Chapter 2: Space Curves
Section 2.2: Arc Length as Parameter
Obtain the arc-length function for the curve x=p2−p/2,y=4/3 p3/2, where p∈0,∞.
Invert s=sp to obtain p=ps and reparametrize the curve with the arc length s as the parameter.
Show that dRds=1.
Figure 2.2.6(a) provides a graph of the given curve.
If the position-vector description of a curve is given by Rt=xt i+yt j, then R.=x. i+y. j, where the over-dot notation represents differentiation with respect to t. Hence, the integrand in the arc-length integral for R is x.2+y.2 = R., which becomes
Figure 2.2.6(a) Graph of the given plane curve
=4 p2+2 p+1/4
The arc-length function is then sp=∫0p2 u+1/2 ⅆu=p2+p/2, where the variable of integration is chosen as u because the upper limit of integration is itself p.
To obtain p=ps from sp=p2+p/2, solve for p by the quadratic formula.
The result is p=−1 ±1+16 s/4, but the restriction p∈0,∞ means that ps=−1+1+16 s/4.
Obtain Rs=Rps as follows.
The relevant calculations are as follows.
=41+16 s+1−41+16 s+4−1+1+16 s1+16 s
=41+16 s+1−41+16 s−41+16 s+41+16 s
Maple Solution - Interactive
Within the Student MultivariateCalculus package, the differentiation operator automatically maps onto the components of vectors. Also, in this package, the norm of a vector defaults to the Euclidean norm.
Tools≻Load Package: Student Multivariate Calculus
Define the curve as the position vector R
Enter R as per Table 1.1.1.
Context Panel: Assign to a Name≻R
p2−p/2,4/3 p3/2→assign to a nameR
Write and evaluate the arc-length integral
Using vertical strokes for norm bars and the Calculus palette for the differentiation operator, write the norm of R′.
Context Panel: Simplify≻Assuming Positive
Context Panel: Constructions≻Definite Integral≻p ≻ Set range from 0 to t
Context Panel: Evaluate Integral
ⅆⅆ p R→assuming positive12+2⁢p→integrate w.r.t. p∫0t12+2⁢pⅆp=12⁢t+t2
Interpret the result st=t2+t/2 as sp=p2+p/2.
Write the equation s=…
Solve≻Obtain Solutions for≻p
s=p2+p/2→solutions for p−14+1+16⁢s4,−14−1+16⁢s4
Control-drag the solution with the positive radical.
Context Panel: Assign to a Name≻P
−14+14⁢1+16⁢s→assign to a nameP
Re-parametrize R by making the substitution p=ps=P
Write R, the name of the position vector Rp, and press the Enter key.
Context Panel: Evaluate at a Point≻p→P
Context Panel: Simplify≻Simplify
Context Panel: Assign to a Name≻Rs
→evaluate at point
→assign to a name
Calculus palette: Differentiation operator
Apply to Rs
Context Panel: Evaluate and Display Inline
ⅆⅆ s Rs = −21+16⁢s+12+4⁢−1+1+16⁢s1+16⁢s= simplify 1
Maple Solution - Coded
Install the Student MultivariateCalculus package.
Define the curve as the position vector R.
Apply the int, Norm, and diff commands, imposing the positivity assumption on t.
intNormdiffR,p,p=0..t assuming t>0 = 12⁢t+t2
The result of these calculations is st=t2+t/2, or sp=p2+p/2.
Use the solve command to obtain p=ps from the equation s=sp.
Assign the sequence of solutions to the name P.
Apply the eval command to obtain Rps=Rs.
In addition, apply the simplify command.
Apply simplify to the Norm command applied to ddsRs, obtained in turn by an application of the diff command.
simplifyNormdiffRs,s = 1
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