Chapter 2: Space Curves
Section 2.6: Binormal and Torsion
Example 2.6.2
For C, the curve defined by Rp=p i+3 p2 j+p3 k in Example 2.5.7,
Obtain the TNB-frame.
Calculate the torsion τ by both formulas on the right in Table 2.6.1.
Verify the equality T.p·B.p=−κ τ ρ2.
Graph C, along with the TNB-frame at p=1.
Solution
Mathematical Solution
Part (a)
The vectors T, N, and B are respectively
19⁢p4+36⁢p2+16⁢p9⁢p4+36⁢p2+13⁢p29⁢p4+36⁢p2+1,−3⁢p⁢p2+29⁢p4+36⁢p2+1⁢9⁢p4+p2+1−9⁢p4−19⁢p4+36⁢p2+1⁢9⁢p4+p2+1p⁢18⁢p2+19⁢p4+36⁢p2+1⁢9⁢p4+p2+1,3⁢p29⁢p4+p2+1−p9⁢p4+p2+119⁢p4+p2+1
Table 2.6.2(a) provides a path through the manual calculations of the TNB-frame. The overdot represents differentiation with respect to p; the prime, with respect to arc length s. The calculations are done in the following order: three down the left-hand column then three down the right-hand column, and finally, the calculation across the bottom.
R.=16 p3 p2
T′=T./ρ=−18⁢p⁢p2+29⁢p4+36⁢p2+12−6⁢9⁢p4−19⁢p4+36⁢p2+126⁢p⁢18⁢p2+19⁢p4+36⁢p2+12
ρ=R.=1/9⁢p4+36⁢p2+1
κ=∥T′∥ = 6⁢9⁢p4+p2+19⁢p4+36⁢p2+13/2
T=R./ρ=19⁢p4+36⁢p2+16⁢p9⁢p4+36⁢p2+13⁢p29⁢p4+36⁢p2+1
N=T′/κ=−3⁢p⁢p2+29⁢p4+36⁢p2+1⁢9⁢p4+p2+1−9⁢p4−19⁢p4+36⁢p2+1⁢9⁢p4+p2+1p⁢18⁢p2+19⁢p4+36⁢p2+1⁢9⁢p4+p2+1
B=T×N=ijk16 p3 p2−3⁢p⁢p2+2−9⁢p4−1p⁢18⁢p2+19⁢p4+36⁢p2+19⁢p4+p2+1 = 3⁢p29⁢p4+p2+1−p9⁢p4+p2+119⁢p4+p2+1
Table 2.6.2(a) Manual calculation of the TNB-frame
There are other ways to obtain the TNB-frame. The student taught a different path might want to modify Table 2.6.2(a) to reflect one of those different methods.
Part (b)
Torsion by first formula:
τ= −B./ρ·N
=−3⁢p⁢p2+29⁢p4−1−p⁢18⁢p2+19⁢p4+36⁢p2+19⁢p4+p2+13/2·−3⁢p⁢p2+2−(9 p4−1)p⁢18⁢p2+19⁢p4+36⁢p2+1⁢9⁢p4+p2+1
=−−9⁢p4+p2+1⁢9⁢p4+36⁢p2+19⁢p4+36⁢p2+19⁢p4+p2+12
=19⁢p4+p2+1
Torsion by second formula:
R.R..R...=R.·R..×R... = 16 p3 p2066 p006 = 36
R.×R.. = |ijk16 p3 p2066 p| = 18 p2−6 p6 ⇒ ∥R.×R..∥2 = 369 p4+p2+1
τ=R.·R..×R...R.×R..2 = 36369 p4+p2+1=19 p4+p2+1
Part (c)
Left-hand side:
T.p·B.p
=−18⁢p⁢p2+29⁢p4+36⁢p2+13/2−6⁢9⁢p4−19⁢p4+36⁢p2+13/26⁢p⁢18⁢p2+19⁢p4+36⁢p2+13/2·3⁢p⁢p2+29⁢p4+p2+13/29⁢p4−19⁢p4+p2+13/2−p⁢18⁢p2+19⁢p4+p2+13/2
=−69⁢p4+p2+1⁢9⁢p4+36⁢p2+1
Right-hand side:
−κ τ ρ2
=−6⁢9⁢p4+p2+19⁢p4+36⁢p2+13/2 19⁢p4+p2+1 9⁢p4+36⁢p2+12
=−69⁢p4+36⁢p2+1⁢9⁢p4+p2+1
Part (d)
Figure 2.6.2(a) contains a graph of C, along with the TNB-frame at p=1.
T1 is represented by the black arrow.