Section 13.2 Derivatives and Integrals of Vector Functions

Section 13.2 Derivatives and Integrals of Vector Functions

24. If r(t) =< e^2t, e^−3t, t >, find r⁰(0), T(0), r⁰⁰(0), and r⁰(0) × r⁰⁰(0).

Solution:

1288 ¤ CHAPTER 13 VECTOR FUNCTIONS

19. r() = cos  i + 3 j + 2 sin 2 k ⇒ r⁰() = − sin  i + 3 j + 4 cos 2 k ⇒ r⁰(0) = 3 j + 4 k. So

|r⁰(0)| =√

0²+ 3²+ 4² =√

25 = 5and T(0) = r⁰(0)

|r⁰(0)| = ¹₅(3 j + 4 k) = ³₅j+ ⁴₅k.

20. r() = sin² i + cos² j + tan² k ⇒ 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. Sor⁰_

4=√

1²+ 1²+ 4²=√

18 = 3√ 2 and T

4

= r⁰_

4



r⁰

4 = 1 3√

2(i − j + 4 k) = 1 3√

2i− 1 3√

2j+ 4 3√

2k.

21. The point (2 −2 4) corresponds to  = 1 [note that 4 = 4]. Then r() =

³+ 1 3 − 5 4

⇒ r⁰() =

3² 3 −4²

⇒ r⁰(1) = h3 3 −4i

So |r⁰(1)| =

3²+ 3²+ (−4)²=√ 34

and T(1) = r⁰(1)

|r⁰(1)| = 1

√34h3 3 −4i =

 3

√34 3

√34 − 4

√34



22. The point (0 0 1) corresponds to  = 0 [note that 5 = 0]. Then

r() = sin  i + 5 j + cos  k ⇒ r⁰() = cos  i + 5 j − sin  k ⇒ r⁰(0) = i + 5 j

So |r⁰(0)| =√

1²+ 5²+ 0²=√ 26

and T(0) = r⁰(0)

|r⁰(0)|= 1

√26(i + 5j) = 1

√26i+ 5

√26j

23. r() =

⁴  ²

⇒ r⁰() =

4³ 1 2. Then r⁰(1) = h4 1 2i, |r⁰(1)| =√

4²+ 1²+ 2²=√ 21, and

T(1) = r⁰(1)

|r⁰(1)| = 1

√21h4 1 2i =

 4

√21 1

√21 2

√21



r⁰⁰() =

12² 0 2 , so

r⁰() × r⁰⁰() =









i j k

4³ 1 2

12² 0 2









=





 1 2

0 2





i−







4³ 2

12² 2





j+





 4³ 1 12² 0





k

= (2 − 0) i − (8³− 24³) j + (0 − 12²) k =

2 16³ −12² 24. r() =

^2 ^−3 

⇒ r⁰() =

2^2 −3^−3 1. Then r⁰(0) = h2 −3 1i, |r⁰(0)| =

2²+ (−3)²+ 1²=√14, and

T(0) = r⁰(0)

|r⁰(0)| = 1

√14h2 −3 1i =

 2

√14 − 3

√14 1

√14



r⁰⁰() =

4^2 9^−3 0

⇒ r⁰⁰(0) = h4 9 0i. Then

r⁰(0) × r⁰⁰(0) =









i j k

2 −3 1

4 9 0









=







−3 1

9 0





i−





 2 1 4 0





j+





 2 −3

4 9





k

= (0 − 9) i − (0 − 4) j + [18 − (−12)] k = h−9 4 30i

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28. Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point.

x =p

t²+ 3, y = ln(t²+ 3), z = t; (2, ln 4, 1) Solution:

SECTION 13.2 DERIVATIVES AND INTEGRALS OF VECTOR FUNCTIONS ¤ 1289 25. 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 + .

26. 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).

27. 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 − .

28. 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 + .

29. 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.

30. 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).

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30. Find the point on the curve r(t) =< 2 cos t, 2 sin t, e^t >, 0 ≤ t ≤ π, where the tangent line is parallel to the plane

√

3x + y = 1.

Solution:

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() = 1 + 2√

 ³−  ³+ , so r⁰() = 1√

 3²− 1 3²+ 1. The point (3 0 2)corresponds to  = 1, so the tangent vector there is r⁰(1) = h1 2 4i. Thus, the tangent line goes through the point (3 0 2)and is parallel to the vector h1 2 4i. Parametric equations are  = 3 + ,  = 2,  = 2 + 4.

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).

° 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

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