Chapter 2: Space Curves
Section 2.2: Arc Length as Parameter
Example 2.2.3
Calculate the length of the curve defined parametrically by xt=t cost, yt=t sint, zt=t3/6 for 0≤ t≤2 π.
Solution
Mathematical Solution
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..
Figure 2.2.3(a) provides a graph of the given curve.
Figure 2.2.3(a) Graph of the given curve
For the given curve,
R.
= ddt t cost2+ddtt sint2+ddtt362
= cos⁡t−t⁢sin⁡t2+sin⁡t+t⁢cos⁡t2+t4/4
=t4+4 t2+4/4
=t2+22/4
=t2+2/2
The arc length is then 12∫02 πt2+2 ⅆt=128 π33+4 π=43π3+π
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 helix as the position vector R
Enter R as per Table 1.1.1.
Context Panel: Assign to a Name≻R
t cost,t sint,t3/6→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
∫02 πⅆⅆ t R ⅆt = 43⁢π3+2⁢π
Maple Solution - Coded
Install the Student MultivariateCalculus package.
withStudent:-MultivariateCalculus:
Define the helix as the position vector R.
R≔t cost,t sint,t3/6:
Apply the int, Norm, and diff commands.
intNormdiffR,t,t=0..2 π = 43⁢π3+2⁢π
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