Chapter 2: Space Curves
Section 2.2: Arc Length as Parameter
Example 2.2.6
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.
Solution
Mathematical Solution
Part (a)
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
dRdp
=ddpp2−p/22+ddp4/3 p3/22
=2 p−1/22+2p2
=4 p2+2 p+1/4
=4 p+12/4
=2 p+1/2
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.
Part (b)
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.
Rs=Rps
= −14+141+16s2+18−181+16s43−14+141+16s3/2
=14−141+16s+s16−1+1+16s3/2
Part (c)
The relevant calculations are as follows.
ddsRs=−21+16s+12−1+1+16s1+16s
ddsRs2
=−21+16s+12+2−1+1+16s1+16s2
=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
=1
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
Loading Student:-MultivariateCalculus
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+2p→integrate w.r.t. p∫0t12+2pⅆp=12t+t2
Interpret the result st=t2+t/2 as sp=p2+p/2.
Write the equation s=…
Context Panel: Solve≻Obtain Solutions for≻p
s=p2+p/2→solutions for p−14+1+16s4,−14−1+16s4
Control-drag the solution with the positive radical.
Context Panel: Assign to a Name≻P
−14+141+16s→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
R
p2−12p4p323
→evaluate at point
−14+1+16s42+18−1+16s84−14+1+16s4323
= simplify
14−1+16s4+s−1+1+16s326
→assign to a name
Rs
Calculus palette: Differentiation operator Apply to Rs
Context Panel: Evaluate and Display Inline
ⅆⅆ s Rs = −21+16s+12+4−1+1+16s1+16s= simplify 1
Maple Solution - Coded
Install the Student MultivariateCalculus package.
withStudent:-MultivariateCalculus:
Define the curve as the position vector R.
R≔p2−p/2,4/3 p3/2:
Apply the int, Norm, and diff commands, imposing the positivity assumption on t.
intNormdiffR,p,p=0..t assuming t>0 = 12t+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.
P≔solves=p2+p/2,p
P≔−14+1+16s4,−14−1+16s4
Apply the eval command to obtain Rps=Rs. In addition, apply the simplify command.
Rs≔simplifyevalR,p=P1
Rs≔14−1+16s4+s−1+1+16s326
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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