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Chapter 2: Space Curves

Section 2.3: Tangent Vectors

Example 2.3.7

Given the plane curve $C$ defined by $y (x) = \frac{1 - x^{2}}{1 + x^{2}}$ ,

a)	Obtain $R (p)$ , the radius-vector form of the curve, by the parametrization $x = p, y = y (p)$ .

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

c)	Graph $C$ and the vectors $R (1)$ and $T (1)$ .

d)	Graph $ρ (p), p \in [0, 3]$ , and determine the point $[x, y]$ at which $ρ$ is a maximum in this interval.

Solution

Mathematical Solution

If $R = [\begin{array}{c} p \\ \frac{1 - p^{2}}{1 + p^{2}} \end{array}]$ , then $R' = [\begin{array}{c} 1 \\ - \frac{4 p}{{(p^{2} + 1)}^{2}} \end{array}]$ , $ρ = ∥ R' ∥$ = $\frac{\sqrt{p^{8} + 4 p^{6} + 6 p^{4} + 20 p^{2} + 1}}{{(p^{2} + 1)}^{2}}$ ,

$T = [\begin{array}{c} \frac{{(p^{2} + 1)}^{2}}{\sqrt{p^{8} + 4 p^{6} + 6 p^{4} + 20 p^{2} + 1}} \\ - \frac{4 p}{\sqrt{p^{8} + 4 p^{6} + 6 p^{4} + 20 p^{2} + 1}} \end{array}]$ , $R (1) = [\begin{array}{c} 1 \\ 0 \end{array}]$ , and $T (1) = \frac{1}{\sqrt{2}} [\begin{array}{c} 1 \\ - 1 \end{array}]$ .

Figure 2.3.7(a) displays a graph of the curve $C$ , the vector $R (1)$ (green arrow), and the vector $T (1)$ (black arrow). Figure 2.3.7(b) shows a graph of $ρ$ , from which is inferred the existence of a single maximum in the interval $0 \leq p \leq 3$ .

use plots, Student:-VectorCalculus in
module()
local p1,p2,p3,R,T,R1,T1;
R:=<p,(-p^2+1)/(p^2+1)>;
T:=TangentVector(R,p,normalized);
R1:=eval(R,p=1);
T1:=RootedVector(root=[1,0],<1,-1>/sqrt(2));
p1:=SpaceCurve(R,p=0..3,caption="");
p2:=PlotVector([R1,T1],color=[green,black]);
p3:=display(p1,p2,scaling=constrained,labels=[x,y],tickmarks=[3,3],view =[0..3,-1..1]);
print(p3);
end module:
end use:

Figure 2.3.7(a) $C$ ; $R (1)$ (green), $T (1)$ (black)

Figure 2.3.7(b) Graph of $ρ$

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

$ρ' = \frac{16 p (1 - 3 p^{2})}{{(p^{2} + 1)}^{3} \sqrt{p^{8} + 4 p^{6} + 6 p^{4} + 20 p^{2} + 1}} = 0$

and obtaining $\overset{&Hat;}{ρ} = 1 / \sqrt{3}$ , in which case $R (1 / \sqrt{3})$ determines the point $(1 / \sqrt{3}, 1 / 2)$ .

Maple Solution - Interactive

Initialize

•	Tools≻Load Package: Student Vector Calculus

Loading Student:-MultivariateCalculus

Part (a)

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

•	Context Panel: Assign Name

$R = ⟨p, (1 - p^{2}) / (1 + p^{2})⟩$ $\overset{assign}{\to}$

Part (b)

•	Calculus palette: Differentiation operator

•	Context Panel: Evaluate and Display Inline

•	Context Panel: Simplify≻Simplify

•	Context Panel: Assign to a Name≻dR

$\frac{&DifferentialD;}{&DifferentialD; p} R$ = $[\begin{array}{c} 1 \\ - \frac{2 p}{p^{2} + 1} - \frac{2 (- p^{2} + 1) p}{{(p^{2} + 1)}^{2}} \end{array}]$ $\overset{simplify}{=}$ $[\begin{array}{c} 1 \\ - \frac{4 p}{{(p^{2} + 1)}^{2}} \end{array}]$ $\overset{assign to a name}{\to}$ $dR$

•	Keyboard the norm bars.

•	Context Panel: Evaluate and Display Inline

•	Context Panel: Simplify≻Assuming Positive

•	Context Panel: Assign to a Name≻rho

$∥dR∥$ = $\sqrt{1 + \frac{16 p^{2}}{{(p^{2} + 1)}^{4}}}$ $\overset{assuming positive}{\to}$ $\frac{\sqrt{p^{8} + 4 p^{6} + 6 p^{4} + 20 p^{2} + 1}}{{(p^{2} + 1)}^{2}}$ $\overset{assign to a name}{\to}$ $ρ$

•	Context Panel: Evaluate and Display Inline

•	Context Panel: Assign to a Name≻T

$\frac{dR}{ρ}$ = $[\begin{array}{c} \frac{{(p^{2} + 1)}^{2}}{\sqrt{p^{8} + 4 p^{6} + 6 p^{4} + 20 p^{2} + 1}} \\ - \frac{4 p}{\sqrt{p^{8} + 4 p^{6} + 6 p^{4} + 20 p^{2} + 1}} \end{array}]$ $\overset{assign to a name}{\to}$ $T$

Part (c)

•	For $R (1)$ and $T (1)$ : Expression palette: Evaluation template Context Panel: Evaluate and Display Inline Context Panel: Plots≻Arrow from origin (for $R (1)$ ) and Arrow from point≻ $(1, 0)$ (for $T (1)$ )

•	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 $C$

$\binom{R}{} | \binom{}{p = 1}$ = $[\begin{array}{c} 1 \\ 0 \end{array}]$ $\overset{plot arrow}{\to}$

$\binom{T}{} | \binom{}{p = 1}$ = $[\begin{array}{c} \frac{\sqrt{32}}{8} \\ - \frac{\sqrt{32}}{8} \end{array}]$ $\overset{plot arrow}{\to}$

$R$ = $[\begin{array}{c} p \\ \frac{- p^{2} + 1}{p^{2} + 1} \end{array}]$ $\overset{to list}{\to}$ $[p, \frac{- p^{2} + 1}{p^{2} + 1}]$ $\to$

Part (d)

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

$ρ$ = $\frac{\sqrt{p^{8} + 4 p^{6} + 6 p^{4} + 20 p^{2} + 1}}{{(p^{2} + 1)}^{2}}$ $\to$

•	Write $ρ$ and press the Enter key.

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

•	Context Panel: Conversions≻Equate to 0

•	Context Panel: Solve≻Solve

$ρ$

$\frac{\sqrt{p^{8} + 4 p^{6} + 6 p^{4} + 20 p^{2} + 1}}{{(p^{2} + 1)}^{2}}$

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

$- \frac{4 \sqrt{p^{8} + 4 p^{6} + 6 p^{4} + 20 p^{2} + 1} p}{{(p^{2} + 1)}^{3}} + \frac{8 p^{7} + 24 p^{5} + 24 p^{3} + 40 p}{2 {(p^{2} + 1)}^{2} \sqrt{p^{8} + 4 p^{6} + 6 p^{4} + 20 p^{2} + 1}}$

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

$- \frac{4 \sqrt{p^{8} + 4 p^{6} + 6 p^{4} + 20 p^{2} + 1} p}{{(p^{2} + 1)}^{3}} + \frac{8 p^{7} + 24 p^{5} + 24 p^{3} + 40 p}{2 {(p^{2} + 1)}^{2} \sqrt{p^{8} + 4 p^{6} + 6 p^{4} + 20 p^{2} + 1}} = 0$

$\overset{solve}{\to}$

$\{p = 0\}, \{p = \frac{\sqrt{3}}{3}\}, \{p = - \frac{\sqrt{3}}{3}\}$

•	Expression palette: Evaluation template

•	Context Panel: Evaluate and Display Inline

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

$\binom{R}{} | \binom{}{p = 1 / \sqrt{3}}$ = $[\begin{array}{c} \frac{\sqrt{3}}{3} \\ \frac{1}{2} \end{array}]$ $\overset{to list}{\to}$ $[\frac{\sqrt{3}}{3}, \frac{1}{2}]$

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.

$R ≔ ⟨p, (1 - p^{2}) / (1 + p^{2})⟩ &colon;$

Part (b)

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

$R' ≔ simplify (diff (R, p))$

$R' ≔ [\begin{array}{c} 1 \\ - \frac{4 p}{{(p^{2} + 1)}^{2}} \end{array}]$

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

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

$ρ ≔ \frac{\sqrt{p^{8} + 4 p^{6} + 6 p^{4} + 20 p^{2} + 1}}{{(p^{2} + 1)}^{2}}$

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

$T ≔ R' / ρ$

$T ≔ [\begin{array}{c} \frac{{(p^{2} + 1)}^{2}}{\sqrt{p^{8} + 4 p^{6} + 6 p^{4} + 20 p^{2} + 1}} \\ - \frac{4 p}{\sqrt{p^{8} + 4 p^{6} + 6 p^{4} + 20 p^{2} + 1}} \end{array}]$

Part (c)

•	Use the eval command to make the substitution $p = 1$ in each of R and T.

•	Use the ConvertVector command to convert $T (1)$ to a rooted vector.

$R1 ≔ eval (R, p = 1) &colon; temp ≔ eval (T, p = 1) &colon; T1 ≔ ConvertVector (temp, rooted, [1, 0]) &colon; R1, T1$

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

•	To graph the vectors $R (1)$ and $T (1)$ , use the PlotVector command.

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

$q_{1} ≔ SpaceCurve (R, p = 0 .. 3, caption =) &colon; q_{2} ≔ PlotVector ([R1, T1], color = [green, black]) &colon; plots :- display (q_{1}, q_{2}, scaling = constrained, labels = [x, y], tickmarks = [3, 3], view = [0 .. 3, - 1 .. 1])$

Part (d)

•	Apply the solve and diff commands to obtain the critical values for $ρ$ .

$solve (diff (ρ, p) = 0)$

$0, \frac{\sqrt{3}}{3}, - \frac{\sqrt{3}}{3}$

•	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.

$convert (eval (R, p = 1 / \sqrt{3}), list)$

$[\frac{\sqrt{3}}{3}, \frac{1}{2}]$

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