Chapter 2: Space Curves
Section 2.2: Arc Length as Parameter
Obtain st, the arc-length function for the curve in Example 2.1.3.
If the position-vector description of a curve is given by Rt=xt i+yt j+zt k, then R.=x. i+y. j+z. k, 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+z.2 = R..
For the given curve,
= ddtt cost2+ddtt sint2+ddtt362
Hence, the arc-length function is st=12∫0tu2+2 ⅆu=t36+t. (Note that because the upper limit of integration is t, the integration variable itself must be some other variable, here chosen to be u.)
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
Using p as the parameter on the helix, define this curve as the position vector R
Enter R as per Table 1.1.1.
Context Panel: Assign to a Name≻R
p cosp,p sinp,p3/6→assign to a nameR
Write and evaluate the arc-length integral
Calculus palette: Definite integral template
Calculus palette: Differentiation operator
Context Panel: Evaluate and Display Inline
Context Panel: Simplify≻Assuming Positive
∫0tⅆⅆ p R ⅆp = 16⁢csgn⁡t2+2⁢t⁢t2+6→assuming positive16⁢t⁢t2+6
The arc-length function is therefore st=t3/6+t. (Note that because the upper limit of integration is t, the integration variable itself must be some other variable, here chosen to be p.)
Maple Solution - Coded
Install the Student MultivariateCalculus package.
Define the helix as the position vector R.
R≔p cosp,p sinp,p3/6:
Apply the int, Norm, and diff commands.
Note the need for a positivity assumption on t.
intNormdiffR,p,p=0..t assuming t>0 = 16⁢t3+t
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