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Chapter 2: Space Curves

Section 2.3: Tangent Vectors

Example 2.3.8

If $R (p)$ is the position-vector form of the curve $C$ defined parametrically by the equations $x (p) = 2 + 3 p - p^{2}, y (p) = \cos (p)$ , $p \in [0, π]$ ,

a)	Obtain $R' (p), ρ (p)$ and $T (p)$ , where $ρ = ∥R'∥$ and $T = R' / ρ$ .

b)	Graph $C$ and the vectors $T (0), T (1), T (2)$ .

c)	On the given interval, graph $ρ$ and determine its absolute minimum and the point on the curve where this minimum occurs.

Solution

Mathematical Solution

If $R = [\begin{array}{c} 2 + 3 p - p^{2} \\ \cos (p) \end{array}]$ , then

$R' = [\begin{array}{c} 3 - 2 p \\ - \sin (p) \end{array}]$ , $ρ = ∥ R' ∥$ = $\sqrt{{(- 2 p + 3)}^{2} + {\sin (p)}^{2}}$ , $T = [\begin{array}{c} \frac{- 2 p + 3}{\sqrt{{(- 2 p + 3)}^{2} + {\sin (p)}^{2}}} \\ - \frac{\sin (p)}{\sqrt{{(- 2 p + 3)}^{2} + {\sin (p)}^{2}}} \end{array}]$ ,

$T (0) = [\begin{array}{c} 1 \\ 0 \end{array}]$ , $T (1) = [\begin{array}{c} \frac{1}{\sqrt{1 + {\sin (1)}^{2}}} \\ - \frac{\sin (1)}{\sqrt{1 + {\sin (1)}^{2}}} \end{array}]$ , $T (2) =$ $[\begin{array}{c} - \frac{1}{\sqrt{1 + {\sin (2)}^{2}}} \\ - \frac{\sin (2)}{\sqrt{1 + {\sin (2)}^{2}}} \end{array}]$ .

Figure 2.3.8(a) displays a graph of the curve $C$ , and the vectors $T (0), T (1), T (2)$ as the green, black, and red vectors, respectively. Figure 2.3.8(b) shows a graph of $ρ$ , from which is inferred the existence of a single minimum in the interval $0 \leq p \leq π$ .

use plots, Student:-VectorCalculus in
module()
local p1,p2,p3,R,T,T0,T1,T2;
R:=<-p^2+3*p+2,cos(p)>;
T:=TangentVector(R,p,normalized);
T0 := simplify(ConvertVector(eval(T, p = 0), rooted, eval(R, p = 0))); T1 := ConvertVector(eval(T, p = 1), rooted, eval(R, p = 1));
T2 := ConvertVector(eval(T, p = 2), rooted, eval(R, p = 2));
p1:=SpaceCurve(R,p=0..Pi,caption="");
p2:=PlotVector([T0,T1,T2],color=[green,black,red]);
p3:=display(p1,p2,scaling=constrained,labels=[x,y],tickmarks=[3,3]);
print(p3);
end module:
end use:

Figure 2.3.8(a) $C$ ; $T (k), k = 0, 1, 2$

>	plot(sqrt((-2*p+3)^2+sin(p)^2),p=0..Pi);

Figure 2.3.8(b) Graph of $ρ$

The critical values for $ρ$ are found by solving the equation

$ρ' = \frac{\sin (p) \cos (p) + 4 p - 6}{\sqrt{- {\cos (p)}^{2} + 4 p^{2} - 12 p + 10}} = 0$

and obtaining $\overset{&Hat;}{ρ} = 1.476586707$ , in which case $R (\overset{&Hat;}{ρ})$ determines the point $(4.25, 0.094)$ . (A moment's thought will show that there is no exact solution to this equation - it has to be solved numerically.)

Maple Solution - Interactive

Initialize

•	Tools≻Load Package: Student Multivariate Calculus

Loading Student:-MultivariateCalculus

Part (a)

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

•	Context Panel: Assign Name

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

•	Calculus palette: Differentiation operator

•	Context Panel: Evaluate and Display Inline

•	Context Panel: Assign to a Name≻dR

$\frac{&DifferentialD;}{&DifferentialD; p} R$ = $[\begin{array}{c} - 2 p + 3 \\ - \sin (p) \end{array}]$ $\overset{assign to a name}{\to}$ $dR$

•	Keyboard the norm bars.

•	Context Panel: Evaluate and Display Inline

•	Context Panel: Simplify≻Assuming Positive

•	Context Panel: Assign to a Name≻rho

$∥dR∥$ = $\sqrt{{(- 2 p + 3)}^{2} + {\sin (p)}^{2}}$ $\overset{assign to a name}{\to}$ $ρ$

•	Context Panel: Evaluate and Display Inline

•	Context Panel: Assign to a Name≻T

$\frac{dR}{ρ}$ = $[\begin{array}{c} \frac{- 2 p + 3}{\sqrt{{(- 2 p + 3)}^{2} + {\sin (p)}^{2}}} \\ - \frac{\sin (p)}{\sqrt{{(- 2 p + 3)}^{2} + {\sin (p)}^{2}}} \end{array}]$ $\overset{assign to a name}{\to}$ $T$

Part (b)

•	For $T (k), k = 0, 1, 2$ : Expression palette: Evaluation template Context Panel: Evaluate and Display Inline Context Panel: Plots≻Arrow from point≻ $(2, 1)$ for $T (1)$ , $(4, \cos (1))$ for $T (1)$ , $(4, \cos (2))$ for $T (2)$

•	For $C$ : Write R Context Panel: Evaluate and Display Inline Context Panel: Student Vector Calculus≻Conversions≻To List Context Panel: Plots≻Plot Builder

•	Copy and paste the arrows onto the graph of $C$

$\binom{T}{} | \binom{}{p = 0}$ = $[\begin{array}{c} \frac{\sqrt{9}}{3} \\ 0 \end{array}]$ $\overset{plot arrow}{\to}$

$\binom{T}{} | \binom{}{p = 1}$ = $[\begin{array}{c} \frac{1}{\sqrt{1 + {\sin (1)}^{2}}} \\ - \frac{\sin (1)}{\sqrt{1 + {\sin (1)}^{2}}} \end{array}]$ $\overset{plot arrow}{\to}$

$\binom{T}{} | \binom{}{p = 2}$ = $[\begin{array}{c} - \frac{1}{\sqrt{1 + {\sin (2)}^{2}}} \\ - \frac{\sin (2)}{\sqrt{1 + {\sin (2)}^{2}}} \end{array}]$ $\overset{plot arrow}{\to}$

•	For $C$ : Write R Context Panel: Evaluate and Display Inline Context Panel: Student Vector Calculus≻Conversions≻To List Context Panel: Plots≻Plot Builder

•	Copy and paste the arrows onto the graph of $C$

$R$ = $[\begin{array}{c} - p^{2} + 3 p + 2 \\ \cos (p) \end{array}]$ $\overset{to list}{\to}$ $[- p^{2} + 3 p + 2, \cos (p)]$ $\to$

Part (c)

•	Write $ρ$ . Context Panel: Plots≻Plot Builder

$ρ$ = $\sqrt{{(- 2 p + 3)}^{2} + {\sin (p)}^{2}}$ $\to$

•	Write $ρ$ and press the Enter key.

•	Context Panel: Differentiate≻With Respect To≻ $ρ$

•	Context Panel: Conversions≻Equate to 0

•	Context Panel: Solve≻Numerically Solve

•	Context Panel: Assign to a Name≻P

$ρ$

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

$\overset{differentiate w.r.t. p}{\to}$

$\frac{8 p - 12 + 2 \sin (p) \cos (p)}{2 \sqrt{{(- 2 p + 3)}^{2} + {\sin (p)}^{2}}}$

$\overset{equate to 0}{\to}$

$\frac{8 p - 12 + 2 \sin (p) \cos (p)}{2 \sqrt{{(- 2 p + 3)}^{2} + {\sin (p)}^{2}}} = 0$

$\overset{solve}{\to}$

$1.476586707$

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

$P$

•	Expression palette: Evaluation template

•	Context Panel: Evaluate and Display Inline

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

$\binom{R}{} | \binom{}{p = P}$ = $[\begin{array}{c} 4.249451818 \\ 0.09407032279 \end{array}]$ $\overset{to list}{\to}$ $[4.249451818, 0.09407032279]$

Maple Solution - Coded

Initialize

•	Install the Student VectorCalculus package.

•	Apply the BasisFormat command.

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

Part (a)

•	Define $C$ as the position vector R.

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

•	Apply the diff and simplify commands. Note that the name $R'$ is an Atomic Identifier.

$R' ≔ simplify (diff (R, p))$

$R' ≔ [\begin{array}{c} - 2 p + 3 \\ - \sin (p) \end{array}]$

•	Apply the Norm and simplify commands. Note that the name $R'$ is an Atomic Identifier.

$ρ ≔ Norm (R')$

$ρ ≔ \sqrt{{(- 2 p + 3)}^{2} + {\sin (p)}^{2}}$

•	The name $R'$ is an Atomic Identifier.

$T ≔ R' / ρ$

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

Part (b)

•	Use the eval command to make the substitutions $p = 0, 1, 2$ in T.

•	Use the ConvertVector command to convert each vector to a rooted vector.

$T0 ≔ simplify (ConvertVector (eval (T, p = 0), rooted, eval (R, p = 0))) &colon; T1 ≔ ConvertVector (eval (T, p = 1), rooted, eval (R, p = 1)) &colon; T2 ≔ ConvertVector (eval (T, p = 2), rooted, eval (R, p = 2)) &colon; T0, T1, T2$

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

•	To graph $C$ , use the SpaceCurve command.

•	To graph the vectors $T (0), T (1), T (2)$ , use the PlotVector command.

•	Use the display command (plots package) to join the graph of $C$ with the graph of the vectors.

$q_{1} ≔ SpaceCurve (R, p = 0 .. π, caption =) &colon; q_{2} ≔ PlotVector ([T0, T1, T2], color = [green, black, red]) &colon; plots :- display (q_{1}, q_{2}, scaling = constrained, labels = [x, y])$

Part (c)

•	To $ρ$ , apply the plot command.

$plot (ρ, p = 0 .. π)$

•	Apply the fsolve and diff commands to obtain $c$ , the critical value for $ρ$ .

$c ≔ fsolve (diff (ρ, p) = 0)$

$c ≔ 1.476586707$

•	Apply the eval command to R to obtain the extreme point as a vector.

•	Apply the convert command (with option list) to change the column vector to a list.

$convert (eval (R, p = c), list)$

$[4.249451818, 0.09407032279]$

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