Section 13.2 Derivatives and Integrals of Vector Functions

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Section 13.2 Derivatives and Integrals of Vector Functions

SECTION 13.2 DERIVATIVES AND INTEGRALS OF VECTOR FUNCTIONS ¤ 327

6.r() = ^i+ 2 j, r(0) = i.

Since  = ^ ⇔  = ln  and

 = 2 = 2 ln , the curve is the graph of  = 2 ln .

(a), (c) (b) r⁰() = ^i+ 2 j,

r⁰(0) = i + 2 j

7.r() = 4 sin  i − 2 cos  j, r(34) = 4(√

22) i − 2(−√

22) j = 2√ 2 i +√

2 j.

Here (4)²+ (2)²= sin² + cos² = 1, so the curve is the ellipse² 16+²

4 = 1.

(a), (c) (b) r⁰() = 4 cos  i + 2 sin  j,

r⁰(34) = −2√ 2 i +√

2 j.

8.r() = (cos  + 1) i + (sin  − 1) j, r(−3) =₁

2+ 1 i+

−^√2³− 1

j=³₂i+

−^√2³− 1

j≈ 15 i − 187 j.

Here ( − 1)²+ ( + 1)²= cos² + sin² = 1, so the curve is a circle of radius 1 with center (1 −1).

(a), (c) (b) r⁰() = − sin  i + cos  j,

r⁰(−3) = ^√2³i+¹₂j≈ 087 i + 05 j

9.r() =√

 − 2 3 1²

⇒ r⁰() =





√ − 2

 

[3]  



1²

=

1

2( − 2)^−12 0 −2⁻³

=

 1

2√

 − 2 0 −2

³



10.r() =

^−  − ³ ln 

⇒ r⁰() =

−^− 1 − 3² 1 11. r() = ²i+ cos

²

j+ sin² k ⇒ r⁰() = 2 i +

− sin(²) · 2

j+ (2 sin  · cos ) k = 2 i − 2 sin(²) j + 2 sin  cos  k

12.r() = 1

1 + i+ 

1 + j+ ² 1 + k ⇒ r⁰() = 0 − 1(1)

(1 + )² i+(1 + ) · 1 − (1)

(1 + )² j+(1 + ) · 2 − ²(1)

(1 + )² k= − 1

(1 + )² i+ 1

(1 + )² j+ ²+ 2

(1 + )²k 13.r() =  sin  i + ^cos  j + sin  cos  k ⇒

r⁰() = [ · cos  + (sin ) · 1] i +

^(− sin ) + (cos )^

j+ [(sin )(− sin ) + (cos )(cos )] k

= ( cos  + sin ) i + ^(cos  − sin ) j +

cos² − sin² k

328 ¤ CHAPTER 13 VECTOR FUNCTIONS

14. r() = sin² i + ^j+ cos² k ⇒ r⁰() = [2(sin ) · (cos )()] i +

 · ^() + ^· 1

j+ [2(cos ) · (− sin )()] k

= 2 sin  cos  i + ^( + 1) j − 2 sin  cos  k 15. r⁰() = 0 + b + 2 c = b + 2 cby Formulas 1 and 3 of Theorem 3.

16. To ﬁnd r⁰(), we ﬁrst expand r() =  a × (b +  c) = (a × b) + ²(a × c), so r⁰() = a × b + 2(a × c).

17. r() =

²− 2 1 + 3¹3³+¹₂²

⇒ r⁰() =

2 − 2 3 ²+ 

⇒ r⁰(2) = h2 3 6i.

So |r⁰(2)| =√

2²+ 3²+ 6²=√

49 = 7and T(2) = r⁰(2)

|r⁰(2)|= ¹₇h2 3 6i =2 7³₇⁶₇. 18. r() =

tan⁻¹ 2^2 8^

⇒ r⁰() =

1(1 + ²) 4^2 8^+ 8^

⇒ r⁰(0) = h1 4 8i.

So |r⁰(0)| =√

1²+ 4²+ 8²=√

81 = 9and T(0) = r⁰(0)

|r⁰(0)|= ¹₉h1 4 8i =1 9⁴₉⁸₉. 19. r⁰() = − sin  i + 3 j + 4 cos 2 k ⇒ r⁰(0) = 3 j + 4 k. Thus

T(0) = r⁰(0)

|r⁰(0)|= 1

√0²+ 3²+ 4² (3 j + 4 k) =¹₅(3 j + 4 k) =³₅j+⁴₅k.

20. r⁰() = 2 sin  cos  i − 2 cos  sin  j + 2 tan  sec² k ⇒ r⁰

4

= 2 ·^√2²·^√2²i− 2 ·^√2²·^√2²j+ 2 · 1 · (√

2)²k= i − j + 4 k andr⁰ 4=√

1 + 1 + 16 =√

18 = 3√ 2. Thus

T 4

= r⁰ 4



r⁰ 4 = 1

3√

2(i − j + 4 k) = 1 3√

2i− 1 3√

2j+ 4 3√

2k.

21. r() =

 ² ³

⇒ r⁰() =

1 2 3². Then r⁰(1) = h1 2 3i and |r⁰(1)| =√

1²+ 2²+ 3²=√ 14, so

T(1) = r⁰(1)

|r⁰(1)|= ^√¹₁₄h1 2 3i =

√1 14√²

14√³ 14

. r⁰⁰() = h0 2 6i, so

r⁰() × r⁰⁰() =





i j k

1 2 3² 0 2 6





=



 2 3²

2 6



i −



 1 3² 0 6



j +



 1 2

0 2



k

= (12²− 6²) i − (6 − 0) j + (2 − 0) k =

6² −6 2 22. r() =

^2 ^−2 ^2

⇒ r⁰() =

2^2 −2^−2 (2 + 1)^2

⇒ r⁰(0) =

2⁰ −2⁰ (0 + 1)⁰

= h2 −2 1i and |r⁰(0)| =

2²+ (−2)²+ 1²= 3. Then T(0) = r⁰(0)

|r⁰(0)|= ¹₃h2 −2 1i =2

3 −²3¹₃. r⁰⁰() =

4^2 4^−2 (4 + 4)^2

⇒ r⁰⁰(0) =

4⁰ 4⁰ (0 + 4)⁰

= h4 4 4i.

r⁰() · r⁰⁰() =

2^2 −2^−2 (2 + 1)^2

·

4^2 4^−2 (4 + 4)^2

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

= 8^4− 8^−4+ (8²+ 12 + 4)^4= (8²+ 12 + 12)^4− 8^−4

14. r() = sin² i + ^j+ cos² k ⇒ r⁰() = [2(sin ) · (cos )()] i +

 · ^() + ^· 1

j+ [2(cos ) · (− sin )()] k

= 2 sin  cos  i + ^( + 1) j − 2 sin  cos  k 15. r⁰() = 0 + b + 2 c = b + 2 cby Formulas 1 and 3 of Theorem 3.

16. To ﬁnd r⁰(), we ﬁrst expand r() =  a × (b +  c) = (a × b) + ²(a × c), so r⁰() = a × b + 2(a × c).

17. r() =

²− 2 1 + 3¹3³+¹₂²

⇒ r⁰() =

2 − 2 3 ²+ 

⇒ r⁰(2) = h2 3 6i.

So |r⁰(2)| =√

2²+ 3²+ 6²=√

49 = 7and T(2) = r⁰(2)

|r⁰(2)|= ¹₇h2 3 6i =2 7³₇⁶₇. 18. r() =

tan⁻¹ 2^2 8^

⇒ r⁰() =

1(1 + ²) 4^2 8^+ 8^

⇒ r⁰(0) = h1 4 8i.

So |r⁰(0)| =√

1²+ 4²+ 8²=√

81 = 9and T(0) = r⁰(0)

|r⁰(0)|= ¹₉h1 4 8i =1 9⁴₉⁸₉. 19. r⁰() = − sin  i + 3 j + 4 cos 2 k ⇒ r⁰(0) = 3 j + 4 k. Thus

T(0) = r⁰(0)

|r⁰(0)|= 1

√0²+ 3²+ 4² (3 j + 4 k) =¹₅(3 j + 4 k) =³₅j+⁴₅k.

20. r⁰() = 2 sin  cos  i − 2 cos  sin  j + 2 tan  sec² k ⇒ r⁰

4

= 2 ·^√2²·^√2²i− 2 ·^√2²·^√2²j+ 2 · 1 · (√

2)²k= i − j + 4 k andr⁰ 4=√

1 + 1 + 16 =√

18 = 3√ 2. Thus

T 4

= r⁰ 4



r⁰_

4 = 1 3√

2(i − j + 4 k) = 1 3√

2i− 1 3√

2j+ 4 3√

2k.

21. r() =

 ² ³

⇒ r⁰() =

1 2 3². Then r⁰(1) = h1 2 3i and |r⁰(1)| =√

1²+ 2²+ 3²=√ 14, so

T(1) = r⁰(1)

|r⁰(1)|= ^√¹₁₄h1 2 3i =

√1 14√²

14√³ 14

. r⁰⁰() = h0 2 6i, so

r⁰() × r⁰⁰() =





i j k

1 2 3² 0 2 6





=



 2 3²

2 6



i −



 1 3² 0 6



j +



 1 2

0 2



k

= (12²− 6²) i − (6 − 0) j + (2 − 0) k =

6² −6 2 22. r() =

^2 ^−2 ^2

⇒ r⁰() =

2^2 −2^−2 (2 + 1)^2

⇒ r⁰(0) =

2⁰ −2⁰ (0 + 1)⁰

= h2 −2 1i and |r⁰(0)| =

2²+ (−2)²+ 1²= 3. Then T(0) = r⁰(0)

|r⁰(0)|= ¹₃h2 −2 1i =2

3 −²3¹₃. r⁰⁰() =

4^2 4^−2 (4 + 4)^2

⇒ r⁰⁰(0) =

4⁰ 4⁰ (0 + 4)⁰

= h4 4 4i.

r⁰() · r⁰⁰() =

2^2 −2^−2 (2 + 1)^2

·

4^2 4^−2 (4 + 4)^2

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

= 8^4− 8^−4+ (8²+ 12 + 4)^4= (8²+ 12 + 12)^4− 8^−4

SECTION 13.2 DERIVATIVES AND INTEGRALS OF VECTOR FUNCTIONS ¤ 329 23. The vector equation for the curve is r() =

²+ 1 4√

 ^²^−

, so r⁰() = 2 2√

 (2 − 1)^²^−

. The point (2 4 1) corresponds to  = 1, so the tangent vector there is r⁰(1) = h2 2 1i. Thus, the tangent line goes through the point (2 4 1) and is parallel to the vector h2 2 1i. Parametric equations are  = 2 + 2,  = 4 + 2,  = 1 + .

24. The vector equation for the curve is r() =

ln( + 1)  cos 2 2^, so r⁰() =

1( + 1) cos 2 − 2 sin 2 2^ln 2. The point (0 0 1) corresponds to  = 0, so the tangent vector there is r⁰(0) = h1 1 ln 2i. Thus, the tangent line goes through the point (0 0 1) and is parallel to the vector h1 1 ln 2i. Parametric equations are  = 0 + 1 ·  = ,  = 0 + 1 ·  = ,

 = 1 + (ln 2).

25. The vector equation for the curve is r() =

^−cos  ^−sin  ^−, so

r⁰() =^−(− sin ) + (cos )(−^−), ^−cos  + (sin )(−^−), (−^−)

=

−^−(cos  + sin ) ^−(cos  − sin ) −^− The point (1 0 1) corresponds to  = 0, so the tangent vector there is

r⁰(0) =

−⁰(cos 0 + sin 0) ⁰(cos 0 − sin 0) −⁰

= h−1 1 −1i. Thus, the tangent line is parallel to the vector h−1 1 −1i and parametric equations are  = 1 + (−1) = 1 − ,  = 0 + 1 ·  = ,  = 1 + (−1) = 1 − .

26. The vector equation for the curve is r() =√

²+ 3 ln(²+ 3) , so r⁰() =

√

²+ 3 2(²+ 3) 1. At (2 ln 4 1),

 = 1and r⁰(1) =1

2¹₂ 1. Thus, parametric equations of the tangent line are  = 2 +¹₂,  = ln 4 +¹₂,  = 1 + .

27. First we parametrize the curve  of intersection. The projection of  onto the -plane is contained in the circle

²+ ²= 25,  = 0, so we can write  = 5 cos ,  = 5 sin .  also lies on the cylinder ²+ ²= 20, and  ≥ 0 near the point (3 4 2), so we can write  =

20 − ²=

20 − 25 sin². A vector equation then for  is r() =

5 cos  5 sin 

20 − 25 sin²

⇒ r⁰() =

−5 sin  5 cos ¹2(20 − 25 sin²)^−12(−50 sin  cos ) . The point (3 4 2) corresponds to  = cos⁻¹₃

5

, so the tangent vector there is

r⁰ cos⁻¹₃

5

=



−5₄

5

 5₃

5

¹₂

20 − 25₄

5

2_−12

−50₄

5

₃

5



= h−4 3 −6i.

The tangent line is parallel to this vector and passes through (3 4 2), so a vector equation for the line is r() = (3 − 4)i + (4 + 3)j + (2 − 6)k.

28. r() =

2 cos  2 sin  ^

⇒ r⁰() =

−2 sin  2 cos  ^. The tangent line to the curve is parallel to the plane when the curve’s tangent vector is orthogonal to the plane’s normal vector. Thus we require

−2 sin  2 cos  ^

·√

3 1 0

= 0 ⇒

−2√

3 sin  + 2 cos  + 0 = 0 ⇒ tan  = ^√¹₃ ⇒  =^6 [since 0 ≤  ≤ ].

r 6

=√

3 1 ^6

, so the point is (√

3 1 ^6).

SECTION 13.2 DERIVATIVES AND INTEGRALS OF VECTOR FUNCTIONS ¤ 331

r⁰2() = hcos  2 cos 2 1i and since r²(0) = h0 0 0i, r⁰²(0) = h1 2 1i is a tangent vector to r²at (0 0 0). If  is the angle between these two tangent vectors, then cos  =√¹

1√

6h1 0 0i · h1 2 1i = ^√¹₆and  = cos⁻¹

√1 6



≈ 66^◦.

34. To ﬁnd the point of intersection, we must ﬁnd the values of  and  which satisfy the following three equations simultaneously:

 = 3 − , 1 −  =  − 2, 3 + ²= ². Solving the last two equations gives  = 1,  = 2 (check these in the ﬁrst equation).

Thus the point of intersection is (1 0 4). To ﬁnd the angle  of intersection, we proceed as in Exercise 33. The tangent vectors to the respective curves at (1 0 4) are r⁰1(1) = h1 −1 2i and r⁰2(2) = h−1 1 4i. So

cos  = ^√₆¹^√₁₈(−1 − 1 + 8) =₆^√⁶₃ =^√¹₃ and  = cos⁻¹

√1 3



≈ 55^◦.

Note: In Exercise 33, the curves intersect when the value of both parameters is zero. However, as seen in this exercise, it is not necessary for the parameters to be of equal value at the point of intersection.

35. 2

0 ( i − ³j+ 3⁵k)  =2 0  

i−2 0 ³

j+2 0 3⁵

k

=₁

2²2 0i−₁

4⁴2 0 j+₁

2⁶2 0 k

=¹₂(4 − 0) i −¹4(16 − 0) j +¹2(64 − 0) k = 2 i − 4 j + 32 k

36.

 4 1



2^32i+ ( + 1)√

 k

 =4

1 2^32 i+4

1 (^32+ ^12)  k

=

4 5^524

1 i+

2

5^52+²₃^324 1k

= ⁴₅(4^52− 1) i +

2

5(4)^52+²₃(4)^32−²5−²3

 k

= ⁴₅(31) i +2

5(32) +²₃(8) −²5 −²3

k=¹²⁴₅ i+²⁵⁶₁₅ k

37.

 1 0

 1

 + 1 i+ 1

²+ 1 j+ 

²+ 1k



 =

 1 0

1

 + 1

 i+

 1 0

1

²+ 1

 j+

 1 0



²+ 1

 k

= [ ln | + 1| ]¹0 i+

tan⁻¹1 0j+₁

2ln(²+ 1)1 0 k

= (ln 2 − ln 1) i + (^4 − 0) j +¹2(ln 2 − ln 1) k = ln 2 i +^4j+¹₂ln 2 k 38. 4

0 (sec  tan  i +  cos 2 j + sin²2 cos 2 k) 

=4

0 sec  tan  

i+4

0  cos 2 

j+4

0 sin²2 cos 2  k

= sec 4

0 i+₁

2 sin 24

0 −4

0 1

2sin 2  j+₁

6sin³24

0 k

[For the -component, integrate by parts with  = ,  = cos 2 .]

= (sec^₄ − sec 0) i +



8 sin^₂ − 0 −

−¹4cos 24 0

 j+¹₆

sin^{3 }₂ − sin³0 k

= (√

2 − 1) i +

8 +¹₄cos^₂ −¹4cos 0

j+¹₆(1 − 0) k = (√

2 − 1) i + 8 −¹4

j+¹₆k

39. 

(sec² i + (²+ 1)³j+ ²ln  k)  =

sec²  i+

(²+ 1)³ j+

²ln   k

= tan  i +¹₈(²+ 1)⁴j+1

3³ln  −¹9³ k+ C,

where C is a vector constant of integration. [For the -component, integrate by parts with  = ln ,  = ².]

52. From Exercise 51, r⁰() = −a  sin  + b  cos  ⇒ r⁰⁰() = −a ²cos  − b ²sin . Then r⁰⁰() + ²r() =

−a ²cos  − b ²sin 

+ ²(a cos  + b sin )

= −a ²cos  − b ²sin  + a ²cos  + b ²sin  = 0

53. 

[r() × r⁰()] = r⁰() × r⁰() + r() × r⁰⁰()by Formula 5 of Theorem 3. But r⁰() × r⁰() = 0 (by Example 12.4.2).

Thus, 

[r() × r⁰()] = r() × r⁰⁰().

54. 

(u() · [v () × w()])= u⁰() · [v() × w()] + u() · 

[v() × w ()]

= u⁰() · [v() × w()] + u() · [v⁰() × w() + v() × w⁰()]

= u⁰() · [v() × w()] + u() · [v⁰() × w()] + u() · [v() × w⁰()]

= u⁰() · [v() × w()] − v⁰() · [u() × w()] + w⁰() · [u() × v()]

55. 

|r()| = 

[r() · r()]^12=¹₂[r() · r()]^−12[2r() · r⁰()] = 1

|r()|r() · r⁰()

56. Since r() · r⁰() = 0, we have 0 = 2r() · r⁰() = 

[r() · r()] = 

|r()|². Thus |r()|², and so |r()|, is a constant, and hence the curve lies on a sphere with center the origin.

57. Since u() = r() · [r⁰() × r⁰⁰()],

u⁰() = r⁰() · [r⁰() × r⁰⁰()] + r() · 

[r⁰() × r⁰⁰()]

= 0 + r() · [r⁰⁰() × r⁰⁰() + r⁰() × r⁰⁰⁰()] [since r⁰() ⊥ r⁰() × r⁰⁰()]

= r() · [r⁰() × r⁰⁰⁰()] [since r⁰⁰() × r⁰⁰() = 0]

58. The tangent vector r⁰()is deﬁned as lim

→0

r( + ) − r()

 . Here we assume that this limit exists and r⁰() 6= 0; then we know that this vector lies on the tangent line to the curve. As in Figure 1, let points  and  have position vectors r() and r( + ).

The vector r( + ) − r() points from  to , so r( + ) − r() =−−→ . If   0 then    + , so  lies “ahead”

of  on the curve. If  is sufﬁciently small (we can take  to be as small as we like since  → 0) then−−→

 approximates the curve from  to  and hence points approximately in the direction of the curve as  increases. Since  is positive,

1



−−→  = r( + ) − r()

 points in the same direction. If   0, then    +  so  lies “behind”  on the curve. For  sufﬁciently small, −−→ approximates the curve but points in the direction of decreasing . However,  is negative, so

1



−−→  = r( + ) − r()

 points in the opposite direction, that is, in the direction of increasing . In both cases, the difference quotientr( + ) − r()

 points in the direction of increasing . The tangent vector r⁰()is the limit of this difference quotient, so it must also point in the direction of increasing .

52. From Exercise 51, r⁰() = −a  sin  + b  cos  ⇒ r⁰⁰() = −a ²cos  − b ²sin . Then r⁰⁰() + ²r() =

−a ²cos  − b ²sin 

+ ²(a cos  + b sin )

= −a ²cos  − b ²sin  + a ²cos  + b ²sin  = 0

53. 

[r() × r⁰()] = r⁰() × r⁰() + r() × r⁰⁰()by Formula 5 of Theorem 3. But r⁰() × r⁰() = 0 (by Example 12.4.2).

Thus, 

[r() × r⁰()] = r() × r⁰⁰().

54. 

(u() · [v () × w()])= u⁰() · [v() × w()] + u() · 

[v() × w ()]

= u⁰() · [v() × w()] + u() · [v⁰() × w() + v() × w⁰()]

= u⁰() · [v() × w()] + u() · [v⁰() × w()] + u() · [v() × w⁰()]

= u⁰() · [v() × w()] − v⁰() · [u() × w()] + w⁰() · [u() × v()]

55. 

|r()| = 

[r() · r()]^12=¹₂[r() · r()]^−12[2r() · r⁰()] = 1

|r()|r() · r⁰()

56. Since r() · r⁰() = 0, we have 0 = 2r() · r⁰() = 

[r() · r()] = 

|r()|². Thus |r()|², and so |r()|, is a constant, and hence the curve lies on a sphere with center the origin.

57. Since u() = r() · [r⁰() × r⁰⁰()],

u⁰() = r⁰() · [r⁰() × r⁰⁰()] + r() · 

[r⁰() × r⁰⁰()]

= 0 + r() · [r⁰⁰() × r⁰⁰() + r⁰() × r⁰⁰⁰()] [since r⁰() ⊥ r⁰() × r⁰⁰()]

= r() · [r⁰() × r⁰⁰⁰()] [since r⁰⁰() × r⁰⁰() = 0]

58. The tangent vector r⁰()is deﬁned as lim

→0

r( + ) − r()

 . Here we assume that this limit exists and r⁰() 6= 0; then we know that this vector lies on the tangent line to the curve. As in Figure 1, let points  and  have position vectors r() and r( + ).

The vector r( + ) − r() points from  to , so r( + ) − r() =−−→ . If   0 then    + , so  lies “ahead”

of  on the curve. If  is sufﬁciently small (we can take  to be as small as we like since  → 0) then−−→ approximates the curve from  to  and hence points approximately in the direction of the curve as  increases. Since  is positive,

1



−−→  = r( + ) − r()

 points in the same direction. If   0, then    +  so  lies “behind”  on the curve. For  sufﬁciently small, −−→ approximates the curve but points in the direction of decreasing . However,  is negative, so

1



−−→  = r( + ) − r()

 points in the opposite direction, that is, in the direction of increasing . In both cases, the difference quotientr( + ) − r()

 points in the direction of increasing . The tangent vector r⁰()is the limit of this difference quotient, so it must also point in the direction of increasing .