Example 2-7-14 - Maple Help

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Home : Support : Online Help : Study Guides : MultivariateCalculus : Chapter 2 : Examples : Section 2-7 : Example 2-7-14

Chapter 2: Space Curves

Section 2.7: Frenet-Serret Formalism

Example 2.7.14

If $C$ is the curve given by $R (p) = \ln (\cos (p)) i + \ln (\sin (p)) j + \sqrt{2} p k$ , $p \in (0, π / 2)$ , in Example 2.6.3,

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 2.7.14(a).

$T = [\begin{array}{c} - {\sin (p)}^{2} \\ {\cos (p)}^{2} \\ \sqrt{2} \cos (p) \sin (p) \end{array}]$

$N = [\begin{array}{c} - \sqrt{2} \cos (p) \sin (p) \\ - \sqrt{2} \cos (p) \sin (p) \\ 2 {\cos (p)}^{2} - 1 \end{array}]$

$B = [\begin{array}{c} {\cos (p)}^{2} \\ - {\sin (p)}^{2} \\ \sqrt{2} \cos (p) \sin (p) \end{array}]$

$ρ = \frac{1}{\cos (p) \sin (p)}$

$κ = \sqrt{2} \cos (p) \sin (p)$

$τ = - \sqrt{2} \cos (p) \sin (p)$

Table 2.7.14(a) Frenet formalism

The Darboux vector is then

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

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, $\frac{d}{ds} = \frac{1}{ρ} \frac{d}{dp}$ .

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

$\frac{d N}{dp} \frac{1}{ρ}$ = $[\begin{array}{c} - \sin (p) \cos (p) \sqrt{2} (2 {\cos (p)}^{2} - 1) \\ - \sin (p) \cos (p) \sqrt{2} (2 {\cos (p)}^{2} - 1) \\ - 4 {\cos (p)}^{2} {\sin (p)}^{2} \end{array}]$ = $d \times N$

$\frac{d B}{dp} \frac{1}{ρ}$ = $[\begin{array}{c} - 2 {\cos (p)}^{2} {\sin (p)}^{2} \\ - 2 {\cos (p)}^{2} {\sin (p)}^{2} \\ \sin (p) \cos (p) \sqrt{2} (2 {\cos (p)}^{2} - 1) \end{array}]$ = $d \times B$

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, $\frac{d}{ds} = \frac{1}{ρ} \frac{d}{dp}$ .

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

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: $ρ, κ, τ, T, N, B$

•	Contest Menu: Assign Name

$R = ⟨\ln (\cos (p)), \ln (\sin (p)), \sqrt{2} p⟩$ $\overset{assign}{\to}$

•	Keyboard the norm bars.

•	Calculus palette: Differentiation operator

•	Context Panel: Evaluate and Display Inline

•	Context Panel: Simplify≻Assuming Real Range≻ $p \in (0, π / 2)$

•	Context Panel: Assign to a Name≻rho

$∥\frac{&DifferentialD;}{&DifferentialD; p} R∥$ = $\sqrt{\frac{1}{{\cos (p)}^{2} {\sin (p)}^{2}}}$ $\overset{assuming real range}{\to}$ $\frac{1}{\cos (p) \sin (p)}$ $\overset{assign to a name}{\to}$ $ρ$

•	Write R and press the Enter key.

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

•	Context Panel: Simplify≻Assuming Real Range≻ $p \in (0, π / 2)$

•	Context Panel: Assign to a Name≻kappa

$R$

$[\begin{array}{c} \ln (\cos (p)) \\ \ln (\sin (p)) \\ \sqrt{2} p \end{array}]$

$\overset{curvature}{\to}$

$\frac{1}{2} \frac{\sqrt{4 {(\frac{1}{2} \frac{\sin (p) (\frac{2}{{\cos (p)}^{3} \sin (p)} - \frac{2}{\cos (p) {\sin (p)}^{3}})}{{(\frac{1}{{\cos (p)}^{2} {\sin (p)}^{2}})}^{3 / 2} \cos (p)} - \frac{1}{\sqrt{\frac{1}{{\cos (p)}^{2} {\sin (p)}^{2}}}} - \frac{{\sin (p)}^{2}}{\sqrt{\frac{1}{{\cos (p)}^{2} {\sin (p)}^{2}}} {\cos (p)}^{2}})}^{2} + 4 {(- \frac{1}{2} \frac{\cos (p) (\frac{2}{{\cos (p)}^{3} \sin (p)} - \frac{2}{\cos (p) {\sin (p)}^{3}})}{{(\frac{1}{{\cos (p)}^{2} {\sin (p)}^{2}})}^{3 / 2} \sin (p)} - \frac{1}{\sqrt{\frac{1}{{\cos (p)}^{2} {\sin (p)}^{2}}}} - \frac{{\cos (p)}^{2}}{\sqrt{\frac{1}{{\cos (p)}^{2} {\sin (p)}^{2}}} {\sin (p)}^{2}})}^{2} + 2 {\cos (p)}^{6} {\sin (p)}^{6} {(\frac{2}{{\cos (p)}^{3} \sin (p)} - \frac{2}{\cos (p) {\sin (p)}^{3}})}^{2}}}{\sqrt{\frac{1}{{\cos (p)}^{2} {\sin (p)}^{2}}}}$

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

$\sqrt{2} \cos (p) \sin (p)$

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

$κ$

•	Write R. Context Panel: Evaluate and Display Inline

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

•	Context Panel: Assign to a Name≻tau

$R$ = $[\begin{array}{c} \ln (\cos (p)) \\ \ln (\sin (p)) \\ \sqrt{2} p \end{array}]$ $\overset{torsion}{\to}$ $- \sqrt{2} \cos (p) \sin (p)$ $\overset{assign to a name}{\to}$ $τ$

•	Write R and press the Enter key.

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

•	Context Panel: Simplify≻Symbolic

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

$R$

$[\begin{array}{c} \ln (\cos (p)) \\ \ln (\sin (p)) \\ \sqrt{2} p \end{array}]$

$\overset{TNB frame}{\to}$

$[\begin{array}{c} - \frac{\sin (p)}{\sqrt{\frac{1}{{\cos (p)}^{2} {\sin (p)}^{2}}} \cos (p)} \\ \frac{\cos (p)}{\sqrt{\frac{1}{{\cos (p)}^{2} {\sin (p)}^{2}}} \sin (p)} \\ \frac{\sqrt{2}}{\sqrt{\frac{1}{{\cos (p)}^{2} {\sin (p)}^{2}}}} \end{array}], [\begin{array}{c} - \frac{\sqrt{2}}{\sqrt{\frac{1}{{\cos (p)}^{2} {\sin (p)}^{2}}}} \\ - \frac{\sqrt{2}}{\sqrt{\frac{1}{{\cos (p)}^{2} {\sin (p)}^{2}}}} \\ \frac{2 {\cos (p)}^{2} - 1}{\cos (p) \sin (p) \sqrt{\frac{1}{{\cos (p)}^{2} {\sin (p)}^{2}}}} \end{array}], [\begin{array}{c} {\cos (p)}^{2} \\ - {\sin (p)}^{2} \\ \sqrt{2} \cos (p) \sin (p) \end{array}]$

$\overset{simplify symbolic}{\to}$

$[\begin{array}{c} - {\sin (p)}^{2} \\ {\cos (p)}^{2} \\ \sqrt{2} \cos (p) \sin (p) \end{array}], [\begin{array}{c} - \sqrt{2} \cos (p) \sin (p) \\ - \sqrt{2} \cos (p) \sin (p) \\ 2 {\cos (p)}^{2} - 1 \end{array}], [\begin{array}{c} {\cos (p)}^{2} \\ - {\sin (p)}^{2} \\ \sqrt{2} \cos (p) \sin (p) \end{array}]$

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

$Q$

•	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 the Darboux vector

•	Context Panel: Evaluate and Display Inline

•	Context Panel: Simplify≻Symbolic

•	Context Panel: Assign to a Name≻d

$τ T + κ B$ = $[\begin{array}{c} \cos (p) {\sin (p)}^{3} \sqrt{2} + {\cos (p)}^{3} \sin (p) \sqrt{2} \\ - {\cos (p)}^{3} \sin (p) \sqrt{2} - \cos (p) {\sin (p)}^{3} \sqrt{2} \\ 0 \end{array}]$ $\overset{simplify symbolic}{\to}$ $[\begin{array}{c} \sqrt{2} \cos (p) \sin (p) \\ - \sqrt{2} \cos (p) \sin (p) \\ 0 \end{array}]$ $\overset{assign to a name}{\to}$ $d$

Part (b)

•	Calculus palette: Differentiation operator or cross-product operator

•	Context Panel: Evaluate and Display Inline

•	Context Panel: Simplify≻Symbolic

$T' = d \times T$

$(\frac{&DifferentialD;}{&DifferentialD; p} T) / ρ$ = $[\begin{array}{c} - 2 {\sin (p)}^{2} {\cos (p)}^{2} \\ - 2 {\sin (p)}^{2} {\cos (p)}^{2} \\ \sin (p) \cos (p) (- \sqrt{2} {\sin (p)}^{2} + \sqrt{2} {\cos (p)}^{2}) \end{array}]$ $\overset{simplify symbolic}{\to}$ $[\begin{array}{c} - 2 {\sin (p)}^{2} {\cos (p)}^{2} \\ - 2 {\sin (p)}^{2} {\cos (p)}^{2} \\ \sqrt{2} \cos (p) \sin (p) (2 {\cos (p)}^{2} - 1) \end{array}]$

$d \times T$ = $[\begin{array}{c} - 2 {\sin (p)}^{2} {\cos (p)}^{2} \\ - 2 {\sin (p)}^{2} {\cos (p)}^{2} \\ {\cos (p)}^{3} \sin (p) \sqrt{2} - \cos (p) {\sin (p)}^{3} \sqrt{2} \end{array}]$ $\overset{simplify symbolic}{\to}$ $[\begin{array}{c} - 2 {\sin (p)}^{2} {\cos (p)}^{2} \\ - 2 {\sin (p)}^{2} {\cos (p)}^{2} \\ \sqrt{2} \cos (p) \sin (p) (2 {\cos (p)}^{2} - 1) \end{array}]$

$N' = d \times N$

$(\frac{&DifferentialD;}{&DifferentialD; p} N) / ρ$ = $[\begin{array}{c} \sin (p) \cos (p) (\sqrt{2} {\sin (p)}^{2} - \sqrt{2} {\cos (p)}^{2}) \\ \sin (p) \cos (p) (\sqrt{2} {\sin (p)}^{2} - \sqrt{2} {\cos (p)}^{2}) \\ - 4 {\sin (p)}^{2} {\cos (p)}^{2} \end{array}]$ $\overset{simplify symbolic}{\to}$ $[\begin{array}{c} - \sqrt{2} \cos (p) \sin (p) (2 {\cos (p)}^{2} - 1) \\ - \sqrt{2} \cos (p) \sin (p) (2 {\cos (p)}^{2} - 1) \\ - 4 {\sin (p)}^{2} {\cos (p)}^{2} \end{array}]$

$d \times N$ = $[\begin{array}{c} - \sqrt{2} \cos (p) \sin (p) (2 {\cos (p)}^{2} - 1) \\ - \sqrt{2} \cos (p) \sin (p) (2 {\cos (p)}^{2} - 1) \\ - 4 {\sin (p)}^{2} {\cos (p)}^{2} \end{array}]$ $\overset{simplify symbolic}{\to}$ $[\begin{array}{c} - \sqrt{2} \cos (p) \sin (p) (2 {\cos (p)}^{2} - 1) \\ - \sqrt{2} \cos (p) \sin (p) (2 {\cos (p)}^{2} - 1) \\ - 4 {\sin (p)}^{2} {\cos (p)}^{2} \end{array}]$

$B' = d \times B$

$(\frac{&DifferentialD;}{&DifferentialD; p} B) / ρ$ = $[\begin{array}{c} - 2 {\sin (p)}^{2} {\cos (p)}^{2} \\ - 2 {\sin (p)}^{2} {\cos (p)}^{2} \\ \sin (p) \cos (p) (- \sqrt{2} {\sin (p)}^{2} + \sqrt{2} {\cos (p)}^{2}) \end{array}]$ $\overset{simplify symbolic}{\to}$ $[\begin{array}{c} - 2 \end{array}]$

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