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Chapter 2: Space Curves

Section 2.6: Binormal and Torsion

Example 2.6.1

For $C$ , the helix defined by $R (p) = \cos (p) i + \sin (p) j + p k$ in Example 2.5.6,

a)	Obtain the TNB-frame.

b)	Calculate the torsion $τ$ by both formulas on the right in Table 2.6.1.

c)	Verify the equality $\overset{&period;}{T} (p) \cdot \overset{&period;}{B} (p) = - κ τ ρ^{2}$ .

d)	Graph $C$ , along with the TNB-frame at $p = π / 3$ .

Solution

Mathematical Solution

Part (a)

The vectors T, N, and B are respectively

$\frac{1}{\sqrt{2}} [\begin{array}{c} - \sin (p) \\ \cos (p) \\ 1 \end{array}]$ , $- [\begin{array}{c} \cos (p) \\ \sin (p) \\ 0 \end{array}]$ , $\frac{1}{\sqrt{2}} [\begin{array}{c} \sin (p) \\ - \cos (p) \\ 1 \end{array}]$

Table 2.6.1(a) provides a path through the manual calculations of the TNB-frame. The overdot represents differentiation with respect to $p$ ; the prime, with respect to arc length $s$ . The calculations are done in the following order: three down the left-hand column then three down the right-hand column, and finally, the calculation across the bottom.

$\overset{&period;}{R} = [\begin{array}{c} - \sin (p) \\ \cos (p) \\ 1 \end{array}]$	$T' = \overset{&period;}{T} / ρ = - \frac{1}{2} [\begin{array}{c} \cos (p) \\ \sin (p) \\ 0 \end{array}]$
$ρ = ∥\overset{&period;}{R}∥ = \sqrt{2}$	$κ = ∥ T' ∥$ = $1 / 2$
$T = \overset{&period;}{R} / ρ = \frac{1}{\sqrt{2}} [\begin{array}{c} - \sin (p) \\ \cos (p) \\ 1 \end{array}]$	$N = T' / κ = - [\begin{array}{c} \cos (p) \\ \sin (p) \\ 0 \end{array}]$
$B = T \times N = \|\begin{array}{c} i & j & k \\ - \sin (p) / \sqrt{2} & \cos (p) / \sqrt{2} & 1 / \sqrt{2} \\ - \cos (p) & - \sin (p) & 0 \end{array}\|$ = $\frac{1}{\sqrt{2}} [\begin{array}{c} \sin (p) \\ - \cos (p) \\ 1 \end{array}]$
Table 2.6.1(a) Manual calculation of the TNB-frame

There are other ways to obtain the TNB-frame. The student taught a different path might want to modify Table 2.6.1(a) to reflect one of those different methods.

Part (b)

Torsion by first formula:

$τ = - (\overset{&period;}{B} / ρ) \cdot N$ = $- \frac{1}{\sqrt{2}} (\frac{1}{\sqrt{2}} [\begin{array}{c} \cos (p) \\ \sin (p) \\ 0 \end{array}]) \cdot (- [\begin{array}{c} \cos (p) \\ \sin (p) \\ 0 \end{array}])$ = $\frac{1}{2} (\cos^{2} (p) + \sin^{2} (p)) = \frac{1}{2}$

Torsion by second formula:

$[\overset{&period;}{R} \overset{..}{R} \overset{.. &period;}{R}] = \overset{&period;}{R} \cdot (\overset{..}{R} \times \overset{.. &period;}{R})$ = $|\begin{array}{c} - \sin (p) & \cos (p) & 1 \\ - \cos (p) & - \sin (p) & 0 \\ \sin (p) & - \cos (p) & 0 \end{array}|$ = 1

$\overset{&period;}{R} \times \overset{..}{R}$ = $| \begin{array}{c} i & j & k \\ - \sin (p) & \cos (p) & 1 \\ - \cos (p) & - \sin (p) & 0 \end{array} |$ = $[\begin{array}{c} \sin (p) \\ - \cos (p) \\ 1 \end{array}]$ ⇒ ${∥ \overset{&period;}{R} \times \overset{..}{R} ∥}^{2}$ = 2 ⇒ $τ = \frac{\overset{&period;}{R} \cdot (\overset{..}{R} \times \overset{.. &period;}{R})}{{∥\overset{&period;}{R} \times \overset{..}{R}∥}^{2}}$ = $\frac{1}{2}$

Part (c)

Left-hand side:

$\overset{&period;}{T} (p) \cdot \overset{&period;}{B} (p)$ = $(\frac{1}{\sqrt{2}} [\begin{array}{c} - \cos (p) \\ - \sin (p) \\ 0 \end{array}]) \cdot (\frac{1}{\sqrt{2}} [\begin{array}{c} \cos (p) \\ \sin (p) \\ 0 \end{array}])$ = $- \frac{1}{2} (\cos^{2} (p) + \sin^{2} (p))$ = $- \frac{1}{2}$

Right-hand side:

$- κ τ ρ^{2} = - (1 / 2) (1 / 2) {(\sqrt{2})}^{2} = - \frac{1}{2}$

Part (d)

•	Figure 2.6.1(a) contains a graph of the helix, along with the TNB-frame at $p = π / 3$ .

•	$T (π / 3)$ is represented by the black arrow.

•	$N (π / 3)$ is represented by the red arrow.

•	$B (π / 3)$ is represented by the green arrow.

•	The animation in the tutor can be used to verify that as T advances with increasing arclength, the osculating plane twists about the tangent line clockwise (as viewed in the direction of the advance of T), a twist consistent with the positive value of the torsion.

use plots, Student:-VectorCalculus in
module()
local p1,p2,p3,R,V,TNB;
R:=<cos(p),sin(p),p>;
TNB:=TNBFrame(R,p);
V:=map(ConvertVector,eval([TNB],p=(1/3)*Pi),rooted,[1/2,(1/2)*sqrt(3), (1/3)*Pi]);
p1:=PlotVector(V,color=[black,red,green],width=.1);
p2:=PlotPositionVector(ConvertVector(R,position),p=0..2*Pi, curveoptions=[color=blue]);
p3:=display(p1,p2,scaling=constrained,axes=frame,labels=[x,y,z], tickmarks=[[0,1],[0,1],8],orientation=[-65,70,0]);
print(p3);
end module:
end use:

Figure 2.6.1(a) $C$ and TNB-frame at $p = π / 3$

Maple Solution - Interactive

Part (a)

•	Tools≻Load Package: Student Vector Calculus

Loading Student:-VectorCalculus

•	Execute the BasisFormat command at the right, or use the task template.

$BasisFormat (false) &colon;$

•	Enter the vector notation for $C$ as per Table 1.1.1. Context Panel: Assign Name

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

•	Write R Context Panel: Evaluate and Display Inline

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

•	Context Panel: Assign to a Name≻TNB

$R$ = $\overset{TNB frame}{\to}$ $\overset{assign to a name}{\to}$ $TNB$

•	Figure 2.6.1(a) is a screenshot of the tutor adjusted to display an animation of a single TNB-frame traversing one loop of the helix.

•	The Plot Options button is used to impose constrained scaling and the frame style for the axes.

•	The default number of frames, 5, is changed to 30, and the Animate button is pressed.

•	The Display Options drop-down box provides other options: The individual vectors of the TNB-frame can be separately graphed, and graphs of the curvature and torsion can be displayed.

•

In Figure 2.6.1(b), the animation of a single TNB-frame traversing the curve was generated by the TNBFrame command, which allows for greater control of all aspects of the animation. (The animation shown in Figure 2.6.1(a) would be written to the worksheet upon pressing the Close button in the tutor.) The colors black, red, and green are used respectively for T, N, and B.

Figure 2.6.1(a) Space Curves tutor

Student:-VectorCalculus:-TNBFrame(<cos(p),sin(p),p>,range=0..2*Pi,output=animation,tangentoptions=[color=black,width=.1],normaloptions=[color=red,width=.1],binormaloptions=[color=green,width=.1],curveoptions=[scaling=constrained,labels=[x,y,z],axes=frame,orientation=[-40,80,0],tickmarks=[[0],[0],6],lightmodel=none],caption="",frames=30);

Figure 2.6.1(b) TNB-frame animation

The tutor is based on the TNBFrame command whose output can be a graph of the curve along with a specified number of TNB-frames, an animation such as shown in Figure 2.6.1(b), or the algebraic representation of the vectors of the TNB-frame.

The Display Options in the tutor provides for graphs of the individual tangent, principal normal, and binormal vectors, and normalized versions of these vectors. That is because the underlying commands TangentVector, PrincipalNormal, and Binormal, originally did not have options for returning normalized vectors. Surprisingly, the vectors in the TNB-frame returned by the TNBFrame command are normalized by default.

Part (b)

Obtain the torsion via the Context Panel system

•	Write R Context Panel: Evaluate and Display Inline

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

$R$ = $\overset{torsion}{\to}$ $\frac{1}{2}$

Obtain the torsion by the upper-right formula in Table 2.6.1

•	Keyboard the norm bars. Calculus palette: Differentiation operator

•	Context Panel: Evaluate and Display Inline

•	Context Panel: Assign to a Name≻rho

$∥\frac{&DifferentialD;}{&DifferentialD; p} R∥$ = $\sqrt{2}$ $\overset{assign to a name}{\to}$ $ρ$

•	Extract B and N from the TNB-frame obtained in Part (a).

•	Calculus palette: Differentiation operator Common Symbols palette: Dot product operator

•	Context Panel: Evaluate and Display Inline

•	Context Panel: Simplify≻Simplify

$- (\frac{&DifferentialD;}{&DifferentialD; p} {TNB}_{3}) \cdot {TNB}_{2} / ρ$ = $- \frac{1}{2} (- \frac{1}{2} \sqrt{2} {\cos (p)}^{2} - \frac{1}{2} \sqrt{2} {\sin (p)}^{2}) \sqrt{2}$ $\overset{simplify}{=}$ $\frac{1}{2}$

Obtain the torsion by the lower-right formula in Table 2.6.1

•	Form the name $\overset{&period;}{R}$ as an Atomic Identifier.

•	Calculus palette: Differentiation operator

•	Context Panel: Assign Name

$\overset{&period;}{R} = \frac{&DifferentialD;}{&DifferentialD; p} R$ $\overset{assign}{\to}$

•	Form the name $\overset{..}{R}$ as an Atomic Identifier.

•	Calculus palette: Differentiation operator

•	Context Panel: Assign Name

$\overset{..}{R} = \frac{{&DifferentialD;}^{2}}{{&DifferentialD;}^{} p^{2}} R$ $\overset{assign}{\to}$

•	Form the name $\overset{.. &period;}{R}$ as an Atomic Identifier.

•	Calculus palette: Differentiation operator

•	Context Panel: Assign Name

$\overset{.. &period;}{R} = \frac{{&DifferentialD;}^{3}}{{&DifferentialD;}^{} p^{3}} R$ $\overset{assign}{\to}$

•	Keyboard the norm bars, and use Atomic Identifiers when referencing the derivatives of R.

•	Common Symbols palette: Dot and cross product operators

•	Context Panel: Evaluate and Display Inline

•	Context Panel: Assign to a Name≻tau

$\frac{\overset{&period;}{R} \cdot (\overset{..}{R} \times \overset{.. &period;}{R})}{{&par; \overset{&period;}{R} \times \overset{..}{R} &par;}^{2}}$ = $\frac{1}{2} {\cos (p)}^{2} + \frac{1}{2} {\sin (p)}^{2}$ $\overset{simplify}{=}$ $\frac{1}{2}$ $\overset{assign to a name}{\to}$ $τ$

Part (c)

Evaluate the right-hand side of the given identity

•	Write R Context Panel: Evaluate and Display Inline

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

•	Context Panel: Simplify≻Simplify

•	Context Panel: Assign to a Name≻kappa

$R$ = $\overset{curvature}{\to}$ $\frac{1}{4} \sqrt{2 {\cos (p)}^{2} + 2 {\sin (p)}^{2}} \sqrt{2}$ $\overset{simplify}{=}$ $\frac{1}{2}$ $\overset{assign to a name}{\to}$ $κ$

•	Write the expression. Context Panel: Evaluate and Display Inline

$- κ τ ρ^{2}$ = $- \frac{1}{2}$

Evaluate the left-hand side of the given identity

•	Calculus palette: Differentiation operator; extract T and B from the TNB-frame in Part (a)

•	Common Symbols palette: Dot product operator

•	Context Panel: Evaluate and Display Inline

•	Context Panel: Simplify≻Simplify

$(\frac{&DifferentialD;}{&DifferentialD; p} {TNB}_{1}) \cdot (\frac{&DifferentialD;}{&DifferentialD; p} {TNB}_{3})$ = $- \frac{1}{2} {\cos (p)}^{2} - \frac{1}{2} {\sin (p)}^{2}$ $\overset{simplify}{=}$ $- \frac{1}{2}$

Part (d)

•	As a first recourse, use the tutor to explore the behavior of the TNB-frame along the helix.

•	Figure 2.6.1(b) shows the state of the tutor after it has been applied to the given helix, the number of frames changed from the default 5 to 6, the axes changed to frame-style, and scaling set to constrained.

•	If the Animate option is selected, then the selected number of frames will be seen to traverse the curve.

Figure 2.6.1(b) Space Curve tutor

The following interactive construction results in an approximation to Figure 2.6.1(a).

Evaluate the TNB-frame at $p = π / 3$

•	Expression palette: Evaluation template

•	Context Panel: Evaluate and Display Inline

•	Context Panel: Assign to a Name≻TNBp

$\binom{TNB}{} | \binom{}{p = π / 3}$ = $\overset{assign to a name}{\to}$ $TNBp$

Graph each vector in the evaluated TNB-frame

•	Type $TNBp [k], k = 1, 2, 3$ , to reference respectively $T (π / 3), N (π / 3), B (π / 3)$

•	Context Panel: Evaluate and Display Inline

•	Context Panel: Student Vector Calculus≻Conversions≻To Free Vector

•	Context Panel: Plots≻Arrow from point≻ $1 / 2, \sqrt{3 / 2}, π / 3$ (for $x, y, z$ , respectively) Context Panel: Color≻black, red, green, respectively

$TNBp [1]$ = $\overset{to free Vector}{\to}$ $\overset{plot arrow}{\to}$

$TNBp [2]$ = $\overset{to free Vector}{\to}$ $\overset{plot arrow}{\to}$

$TNBp [3]$ = $\overset{to free Vector}{\to}$ $\overset{plot arrow}{\to}$

Graph R and add the vectors of the TNB-frame

•	Write R Context Panel: Evaluate and Display Inline

•	Context Panel: Student Vector Calculus: Conversions≻To List

•	Context Panel: Plots≻Plot Builder Set $p \in [0, 2 π]$ Options: Constrained Scaling, color = blue, Axes = frame

•	Copy/paste the graphs of the three vectors

$R$ = $\overset{to list}{\to}$ $[\cos (p), \sin (p), p]$ $\to$

Unfortunately, there is no provision as yet to vary the "heft" of vectors graphed interactively.

Maple Solution - Coded

Part (a)

•	Install the Student VectorCalculus package.

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

•	Execute the BasisFormat command.

$BasisFormat (false) &colon;$

•	Define $C$ as the position vector R.

$R ≔ ⟨\cos (p), \sin (p), p⟩ &colon;$

•	Invoke the TNBFrame command.

$TNB ≔ TNBFrame (R)$

Part (b)

•	Apply the Torsion command.

$τ ≔ Torsion (R)$ = $\frac{1}{2}$

Obtain the torsion by the upper-right formula in Table 2.6.1

•	Use the diff and Norm commands to obtain $ρ$ .

$ρ ≔ Norm (diff (R, p))$ = $\sqrt{2}$

•	Apply the diff, DotProduct, and simplify commands, extracting B and N from the TNB-frame.

$simplify (- \frac{1}{ρ} DotProduct (diff ({TNB}_{3}, p), {TNB}_{2}))$ = $\frac{1}{2}$

Obtain the torsion by the lower-right formula in Table 2.6.1

•	Apply the diff command; $\overset{&period;}{R}$ is an Atomic Identifier.

$\overset{&period;}{R} ≔ diff (R, p) &colon;$

•	Apply the diff command; $\overset{..}{R}$ is an Atomic Identifier.

$\overset{..}{R} ≔ diff (R, p &dollar; 2) &colon;$

•	Apply the diff command; $\overset{.. &period;}{R}$ is an Atomic Identifier.

$\overset{.. &period;}{R} ≔ diff (R, p &dollar; 3) &colon;$

Apply the DotProduct, CrossProduct, Norm, and simplify commands as appropriate.

$simplify (\frac{DotProduct (\overset{&period;}{R}, CrossProduct (\overset{..}{R}, \overset{.. &period;}{R}))}{Norm {(CrossProduct (\overset{&period;}{R}, \overset{..}{R}))}^{2}})$ = $\frac{1}{2}$

Part (c)

•	Obtain $κ$ by applying the Curvature and simplify commands to R.

$κ ≔ simplify (Curvature (R))$ = $\frac{1}{2}$

•	Form the product on the right-hand side.

$- κ τ ρ^{2}$ = $- \frac{1}{2}$

•	Obtain the left-hand side by applying the diff, DotProduct, and simplify commands, extracting T and B from the TNB-frame obtained in Part (a).

$simplify (DotProduct (diff ({TNB}_{1}, p), diff ({TNB}_{3}, p)))$ = $- \frac{1}{2}$

Part (d)

The TNBFrame command can return the TNB-frame, a graph of the curve along with a specified number of individual frames, or an animation of a frame moving along the curve. The following implementation of the command returns six frames so that the first one must occur at the point corresponding to $p = π / 3$ . Unfortunately, there is no way to specify the exact values of $p$ at which TNB-frames are to be graphed.

$TNBFrame (R, range = 0 .. 2 π, output = plot, caption =, frames = 6, curveoptions = [scaling = constrained, axes = frame, color = blue, labels = [x, y, z], orientation = [- 65, 70, 0]], tangentoptions = [color = black, width = .1], normaloptions = [color = red, width = .1], binormaloptions = [color = green, width = .1])$

The following implementation of the TNBFrame command returns an animation of a single frame moving along the helix.

$TNBFrame (R, range = 0 .. 2 π, output = animation, caption =, frames = 30, curveoptions = [scaling = constrained, axes = frame, color = blue, labels = [z, y, z], orientation = [- 65, 70, 0]], tangentoptions = [color = black, width = .1], normaloptions = [color = red, width = .1], binormaloptions = [color = green, width = .1])$

A graph of the helix, along with the TNB-frame at $p = π / 3$ , can be constructed with the following commands. The TNB-frame is evaluated at $p = π / 3$ , but the resulting vectors are rooted at the symbolic point $[\cos (p), \sin (p), p]$ . Hence, these vectors are "converted" to vectors rooted at the contact point determined by $R (π / 3)$ . Then they can be graphed with the PlotVector command. The PlotPositionVector is used to graph the helix (after R is converted to a position vector with the ConvertVector command), and the display command merges the two graphs.

$V ≔ map (ConvertVector, eval ([TNB], p = π / 3), rooted, [1 / 2, \sqrt{3} / 2, π / 3]) &colon; p_{1} ≔ PlotVector (V, color = [black, red, green], width = .1) &colon; p_{2} ≔ PlotPositionVector (ConvertVector (R, position), p = 0 .. 2 π, curveoptions = [color = blue]) &colon; plots :- display (p_{1}, p_{2}, scaling = constrained, axes = frame, labels = [x, y, z], tickmarks = [[0, 1], [0, 1], 8], orientation = [- 65, 70, 0])$

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