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Chapter 2: Space Curves

Section 2.3: Tangent Vectors

Example 2.3.2

If $R (p)$ is the position-vector representation of $C$ , the parametric curve $x = p \cos (p)$ , $y = p \sin (p)$ , $p \in [0, 3 π]$ ,

a)	Obtain $ρ = ∥R' (p)∥$ and the unit tangent vector $T = R' / ρ$ .

b)	Graph R and the vectors $R' (1), R' (5), R' (9)$ along $C$ .

c)	Graph R and the vectors $T (1), T (5), T (9)$ along $C$ .

d)	Show that $T \cdot T' (p) = 0$ , thus verifying that a unit vector is necessarily orthogonal to its derivative.

e)	To the graph in Part (c), add the vectors $T' (1), T' (5), T' (9)$ .

Solution

Mathematical Solution

Part (a)

$ρ = ∥R' (p)∥$	$= ∥(p \cos (p))' i + (p \sin (p))' j∥$
	$= \sqrt{{(\cos (p) - p \sin (p))}^{2} + {(\sin (p) + p \cos (p))}^{2}}$
	$= \sqrt{1 + p^{2}}$

$T = \frac{R'}{ρ} = \frac{1}{\sqrt{1 + p^{2}}} [\begin{array}{c} \cos (p) - p \sin (p) \\ \sin (p) + p \cos (p) \end{array}]$

Part (b)

See Figure 2.3.2(a) where the points corresponding to $p = 1, 5$ , and 9 are indicated by red dots, and the vectors $R' (p)$ at these points are shown as green arrows.

Part (c)

The vectors $T (p)$ for $p \in [1, 5, 9]$ are shown as the black arrows in Figure 2.3.2(b).

Part (d)

$T \cdot T' (p)$	$= \frac{1}{\sqrt{1 + p^{2}}} [\begin{array}{c} \cos (p) - p \sin (p) \\ \sin (p) + p \cos (p) \end{array}] \cdot \frac{1}{{(1 + p^{2})}^{3 / 2}} [\begin{array}{c} - ((p^{3} + 2 p) \cos (p) + (p^{2} + 2) \sin (p)) \\ (p^{2} + 2) \cos (p) - (p^{3} + 2 p) \sin (p) \end{array}]$
	$= \frac{1}{{(1 + p^{2})}^{2}} ([\begin{array}{c} \cos (p) - p \sin (p) \\ \sin (p) + p \cos (p) \end{array}] \cdot [\begin{array}{c} - ((p^{3} + 2 p) \cos (p) + (p^{2} + 2) \sin (p)) \\ (p^{2} + 2) \cos (p) - (p^{3} + 2 p) \sin (p) \end{array}])$
	$= \frac{(\cos (p) - p \sin (p)) (- (p^{3} + 2 p) \cos (p) - (p^{2} + 2) \sin (p)) + (\sin (p) + p \cos (p)) ((p^{2} + 2) \cos (p) - (p^{3} + 2 p) \sin (p))}{{(1 + p^{2})}^{2}}$
	$= 0$

Part (e)

The vectors $T' (p), p \in [1, 5, 9]$ , are shown as the gold arrows in Figure 2.3.2(b).

use plots, Student:-VectorCalculus in
module()
local p1,R,RR,RP;
R:=PositionVector([p*cos(p),p*sin(p)]);
RR:=VectorField(<p*cos(p),p*sin(p)>);
RP:=diff(RR,p);
p1:=PlotPositionVector(R,p=0..3*Pi,curveoptions=[scaling=constrained,labels=[x,y]],vectorfield=[RP],vectorfieldoptions=[width = .15, color = green, head_length = 1],points=[1,5,9]);
print(p1);
end module:
end use:

Figure 2.3.2(a) Graph for Part (b)

use plots, Student:-VectorCalculus in
module()
local p1,R,T,T1;
R:=PositionVector([p*cos(p),p*sin(p)]);
T:=VectorField(ConvertVector(TangentVector(R,normalized),free));
T1:=diff(T,p);
p1:=PlotPositionVector(R,p=0..3*Pi,curveoptions=[scaling=constrained,labels=[x,y]],vectorfield=[T1],vectorfieldoptions=[width = .15, color = gold, head_length = .5],points=[1,5,9],tangent=true,tangentoptions=[width=.1,color=black]);
print(p1);
end module:
end use:

Figure 2.3.2(b) Graph for Parts (c) and (e)

Figure 2.3.2(b) provides visual confirmation that the vectors T and $T'$ are orthogonal, as the calculations in Part (d) assert.

Maple Solution - Interactive

In the Student MultivariateCalculus package, differentiation maps onto the components of a vector and the Norm command defaults to the Euclidean norm.

Initialize

•	Tools≻Load Package: Student Multivariate Calculus

Loading Student:-MultivariateCalculus

•	Write $R = \dots$ as per Table 1.1.1.

•	Context Panel: Assign Name

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

Part (a)

•	Calculus palette: Differentiation operator Keyboard the norm-bars: two vertical strokes.

•	Context Panel: Evaluate and Display Inline

•	Context Panel: Assign to a Name≻rho

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

•	Calculus palette: Differentiation operator

•	Context Panel: Assign Name

$T = \frac{1}{ρ} (\frac{&DifferentialD;}{&DifferentialD; p} R)$ $\overset{assign}{\to}$

•	To display the vector T: Write T Context Panel: Evaluate and Display Inline

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

Part (b)

•	Expression palette: Evaluation template

•	Calculus palette: Differentiation operator

•	Press the Enter key.

$\binom{\frac{&DifferentialD;}{&DifferentialD; p} R}{} | \binom{}{p = 1}$

$[\begin{array}{c} \cos (1) - \sin (1) \\ \sin (1) + \cos (1) \end{array}]$

$\binom{\frac{&DifferentialD;}{&DifferentialD; p} R}{} | \binom{}{p = 5}$

$[\begin{array}{c} \cos (5) - 5 \sin (5) \\ \sin (5) + 5 \cos (5) \end{array}]$

$\binom{\frac{&DifferentialD;}{&DifferentialD; p} R}{} | \binom{}{p = 9}$

$[\begin{array}{c} \cos (9) - 9 \sin (9) \\ \sin (9) + 9 \cos (9) \end{array}]$

Part (c)

•	Expression palette: Evaluation template

•	Press the Enter key.

$\binom{T}{} | \binom{}{p = 1}$

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

$\binom{T}{} | \binom{}{p = 5}$

$[\begin{array}{c} \frac{\sqrt{26} (\cos (5) - 5 \sin (5))}{26} \\ \frac{\sqrt{26} (\sin (5) + 5 \cos (5))}{26} \end{array}]$

$\binom{T}{} | \binom{}{p = 9}$

$[\begin{array}{c} \frac{\sqrt{82} (\cos (9) - 9 \sin (9))}{82} \\ \frac{\sqrt{82} (\sin (9) + 9 \cos (9))}{82} \end{array}]$

Part (d)

•	Common Symbols palette: Dot product operator

•	Calculus palette: Differentiation operator

•	Press the Enter key.

•	Context Panel: Simplify≻Simplify

$T \cdot (\frac{&DifferentialD;}{&DifferentialD; p} T)$

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

$\overset{simplify}{=}$

$0$

Part (e)

•	Expression palette: Evaluation template

•	Calculus palette: Differentiation operator

•	Context Panel: Evaluate and Display Inline

$\binom{\frac{&DifferentialD;}{&DifferentialD; p} T}{} | \binom{}{p = 1}$ = $[\begin{array}{c} - \frac{\sqrt{2} (\cos (1) - \sin (1))}{4} + \frac{\sqrt{2} (- 2 \sin (1) - \cos (1))}{2} \\ - \frac{\sqrt{2} (\sin (1) + \cos (1))}{4} + \frac{\sqrt{2} (2 \cos (1) - \sin (1))}{2} \end{array}]$

$\binom{\frac{&DifferentialD;}{&DifferentialD; p} T}{} | \binom{}{p = 5}$ = $[\begin{array}{c} - \frac{5 \sqrt{26} (\cos (5) - 5 \sin (5))}{676} + \frac{\sqrt{26} (- 2 \sin (5) - 5 \cos (5))}{26} \\ - \frac{5 \sqrt{26} (\sin (5) + 5 \cos (5))}{676} + \frac{\sqrt{26} (2 \cos (5) - 5 \sin (5))}{26} \end{array}]$

$\binom{\frac{&DifferentialD;}{&DifferentialD; p} T}{} | \binom{}{p = 9}$ = $[\begin{array}{c} - \frac{9 \sqrt{82} (\cos (9) - 9 \sin (9))}{6724} + \frac{\sqrt{82} (- 2 \sin (9) - 9 \cos (9))}{82} \\ - \frac{9 \sqrt{82} (\sin (9) + 9 \cos (9))}{6724} + \frac{\sqrt{82} (2 \cos (9) - 9 \sin (9))}{82} \end{array}]$

The graphs in Parts (b), (c), and (e) are tedious to create interactively. The curve $C$ has to be drawn first. Then, each vector has to be graphed separately, and dragged onto the graph of $C$ . It is far easier to use the PlotVector and PlotPositionVector commands in the Student VectorCalculus package, as detailed below in the Coded Solution.

Maple Solution - Coded

In the Student VectorCalculus package, the diff command maps onto the components of vectors, the TangentVector command returns a tangent vector, and the PlotPositionVector command graphs a curve defined by the PositionVector command, and adds to the graph of the curve different vectors and vector fields with appropriate changes in syntax. The BasisFormat command is used to change the default display of vectors to "column vectors."

•	Install the Student VectorCalculus package.

•	Apply the BasisFormat command.

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

Part (a)

Use the PositionVector command to define $C$ .

$R ≔ PositionVector ([p \cos (p), p \sin (p)]) &colon;$

•	Use the diff command to obtain $R'$ , and apply the Norm command to obtain $∥ R' ∥$ .

$ρ ≔ Norm (diff (R, p))$ = $\sqrt{p^{2} + 1}$

•	Apply the diff command to obtain $T'$ .

$T ≔ \frac{diff (R, p)}{ρ}$ = $[\begin{array}{c} \frac{\cos (p) - p \sin (p)}{\sqrt{p^{2} + 1}} \\ \frac{\sin (p) + p \cos (p)}{\sqrt{p^{2} + 1}} \end{array}]$

•	Alternatively, to the position vector R, apply the TangentVector command with the normalized option to obtain a unit tangent vector along $C$ .

$TangentVector (R, normalized)$

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

Part (d): Show $T \cdot T' (p) = 0$

•	To the result of the DotProduct command, apply the simplify command.

$simplify (DotProduct (T, diff (T, p)))$ = $0$

Figure 2.3.2(a), the graph required in Part (b), is obtained with the following commands.

$F ≔ diff (VectorField (⟨p \cos (p), p \sin (p)⟩), p) &colon; PlotPositionVector (R, p = 0 .. 3 π, curveoptions = [scaling = constrained, labels = [x, y]], points = [1, 5, 9], vectorfield = [F], vectorfieldoptions = [width = .15, color = green, head_length = 1]) &colon;$

Differentiating the position vector R results in a rooted vector, a construct that the PlotPositionVector command will not graph as a vector field. Hence, the vector $F = R'$ is constructed separately as a vector field.

Figure 2.3.2(b), combining the graphs required in Parts (c) and (e), is obtained with the following commands.

$TP ≔ diff (VectorField (ConvertVector (TangentVector (R, normalized), free)), p) &colon; PlotPositionVector (R, p = 0 .. 3 π, curveoptions = [scaling = constrained, labels = [x, y]], vectorfield = [TP], vectorfieldoptions = [width = .15, color = gold, head_length = .5], points = [1, 5, 9], tangent = true, tangentoptions = [width = .1, color = black]) &colon;$

The TangentVector command returns a rooted vector, and its derivative is still a rooted vector. Hence, the tangent vector is converted to a free vector, then a vector field, with its derivative remaining a vector field. The tangent option in the PlotPositionVector command produces the black tangent vectors in Figure 2.3.2(b), while the vectorfield option produces the gold vectors corresponding to values of $T'$ .

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