ), r \cos (θ))) r \sin (θ)}{\sqrt{r^{2} {\cos (θ)}^{2} + r^{2} {\sin (θ)}^{2}}} + \frac{D_{2} (f) (\sqrt{r^{2} {\cos (θ)}^{2} + r^{2} {\sin (θ)}^{2}}, \arctan (r \sin (θ), r \cos (θ)))}{r \cos (θ) (1 + \frac{{\sin (θ)}^{2}}{{\cos (θ)}^{2}})}$

$\overset{assuming positive}{\to}$

$\frac{D_{1} (f) (r, \arctan (\sin (θ), \cos (θ))) r \sin (θ) + \cos (θ) D_{2} (f) (r, \arctan (\sin (θ), \cos (θ)))}{r}$

$\overset{assuming real range}{\to}$

$\frac{D_{1} (f) (r, θ) r \sin (θ) + \cos (θ) D_{2} (f) (r, θ)}{r}$

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

$Fy$

Complete the derivation of the gradient in polar coordinates.

•	Expression palette: Evaluation template Press the Enter key.

•	Context Panel: Simplify≻Simplify

•	Context Panel: Expand≻Expand

•	Context Panel: Apply a Command (Complete the dialog as per figure to the right.)

$\binom{Fx i + Fy j}{} | \binom{}{S}$

$\frac{(D_{1} (f) (r, θ) r \cos (θ) - D_{2} (f) (r, θ) \sin (θ)) (\cos (θ) e_{r} - \sin (θ) e_{θ})}{r} + \frac{(D_{1} (f) (r, θ) r \sin (θ) + \cos (θ) D_{2} (f) (r, θ)) (\sin (θ) e_{r} + \cos (θ) e_{θ})}{r}$

$\overset{simplify}{=}$

$\frac{D_{1} (f) (r, θ) r e_{r} + D_{2} (f) (r, θ) e_{θ}}{r}$

$\overset{expand}{=}$

$D_{1} (f) (r, θ) e_{r} + \frac{D_{2} (f) (r, θ) e_{θ}}{r}$

$\to$

$(\frac{\partial}{\partial r} f (r, θ)) e_{r} + \frac{(\frac{\partial}{\partial θ} f (r, θ)) e_{θ}}{r}$

Unfortunately, the Context Panel does not provide for a conversion of the differential operator notation, so the last step is obtained by selecting from the Context Panel the option "Apply a Command", and applying the convert command as per the figure.

Maple Solution - Coded

Initialize

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

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

Define $F = f (r (x, y), θ (x, y))$

$F ≔ f (\sqrt{x^{2} + y^{2}}, \arctan (y, x)) &colon;$

Differentiate with respect to $x$ , then with respect to $y$

•	Apply the diff command.

•	Apply the eval command to replace $x$ and $y$ with their equivalents in $r$ and $θ$ .

•	Apply the simplify command with appropriate assumptions on $r$ and $θ$ .

•	Finally, convert the differential operator notation to partial-derivative notation via convert.

$temp ≔ diff (F, x)$

$\frac{D_{1} (f) (\sqrt{x^{2} + y^{2}}, \arctan (y, x)) x}{\sqrt{x^{2} + y^{2}}} - \frac{D_{2} (f) (\sqrt{x^{2} + y^{2}}, \arctan (y, x)) y}{x^{2} (\frac{y^{2}}{x^{2}} + 1)}$

$Temp ≔ simplify (eval (temp, [x = r \cos (θ), y = r \sin (θ)])) assuming r > 0, θ ∷ RealRange (Open (- π), π)$

$\frac{D_{1} (f) (r, θ) r \cos (θ) - D_{2} (f) (r, θ) \sin (θ)}{r}$

$Fx ≔ convert (Temp, diff)$

$\frac{(\frac{\partial}{\partial r} f (r, θ)) r \cos (θ) - (\frac{\partial}{\partial θ} f (r, θ)) \sin (θ)}{r}$

$temp ≔ diff (F, y)$

$\frac{D_{1} (f) (\sqrt{x^{2} + y^{2}}, \arctan (y, x)) y}{\sqrt{x^{2} + y^{2}}} + \frac{D_{2} (f) (\sqrt{x^{2} + y^{2}}, \arctan (y, x))}{x (\frac{y^{2}}{x^{2}} + 1)}$

$Temp ≔ simplify (eval (temp, [x = r \cos (θ), y = r \sin (θ)])) assuming r > 0, θ ∷ RealRange (Open (- π), π)$

$\frac{D_{1} (f) (r, θ) r \sin (θ) + D_{2} (f) (r, θ) \cos (θ)}{r}$

$Fy ≔ convert (Temp, diff)$

$\frac{(\frac{\partial}{\partial r} f (r, θ)) r \sin (θ) + (\frac{\partial}{\partial θ} f (r, θ)) \cos (θ)}{r}$

From Example 9.2.8, define i and j in terms of $e_{r}$ and $e_{θ}$

$S ≔ \{i = \cos (θ) e_{r} - \sin (θ) e_{θ}, j = \sin (θ) e_{r} + \cos (θ) e_{θ}\} &colon;$

Apply the chain rule, writing $F_{x} i + F_{y} j$ and simplifying

•	Make the substitutions in $S$ via the eval command.

•	Apply the collect command to group the coefficients of $e_{r}$ and $e_{θ}$ .

•	Use the collect command to apply both the simplify and expand commands to the expression.

$collect (eval (Fx i + Fy j, S), [e_{r}, e_{θ}], expand &commat; simplify)$

$(\frac{\partial}{\partial r} f (r, θ)) e_{r} + \frac{(\frac{\partial}{\partial θ} f (r, θ)) e_{θ}}{r}$

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