James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Proof Since T − r9y| r9 | and | r9 |
Section 13.3 Arc Length and Curvature 865
− dsydt, we have
r9 − | r9 | T − ds
dt T
so the Product Rule (Theorem 13.2.3, Formula 3) gives
r0 − d 2 s
dt 2 T 1 ds
dt T9
Using the fact that T 3 T − 0 (see Example 12.4.2), we have
r9 3 r0 −S
dtD
ds 2
sT 3 T9d
− 1 for all t, so T and T9 are orthogonal by Example 13.2.4. Therefore, by
Theorem 12.4.9,
| r9 3 r0 | S
dtD − ds 2
| T 3 T9 | S
dtD − ds 2
| T | | T9 | S
dtD − ds 2
| T9 |
Now | Tstd |
Thus | T9 | − | r9 3 r0 |
sdsydtd 2 − | r9 3 r0 |
| r9 | 2
and − | T9 |
| r9 |
− | r9 3 r0 |
| r9 | 3 ■
ExamplE 4 Find the curvature of the twisted cubic rstd − kt, t 2 , t 3 l at a general point
and at s0, 0, 0d.
Solution We first compute the required ingredients:
r9std − k1, 2t, 3t 2 l
r0std − k0, 2, 6tl
Theorem 10 then gives
| r9std | − s1 1 4t 2 1 9t 4
Z
i
r9std 3 r0std − 1
0
j
2t
2
k
3t 2
6t
Z − 6t 2 i 2 6t j 1 2 k
| r9std 3 r0std | − s36t 4 1 36t 2 1 4 − 2s9t 4 1 9t 2 1 1
std − | r9std 3 r0std |
| r9std | 3 − 2s1 1 9t 2 1 9t 4
s1 1 4t 2 1 9t 4 d 3y2
At the origin, where t − 0, the curvature is s0d − 2.
For the special case of a plane curve with equation y − f sxd, we choose x as the
parameter and write rsxd − x i 1 f sxd j. Then r9sxd − i 1 f 9sxd j and r0sxd − f 0sxd j.
Since i 3 j − k and j 3 j − 0, it follows that r9sxd 3 r0sxd − f 0sxd k. We also have
| r9sxd | − s1 1 f f 9sxdg2 and so, by Theorem 10,
11 sxd −
| f 0sxd |
f1 1 s f 9sxdd 2 g 3y2
■
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.