Section 13.3 Arc Length and Curvature

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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 ²¹₃³

⇒ r⁰() = 2 2 ²

⇒

|r⁰()| =

2²+ (2)²+ (²)² =√

4 + 4²+ ⁴ =

(2 + ²)²= 2 + ²for 0 ≤  ≤ 1. Then using Formula 3, we have

 =1

0 |r⁰()|  =1

0(2 + ²)  = 2 +¹₃³1 0=⁷₃. 3.r() =√

2  i + ^j+ ^−k ⇒ r⁰() =√

2 i + ^j− ^−k ⇒

|r⁰()| =√

22

+ (^)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 = lnsec^₄ + tan^₄

 − ln |sec 0 + tan 0|

= ln√2 + 1

 − ln |1 + 0| = ln(√ 2 + 1)

5.r() = i + ²j+ ³k ⇒ r⁰() = 2 j + 3²k ⇒ |r⁰()| =√

4²+ 9⁴= √

4 + 9² [since  ≥ 0].

Then  =1

0 |r⁰()|  =1 0 √

4 + 9² = ₁₈¹ ·²3(4 + 9²)^321

0= ₂₇¹(13^32− 4^32) = ₂₇¹(13^32− 8).

6.r() = ²i+ 9 j + 4^32k ⇒ r⁰() = 2 i + 9 j + 6√

 k ⇒

|r⁰()| =√

4²+ 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+ 94

1= 52 − 10 = 42.

7.r() =

² ³ ⁴

⇒ r⁰() =

2 3² 4³

⇒ |r⁰()| =



(2)²+ (3²)²+ (4³)² =√

4²+ 9⁴+ 16⁶, so

 =2

0 |r⁰()|  =2 0

√4²+ 9⁴+ 16⁶ ≈ 186833.

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 ≈ 20454.

336 ¤ CHAPTER 13 VECTOR FUNCTIONS

9. r() = hcos  2 sin 2i ⇒ r⁰() = h− sin  2 2 cos 2i ⇒ |r⁰()| =

²sin² + 4 + 4²cos²2.

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 + 4²cos²2  ≈ 103311.

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 007 around the curve.

The complete curve is given by the parameter interval [0 2] and we have r⁰() = hcos  2 cos 2 3 cos 3i ⇒

|r⁰()| =√

cos² + 4 cos²2 + 9 cos²3, so  =2

0 |r⁰()|  =2

0

√cos² + 4 cos²2 + 9 cos²3  ≈ 160264.

11. The projection of the curve  onto the -plane is the curve ²= 2or  =¹₂²,  = 0. Then we can choose the parameter

 =  ⇒  = ¹2². Since  also lies on the surface 3 = , we have  =¹₃ =¹₃()(¹₂²) = ¹₆³. Then parametric equations for  are  = ,  =¹₂²,  =¹₆³and the corresponding vector equation is r() =

¹₂²¹₆³. The origin corresponds to  = 0 and the point (6 18 36) corresponds to  = 6, so

 =6

0 |r⁰()|  =6 0

1 ¹₂² =6 0



1²+ ²+₁

2²2

 =6 0



1 + ²+¹₄⁴

=6 0



(1 +¹₂²)² =6

0(1 +¹₂²)  =

 +¹₆³6

0= 6 + 36 = 42

12. Let  be the curve of intersection. The projection of  onto the -plane is the ellipse 4²+ ²= 4or ²+ ²4 = 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   ≈ 135191.

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

() =

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+ 2^2

=

^2

2(cos² + sin²) + 2 cos  sin  − 2 cos  sin  + 2

=√

4^2= 2^ The point 

0 1√

2corresponds to  = 0, so the arc length function from  is

() =

0 |r⁰()|  =

0 2^ = 2^|^0 = 2(^− 1). Since  = 2(^− 1), we have ^= 

2+ 1 ⇔

 = ln₁

2 + 1. Substituting for  in the original equation, the reparametrization of the curve with respect to arc length is r(()) =₁

2 + 1 sin

ln₁

2 + 1

i+₁

2 + 1 cos

ln₁

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

ln₁

2(4) + 1

i+₁

2(4) + 1 cos

ln₁

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+ −2²+ 2 (²+ 1)² j,



 = |r⁰()| =

 −4

(²+ 1)²

2

+

−2²+ 2 (²+ 1)²

2

=



4⁴+ 8²+ 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

r(()) =

 2

tan²₁

2 + 1− 1



i+ 2 tan1 2 tan²₁

2

+ 1j=1 − tan²1 2 1 + tan²₁

2 i +2 tan1 2 sec²₁

2 j

=1 − tan²₁

2 sec²₁

2 i+ 2 tan₁

2 cos²₁

2 j=

cos²₁

2

− sin²₁

2

i+ 2 sin₁

2 cos₁

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  = tan1

2is 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()| =

4²+ ²sin² + ²cos² =

4²+ ²(cos² + sin²) =√ 5²=√

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 ^ 2^2 0

− 2^2

(^2+ 1)²

√2 ^ ^2 −1

= 1

(^2+ 1)²

(^2+ 1)√

2 ^ 2^2 0

− 2^2√

2 ^ ^2 −1

= 1

(^2+ 1)²

√2 ^ 1 − ^2

 2^2 2^2 Then

|T⁰()| = 1 (^2+ 1)²

2^2(1 − 2^2+ ^4) + 4^4+ 4^4= 1 (^2+ 1)²

2^2(1 + 2^2+ ^4)

= 1

(^2+ 1)²



2^2(1 + ^2)²=

√2 ^(1 + ^2) (^2+ 1)² =

√2 ^

^2+ 1

1

(2)

r(()) =

 2

tan²₁

2 + 1− 1



i+ 2 tan₁

2 tan²₁

2

+ 1j=1 − tan²₁

2 1 + tan²₁

2 i +2 tan₁

2 sec²₁

2 j

=1 − tan²₁

2 sec²₁

2 i+ 2 tan₁

2 cos²₁

2 j=

cos²₁

2

− sin²₁

2

i+ 2 sin₁

2 cos₁

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  = tan1

2is 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()| =

4²+ ²sin² + ²cos² =

4²+ ²(cos² + sin²) =√ 5²=√

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 ^ 2^2 0

− 2^2

(^2+ 1)²

√2 ^ ^2 −1

= 1

(^2+ 1)²

(^2+ 1)√

2 ^ 2^2 0

− 2^2√

2 ^ ^2 −1

= 1

(^2+ 1)²

√2 ^ 1 − ^2

 2^2 2^2 Then

|T⁰()| = 1 (^2+ 1)²

2^2(1 − 2^2+ ^4) + 4^4+ 4^4= 1 (^2+ 1)²

2^2(1 + 2^2+ ^4)

= 1

(^2+ 1)²



2^2(1 + ^2)²=

√2 ^(1 + ^2) (^2+ 1)² =

√2 ^

^2+ 1

Therefore

N() = T⁰()

|T⁰()|=^2+ 1

√2 ^ 1 (^2+ 1)²

√2 ^(1 − ^2) 2^2 2^2

= 1

√2 ^(^2+ 1)

√2 ^(1 − ^2) 2^2 2^2

= 1

^2+ 1

1 − ^2√ 2 ^√

2 ^

(b) () = |T⁰()|

|r⁰()| =

√2 ^

^2+ 1· 1

^+ ^− =

√2 ^

^3+ 2^+ ^− =

√2 ^2

^4+ 2^2+ 1=

√2 ^2

(^2+ 1)² 20. (a) r() =

¹₂² ²

⇒ r⁰() = h1  2i ⇒ |r⁰()| =√

1 + ²+ 4² =√

1 + 5². Then

T() = r⁰()

|r⁰()|= 1

√1 + 5² h1  2i.

T⁰() = −5

(1 + 5²)^32 h1  2i + 1

√1 + 5² h0 1 2i [by Formula 3 of Theorem 13.2.3]

= 1

(1 + 5²)^32

−5 −5² −10² +

0 1 + 5² 2 + 10²

= 1

(1 + 5²)^32 h−5 1 2i

|T⁰()| = 1 (1 + 5²)^32

√25²+ 1 + 4 = 1 (1 + 5²)^32

√25²+ 5 =

√5√ 5²+ 1 (1 + 5²)^32 =

√5 1 + 5² Thus N() = T⁰()

|T⁰()| =1 + 5²

√5 · 1

(1 + 5²)^32 h−5 1 2i = 1

√5 + 25² h−5 1 2i.

(b) () = |T⁰()|

|r⁰()| =

√5(1 + 5²)

√1 + 5² =

√5 (1 + 5²)^32

21. r() = ³j+ ²k ⇒ r⁰() = 3²j+ 2 k, r⁰⁰() = 6 j + 2 k, |r⁰()| =

0²+ (3²)²+ (2)²=√

9⁴+ 4²,

r⁰() × r⁰⁰() = −6²i, |r⁰() × r⁰⁰()| = 6². Then () = |r⁰() × r⁰⁰()|

|r⁰()|³ = 6²

√9⁴+ 4²³ = 6² (9⁴+ 4²)^32.

22. r() =  i + ²j+ ^k ⇒ r⁰() = i + 2 j + ^k, r⁰⁰() = 2 j + ^k,

|r⁰()| =

1²+ (2)²+ (^)²=√

1 + 4²+ ^2, r⁰() × r⁰⁰() = (2 − 2)^i− ^j+ 2 k,

|r⁰() × r⁰⁰()| =

[(2 − 2)^]²+ (−^)²+ 2²=

(2 − 2)²^2+ ^2+ 4 =

(4²− 8 + 5)^2+ 4.

Then () = |r⁰() × r⁰⁰()|

|r⁰()|³ =

(4²− 8 + 5)^2+ 4

√1 + 4²+ ^2³ =

(4²− 8 + 5)^2+ 4 (1 + 4²+ ^2)^32 . 23. r() =√

6 ²i+ 2 j + 2³k ⇒ r⁰() = 2√

6  i + 2 j + 6²k, r⁰⁰() = 2√6 i + 12 k,

|r⁰()| =√

24²+ 4 + 36⁴=

4(9⁴+ 6²+ 1) =

4(3²+ 1)² = 2(3²+ 1), r⁰() × r⁰⁰() = 24 i − 12√

6 ²j− 4√ 6 k,

|r⁰() × r⁰⁰()| =√

576²+ 864⁴+ 96 =

96(9⁴+ 6²+ 1) =

96(3²+ 1)²= 4√

6 (3²+ 1).

Then () = |r⁰() × r⁰⁰()|

|r⁰()|³ =4√

6 (3²+ 1) 8(3²+ 1)³ =

√6 2(3²+ 1)².

Therefore

N() = T⁰()

|T⁰()|=^2+ 1

√2 ^ 1 (^2+ 1)²

√2 ^(1 − ^2) 2^2 2^2

= 1

√2 ^(^2+ 1)

√2 ^(1 − ^2) 2^2 2^2

= 1

^2+ 1

1 − ^2√ 2 ^√

2 ^

(b) () = |T⁰()|

|r⁰()| =

√2 ^

^2+ 1· 1

^+ ^− =

√2 ^

^3+ 2^+ ^− =

√2 ^2

^4+ 2^2+ 1=

√2 ^2

(^2+ 1)² 20. (a) r() =

¹₂² ²

⇒ r⁰() = h1  2i ⇒ |r⁰()| =√

1 + ²+ 4² =√

1 + 5². Then

T() = r⁰()

|r⁰()|= 1

√1 + 5² h1  2i.

T⁰() = −5

(1 + 5²)^32 h1  2i + 1

√1 + 5² h0 1 2i [by Formula 3 of Theorem 13.2.3]

= 1

(1 + 5²)^32

−5 −5² −10² +

0 1 + 5² 2 + 10²

= 1

(1 + 5²)^32 h−5 1 2i

|T⁰()| = 1 (1 + 5²)^32

√25²+ 1 + 4 = 1 (1 + 5²)^32

√25²+ 5 =

√5√ 5²+ 1 (1 + 5²)^32 =

√5 1 + 5² Thus N() = T⁰()

|T⁰()| =1 + 5²

√5 · 1

(1 + 5²)^32 h−5 1 2i = 1

√5 + 25² h−5 1 2i.

(b) () = |T⁰()|

|r⁰()| =

√5(1 + 5²)

√1 + 5² =

√5 (1 + 5²)^32

21. r() = ³j+ ²k ⇒ r⁰() = 3²j+ 2 k, r⁰⁰() = 6 j + 2 k, |r⁰()| =

0²+ (3²)²+ (2)²=√

9⁴+ 4²,

r⁰() × r⁰⁰() = −6²i, |r⁰() × r⁰⁰()| = 6². Then () = |r⁰() × r⁰⁰()|

|r⁰()|³ = 6²

√9⁴+ 4²3 = 6² (9⁴+ 4²)^32.

22. r() =  i + ²j+ ^k ⇒ r⁰() = i + 2 j + ^k, r⁰⁰() = 2 j + ^k,

|r⁰()| =

1²+ (2)²+ (^)²=√

1 + 4²+ ^2, r⁰() × r⁰⁰() = (2 − 2)^i− ^j+ 2 k,

|r⁰() × r⁰⁰()| =

[(2 − 2)^]²+ (−^)²+ 2²=

(2 − 2)²^2+ ^2+ 4 =

(4²− 8 + 5)^2+ 4.

Then () = |r⁰() × r⁰⁰()|

|r⁰()|³ =

(4²− 8 + 5)^2+ 4

√1 + 4²+ ^23 =

(4²− 8 + 5)^2+ 4 (1 + 4²+ ^2)^32 .

23. r() =√

6 ²i+ 2 j + 2³k ⇒ r⁰() = 2√

6  i + 2 j + 6²k, r⁰⁰() = 2√

6 i + 12 k,

|r⁰()| =√

24²+ 4 + 36⁴=

4(9⁴+ 6²+ 1) =

4(3²+ 1)² = 2(3²+ 1), r⁰() × r⁰⁰() = 24 i − 12√

6 ²j− 4√ 6 k,

|r⁰() × r⁰⁰()| =√

576²+ 864⁴+ 96 =

96(9⁴+ 6²+ 1) =

96(3²+ 1)²= 4√

6 (3²+ 1).

Then () = |r⁰() × r⁰⁰()|

|r⁰()|³ =4√

6 (3²+ 1) 8(3²+ 1)³ =

√6 2(3²+ 1)².

and ⁰()  0for  √¹

2, () attains its maximum at  =√¹

2. Thus, the maximum curvature occurs at

√1 2 ln√¹

2

.

Since lim

→∞



(²+ 1)^32 = 0, () approaches 0 as  → ∞.

31. Since ⁰= ⁰⁰= ^, the curvature is () = |⁰⁰()|

[1 + (⁰())²]^32 = ^

(1 + ^2)^32 = ^(1 + ^2)^−32. To find the maximum curvature, we first find the critical numbers of ():

⁰() = ^(1 + ^2)^−32+ ^

−³2

(1 + ^2)^−52(2^2) = ^1 + ^2− 3^2

(1 + ^2)^52 = ^ 1 − 2^2

(1 + ^2)^52.

⁰() = 0when 1 − 2^2= 0, so ^2= ¹₂or  = −¹2ln 2. And since 1 − 2^2 0for   −¹2ln 2and 1 − 2^2 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)^−32= 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]^32 = |2|

[1 + (2)²]^32 = 2

(1 + 4²²)^32, thus (0) = 2. We want (0) = 4, so

 = 2and the equation is  = 2².

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 08 units, thus  =1

 ⇒

 =1

 ≈ 1

08≈ 13. Similarly, we estimate the radius of the osculating circle at  to be 14 units, so  =1

 ≈ 1 14 ≈ 07.

34.  = ⁴− 2² ⇒ ⁰= 4³− 4, ⁰⁰= 12²− 4, and

() = |⁰⁰|

1 + (⁰)²32 =

12²− 4

1 + (4³− 4)²32. The graph of the

curvature here is what we would expect. The graph of  = ⁴− 2² appears to be bending most sharply at the origin and near  = ±1.

35.  = ⁻² ⇒ ⁰= −2⁻³, ⁰⁰= 6⁻⁴, and

() = |⁰⁰|

1 + (⁰)²32 =

6⁻⁴

1 + (−2⁻³)²32 = 6

⁴(1 + 4⁻⁶)^32.

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.]

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)^32. (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 )^32 . (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() = h⁰() ⁰()i, r⁰⁰() = h⁰⁰() ⁰⁰()i,

|r⁰()|³=

(⁰())²+ (⁰())²3

= [(⁰())²+ (⁰())²]^32= ( ˙²+ ˙²)^32, and

|r⁰() × r⁰⁰()| = |h0 0 ⁰() ⁰⁰() − ⁰⁰() ⁰()i| =

( ˙¨ − ¨ ˙)²12

= | ˙¨ − ˙¨|. Thus () = | ˙¨ − ˙¨|

[ ˙²+ ˙²]^32. 43.  = ² ⇒ ˙ = 2 ⇒ ¨ = 2,  = ³ ⇒  = 3˙ ² ⇒ ¨ = 6.

Then () = |¨˙ − ˙¨|

[ ˙²+ ˙²]^32 =

(2)(6) − (3²)(2) [(2)²+ (3²)²]^32 =

12²− 6²

(4²+ 9⁴)^32 = 6² (4²+ 9⁴)^32.

44.  =  cos  ⇒ ˙ = − sin  ⇒ ¨ = −²cos ,

 =  sin  ⇒  =  cos ˙ ⇒ ¨ = −²sin . Then

() = | ˙¨ − ˙¨|

[ ˙²+ ˙²]^32 =

(− sin )(−²sin ) − ( cos )(−²cos ) [(− sin )²+ ( cos )²]^32

=

³sin² + ³cos² (²²sin² + ²²cos²)^32 =

³

(²²sin² + ²²cos²)^32

45.  = ^cos  ⇒ ˙ = ^(cos  − sin ) ⇒ ¨ = ^(− sin  − cos ) + ^(cos  − sin ) = −2^sin ,

 = ^sin  ⇒  = ˙ ^(cos  + sin ) ⇒ ¨ = ^(− sin  + cos ) + ^(cos  + sin ) = 2^cos . Then

() = | ˙¨ − ˙¨|

[ ˙²+ ˙²]^32 =

^(cos  − sin )(2^cos ) − ^(cos  + sin )(−2^sin ) ([^(cos  − sin )]²+ [^(cos  + sin )]²)^32

=

2^2(cos² − sin  cos  + sin  cos  + sin²)

^2(cos² − 2 cos  sin  + sin² + cos² + 2 cos  sin  + sin²)32 =

2^2(1)

[^2(1 + 1)]^32 = 2^2

^3(2)^32 = 1

√2 ^

46.  () = ^, ⁰() = ^, ⁰⁰() = ²^. Using Formula 11 we have

() = |⁰⁰()|

[1 + (⁰())²]^32 =

²^

[1 + (^)²]^32 = ²^

(1 + ²^2)^32 so the curvature at  = 0 is

(0) = ²

(1 + ²)^32. To determine the maximum value for (0), let () = ²

(1 + ²)^32. Then

⁰() = 2 · (1 + ²)^32− ²·³2(1 + ²)^12(2)

[(1 + ²)^32]² =(1 + ²)^12

2(1 + ²) − 3³ (1 + ²)³ =

2 − ³

(1 + ²)^52. 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

3^32, so the members of the family with the largest value of (0) are () = ^√^2and () = ⁻^√^2.

47. 

1²₃ 1corresponds to  = 1. T() = r⁰()

|r⁰()|=

2 2² 1

√4²+ 4⁴+ 1 =

2 2² 1

2²+ 1 , so T(1) =₂

3²₃¹₃. T⁰() = −4(2²+ 1)⁻²

2 2² 1

+ (2²+ 1)⁻¹h2 4 0i [by Formula 3 of Theorem 13.2.3]

= (2²+ 1)⁻²

−8²+ 4²+ 2 −8³+ 8³+ 4 −4

= 2(2²+ 1)⁻²

1 − 2² 2 −2

N() = T⁰()

|T⁰()|= 2(2²+ 1)⁻²

1 − 2² 2 −2 2(2²+ 1)⁻²

(1 − 2²)²+ (2)²+ (−2)² =

1 − 2² 2 −2

√1 − 4²+ 4⁴+ 8² =

1 − 2² 2 −2 1 + 2² N(1) =

−¹3²₃ −²3

and B(1) = T(1) × N(1) =

−⁴9 −²9 −

−⁴9+¹₉

⁴₉ +²₉

=

−²3¹₃²₃.

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 .