Section 13.3 Arc Length and Curvature 5. Find the length of the curve.

Section 13.3 Arc Length and Curvature

5. Find the length of the curve. r(t) =√

2ti + e^tj + e^−tk, 0 ≤ t ≤ 1.

Solution:

SECTION 13.3 ARC LENGTH AND CURVATURE ¤ 335

13.3 Arc Length and Curvature

1. r() =

2 cos √

5  2 sin 

⇒ r⁰() = 

−2 sin √

5 2 cos 

⇒

|r⁰()| =



(−2 sin )²+√

5²

+ (2 cos )²=

4 sin² + 5 + 4 cos² =√ 9 = 3.

Then using Formula 3, we have  =2

−23  = 32

−2= 6 − (−6) = 12.

2. r() = 

³ 6 3²

⇒ r⁰() =

3² 6 6

⇒

|r⁰()| =

(3²)²+ 6²+ (6)²=√

9⁴+ 36²+ 36 = 3√

⁴+ 4 + 4 = 3

(²+ 2)²= 3(²+ 2).

Then using Formula 3, we have  =3

0 3(²+ 2)  =

³+ 63

0= (27 + 18) − (0 + 0) = 45.

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() = 12 i + 8^32j+ 3²k ⇒ r⁰() = 12 i + 12√

 j + 6 k ⇒

|r⁰()| =√

144 + 144 + 36² =

36( + 2)² = 6 | + 2| = 6( + 2) for 0 ≤  ≤ 1. Then

 =1

0 |r⁰()|  =1

0 6( + 2)  =

3²+ 121 0= 15.

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() =√

  ²

⇒ r⁰() =

 1 2√

 1 2



⇒ |r⁰()| =

1 2√



2

+ 1²+ (2)²=

1

4+ 1 + 4², so

 =4

1 |r⁰()|  =4 1

1

4 + 1 + 4² ≈ 153841.

° 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

24. (a) Find the unit tangent and unit normal vectors T(t) and N(t).

(b) Use Formula 9 to find the curvature.

r(t) =<√

2t, e^t, e^−t>.

Solution:

338 ¤ CHAPTER 13 VECTOR FUNCTIONS 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()| = √¹

10h1 −3 sin  3 cos i or ₁

√10 −^√3₁₀sin √³ 10cos 

.

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 = ^√1₅. 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

° 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

SECTION 13.3 ARC LENGTH AND CURVATURE ¤ 339

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 +  j + (1 + ²) k ⇒ r⁰() = i + j + 2 k, r⁰⁰() = 2 k, |r⁰()| =

1²+ 1²+ (2)²=√ 4²+ 2, r⁰() × r⁰⁰() = 2 i − 2 j, |r⁰() × r⁰⁰()| =√

2²+ 2²+ 0²=√

8 = 2√2.

Then () = |r⁰() × r⁰⁰()|

|r⁰()|³ = 2√

√ 2

4²+ 23 = 2√

√ 2 2√

2²+ 1³ = 1 (2²+ 1)^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.

° 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

26. Use Theorem 10 to find the curvature. r(t) = ti + t²j + e^tk Solution:

1300 ¤ CHAPTER 13 VECTOR FUNCTIONS T⁰() = 1

^2+ 1

√2 ^ 2^2 0

− 2^2

(^2+ 1)²

√2 ^ ^2 −1 [by Formula 3 of Theorem 13.2.3]

= 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

|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 Thus, 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) By Formula 9, the curvature is

() = |T⁰()|

|r⁰()| =

√2 ^

^2+ 1· 1

^+ ^− =

√2 ^

^3+ 2^+ ^− =

√2 ^2

^4+ 2^2+ 1 =

√2 ^2

(^2+ 1)². 25. 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.

26. 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 .

27. 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. 28. r() =

² ln   ln 

⇒ r⁰() = h2 1 1 + ln i, r⁰⁰() =

2 −1² 1. The point (1 0 0) corresponds to  = 1, and r⁰(1) = h2 1 1i, |r⁰(1)| =√

2²+ 1²+ 1²=√6, r⁰⁰(1) = h2 −1 1i, r⁰(1) × r⁰⁰(1) = h2 0 −4i,

|r⁰(1) × r⁰⁰(1)| =

2²+ 0²+ (−4)²=√

20 = 2√5. Then (1) = |r⁰(1) × r⁰⁰(1)|

|r⁰(1)|³ = 2√

√ 5

6³ = 2√ 5 6√

6 or

√30 18 .

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1

52. Find the vectors T, N, and B at the given point.

r(t) =< cos t, sin t, ln cos t >, (1, 0, 0).

Solution:

344 ¤ CHAPTER 13 VECTOR FUNCTIONS

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 .

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SECTION 13.3 ARC LENGTH AND CURVATURE ¤ 345 T() = r⁰()

|r⁰()| = h−sin  cos  − tan i

sec  =

− sin  cos  cos² − sin 

and T(0) = h0 1 0i.

T⁰() = h−[(sin )(− sin ) + (cos )(cos )] 2(cos )(− sin ) − cos i =

sin² − cos² −2 sin  cos  − cos , so

N(0) = T⁰(0)

|T⁰(0)| = h−√1 0 −1i 1 + 0 + 1 = 1

√2h−1 0 −1i =

−^√1₂ 0 −^√1₂ .

Finally, B(0) = T(0) × N(0) = h0 1 0i ×

−^√1₂ 0 −^√1₂

=

−^√1₂ 0^√1₂ .

49. r() = hsin 2 − cos 2 4i ⇒ r⁰() = h2 cos 2 2 sin 2 4i. The point (0 1 2) corresponds to  = 2, and the normal plane there has normal vector r⁰(2) = h−2 0 4i. An equation for the normal plane is

−2( − 0) + 0( − 1) + 4( − 2) = 0 or −2 + 4 = 8 or  − 2 = −4.

T() = r⁰()

|r⁰()| =  h2 cos 2 2 sin 2 4i 4 cos²2 + 4 sin²2 + 16

= 1

2√

5h2 cos 2 2 sin 2 4i = 1

√5hcos 2 sin 2 2i ⇒

T⁰() = √¹

5h−2 sin 2 2 cos 2 0i ⇒ |T⁰()| = ^√1₅

4 sin²2 + 4 cos²2 = √² 5, and N() = T⁰()

|T⁰()| = h− sin 2 cos 2 0i. Then T(2) = ^√1₅h−1 0 2i, N(2) = h0 −1 0i, and

B(2) = T(2) × N(2) = ^√1₅h2 0 1i. Since B(2) is normal to the osculating plane, so is h2 0 1i, and an equation of the plane is 2( − 0) + 0( − 1) + 1( − 2) = 0 or 2 +  = 2.

50. r() =

ln  2 ²

⇒ r⁰() = h1 2 2i. The point (0 2 1) corresponds to  = 1, and the normal plane there has normal vector r⁰(1) = h1 2 2i. An equation for the normal plane is 1( − 0) + 2( − 2) + 2( − 1) = 0 or

 + 2 + 2 = 6.

|r⁰()| =

1²+ 4 + 4²=

[(1) + 2]²= (1) + 2 [since   0] and then

T() = r⁰()

|r⁰()| = h1 2 2i

(1) + 2 = 1 1 + 2²

1 2 2² 

after multiplying by 





. By Formula 3 of Theorem 13.2.3,

T⁰() = − 4

(1 + 2²)²

1 2 2²

+ 1

1 + 2² h0 2 4i

= 1

(1 + 2²)²

−4 −8²+ 2(1 + 2²) −8³+ 4(1 + 2²)

= 1

(1 + 2²)²

−4 2 − 4² 4

Then

|T⁰()| = 1 (1 + 2²)²

16²+ (2 − 4²)²+ 16²= 1 (1 + 2²)²

√16²+ 4 + 16⁴

= 1

(1 + 2²)² · 2

(1 + 2²)²= 2 1 + 2²

and N() = T⁰()

|T⁰()| = 1 2(1 + 2²)

−4 2 − 4² 4

= 1

1 + 2²

−2 1 − 2² 2 .

° 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

54. Find equations of the normal plane and osculating plane of the curve at the given point.

x = ln t, y = 2t, z = t²; (0, 2, 1).

Solution:

SECTION 13.3 ARC LENGTH AND CURVATURE ¤ 345 T() = r⁰()

|r⁰()| = h−sin  cos  − tan i

sec  =

− sin  cos  cos² − sin 

and T(0) = h0 1 0i.

T⁰() = h−[(sin )(− sin ) + (cos )(cos )] 2(cos )(− sin ) − cos i =

sin² − cos² −2 sin  cos  − cos , so

N(0) = T⁰(0)

|T⁰(0)| = h−√1 0 −1i 1 + 0 + 1 = 1

√2h−1 0 −1i =

−^√1₂ 0 −^√1₂ .

Finally, B(0) = T(0) × N(0) = h0 1 0i ×

−^√1₂ 0 −^√1₂

=

−^√1₂ 0√¹ 2

.

49. r() = hsin 2 − cos 2 4i ⇒ r⁰() = h2 cos 2 2 sin 2 4i. The point (0 1 2) corresponds to  = 2, and the normal plane there has normal vector r⁰(2) = h−2 0 4i. An equation for the normal plane is

−2( − 0) + 0( − 1) + 4( − 2) = 0 or −2 + 4 = 8 or  − 2 = −4.

T() = r⁰()

|r⁰()| =  h2 cos 2 2 sin 2 4i

4 cos²2 + 4 sin²2 + 16 = 1 2√

5h2 cos 2 2 sin 2 4i = 1

√5hcos 2 sin 2 2i ⇒

T⁰() = √¹

5h−2 sin 2 2 cos 2 0i ⇒ |T⁰()| = ^√1₅

4 sin²2 + 4 cos²2 = √² 5, and N() = T⁰()

|T⁰()| = h− sin 2 cos 2 0i. Then T(2) = ^√1₅h−1 0 2i, N(2) = h0 −1 0i, and

B(2) = T(2) × N(2) = ^√1₅h2 0 1i. Since B(2) is normal to the osculating plane, so is h2 0 1i, and an equation of the plane is 2( − 0) + 0( − 1) + 1( − 2) = 0 or 2 +  = 2.

50. r() =

ln  2 ²

⇒ r⁰() = h1 2 2i. The point (0 2 1) corresponds to  = 1, and the normal plane there has normal vector r⁰(1) = h1 2 2i. An equation for the normal plane is 1( − 0) + 2( − 2) + 2( − 1) = 0 or

 + 2 + 2 = 6.

|r⁰()| =

1²+ 4 + 4²=

[(1) + 2]²= (1) + 2 [since   0] and then

T() = r⁰()

|r⁰()| = h1 2 2i

(1) + 2 = 1 1 + 2²

1 2 2² 

after multiplying by 





. By Formula 3 of Theorem 13.2.3,

T⁰() = − 4

(1 + 2²)²

1 2 2²

+ 1

1 + 2²h0 2 4i

= 1

(1 + 2²)²

−4 −8²+ 2(1 + 2²) −8³+ 4(1 + 2²)

= 1

(1 + 2²)²

−4 2 − 4² 4

Then

|T⁰()| = 1 (1 + 2²)²

16²+ (2 − 4²)²+ 16²= 1 (1 + 2²)²

√16²+ 4 + 16⁴

= 1

(1 + 2²)² · 2

(1 + 2²)²= 2 1 + 2²

and N() = T⁰()

|T⁰()| = 1 2(1 + 2²)

−4 2 − 4² 4

= 1

1 + 2²

−2 1 − 2² 2.

° 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

346 ¤ CHAPTER 13 VECTOR FUNCTIONS

Thus T(1) = ¹₃h1 2 2i, N(1) = ¹3h−2 −1 2i, and B(1) = T(1) × N(1) = ¹9h6 −6 3i is normal to the osculating plane.

We can take the parallel vector h2 −2 1i as a normal vector for the plane, so an equation is 2( − 0) − 2( − 2) + 1( − 1) = 0 or 2 − 2 +  = −3.

Note: Since r⁰(1)is parallel to T(1) and T⁰(1)is parallel to N(1), we could have taken r⁰(1) × T⁰(1)as a normal vector for the plane.

51. The ellipse is given by the parametric equations  = 2 cos ,  = 3 sin , so using the result from Exercise 42,

() = | ˙¨ − ¨ ˙|

[ ˙²+ ˙²]^32 = |(−2 sin )(−3 sin ) − (3 cos )(−2 cos )|

(4 sin² + 9 cos²)^32 = 6

(4 sin² + 9 cos²)^32. At (2 0),  = 0. Now (0) =₂₇⁶ = ²₉, so the radius of the osculating circle is

1(0) =⁹₂ and its center is

−⁵2 0. Its equation is therefore

 +⁵₂2

+ ²= ⁸¹₄. At (0 3),  = ^₂, and _

2

= ⁶₈ = ³₄. So the radius of the osculating circle is⁴₃ and

its center is

0⁵₃. Hence its equation is ²+

 −⁵3

2

= ¹⁶₉.

52.  = ¹₂² ⇒ ⁰= and ⁰⁰= 1, so Formula 11 gives () = 1

(1 + ²)^32. So the curvature at (0 0) is (0) = 1 and the osculating circle has radius 1 and center (0 1), and hence equation ²+ ( − 1)²= 1. The curvature at

1¹₂ is (1) = 1

(1 + 1²)^32 = 1 2√

2. The tangent line to the parabola at 1¹₂ has slope 1, so the normal line has slope −1. Thus the center of the osculating circle lies in the direction of the unit vector

−^√1₂√¹ 2

.

The circle has radius 2√2, so its center has position vector

1¹₂ + 2√

2

−^√1₂^√1₂

=

−1⁵2

. So the equation of the circle

is ( + 1)²+

 −⁵2

2

= 8.

53. Here r() =

³ 3 ⁴, and r⁰() =

3² 3 4³is normal to the normal plane for any . The given plane has normal vector h6 6 −8i, and the planes are parallel when their normal vectors are parallel. Thus we need to find a value for  where

3² 3 4³

=  h6 6 −8i for some  6= 0. From the -component we see that  = ¹2, and

3² 3 4³

= ¹₂h6 6 −8i = h3 3 −4i for  = −1. Thus the planes are parallel at the point (−1 −3 1).

54. To find the osculating plane, we first calculate the unit tangent and normal vectors.

In Maple, we use the VectorCalculus package and set r:= tˆ3,3*t,tˆ4;. After differentiating, the Normalize command converts the tangent vector to the unit tangent vector: T:=Normalize(diff(r,t));. After

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