Example 2-7-15 - Maple Help

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Chapter 2: Space Curves

Section 2.7: Frenet-Serret Formalism

Example 2.7.15

If $C$ is the curve given by $R (p) = (3 p - p^{3}) i + 3 p^{2} j + (3 p + p^{3}) k$ in Example 2.6.4,

a)	Obtain its Darboux vector $d$ .

b)	Verify the three identities in Table 2.7.3.

c)	Show that $T' \times T ″ = κ^{2} d$ , where primes denote differentiation with respect to arc length $s$ .

Solution

Mathematical Solution

Part (a)

By the usual techniques, obtain the items in Table 2.7.15(a).

$T = [\begin{array}{c} - \frac{1}{2} \frac{\sqrt{2} (p^{2} - 1)}{p^{2} + 1} \\ \frac{\sqrt{2} p}{p^{2} + 1} \\ \frac{1}{2} \sqrt{2} \end{array}]$

$N = [\begin{array}{c} - \frac{2 p}{p^{2} + 1} \\ - \frac{p^{2} - 1}{p^{2} + 1} \\ 0 \end{array}]$

$B = [\begin{array}{c} \frac{1}{2} \frac{\sqrt{2} (p^{2} - 1)}{p^{2} + 1} \\ - \frac{\sqrt{2} p}{p^{2} + 1} \\ \frac{1}{2} \sqrt{2} \end{array}]$

$ρ = 3 \sqrt{2} (p^{2} + 1)$

$κ = \frac{1}{3 {(p^{2} + 1)}^{2}}$

$τ = \frac{1}{3 {(p^{2} + 1)}^{2}}$

Table 2.7.15(a) Frenet formalism

The Darboux vector is then

$d = τ T + κ B$	$= \frac{1}{3 {(p^{2} + 1)}^{2}} [\begin{array}{c} - \frac{1}{2} \frac{\sqrt{2} (p^{2} - 1)}{p^{2} + 1} \\ \frac{\sqrt{2} p}{p^{2} + 1} \\ \frac{1}{2} \sqrt{2} \end{array}] + \frac{1}{3 {(p^{2} + 1)}^{2}} [\begin{array}{c} \frac{1}{2} \frac{\sqrt{2} (p^{2} - 1)}{p^{2} + 1} \\ - \frac{\sqrt{2} p}{p^{2} + 1} \\ \frac{1}{2} \sqrt{2} \end{array}]$
	$= [\begin{array}{c} 0 \\ 0 \\ \frac{1}{3} \frac{\sqrt{2}}{{(p^{2} + 1)}^{2}} \end{array}]$

Part (b)

•	The derivatives of the vectors in the TNB-frame must be taken with respect to the arc length $s$ . Since R is given in terms of the parameter $p$ , the chain rule must be invoked. Hence, $\frac{d}{ds} = \frac{1}{ρ} \frac{d}{dp}$ .

$\frac{d T}{dp} \frac{1}{ρ}$ = $\frac{1}{3 {(p^{2} + 1)}^{3}} [\begin{array}{c} - 2 p \\ 1 - p^{2} \\ 0 \end{array}]$ = $d \times T$

$\frac{d N}{dp} \frac{1}{ρ}$ = $\frac{\sqrt{2}}{3 {(p^{2} + 1)}^{3}} [\begin{array}{c} p^{2} - 1 \\ - 2 p \\ 0 \end{array}]$ = $d \times N$

$\frac{dB}{dp} \frac{1}{ρ}$ = $\frac{1}{3 {(p^{2} + 1)}^{3}} [\begin{array}{c} 2 p \\ p^{2} - 1 \\ 0 \end{array}]$ = $d \times B$

Part (c)

The derivatives of the vectors in the TNB-frame must be taken with respect to the arc length $s$ . Since R is given in terms of the parameter $p$ , the chain rule must be invoked. Hence, $\frac{d}{ds} = \frac{1}{ρ} \frac{d}{dp}$ .

$\frac{d T}{dp} \frac{1}{ρ} \times \frac{d^{2} T}{{dp}^{2}} \frac{1}{ρ^{2}}$ = $\frac{\sqrt{2}}{27 {(p^{2} + 1)}^{6}} [\begin{array}{c} 0 \\ 0 \\ 1 \end{array}]$ = $κ^{2} d$

Maple Solution - Interactive

Part (a)

Initialize

•	Tools≻Load Package: Student Vector Calculus

Loading Student:-VectorCalculus

•	Execute the BasisFormat command at the right, or use the task template to set the display of vectors as columns.

$BasisFormat (false) &colon;$

Frenet formalism: $ρ, κ, τ, T, N, B$

•	Context Panel: Assign Name

$R = ⟨3 p - p^{3}, 3 p^{2}, 3 p + p^{3}⟩$ $\overset{assign}{\to}$

•	Keyboard the norm bars.

•	Calculus palette: Differentiation operator

•	Context Panel: Evaluate and Display Inline

•	Context Panel: Simplify≻Assuming Positive

•	Context Panel: Assign to a Name≻rho

$∥\frac{&DifferentialD;}{&DifferentialD; p} R∥$ = $3 \sqrt{2} \sqrt{{(p^{2} + 1)}^{2}}$ $\overset{assuming positive}{\to}$ $3 \sqrt{2} (p^{2} + 1)$ $\overset{assign to a name}{\to}$ $ρ$

•	Write R. Context Panel: Evaluate and Display Inline

•	Context Panel: Student Vector Calculus≻Frenet Formalism≻Curvature≻ $p$

•	Context Panel: Simplify≻Assuming Positive

•	Context Panel: Assign to a Name≻kappa

$R$ = $[\begin{array}{c} - p^{3} + 3 p \\ 3 p^{2} \\ p^{3} + 3 p \end{array}]$ $\overset{curvature}{\to}$ $\frac{\sqrt{{(- \frac{\sqrt{2} (- 3 p^{2} + 3) csgn (1, p^{2} + 1)}{6 {csgn (p^{2} + 1)}^{2} (p^{2} + 1)} - \frac{\sqrt{2} (- 3 p^{2} + 3) p}{3 csgn (p^{2} + 1) {(p^{2} + 1)}^{2}} - \frac{\sqrt{2} p}{csgn (p^{2} + 1) (p^{2} + 1)})}^{2} + {(- \frac{\sqrt{2} p csgn (1, p^{2} + 1)}{{csgn (p^{2} + 1)}^{2} (p^{2} + 1)} - \frac{2 \sqrt{2} p^{2}}{csgn (p^{2} + 1) {(p^{2} + 1)}^{2}} + \frac{\sqrt{2}}{csgn (p^{2} + 1) (p^{2} + 1)})}^{2} + {(- \frac{\sqrt{2} (3 p^{2} + 3) csgn (1, p^{2} + 1)}{6 {csgn (p^{2} + 1)}^{2} (p^{2} + 1)} - \frac{\sqrt{2} (3 p^{2} + 3) p}{3 csgn (p^{2} + 1) {(p^{2} + 1)}^{2}} + \frac{\sqrt{2} p}{csgn (p^{2} + 1) (p^{2} + 1)})}^{2}} \sqrt{2}}{6 csgn (p^{2} + 1) (p^{2} + 1)}$ $\overset{assuming positive}{\to}$ $\frac{1}{3 {(p^{2} + 1)}^{2}}$ $\overset{assign to a name}{\to}$ $κ$

•	Write R. Context Panel: Evaluate and Display Inline

•	Context Panel: Student Vector Calculus≻Frenet Formalism≻Torsion≻ $p$

•	Context Panel: Simplify≻Assuming Positive

•	Context Panel: Assign to a Name≻tau

$R$ = $[\begin{array}{c} - p^{3} + 3 p \\ 3 p^{2} \\ p^{3} + 3 p \end{array}]$ $\overset{torsion}{\to}$ $\frac{csgn (p^{2} + 1) \sqrt{2}}{3 \sqrt{\frac{{csgn (1, p^{2} + 1)}^{2} p^{4} + 2 {csgn (1, p^{2} + 1)}^{2} p^{2} + {csgn (1, p^{2} + 1)}^{2} + 2}{{(p^{2} + 1)}^{2}}} {(p^{2} + 1)}^{3}}$ $\overset{assuming positive}{\to}$ $\frac{1}{3 {(p^{2} + 1)}^{2}}$ $\overset{assign to a name}{\to}$ $τ$

•	Write R. Context Panel: Evaluate and Display Inline

•	Context Panel: Student Vector Calculus≻Frenet Formalism≻TNB Frame≻ $p$

•	Context Panel: Simplify≻Assuming Positive

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

$R$ = $[\begin{array}{c} - p^{3} + 3 p \\ 3 p^{2} \\ p^{3} + 3 p \end{array}]$ $\overset{TNB frame}{\to}$ $[\begin{array}{c} - \frac{\sqrt{2} (p^{2} - 1) csgn (p^{2} + 1)}{2 (p^{2} + 1)} \\ \frac{\sqrt{2} csgn (p^{2} + 1) p}{p^{2} + 1} \\ \frac{csgn (p^{2} + 1) \sqrt{2}}{2} \end{array}], [\begin{array}{c} - \frac{\sqrt{2} (csgn (1, p^{2} + 1) p^{4} + 4 p csgn (p^{2} + 1) - csgn (1, p^{2} + 1))}{2 \sqrt{\frac{{csgn (1, p^{2} + 1)}^{2} p^{4} + 2 {csgn (1, p^{2} + 1)}^{2} p^{2} + {csgn (1, p^{2} + 1)}^{2} + 2}{{(p^{2} + 1)}^{2}}} {(p^{2} + 1)}^{2}} \\ \frac{\sqrt{2} (csgn (1, p^{2} + 1) p^{3} - p^{2} csgn (p^{2} + 1) + csgn (1, p^{2} + 1) p + csgn (p^{2} + 1))}{\sqrt{\frac{{csgn (1, p^{2} + 1)}^{2} p^{4} + 2 {csgn (1, p^{2} + 1)}^{2} p^{2} + {csgn (1, p^{2} + 1)}^{2} + 2}{{(p^{2} + 1)}^{2}}} {(p^{2} + 1)}^{2}} \\ \frac{\sqrt{2} csgn (1, p^{2} + 1)}{2 \sqrt{\frac{{csgn (1, p^{2} + 1)}^{2} p^{4} + 2 {csgn (1, p^{2} + 1)}^{2} p^{2} + {csgn (1, p^{2} + 1)}^{2} + 2}{{(p^{2} + 1)}^{2}}}} \end{array}], [\begin{array}{c} \frac{p^{2} - 1}{\sqrt{\frac{{csgn (1, p^{2} + 1)}^{2} p^{4} + 2 {csgn (1, p^{2} + 1)}^{2} p^{2} + {csgn (1, p^{2} + 1)}^{2} + 2}{{(p^{2} + 1)}^{2}}} {(p^{2} + 1)}^{2}} \\ - \frac{2 p}{\sqrt{\frac{{csgn (1, p^{2} + 1)}^{2} p^{4} + 2 {csgn (1, p^{2} + 1)}^{2} p^{2} + {csgn (1, p^{2} + 1)}^{2} + 2}{{(p^{2} + 1)}^{2}}} {(p^{2} + 1)}^{2}} \\ \frac{1}{(p^{2} + 1) \sqrt{\frac{{csgn (1, p^{2} + 1)}^{2} p^{4} + 2 {csgn (1, p^{2} + 1)}^{2} p^{2} + {csgn (1, p^{2} + 1)}^{2} + 2}{{(p^{2} + 1)}^{2}}}} \end{array}]$ $\overset{assuming positive}{\to}$ $[\begin{array}{c} \frac{(- p^{2} + 1) \sqrt{2}}{2 p^{2} + 2} \\ \frac{\sqrt{2} p}{p^{2} + 1} \\ \frac{\sqrt{2}}{2} \end{array}], [\begin{array}{c} - \frac{2 p}{p^{2} + 1} \\ \frac{- p^{2} + 1}{p^{2} + 1} \\ 0 \end{array}], [\begin{array}{c} \frac{\sqrt{2} (p^{2} - 1)}{2 p^{2} + 2} \\ - \frac{\sqrt{2} p}{p^{2} + 1} \\ \frac{\sqrt{2}}{2} \end{array}]$ $\overset{assign to a name}{\to}$ $Q$

•	Context Panel: Assign Name

$T = Q_{1}$ $\overset{assign}{\to}$

•	Context Panel: Assign Name

$N = Q_{2}$ $\overset{assign}{\to}$

•	Context Panel: Assign Name

$B = Q_{3}$ $\overset{assign}{\to}$

Obtain and display the Darboux vector

•	Context Panel: Assign Name

$d = τ T + κ B$ $\overset{assign}{\to}$

•	Write d. Context Panel: Evaluate and Display Inline

$d$ = $[\begin{array}{c} \frac{(- p^{2} + 1) \sqrt{2}}{3 {(p^{2} + 1)}^{2} (2 p^{2} + 2)} + \frac{\sqrt{2} (p^{2} - 1)}{3 {(p^{2} + 1)}^{2} (2 p^{2} + 2)} \\ 0 \\ \frac{\sqrt{2}}{3 {(p^{2} + 1)}^{2}} \end{array}]$

Part (b)

•	Calculus palette: Differentiation operator or cross-product operator

•	Context Panel: Evaluate and Display Inline

•	Context Panel: Simplify≻Assuming Positive (where needed)

$T' = d \times T$

$(\frac{&DifferentialD;}{&DifferentialD; p} T) / ρ$ = $[\begin{array}{c} \frac{\sqrt{2} (- \frac{2 p \sqrt{2}}{2 p^{2} + 2} - \frac{4 (- p^{2} + 1) \sqrt{2} p}{{(2 p^{2} + 2)}^{2}})}{6 (p^{2} + 1)} \\ \frac{\sqrt{2} (- \frac{2 \sqrt{2} p^{2}}{{(p^{2} + 1)}^{2}} + \frac{\sqrt{2}}{p^{2} + 1})}{6 (p^{2} + 1)} \\ 0 \end{array}]$ $\overset{assuming positive}{\to}$ $[\begin{array}{c} - \frac{2 p}{3 {(p^{2} + 1)}^{3}} \\ \frac{- p^{2} + 1}{3 {(p^{2} + 1)}^{3}} \\ 0 \end{array}]$

$d \times T$ = $[\begin{array}{c} - \frac{2 p}{3 {(p^{2} + 1)}^{3}} \\ - \frac{(\frac{(- p^{2} + 1) \sqrt{2}}{3 {(p^{2} + 1)}^{2} (2 p^{2} + 2)} + \frac{\sqrt{2} (p^{2} - 1)}{3 {(p^{2} + 1)}^{2} (2 p^{2} + 2)}) \sqrt{2}}{2} + \frac{2 (- p^{2} + 1)}{3 {(p^{2} + 1)}^{2} (2 p^{2} + 2)} \\ \frac{(\frac{(- p^{2} + 1) \sqrt{2}}{3 {(p^{2} + 1)}^{2} (2 p^{2} + 2)} + \frac{\sqrt{2} (p^{2} - 1)}{3 {(p^{2} + 1)}^{2} (2 p^{2} + 2)}) \sqrt{2} p}{p^{2} + 1} \end{array}]$

$N' = d \times N$

$(\frac{&DifferentialD;}{&DifferentialD; p} N) / ρ$ = $[\begin{array}{c} \frac{\sqrt{2} (\frac{4 p^{2}}{{(p^{2} + 1)}^{2}} - \frac{2}{p^{2} + 1})}{6 (p^{2} + 1)} \\ \frac{\sqrt{2} (- \frac{2 p}{p^{2} + 1} - \frac{2 (- p^{2} + 1) p}{{(p^{2} + 1)}^{2}})}{6 (p^{2} + 1)} \\ 0 \end{array}]$ = $\overset{assuming positive}{\to}$ $[\begin{array}{c} \frac{\sqrt{2} (p^{2} - 1)}{3 {(p^{2} + 1)}^{3}} \\ - \frac{2 \sqrt{2} p}{3 {(p^{2} + 1)}^{3}} \\ 0 \end{array}]$

$d \times N$ = $[\begin{array}{c} - \frac{\sqrt{2} (- p^{2} + 1)}{3 {(p^{2} + 1)}^{3}} \\ - \frac{2 \sqrt{2} p}{3 {(p^{2} + 1)}^{3}} \\ \frac{(\frac{(- p^{2} + 1) \sqrt{2}}{3 {(p^{2} + 1)}^{2} (2 p^{2} + 2)} + \frac{\sqrt{2} (p^{2} - 1)}{3 {(p^{2} + 1)}^{2} (2 p^{2} + 2)}) (- p^{2} + 1)}{p^{2} + 1} \end{array}]$

$B' = d \times B$

$(\frac{&DifferentialD;}{&DifferentialD; p} B) / ρ$ = $[\begin{array}{c} \frac{\sqrt{2} (\frac{2 p \sqrt{2}}{2 p^{2} + 2} - \frac{4 \sqrt{2} (p^{2} - 1) p}{{(2 p^{2} + 2)}^{2}})}{6 (p^{2} + 1)} \\ \frac{\sqrt{2} (\frac{2 \sqrt{2} p^{2}}{{(p^{2} + 1)}^{2}} - \frac{\sqrt{2}}{p^{2} + 1})}{6 (p^{2} + 1)} \\ 0 \end{array}]$ $\overset{assuming positive}{\to}$ $[\begin{array}{c} \frac{2 p}{3 {(p^{2} + 1)}^{3}} \\ \frac{p^{2} - 1}{3 {(p^{2} + 1)}^{3}} \\ 0 \end{array}]$

$d \times B$ = $[\begin{array}{c} \frac{2 p}{3 {(p^{2} + 1)}^{3}} \\ - \frac{(\frac{(- p^{2} + 1) \sqrt{2}}{3 {(p^{2} + 1)}^{2} (2 p^{2} + 2)} + \frac{\sqrt{2} (p^{2} - 1)}{3 {(p^{2} + 1)}^{2} (2 p^{2} + 2)}) \sqrt{2}}{2} + \frac{2 (p^{2} - 1)}{3 {(p^{2} + 1)}^{2} (2 p^{2} + 2)} \\ - \frac{(\frac{(- p^{2} + 1) \sqrt{2}}{3 {(p^{2} + 1)}^{2} (2 p^{2} + 2)} + \frac{\sqrt{2} (p^{2} - 1)}{3 {(p^{2} + 1)}^{2} (2 p^{2} + 2)}) \sqrt{2} p}{p^{2} + 1} \end{array}]$

Part (c)

•	Calculus palette: Differentiation operators

•	Common Symbols palette: Cross product operator Press the Enter key.

•	Context Panel: Simplify≻Simplify

$\frac{(\frac{&DifferentialD;}{&DifferentialD; p} T)}{ρ} \times \frac{(\frac{{&DifferentialD;}^{2}}{{&DifferentialD;}^{} p^{2}} T)}{ρ^{2}}$ = $[\begin{array}{c} 0 \\ 0 \\ \frac{\sqrt{2} (- \frac{2 p \sqrt{2}}{2 p^{2} + 2} - \frac{4 (- p^{2} + 1) \sqrt{2} p}{{(2 p^{2} + 2)}^{2}}) (\frac{8 \sqrt{2} p^{3}}{{(p^{2} + 1)}^{3}} - \frac{6 \sqrt{2} p}{{(p^{2} + 1)}^{2}})}{108 {(p^{2} + 1)}^{3}} - \frac{\sqrt{2} (- \frac{2 \sqrt{2} p^{2}}{{(p^{2} + 1)}^{2}} + \frac{\sqrt{2}}{p^{2} + 1}) (- \frac{2 \sqrt{2}}{2 p^{2} + 2} + \frac{16 p^{2} \sqrt{2}}{{(2 p^{2} + 2)}^{2}} + \frac{32 (- p^{2} + 1) \sqrt{2} p^{2}}{{(2 p^{2} + 2)}^{3}} - \frac{4 (- p^{2} + 1) \sqrt{2}}{{(2 p^{2} + 2)}^{2}})}{108 {(p^{2} + 1)}^{3}} \end{array}]$ $\overset{simplify}{=}$ $[\begin{array}{c} 0 \\ 0 \\ \frac{\sqrt{2}}{27 {(p^{2} + 1)}^{6}} \end{array}]$

•	Context Panel: Evaluate and Display Inline

$κ^{2} d$ = $[\begin{array}{c} \frac{\frac{(- p^{2} + 1) \sqrt{2}}{3 {(p^{2} + 1)}^{2} (2 p^{2} + 2)} + \frac{\sqrt{2} (p^{2} - 1)}{3 {(p^{2} + 1)}^{2} (2 p^{2} + 2)}}{9 {(p^{2} + 1)}^{4}} \\ 0 \\ \frac{\sqrt{2}}{27 {(p^{2} + 1)}^{6}} \end{array}]$

Maple Solution - Coded

Part (a)

Initialize

•	Install the Student VectorCalculus package.

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

•	Execute the BasisFormat command.

$BasisFormat (false) &colon;$

Define the position vector R and obtain $ρ, k, τ$

•	Define R as per Table 1.1.1.

$R ≔ ⟨3 p - p^{3}, 3 p^{2}, 3 p + p^{3}⟩ &colon;$

•	Apply the simplify, Norm, and diff commands.

$ρ ≔ simplify (Norm (diff (R, p))) assuming p > 0$

$ρ ≔ 3 \sqrt{2} (p^{2} + 1)$

•	Apply the simplify command to the result of the Curvature command.

$κ ≔ simplify (Curvature (R)) assuming p > 0$

$κ ≔ \frac{1}{3 {(p^{2} + 1)}^{2}}$

•	Apply the Torsion command.

$τ ≔ simplify (Torsion (R)) assuming p > 0$

$τ ≔ \frac{1}{3 {(p^{2} + 1)}^{2}}$

Obtain and display the vectors of the TNB-frame

•	Apply the TNBFrame command to R, assigning the result to a temporary name.

•	Use the map command to simplify the three vectors of the TNB-frame.(This two-step process for obtaining the TNB-frame overcomes a deep-seated problem with simplifying a rooted vector carrying assumptions.)

•	Extract and display the individual vectors of the TNB-frame.

$Temp ≔ TNBFrame (R) &colon; Q ≔ map (simplify, [Temp]) assuming p > 0 &colon; T, N, B ≔ Q_{1}, Q_{2}, Q_{3}$

$T, N, B ≔ [\begin{array}{c} \frac{(- p^{2} + 1) \sqrt{2}}{2 p^{2} + 2} \\ \frac{\sqrt{2} p}{p^{2} + 1} \\ \frac{\sqrt{2}}{2} \end{array}], [\begin{array}{c} - \frac{2 p}{p^{2} + 1} \\ \frac{- p^{2} + 1}{p^{2} + 1} \\ 0 \end{array}], [\begin{array}{c} \frac{\sqrt{2} (p^{2} - 1)}{2 p^{2} + 2} \\ - \frac{\sqrt{2} p}{p^{2} + 1} \\ \frac{\sqrt{2}}{2} \end{array}]$

Obtain the Darboux vector

•

Implement the definition of the Darboux vector, assigning it to a temporary name. To this temporary vector, apply the simplify command and call the result the Darboux vector.

(This two-step process for obtaining the Darboux vector overcomes a deep-seated problem with simplifying a rooted vector carrying assumptions.)

$Temp ≔ τ T + κ B &colon; d ≔ simplify (Temp) assuming p > 0$

$d ≔ [\begin{array}{c} 0 \\ 0 \\ \frac{\sqrt{2}}{3 {(p^{2} + 1)}^{2}} \end{array}]$

Part (b)

•	The derivatives of the vectors in the TNB-frame must be taken with respect to the arc length $s$ . Since R is given in terms of the parameter $p$ , the chain rule must be invoked. Hence, $\frac{d}{ds} = \frac{1}{ρ} \frac{d}{dp}$ .

•	In the following calculations, use the diff, the CrossProduct, and simplify commands as needed.

$simplify (diff (T, p) / ρ)$

$[\begin{array}{c} - \frac{2 p}{3 {(p^{2} + 1)}^{3}} \\ \frac{- p^{2} + 1}{3 {(p^{2} + 1)}^{3}} \\ 0 \end{array}]$

$simplify (CrossProduct (d, T))$

$[\begin{array}{c} - \frac{2 p}{3 {(p^{2} + 1)}^{3}} \\ \frac{- p^{2} + 1}{3 {(p^{2} + 1)}^{3}} \\ 0 \end{array}]$

$simplify (diff (N, p) / ρ)$

$[\begin{array}{c} \frac{\sqrt{2} (p^{2} - 1)}{3 {(p^{2} + 1)}^{3}} \\ - \frac{2 \sqrt{2} p}{3 {(p^{2} + 1)}^{3}} \\ 0 \end{array}]$

$CrossProduct (d, N)$

$[\begin{array}{c} - \frac{\sqrt{2} (- p^{2} + 1)}{3 {(p^{2} + 1)}^{3}} \\ - \frac{2 \sqrt{2} p}{3 {(p^{2} + 1)}^{3}} \\ 0 \end{array}]$

$simplify (diff (B, p) / ρ)$

$[\begin{array}{c} \frac{2 p}{3 {(p^{2} + 1)}^{3}} \\ \frac{p^{2} - 1}{3 {(p^{2} + 1)}^{3}} \\ 0 \end{array}]$

$simplify (CrossProduct (d, B))$

$[\begin{array}{c} \frac{2 p}{3 {(p^{2} + 1)}^{3}} \\ \frac{p^{2} - 1}{3 {(p^{2} + 1)}^{3}} \\ 0 \end{array}]$

Part (c)

$simplify (CrossProduct (\frac{diff (T, p)}{ρ}, \frac{diff (T, p, p)}{ρ^{2}}))$

$[\begin{array}{c} 0 \\ 0 \\ \frac{\sqrt{2}}{27 {(p^{2} + 1)}^{6}} \end{array}]$

$κ^{2} d$

$[\begin{array}{c} 0 \\ 0 \\ \frac{\sqrt{2}}{27 {(p^{2} + 1)}^{6}} \end{array}]$

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