Chapter 2: Space Curves
Section 2.7: Frenet-Serret Formalism
Example 2.7.5
If s is arc length and the prime denotes dds, show that R″R‴R⁗=κ5ddsτκ.
Solution
Mathematical Solution
Since R′s=T, it then follows that R″=T′=κ N (via the first equation on the right in Table 2.7.1). So,
R‴
=R″′
=κ N′
=κ N′+κ′N
=κ −κ T+τ B+κ′N
= −κ2T+κ′N+κ τ B
=a3T+b3N+c3B
Note the use, in the fourth line, of the third equation on the right in Table 2.7.1. The last line simplifies the notation in anticipation of the additional calculations needed to evaluate the box product. However, it is first necessary to obtain R⁗.
R⁗
=R‴′
=−κ2T+κ′N+κ τ B′
=−κ2T′+κ′N′+κ τ B′
=−2 κ κ′T−κ2T′+κ″N+κ′N′+κ′τ B+κ τ′B+κ τ B′
=−2 κ κ′T−κ2κ N+κ″N+κ′−κ T+τ B+κ′τ B+κ τ′B+κ τ −τ N
=−2 κ κ′−κ κ′T+−κ3+κ″−κ τ2N+κ′τ+κ′τ+κ τ′B
= −3 κ κ′T+κ″−κ3−κ τ2N+2 κ′τ+κ τ′B
=a4T+b4N+c4B
By invoking the additive property of determinants, x y z+w=x y z+x y w, and by writing R″=κ N as b2N, the box product becomes
R″R‴R⁗=b2a4c3−a3c4NBT = b2a4c3−a3c4BTN = b2a4c3−a3c4
where the minus signs introduced by row interchanges in the determinant cancel, and
BTN=B·T×N=B·B=1
Making the substitutions b2=κ,a3=−κ2,a4=−3 κ κ′,c3=κ τ,c4=2 κ′τ+κ τ′, into b2a4c3−a3c4 leads to
b2a4c3−a3c4
=κ4τ′−κ3κ′τ
=κ3κ τ′−κ′τ
=κ5κ τ′−κ′τκ2
=κ5 ddsτκ
Maple Solution
The only place in Maple where symbolic calculations can be made with vectors is in the Physics:-Vectors package. Unfortunately, linear input for this package works best, although the output reflects notation greatly desired in the realm of physics.
In the following calculations, the basis vectors i,j,k take the place of the vectors T, N, and B, respectively. Within the package, a left-underscore for the letters i, j, or k indicates a unit basis vector ; a right-underscore, a symbolic vector.
Install the package.
withPhysics:-Vectors:
Use the Setup command to implement the notational benefits of the package.
Setupmathematicalnotation=true:
Let R2 represent R″, where the right- underscore indicates that this is a vector in the package.
R2_ := b[2]*_j;
R2→≔b2j∧
Let R3 represent R‴, where the right- underscore indicates that this is a vector in the package.
R3_ := a[3]*_i+b[3]*_j+c[3]*_k;
R3→≔a3i∧+b3j∧+c3k∧
Let R4 represent R⁗, where the right- underscore indicates that this is a vector in the package.
R4_ := a[4]*_i+b[4]*_j+c[4]*_k;
R4→≔a4i∧+b4j∧+c4k∧
Use the period for the dot product and &x for the cross product. Apply the expand command. The result is the value of the box product.
q := expand(R2_ . (R3_ &x R4_));
q≔−a3b2c4+a4b2c3
Apply the simplify command to the result of the substitutions enacted by the eval command. Both κ′ and τ′ must be set as Atomic Identifiers to prevent Maple from actually taking a derivative, which by default, is with respect to x.
simplifyevalq,b2=κ,a3=−κ2,a4=−3 κ κ′,c3=κ τ,c4=2 κ′τ+κ τ′
κ4τ′−κ3κ′τ
Write this final result as κ3κ τ′−κ′τ=κ5κ τ′−κ′τκ2=κ5 ddsτκ.
Slightly more insight into the details of these calculations can be obtained if, instead of using the basis vectors i,j,k, the derivatives of R are written in terms of T, N, and B.
r2_ := b[2]*N_: r3_ := a[3]*T_+b[3]*N_+c[3]*B_: r4_ := a[4]*T_+b[4]*N_+c[4]*B_: expand(r2_ . (r3_ &x r4_));
b2c3a4N→·B→×T→−b2a3c4N→·B→×T→
The simplification of the box product to a new product involving just the vectors of the TNB-frame, and not linear combinations of these vectors, is the reward for this extra effort.
<< Previous Example Section 2.7 Next Example >>
© Maplesoft, a division of Waterloo Maple Inc., 2024. All rights reserved. This product is protected by copyright and distributed under licenses restricting its use, copying, distribution, and decompilation.
For more information on Maplesoft products and services, visit www.maplesoft.com
Download Help Document
