Section 13.4 Motion in Space: Velocity and Acceleration 14. Find the velocity, acceleration, and speed of a particle with the given position function. r

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Section 13.4 Motion in Space: Velocity and Acceleration

14. Find the velocity, acceleration, and speed of a particle with the given position function.

r(t) =< t², sin t − t cos t, cos t + t sin t >, t ≥ 0 Solution:

1318 ¤ CHAPTER 13 VECTOR FUNCTIONS 11. r() =√

2  i + ^j+ ^−k ⇒ v() = r⁰() =√

2 i + ^j− ^−k, a() = v⁰() = ^j+ ^−k,

|v()| =√

2 + ^2+ ^−2=

(^+ ^−)²= ^+ ^−.

12. r() = ²i+ 2 j + ln  k ⇒ v() = r⁰() = 2 i + 2 j + (1) k, a() = v⁰() = 2 i − (1²) k,

|v()| =

4²+ 4 + (1²) =

[2 + (1)]²= |2 + (1)|.

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

v() = r⁰() = [^(− sin ) + (cos )^] i + [^cos  + (sin )^] j + (^+ ^) k

= ^[(cos  − sin ) i + (sin  + cos ) j + ( + 1) k]

a() = v⁰() = [^(− sin  − cos ) + (cos  − sin )^] i + [^(cos  − sin ) + (sin  + cos )^] j + [^· 1 + ( + 1)^] k

= ^[−2 sin  i + 2 cos  j + ( + 2) k]

|v()| =

^2(cos  − sin )²+ ^2(sin  + cos )²+ ^2( + 1)²

=√

^2

cos² + sin² − 2 cos  sin  + sin² + cos² + 2 sin  cos  + ²+ 2 + 1

= ^√

²+ 2 + 3

14. r() =

² sin  −  cos  cos  +  sin 

⇒

v() = r⁰() = h2 cos  − (− sin  + cos ) − sin  +  cos  + sin i = h2  sin   cos i, a() = v⁰() = h2  cos  + sin  − sin  + cos i,

|v()| =

4²+ ²sin² + ²cos² =√

4²+ ²=√

5²=√

5  [since  ≥ 0].

15. a() = 2 i + 2 k ⇒ v() =

a()  =

(2 i + 2 k)  = 2 i + ²k+ C. Then v(0) = C but we were given that v(0) = 3 i − j, so C = 3 i − j and v() = 2 i + ²k+ 3 i − j = (2 + 3) i − j + ²k.

r() =

v()  = 

(2 + 3) i − j + ²k

 = (²+ 3) i −  j +¹3³k+ D. Here r(0) = D and we were given that r(0) = j + k, so D = j + k and r() = (²+ 3) i + (1 − ) j +₁

3³+ 1k.

16. a() = sin  i + 2 cos  j + 6 k ⇒ v() =

a()  =

(sin  i + 2 cos  j + 6 k)  = − cos  i + 2 sin  j + 3²k+ C.

Then v(0) = −i + C but we were given that v(0) = −k, so −i + C = −k ⇒ C = i − k and v() = (1 − cos ) i + 2 sin  j + (3²− 1) k.

r() =

v()  = 

(1 − cos ) i + 2 sin  j + (3²− 1) k

 = ( − sin ) i − 2 cos  j + (³− ) k + D. Here r(0) = −2 j + D and we were given that r(0) = j − 4 k, so D = 3 j − 4 k and

r() = ( − sin ) i + (3 − 2 cos ) j + (³−  − 4) k.

16. Find the velocity and position vectors of a particle that has the given acceleration and the given initial velocity and position.

a(t) = sin ti + 2 cos tj + 6tk, v(0) = −k, r(0) = j − 4k Solution:

354 ¤ CHAPTER 13 VECTOR FUNCTIONS 9. r() =

²+ 1 ³ ²− 1

⇒ v() = r⁰() =

2 3² 2, a() = v⁰() = h2 6 2i,

|v()| =

(2)²+ (3²)²+ (2)²=√

9⁴+ 8²= ||√ 9²+ 8.

10. r() = h2 cos  3 2 sin i ⇒ v() = r⁰() = h−2 sin  3 2 cos i, a() = v⁰() = h−2 cos  0 −2 sin i,

|v()| =

4 sin² + 9 + 4 cos² =√13.

11. r() =√

2  i + ^j+ ^−k ⇒ v() = r⁰() =√

2 i + ^j− ^−k, a() = v⁰() = ^j+ ^−k,

|v()| =√

2 + ^2+ ^−2=

(^+ ^−)²= ^+ ^−.

12. r() = ²i+ 2 j + ln  k ⇒ v() = r⁰() = 2 i + 2 j + (1) k, a() = v⁰() = 2 i − (1²) k,

|v()| =

4²+ 4 + (1²) =

[2 + (1)]²= |2 + (1)|.

13. r() = ^hcos  sin  i ⇒

v() = r⁰() = ^hcos  sin  i + ^h− sin  cos  1i = ^hcos  − sin  sin  + cos   + 1i a() = v⁰() = ^hcos  − sin  − sin  − cos  sin  + cos  + cos  − sin   + 1 + 1i

= ^h−2 sin  2 cos   + 2i

|v()| = ^

cos² + sin² − 2 cos  sin  + sin² + cos² + 2 sin  cos  + ²+ 2 + 1

= ^√

²+ 2 + 3

14. r() =  sin  i +  cos  j + ²k ⇒ v() = r⁰() = (sin  +  cos ) i + (cos  −  sin ) j + 2 k, a() = v⁰() = (2 cos  −  sin ) i + (−2 sin  −  cos ) j + 2 k,

|v()| =

(sin² + 2 sin  cos  + ²cos²) + (cos² − 2 sin  cos  + ²sin²) + 4²=√ 5²+ 1.

15. a() = 2 i + 2 k ⇒ v() =

a()  =

(2 i + 2 k)  = 2 i + ²k+ C. Then v(0) = C but we were given that v(0) = 3 i − j, so C = 3 i − j and v() = 2 i + ²k+ 3 i − j = (2 + 3) i − j + ²k.

r() =

v()  = 

(2 + 3) i − j + ²k

 = (²+ 3) i −  j + ¹3³k+ D. Here r(0) = D and we were given that r(0) = j + k, so D = j + k and r() = (²+ 3) i + (1 − ) j +₁

3³+ 1k.

16. a() = sin  i + 2 cos  j + 6 k ⇒ v() =

a()  =

(sin  i + 2 cos  j + 6 k)  = − cos  i + 2 sin  j + 3²k+ C.

Then v(0) = −i + C but we were given that v(0) = −k, so −i + C = −k ⇒ C = i − k and v() = (1 − cos ) i + 2 sin  j + (3²− 1) k.

r() =

v()  = 

(1 − cos ) i + 2 sin  j + (3²− 1) k

 = ( − sin ) i − 2 cos  j + (³− ) k + D. Here r(0) = −2 j + D and we were given that r(0) = j − 4 k, so D = 3 j − 4 k and

r() = ( − sin ) i + (3 − 2 cos ) j + (³−  − 4) k.

39. Find the tangential and normal components of the acceleration vector.

r(t) = cos ti + sin tj + tk Solution:

SECTION 13.4 MOTION IN SPACE: VELOCITY AND ACCELERATION ¤ 361 38. r() = 2²i+2

3³− 2

j ⇒ r⁰() = 4 i + (2²− 2) j,

|r⁰()| =

16²+ (2²− 2)²=√

4⁴+ 8²+ 4 =



4 (²+ 1)²= 2(²+ 1), r⁰⁰() = 4 i + 4 j, r⁰() × r⁰⁰() = (8²+ 8) k. Then Equation 9 gives

 = r⁰() · r⁰⁰()

|r⁰()| = (4)(4) + (2²− 2)(4)

2(²+ 1) = 8(²+ 1) 2(²+ 1) = 4



or by Equation 8,  = ⁰= 



2(²+ 1)

= 4



and Equation 10 gives = |r⁰() × r⁰⁰()|

|r⁰()| = 8(²+ 1) 2(²+ 1) = 4.

39. r() = cos  i + sin  j +  k ⇒ r⁰() = − sin  i + cos  j + k, |r⁰()| =

sin² + cos² + 1 =√ 2, r⁰⁰() = − cos  i − sin  j, r⁰() × r⁰⁰() = sin  i − cos  j + k.

Then  = r⁰() · r⁰⁰()

|r⁰()| = sin  cos  − sin  cos √

2 = 0 and  = |r⁰() × r⁰⁰()|

|r⁰()| =

sin² + cos² + 1

√2 =

√2

√2 = 1.

40. r() =  i + 2^j+ ^2k ⇒ r⁰() = i + 2^j+ 2^2k, |r⁰()| =√

1 + 4^2+ 4^4=

(1 + 2^2)²= 1 + 2^2, r⁰⁰() = 2^j+ 4^2k, r⁰() × r⁰⁰() = 4^3i− 4^2j+ 2^k,

|r⁰() × r⁰⁰()| =√

16^6+ 16^4+ 4^2 =

4^2(2^2+ 1)² = 2^(2^2+ 1). Then

 = r⁰() · r⁰⁰()

|r⁰()| = 4^2+ 8^4

1 + 2^2 = 4^2(1 + 2^2)

1 + 2^2 = 4^2 and  = |r⁰() × r⁰⁰()|

|r⁰()| = 2^(2^2+ 1) 1 + 2^2 = 2^. 41. r() = ln  i + (²+ 3) j + 4√

 k ⇒ r⁰() = (1) i + (2 + 3) j + (2√

 ) k ⇒ r⁰⁰() = (−1²) i + 2 j − (1^32) k. The point (0 4 4) corresponds to  = 1, where

r⁰(1) = i + 5 j + 2 k, r⁰⁰(1) = −i + 2 j − k, and r⁰(1) × r⁰⁰(1) = −9 i − j + 7 k. Thus at the point (0 4 4),

 = r⁰(1) · r⁰⁰(1)

|r⁰(1)| = −1 + 10 − 2

√1 + 25 + 4 = 7

√30 and  = |r⁰(1) × r⁰⁰(1)|

|r⁰(1)| =

√81 + 1 + 49

√30 =

131 30 .

42. r() = ⁻¹i+ ⁻²j+ ⁻³k ⇒ r⁰() = −⁻²i− 2⁻³j− 3⁻⁴k ⇒ r⁰⁰() = 2⁻³i+ 6⁻⁴j+ 12⁻⁵k. The point (1 1 1) corresponds to  = 1, where r⁰(1) = −i − 2 j − 3 k, r⁰⁰(1) = 2 i + 6 j + 12 k, and

r⁰(1) × r⁰⁰(1) = −6 i + 6 j − 2 k. Thus at the point (1 1 1), ^ = r⁰(1) · r⁰⁰(1)

|r⁰(1)| = −√2 − 12 − 36

1 + 4 + 9 = − 50

√14 and

= |r⁰(1) × r⁰⁰(1)|

|r⁰(1)| =

√36 + 36 + 4

√14 =

76 14 =

38 7 .

43. The tangential component of a is the length of the projection of a onto T, so we sketch the scalar projection of a in the tangential direction to the curve and estimate its length to be 45 (using the fact that a has length 10 as a guide). Similarly, the normal component of ais the length of the projection of a onto N, so we sketch the scalar projection of a in the normal direction to the curve and estimate its length to be 90. Thus  ≈ 45 cms²and

≈ 90 cms².

42. Find the tangential and normal components of the acceleration vector at the given point.

r(t) = 1 ti + 1

t²j + 1

t³k, (1, 1, 1) Solution:

SECTION 13.4 MOTION IN SPACE: VELOCITY AND ACCELERATION ¤ 361 38. r() = 2²i+₂

3³− 2

j ⇒ r⁰() = 4 i + (2²− 2) j,

|r⁰()| =

16²+ (2²− 2)²=√

4⁴+ 8²+ 4 =



4 (²+ 1)²= 2(²+ 1), r⁰⁰() = 4 i + 4 j, r⁰() × r⁰⁰() = (8²+ 8) k. Then Equation 9 gives

 = r⁰() · r⁰⁰()

|r⁰()| = (4)(4) + (2²− 2)(4)

2(²+ 1) = 8(²+ 1) 2(²+ 1) = 4



or by Equation 8,  = ⁰= 



2(²+ 1)

= 4



and Equation 10 gives = |r⁰() × r⁰⁰()|

|r⁰()| = 8(²+ 1) 2(²+ 1) = 4.

39. r() = cos  i + sin  j +  k ⇒ r⁰() = − sin  i + cos  j + k, |r⁰()| =

sin² + cos² + 1 =√2, r⁰⁰() = − cos  i − sin  j, r⁰() × r⁰⁰() = sin  i − cos  j + k.

Then  = r⁰() · r⁰⁰()

|r⁰()| = sin  cos  − sin  cos √

2 = 0 and  = |r⁰() × r⁰⁰()|

|r⁰()| =

sin² + cos² + 1

√2 =

√2

√2 = 1.

40. r() =  i + 2^j+ ^2k ⇒ r⁰() = i + 2^j+ 2^2k, |r⁰()| =√

1 + 4^2+ 4^4=

(1 + 2^2)²= 1 + 2^2, r⁰⁰() = 2^j+ 4^2k, r⁰() × r⁰⁰() = 4^3i− 4^2j+ 2^k,

|r⁰() × r⁰⁰()| =√

16^6+ 16^4+ 4^2 =

4^2(2^2+ 1)² = 2^(2^2+ 1). Then

 = r⁰() · r⁰⁰()

|r⁰()| = 4^2+ 8^4

1 + 2^2 = 4^2(1 + 2^2)

1 + 2^2 = 4^2 and  = |r⁰() × r⁰⁰()|

|r⁰()| = 2^(2^2+ 1) 1 + 2^2 = 2^. 41. r() = ln  i + (²+ 3) j + 4√

 k ⇒ r⁰() = (1) i + (2 + 3) j + (2√

 ) k ⇒ r⁰⁰() = (−1²) i + 2 j − (1^32) k. The point (0 4 4) corresponds to  = 1, where

r⁰(1) = i + 5 j + 2 k, r⁰⁰(1) = −i + 2 j − k, and r⁰(1) × r⁰⁰(1) = −9 i − j + 7 k. Thus at the point (0 4 4),

 = r⁰(1) · r⁰⁰(1)

|r⁰(1)| = −1 + 10 − 2

√1 + 25 + 4 = 7

√30 and  = |r⁰(1) × r⁰⁰(1)|

|r⁰(1)| =

√81 + 1 + 49

√30 =

131 30 .

42. r() = ⁻¹i+ ⁻²j+ ⁻³k ⇒ r⁰() = −⁻²i− 2⁻³j− 3⁻⁴k ⇒ r⁰⁰() = 2⁻³i+ 6⁻⁴j+ 12⁻⁵k. The point (1 1 1) corresponds to  = 1, where r⁰(1) = −i − 2 j − 3 k, r⁰⁰(1) = 2 i + 6 j + 12 k, and

r⁰(1) × r⁰⁰(1) = −6 i + 6 j − 2 k. Thus at the point (1 1 1), ^ = r⁰(1) · r⁰⁰(1)

|r⁰(1)| = −√2 − 12 − 36

1 + 4 + 9 = − 50

√14 and

= |r⁰(1) × r⁰⁰(1)|

|r⁰(1)| =

√36 + 36 + 4

√14 =

76 14 =

38 7 .

43. The tangential component of a is the length of the projection of a onto T, so we sketch the scalar projection of a in the tangential direction to the curve and estimate its length to be 45 (using the fact that a has length 10 as a guide). Similarly, the normal component of ais the length of the projection of a onto N, so we sketch the scalar projection of a in the normal direction to the curve and estimate its length to be 90. Thus  ≈ 45 cms²and

≈ 90 cms².

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