Section 13.3 Arc Length and Curvature
SECTION 13.3 ARC LENGTH AND CURVATURE ¤ 335
13.3 Arc Length and Curvature
1.r() = h 3 cos 3 sin i ⇒ r0() = h1 −3 sin 3 cos i ⇒
|r0()| =
12+ (−3 sin )2+ (3 cos )2=
1 + 9(sin2 + cos2) =√10.
Then using Formula 3, we have =5
−5|r0()| =5
−5
√10 = √ 10 5
−5= 10√
10.
2.r() =
2 2133
⇒ r0() = 2 2 2
⇒
|r0()| =
22+ (2)2+ (2)2 =√
4 + 42+ 4 =
(2 + 2)2= 2 + 2for 0 ≤ ≤ 1. Then using Formula 3, we have
=1
0 |r0()| =1
0(2 + 2) = 2 +1331 0=73. 3.r() =√
2 i + j+ −k ⇒ r0() =√
2 i + j− −k ⇒
|r0()| =√
22
+ ()2+ (−−)2=√
2 + 2+ −2=
(+ −)2= + − [since + − 0].
Then =1
0 |r0()| =1
0(+ −) =
− −1
0= − −1. 4.r() = cos i + sin j + ln cos k ⇒ r0() = − sin i + cos j +− sin
cos k= − sin i + cos j − tan k,
|r0()| =
(− sin )2+ cos2 + (− tan )2=√
1 + tan2 =√
sec2 = |sec |. Since sec 0 for 0 ≤ ≤ 4, here we can say |r0()| = sec . Then
=4
0 sec =
ln |sec + tan |4
0 = lnsec4 + tan4
− ln |sec 0 + tan 0|
= ln√2 + 1
− ln |1 + 0| = ln(√ 2 + 1)
5.r() = i + 2j+ 3k ⇒ r0() = 2 j + 32k ⇒ |r0()| =√
42+ 94= √
4 + 92 [since ≥ 0].
Then =1
0 |r0()| =1 0 √
4 + 92 = 181 ·23(4 + 92)321
0= 271(1332− 432) = 271(1332− 8).
6.r() = 2i+ 9 j + 432k ⇒ r0() = 2 i + 9 j + 6√
k ⇒
|r0()| =√
42+ 81 + 36 =
(2 + 9)2= |2 + 9| = 2 + 9 [since 2 + 9 ≥ 0 for 1 ≤ ≤ 4]. Then
=4
1 |r0()| =4
1(2 + 9) =
2+ 94
1= 52 − 10 = 42.
7.r() =
2 3 4
⇒ r0() =
2 32 43
⇒ |r0()| =
(2)2+ (32)2+ (43)2 =√
42+ 94+ 166, so
=2
0 |r0()| =2 0
√42+ 94+ 166 ≈ 186833.
8.r() =
− −
⇒ r0() =
1 −− (1 − )−
⇒
|r0()| =
12+ (−−)2+ [(1 − )−]2 =
1 + −2+ (1 − )2−2=
1 + (2 − 2 + 2)−2, so
=3
1 |r0()| =3 1
1 + (2 + 2 + 2)−2 ≈ 20454.
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336 ¤ CHAPTER 13 VECTOR FUNCTIONS
9. r() = hcos 2 sin 2i ⇒ r0() = h− sin 2 2 cos 2i ⇒ |r0()| =
2sin2 + 4 + 42cos22.
The point (1 0 0) corresponds to = 0 and (1 4 0) corresponds to = 2, so the length is
=2
0 |r0()| =2 0
2sin2 + 4 + 42cos22 ≈ 103311.
10. We plot two different views of the curve with parametric equations = sin , = sin 2, = sin 3. To help visualize the curve, we also include a plot showing a tube of radius 007 around the curve.
The complete curve is given by the parameter interval [0 2] and we have r0() = hcos 2 cos 2 3 cos 3i ⇒
|r0()| =√
cos2 + 4 cos22 + 9 cos23, so =2
0 |r0()| =2
0
√cos2 + 4 cos22 + 9 cos23 ≈ 160264.
11. The projection of the curve onto the -plane is the curve 2= 2or =122, = 0. Then we can choose the parameter
= ⇒ = 122. Since also lies on the surface 3 = , we have =13 =13()(122) = 163. Then parametric equations for are = , =122, =163and the corresponding vector equation is r() =
122163. The origin corresponds to = 0 and the point (6 18 36) corresponds to = 6, so
=6
0 |r0()| =6 0
1 122 =6 0
12+ 2+1
222
=6 0
1 + 2+144
=6 0
(1 +122)2 =6
0(1 +122) =
+1636
0= 6 + 36 = 42
12. Let be the curve of intersection. The projection of onto the -plane is the ellipse 42+ 2= 4or 2+ 24 = 1,
= 0. Then we can write = cos , = 2 sin , 0 ≤ ≤ 2. Since also lies on the plane + + = 2, we have
= 2 − − = 2 − cos − 2 sin . Then parametric equations for are = cos , = 2 sin , = 2 − cos − 2 sin , 0 ≤ ≤ 2, and the corresponding vector equation is r() = hcos 2 sin 2 − cos − 2 sin i. Differentiating gives r0() = h− sin 2 cos sin − 2 cos i ⇒
|r0()| =
(− sin )2+ (2 cos )2+ (sin − 2 cos )2=
2 sin2 + 8 cos2 − 4 sin cos . The length of is
=2
0 |r0()| =2
0
2 sin2 + 8 cos2 − 4 sin cos ≈ 135191.
13. (a) r() = (5 − ) i + (4 − 3) j + 3 k ⇒ r0() = −i + 4 j + 3 k and = |r0()| =√
1 + 16 + 9 =√
26. The point
(4 1 3)corresponds to = 1, so the arc length function from is
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SECTION 13.3 ARC LENGTH AND CURVATURE ¤ 337
() =
1 |r0()| = 1
√26 = √ 26
1=√
26 ( − 1). Since =√
26 ( − 1), we have =
√26+ 1.
Substituting for in the original equation, the reparametrization of the curve with respect to arc length is
r(()) =
5 −
√26+ 1
i+
4
√26+ 1
− 3
j+ 3
√26+ 1
k
=
4 −
√26
i+
4
√26+ 1
j+
3
√26+ 3
k
(b) The point 4 units along the curve from has position vector
r((4)) =
4 − 4
√26
i+
4(4)
√26+ 1
j+
3(4)
√26+ 3
k, so the point is
4 − 4
√26 16
√26+ 1 12
√26+ 3
.
14. (a) r() = sin i + cos j +√
2 k ⇒ r0() = (cos + sin ) i + (cos − sin ) j +√
2 kand
= |r0()| =
2(cos + sin )2+ 2(cos − sin )2+ 22
=
2
2(cos2 + sin2) + 2 cos sin − 2 cos sin + 2
=√
42= 2 The point
0 1√
2corresponds to = 0, so the arc length function from is
() =
0 |r0()| =
0 2 = 2|0 = 2(− 1). Since = 2(− 1), we have =
2+ 1 ⇔
= ln1
2 + 1. Substituting for in the original equation, the reparametrization of the curve with respect to arc length is r(()) =1
2 + 1 sin
ln1
2 + 1
i+1
2 + 1 cos
ln1
2 + 1
j+√ 2 2 +√
2 k.
(b) The point 4 units along the curve from has position vector r((4)) =1
2(4) + 1 sin
ln1
2(4) + 1
i+1
2(4) + 1 cos
ln1
2(4) + 1
j+√ 2 2 (4) +√
2
k, so the point is
3 sin(ln 3) 3 cos(ln 3) 3√ 2.
15. Here r() = h3 sin 4 3 cos i, so r0() = h3 cos 4 −3 sin i and |r0()| =
9 cos2 + 16 + 9 sin2 =√25 = 5.
The point (0 0 3) corresponds to = 0, so the arc length function beginning at (0 0 3) and measuring in the positive direction is given by () =
0|r0()| =
05 = 5. () = 5 ⇒ 5 = 5 ⇒ = 1, thus your location after moving 5 units along the curve is (3 sin 1 4 3 cos 1).
16. r() =
2
2+ 1− 1
i+ 2
2+ 1j ⇒ r0() = −4
(2+ 1)2 i+ −22+ 2 (2+ 1)2 j,
= |r0()| =
−4
(2+ 1)2
2
+
−22+ 2 (2+ 1)2
2
=
44+ 82+ 4 (2+ 1)4 =
4(2+ 1)2 (2+ 1)4 =
4
(2+ 1)2 = 2
2+ 1. Since the initial point (1 0) corresponds to = 0, the arc length function is
() =
0
r0() = 0
2
2+ 1 = 2 arctan . Then arctan = 12 ⇒ = tan12. Substituting, we have
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338 ¤ CHAPTER 13 VECTOR FUNCTIONS
r(()) =
2
tan21
2 + 1− 1
i+ 2 tan1 2 tan21
2
+ 1j=1 − tan21 2 1 + tan21
2 i +2 tan1 2 sec21
2 j
=1 − tan21
2 sec21
2 i+ 2 tan1
2 cos21
2 j=
cos21
2
− sin21
2
i+ 2 sin1
2 cos1
2
j= cos i + sin j With this parametrization, we recognize the function as representing the unit circle. Note here that the curve approaches, but does not include, the point (−1 0), since cos = −1 for = + 2 ( an integer) but then = tan1
2is undefined.
17. (a) r() = h 3 cos 3 sin i ⇒ r0() = h1 −3 sin 3 cos i ⇒ |r0()| =
1 + 9 sin2 + 9 cos2 =√ 10.
Then T() = r0()
|r0()| = √110h1 −3 sin 3 cos i or
√1
10 −√310sin √310cos .
T0() = √1
10h0 −3 cos −3 sin i ⇒ |T0()| = √110
0 + 9 cos2 + 9 sin2 = √3 10. Thus N() = T0()
|T0()| =1√ 10 3√
10h0 −3 cos −3 sin i = h0 − cos − sin i.
(b) () = |T0()|
|r0()| = 3√
√ 10 10 = 3
10 18. (a) r() =
2 sin − cos cos + sin
⇒
r0() = h2 cos + sin − cos , −sin + cos + sin i = h2 sin cos i ⇒
|r0()| =
42+ 2sin2 + 2cos2 =
42+ 2(cos2 + sin2) =√ 52=√
5 [since 0]. Then T() = r0()
|r0()| = 1
√5 h2 sin cos i = √15h2 sin cos i. T0() = √1
5h0 cos − sin i ⇒
|T0()| = √15
0 + cos2 + sin2 = √1
5. Thus N() = T0()
|T0()|= 1√ 5 1√
5 h0 cos − sin i = h0 cos − sin i.
(b) () = |T0()|
|r0()| = 1√
√ 5 5 = 1
5
19. (a) r() =√
2 −
⇒ r0() =√
2 −−
⇒ |r0()| =√
2 + 2+ −2=
(+ −)2= + −. Then
T() = r0()
|r0()|= 1
+ −
√2 −−
= 1
2+ 1
√2 2 −1
after multiplying by
and
T0() = 1
2+ 1
√2 22 0
− 22
(2+ 1)2
√2 2 −1
= 1
(2+ 1)2
(2+ 1)√
2 22 0
− 22√
2 2 −1
= 1
(2+ 1)2
√2 1 − 2
22 22 Then
|T0()| = 1 (2+ 1)2
22(1 − 22+ 4) + 44+ 44= 1 (2+ 1)2
22(1 + 22+ 4)
= 1
(2+ 1)2
22(1 + 2)2=
√2 (1 + 2) (2+ 1)2 =
√2
2+ 1
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1
338 ¤ CHAPTER 13 VECTOR FUNCTIONS
r(()) =
2
tan21
2 + 1− 1
i+ 2 tan1
2 tan21
2
+ 1j=1 − tan21
2 1 + tan21
2 i +2 tan1
2 sec21
2 j
=1 − tan21
2 sec21
2 i+ 2 tan1
2 cos21
2 j=
cos21
2
− sin21
2
i+ 2 sin1
2 cos1
2
j= cos i + sin j With this parametrization, we recognize the function as representing the unit circle. Note here that the curve approaches, but does not include, the point (−1 0), since cos = −1 for = + 2 ( an integer) but then = tan1
2is undefined.
17. (a) r() = h 3 cos 3 sin i ⇒ r0() = h1 −3 sin 3 cos i ⇒ |r0()| =
1 + 9 sin2 + 9 cos2 =√ 10.
Then T() = r0()
|r0()| = √110h1 −3 sin 3 cos i or
√1
10 −√310sin √310cos .
T0() = √1
10h0 −3 cos −3 sin i ⇒ |T0()| = √110
0 + 9 cos2 + 9 sin2 = √3 10. Thus N() = T0()
|T0()| =1√ 10 3√
10h0 −3 cos −3 sin i = h0 − cos − sin i.
(b) () = |T0()|
|r0()| = 3√
√ 10 10 = 3
10 18. (a) r() =
2 sin − cos cos + sin
⇒
r0() = h2 cos + sin − cos , −sin + cos + sin i = h2 sin cos i ⇒
|r0()| =
42+ 2sin2 + 2cos2 =
42+ 2(cos2 + sin2) =√ 52=√
5 [since 0]. Then T() = r0()
|r0()| = 1
√5 h2 sin cos i = √15h2 sin cos i. T0() = √1
5h0 cos − sin i ⇒
|T0()| = √15
0 + cos2 + sin2 = √1
5. Thus N() = T0()
|T0()|= 1√ 5 1√
5 h0 cos − sin i = h0 cos − sin i.
(b) () = |T0()|
|r0()| = 1√
√ 5 5 = 1
5
19. (a) r() =√
2 −
⇒ r0() =√
2 −−
⇒ |r0()| =√
2 + 2+ −2=
(+ −)2= + −. Then
T() = r0()
|r0()|= 1
+ −
√2 −−
= 1
2+ 1
√2 2 −1
after multiplying by
and
T0() = 1
2+ 1
√2 22 0
− 22
(2+ 1)2
√2 2 −1
= 1
(2+ 1)2
(2+ 1)√
2 22 0
− 22√
2 2 −1
= 1
(2+ 1)2
√2 1 − 2
22 22 Then
|T0()| = 1 (2+ 1)2
22(1 − 22+ 4) + 44+ 44= 1 (2+ 1)2
22(1 + 22+ 4)
= 1
(2+ 1)2
22(1 + 2)2=
√2 (1 + 2) (2+ 1)2 =
√2
2+ 1
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SECTION 13.3 ARC LENGTH AND CURVATURE ¤ 339
Therefore
N() = T0()
|T0()|=2+ 1
√2 1 (2+ 1)2
√2 (1 − 2) 22 22
= 1
√2 (2+ 1)
√2 (1 − 2) 22 22
= 1
2+ 1
1 − 2√ 2 √
2
(b) () = |T0()|
|r0()| =
√2
2+ 1· 1
+ − =
√2
3+ 2+ − =
√2 2
4+ 22+ 1=
√2 2
(2+ 1)2 20. (a) r() =
122 2
⇒ r0() = h1 2i ⇒ |r0()| =√
1 + 2+ 42 =√
1 + 52. Then
T() = r0()
|r0()|= 1
√1 + 52 h1 2i.
T0() = −5
(1 + 52)32 h1 2i + 1
√1 + 52 h0 1 2i [by Formula 3 of Theorem 13.2.3]
= 1
(1 + 52)32
−5 −52 −102 +
0 1 + 52 2 + 102
= 1
(1 + 52)32 h−5 1 2i
|T0()| = 1 (1 + 52)32
√252+ 1 + 4 = 1 (1 + 52)32
√252+ 5 =
√5√ 52+ 1 (1 + 52)32 =
√5 1 + 52 Thus N() = T0()
|T0()| =1 + 52
√5 · 1
(1 + 52)32 h−5 1 2i = 1
√5 + 252 h−5 1 2i.
(b) () = |T0()|
|r0()| =
√5(1 + 52)
√1 + 52 =
√5 (1 + 52)32
21. r() = 3j+ 2k ⇒ r0() = 32j+ 2 k, r00() = 6 j + 2 k, |r0()| =
02+ (32)2+ (2)2=√
94+ 42,
r0() × r00() = −62i, |r0() × r00()| = 62. Then () = |r0() × r00()|
|r0()|3 = 62
√94+ 423 = 62 (94+ 42)32.
22. r() = i + 2j+ k ⇒ r0() = i + 2 j + k, r00() = 2 j + k,
|r0()| =
12+ (2)2+ ()2=√
1 + 42+ 2, r0() × r00() = (2 − 2)i− j+ 2 k,
|r0() × r00()| =
[(2 − 2)]2+ (−)2+ 22=
(2 − 2)22+ 2+ 4 =
(42− 8 + 5)2+ 4.
Then () = |r0() × r00()|
|r0()|3 =
(42− 8 + 5)2+ 4
√1 + 42+ 23 =
(42− 8 + 5)2+ 4 (1 + 42+ 2)32 . 23. r() =√
6 2i+ 2 j + 23k ⇒ r0() = 2√
6 i + 2 j + 62k, r00() = 2√6 i + 12 k,
|r0()| =√
242+ 4 + 364=
4(94+ 62+ 1) =
4(32+ 1)2 = 2(32+ 1), r0() × r00() = 24 i − 12√
6 2j− 4√ 6 k,
|r0() × r00()| =√
5762+ 8644+ 96 =
96(94+ 62+ 1) =
96(32+ 1)2= 4√
6 (32+ 1).
Then () = |r0() × r00()|
|r0()|3 =4√
6 (32+ 1) 8(32+ 1)3 =
√6 2(32+ 1)2.
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SECTION 13.3 ARC LENGTH AND CURVATURE ¤ 339
Therefore
N() = T0()
|T0()|= 2+ 1
√2 1 (2+ 1)2
√2 (1 − 2) 22 22
= 1
√2 (2+ 1)
√2 (1 − 2) 22 22
= 1
2+ 1
1 − 2√ 2 √
2
(b) () = |T0()|
|r0()| =
√2
2+ 1· 1
+ − =
√2
3+ 2+ − =
√2 2
4+ 22+ 1=
√2 2
(2+ 1)2
20. (a) r() =
122 2
⇒ r0() = h1 2i ⇒ |r0()| =√
1 + 2+ 42=√
1 + 52. Then
T() = r0()
|r0()|= 1
√1 + 52 h1 2i.
T0() = −5
(1 + 52)32 h1 2i + 1
√1 + 52 h0 1 2i [by Formula 3 of Theorem 13.2.3]
= 1
(1 + 52)32
−5 −52 −102 +
0 1 + 52 2 + 102
= 1
(1 + 52)32 h−5 1 2i
|T0()| = 1 (1 + 52)32
√252+ 1 + 4 = 1 (1 + 52)32
√252+ 5 =
√5√ 52+ 1 (1 + 52)32 =
√5 1 + 52 Thus N() = T0()
|T0()|= 1 + 52
√5 · 1
(1 + 52)32 h−5 1 2i = 1
√5 + 252 h−5 1 2i.
(b) () = |T0()|
|r0()| =
√5(1 + 52)
√1 + 52 =
√5 (1 + 52)32
21.r() = 3j+ 2k ⇒ r0() = 32j+ 2 k, r00() = 6 j + 2 k, |r0()| =
02+ (32)2+ (2)2=√
94+ 42,
r0() × r00() = −62i, |r0() × r00()| = 62. Then () = |r0() × r00()|
|r0()|3 = 62
√94+ 423 = 62 (94+ 42)32.
22.r() = i + j + (1 + 2) k ⇒ r0() = i + j + 2 k, r00() = 2 k, |r0()| =
12+ 12+ (2)2 =√ 42+ 2, r0() × r00() = 2 i − 2 j, |r0() × r00()| =√
22+ 22+ 02=√ 8 = 2√
2.
Then () = |r0() × r00()|
|r0()|3 = 2√
√ 2
42+ 23 = 2√
√ 2 2√
22+ 13 = 1 (22+ 1)32.
23.r() =√
6 2i+ 2 j + 23k ⇒ r0() = 2√
6 i + 2 j + 62k, r00() = 2√6 i + 12 k,
|r0()| =√
242+ 4 + 364 =
4(94+ 62+ 1) =
4(32+ 1)2 = 2(32+ 1), r0() × r00() = 24 i − 12√
6 2j− 4√ 6 k,
|r0() × r00()| =√
5762+ 8644+ 96 =
96(94+ 62+ 1) =
96(32+ 1)2= 4√
6 (32+ 1).
Then () = |r0() × r00()|
|r0()|3 = 4√
6 (32+ 1) 8(32+ 1)3 =
√6 2(32+ 1)2.
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SECTION 13.3 ARC LENGTH AND CURVATURE ¤ 341
and 0() 0for √1
2, () attains its maximum at =√1
2. Thus, the maximum curvature occurs at
√1
2 ln√12 . Since lim
→∞
(2+ 1)32 = 0, () approaches 0 as → ∞.
31. Since 0= 00= , the curvature is () = |00()|
[1 + (0())2]32 =
(1 + 2)32 = (1 + 2)−32. To find the maximum curvature, we first find the critical numbers of ():
0() = (1 + 2)−32+
−32
(1 + 2)−52(22) = 1 + 2− 32
(1 + 2)52 = 1 − 22
(1 + 2)52.
0() = 0when 1 − 22= 0, so 2= 12or = −12ln 2. And since 1 − 22 0for −12ln 2and 1 − 22 0 for −12ln 2, the maximum curvature is attained at the point
−12ln 2 (− ln 2)2
=
−12ln 2√1 2
.
Since lim
→∞(1 + 2)−32= 0 ()approaches 0 as → ∞.
32. We can take the parabola as having its vertex at the origin and opening upward, so the equation is () = 2 0. Then by Equation 11, () = |00()|
[1 + (0())2]32 = |2|
[1 + (2)2]32 = 2
(1 + 422)32, thus (0) = 2. We want (0) = 4, so
= 2and the equation is = 22.
33. (a) appears to be changing direction more quickly at than , so we would expect the curvature to be greater at . (b) First we sketch approximate osculating circles at and . Using the
axes scale as a guide, we measure the radius of the osculating circle at to be approximately 08 units, thus =1
⇒
=1
≈ 1
08≈ 13. Similarly, we estimate the radius of the osculating circle at to be 14 units, so =1
≈ 1 14 ≈ 07.
34. = 4− 22 ⇒ 0= 43− 4, 00= 122− 4, and
() = |00|
1 + (0)232 =
122− 4
1 + (43− 4)232. The graph of the
curvature here is what we would expect. The graph of = 4− 22 appears to be bending most sharply at the origin and near = ±1.
35. = −2 ⇒ 0= −2−3, 00= 6−4, and
() = |00|
1 + (0)232 =
6−4
1 + (−2−3)232 = 6
4(1 + 4−6)32.
The appearance of the two humps in this graph is perhaps a little surprising, but it is explained by the fact that = −2increases asymptotically at the origin from both directions, and so its graph has very little bend there. [Note that (0) is undefined.]
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SECTION 13.3 ARC LENGTH AND CURVATURE ¤ 343
38. Notice that the curve is highest for the same -values at which curve is turning more sharply, and is 0 or near 0 where is nearly straight. So, must be the graph of = (), and is the graph of = ().
39. Notice that the curve has two inflection points at which the graph appears almost straight. We would expect the curvature to be 0 or nearly 0 at these values, but the curve isn’t near 0 there. Thus, must be the graph of = () rather than the graph of curvature, and is the graph of = ().
40. (a) The complete curve is given by 0 ≤ ≤ 2. Curvature appears to have a local (or absolute) maximum at 6 points. (Look at points where the curve appears to turn more sharply.)
(b) Using a CAS, we find (after simplifying)
() = 3√ 2
(5 sin + sin 5)2
(9 cos 6 + 2 cos 4 + 11)32. (To compute cross products in Maple, use the VectorCalculus or LinearAlgebra package and the
CrossProduct(a,b) command; in Mathematica, use Cross[a,b].) The graph shows 6 local (or absolute) maximum points for 0 ≤ ≤ 2, as observed in part (a).
41. Using a CAS, we find (after simplifying)
() = 6√
4 cos2 − 12 cos + 13
(17 − 12 cos )32 . (To compute cross products in Maple, use the VectorCalculus or
LinearAlgebra package and the CrossProduct(a,b) command; in Mathematica, use Cross[a,b].) Curvature is largest at integer multiples of 2.
42. Here r() = h() ()i, r0() = h0() 0()i, r00() = h00() 00()i,
|r0()|3=
(0())2+ (0())23
= [(0())2+ (0())2]32= ( ˙2+ ˙2)32, and
|r0() × r00()| = |h0 0 0() 00() − 00() 0()i| =
( ˙¨ − ¨ ˙)212
= | ˙¨ − ˙¨|. Thus () = | ˙¨ − ˙¨|
[ ˙2+ ˙2]32. 43. = 2 ⇒ ˙ = 2 ⇒ ¨ = 2, = 3 ⇒ = 3˙ 2 ⇒ ¨ = 6.
Then () = |¨˙ − ˙¨|
[ ˙2+ ˙2]32 =
(2)(6) − (32)(2) [(2)2+ (32)2]32 =
122− 62
(42+ 94)32 = 62 (42+ 94)32.
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344 ¤ CHAPTER 13 VECTOR FUNCTIONS
44. = cos ⇒ ˙ = − sin ⇒ ¨ = −2cos ,
= sin ⇒ = cos ˙ ⇒ ¨ = −2sin . Then
() = | ˙¨ − ˙¨|
[ ˙2+ ˙2]32 =
(− sin )(−2sin ) − ( cos )(−2cos ) [(− sin )2+ ( cos )2]32
=
3sin2 + 3cos2 (22sin2 + 22cos2)32 =
3
(22sin2 + 22cos2)32
45. = cos ⇒ ˙ = (cos − sin ) ⇒ ¨ = (− sin − cos ) + (cos − sin ) = −2sin ,
= sin ⇒ = ˙ (cos + sin ) ⇒ ¨ = (− sin + cos ) + (cos + sin ) = 2cos . Then
() = | ˙¨ − ˙¨|
[ ˙2+ ˙2]32 =
(cos − sin )(2cos ) − (cos + sin )(−2sin ) ([(cos − sin )]2+ [(cos + sin )]2)32
=
22(cos2 − sin cos + sin cos + sin2)
2(cos2 − 2 cos sin + sin2 + cos2 + 2 cos sin + sin2)32 =
22(1)
[2(1 + 1)]32 = 22
3(2)32 = 1
√2
46. () = , 0() = , 00() = 2. Using Formula 11 we have
() = |00()|
[1 + (0())2]32 =
2
[1 + ()2]32 = 2
(1 + 22)32 so the curvature at = 0 is
(0) = 2
(1 + 2)32. To determine the maximum value for (0), let () = 2
(1 + 2)32. Then
0() = 2 · (1 + 2)32− 2·32(1 + 2)12(2)
[(1 + 2)32]2 =(1 + 2)12
2(1 + 2) − 33 (1 + 2)3 =
2 − 3
(1 + 2)52. We have a critical number when 2 − 3= 0 ⇒ (2 − 2) = 0 ⇒ = 0 or = ±√
2. 0()is positive for −√
2, 0 √ 2 and negative elsewhere, so achieves its maximum value when =√
2or −√2. In either case, (0) = 2
332, so the members of the family with the largest value of (0) are () = √2and () = −√2.
47.
123 1corresponds to = 1. T() = r0()
|r0()|=
2 22 1
√42+ 44+ 1 =
2 22 1
22+ 1 , so T(1) =2 32313. T0() = −4(22+ 1)−2
2 22 1
+ (22+ 1)−1h2 4 0i [by Formula 3 of Theorem 13.2.3]
= (22+ 1)−2
−82+ 42+ 2 −83+ 83+ 4 −4
= 2(22+ 1)−2
1 − 22 2 −2
N() = T0()
|T0()|= 2(22+ 1)−2
1 − 22 2 −2 2(22+ 1)−2
(1 − 22)2+ (2)2+ (−2)2 =
1 − 22 2 −2
√1 − 42+ 44+ 82 =
1 − 22 2 −2 1 + 22 N(1) =
−1323 −23
and B(1) = T(1) × N(1) =
−49 −29 −
−49+19
49 +29
=
−231323 .
48. (1 0 0)corresponds to = 0. r() = hcos sin ln cos i, and in Exercise 4 we found that r0() = h− sin cos − tan i and |r0()| = |sec |. Here we can assume −2 2 and then sec 0 ⇒ |r0()| = sec .
° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c