Chapter 2: Space Curves
Section 2.3: Tangent Vectors
Example 2.3.8
If Rp is the position-vector form of the curve C defined parametrically by the equations xp=2+3p− p2,yp=cosp, p∈0,π,
Obtain R′p,ρp and Tp, where ρ=R′ and T=R′/ρ.
Graph C and the vectors T0,T1,T2.
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=2+3 p−p2cos(p), then
R′=3−2 p−sin(p), ρ=∥R′∥ = −2p+32+sinp2, T=−2p+3−2p+32+sinp2−sinp−2p+32+sinp2,
T0=10, T1=11+sin12−sin11+sin12, T2=−11+sin22−sin21+sin22.
Figure 2.3.8(a) displays a graph of the curve C, and the vectors T0,T1,T2 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≤p≤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; Tk,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′=sinpcosp+4p−6−cosp2+4p2−12 p+10=0
and obtaining ρ^=1.476586707, in which case Rρ^ 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=… as per Table 1.1.1.
Context Panel: Assign Name
R=2+3 p−p2,cosp→assign
Calculus palette: Differentiation operator
Context Panel: Evaluate and Display Inline
Context Panel: Assign to a Name≻dR
ⅆⅆ p R = −2p+3−sinp→assign to a namedR
Keyboard the norm bars.
Context Panel: Simplify≻Assuming Positive
Context Panel: Assign to a Name≻rho
dR = −2p+32+sinp2→assign to a nameρ
Context Panel: Assign to a Name≻T
dRρ = −2p+3−2p+32+sinp2−sinp−2p+32+sinp2→assign to a nameT
Part (b)
For Tk,k=0,1,2: Expression palette: Evaluation template Context Panel: Evaluate and Display Inline Context Panel: Plots≻Arrow from point≻2,1 for T1, 4,cos1 for T1, 4,cos2 for T2
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
Tx=a|f(x)p=0 = 930→plot arrow
Tx=a|f(x)p=1 = 11+sin12−sin11+sin12→plot arrow
Tx=a|f(x)p=2 = −11+sin22−sin21+sin22→plot arrow
R = −p2+3p+2cosp→to list−p2+3p+2,cosp→
Part (c)
Write ρ. Context Panel: Plots≻Plot Builder
ρ = −2p+32+sinp2→
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
−2p+32+sinp2
→differentiate w.r.t. p
8p−12+2sinpcosp2−2p+32+sinp2
→equate to 0
8p−12+2sinpcosp2−2p+32+sinp2=0
→solve
1.476586707
→assign to a name
P
Expression palette: Evaluation template
Context Panel: Student Vector Calculus≻Conversions≻To List
Rx=a|f(x)p=P = 4.2494518180.09407032279→to list4.249451818,0.09407032279
Maple Solution - Coded
Install the Student VectorCalculus package.
Apply the BasisFormat command.
withStudent:-VectorCalculus:BasisFormatfalse:
Define C as the position vector R.
R≔2+3 p− p2,cosp:
Apply the diff and simplify commands. Note that the name R′ is an Atomic Identifier.
R′≔simplifydiffR,p
R′≔−2p+3−sinp
Apply the Norm and simplify commands. Note that the name R′ is an Atomic Identifier.
ρ≔NormR′
ρ≔−2p+32+sinp2
The name R′ is an Atomic Identifier.
T≔R′/ρ
T≔−2p+3−2p+32+sinp2−sinp−2p+32+sinp2
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≔simplifyConvertVectorevalT,p=0,rooted,evalR,p=0:T1≔ConvertVectorevalT,p=1,rooted,evalR,p=1:T2≔ConvertVectorevalT,p=2,rooted,evalR,p=2:T0,T1,T2
10,11+sin12−sin11+sin12,−11+sin22−sin21+sin22
To graph C , use the SpaceCurve command.
To graph the vectors T0,T1,T2, use the PlotVector command.
Use the display command (plots package) to join the graph of C with the graph of the vectors.
q1≔SpaceCurveR,p=0..π,caption=:q2≔PlotVectorT0,T1,T2,color=green,black,red:plots:-displayq1,q2,scaling=constrained,labels=x,y
To ρ, apply the plot command.
plotρ,p=0..π
Apply the fsolve and diff commands to obtain c, the critical value for ρ.
c≔fsolvediffρ,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.
convertevalR,p=c,list
4.249451818,0.09407032279
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