Geometry in Space CHAPTER

Geometry in Space CHAPTER

7. P(2, 1, 6), Q(4, 7, 9), R(8, 5, –6)

11.1 Concepts Review

PQ = (2 – 4) 2 + (1 – 7) 2 + (6 – 9) 2 = 7

1. coordinates PR = (2 – 8) 2 +− (1 5) 2 ++ (6 6) 2 = 14

4. plane; 4; –6; 3 triangle formed by joining P, Q, and R is a right

Problem Set 11.1

triangle, since it satisfies the Pythagorean Theorem.

1. A(1, 2, 3), B(2, 0, 1), C(–2, 4, 5), D(0, 3, 0),

E (–1, –2, –3)

8. a. The distance to the xy-plane is 1 since the

point is 1 unit below the plane.

b. The distance is

(2 – 0) 2 + (3 – 3) 2 + (–1 – 0) 2 = 5 since the distance from a point to a line is the

length of the shortest segment joining the point and the line. Using the point (0, 3, 0) on the y-axis clearly minimizes the length.

c. (2 – 0) 2 +− (3 0) 2 + (–1 – 0) 2 = 14 ( )

A 3, – 3, 3 , (0, , – 3), –2, B π C ,2, D (0, 0, e)

9. Since the faces are parallel to the coordinate planes, the sides of the box are in the planes x = 2, y = 3, z = 4, x = 6, y = –1, and z = 0 and the vertices are at the points where 3 of these planes intersect. Thus, the vertices are (2, 3, 4), (2, 3, 0), (2, –1, 4), (2, –1, 0), (6, 3, 4), (6, 3, 0), (6, –1, 4), and (6, –1, 0)

3. x = 0 in the yz-plane. x = 0 and y = 0 on the z-axis.

4. y = 0 in the xz-plane. x = 0 and z = 0 on the y-axis.

5. a. (6 – 1) 2 + (–1 – 2) 2 + (0 – 3) 2 = 43

b. (–2 – 2) 2 + (–2 2) + 2 ++ (0 3) 2 = 5

c. ( e +π+π+ ) 2 ( 4) 2 + ( 0–3 ) ≈ 9.399 were parallel to the x-axis, the y-coordinate could not change, similarly for the z-axis.)

2 10. It is parallel to the y-axis; x = 2 and z = 3. (If it

6. P(4, 5, 3), Q(1, 7, 4), R(2, 4, 6)

2 2 11. a. ( –1) x 2 ( – 2) y 2 2 + + ( – 3) z 2 = 25 PQ = (4 1) − + (5 – 7) + (3 – 4) = 14

2 2 2 b. ( x + 2) + ( y + 3) 2 ++ ( z 6) 2 = PR 5 = (4 – 2) + (5 – 4) + (3 – 6) = 14

QR = (1 – 2) + (7 – 4) + (4 – 6) 2 = 14 c. (–) x π+ (–) y e + ( z –2 ) =π

Since the distances are equal, the triangle formed by joining P, Q, and R is equilateral.

Section 11.1 Instructor’s Resource Manual

12. Since the sphere is tangent to the xy-plane, the point (2, 4, 0) is on the surface of the sphere. Hence, the radius of

the sphere is 5 so the equation is ( – 2) x 2 + ( – 4) y 2 + ( – 5) z 2 = 25.

Center: (6, –7, 4); radius 10

14. ( x 2 + 2 x ++ 1) ( y 2 –6 y ++ 9) ( z 2 –10 z + 25) = –34 1 9 25 +++ ( x + 1) 2 + ( – 3) y 2 + ( – 5) z 2 = 1

Center: (–1, 3, 5); radius 1

x – ⎞++++ ( y 1) 2 ⎜ 17 ⎟ ( z 2) 2 =

Center: ⎜ , – 1, – 2 ; ⎟ radius

16. ( x 2 + 8 x +

+ 16) ( y 2 –4 y ++ 4) ( z 2 – 22 z + 121) = –77 16 4 121 + ++

( x + 4) 2 + ( – 2) y 2 + ( – 11) z 2 = 64

Center: (–4, 2, 11); radius 8

19. x-intercept: y = z = 0 ⇒ x=6 y -intercept: x = z = 0 ⇒ 6y = 12, y = 2

17. x-intercept: y = z = 0 ⇒ 2x = 12, x = 6

y -intercept: x = z = 0 ⇒ 3y = 6, y = 2 z -intercept: x = y = 0 ⇒ 3z = 12, z = 4

z -intercept: x = y = 0 ⇒ –z = 6, z = –6

20. x-intercept: y = z = 0 ⇒ –3x = 6, x = –2

y -intercept: x = z = 0 ⇒ 2y = 6, y = 3 y -intercept: y = z = 0 ⇒ –4y = 24, y = –6

18. x-intercept: y = z = 0 ⇒ 3x = 24, x = 8

z -intercept: x = y = 0 ⇒ z=6 z -intercept: x = y = 0 ⇒ 2z = 24, z = 12

Instructor’s Resource Manual

Section 11.1 669

21. x and y cannot both be zero, so the plane is For problems 25-36, L = b ⎛ dx ⎞ 2 ⎛ dy ⎞ 2 ⎛ dz ⎞ parallel to the z-axis. 2 ∫ a ⎜ dt ⎟ + ⎜ + ⎝ ⎠ ⎝ dt ⎠ ⎟ ⎜ ⎝ dt ⎠ ⎟ dt x -intercept: y = z = 0 ⇒ x=8

8 dx

y -intercept: x = z = 0 ⇒ 3y = 8, y = 25. = 1,

L = 2 ∫ 2 ++ 2 2 2 dt = 0 2 1 1 ∫ 0 6 dt =

L 3 = ⎛⎞ 1 2 ⎛⎞ 1 2 ⎛⎞ 1 2 ∫ 3

1 ⎜⎟ ⎝⎠ 4 + ⎜⎟ ⎝⎠ 3 + ⎜⎟ ⎝⎠ 2 dt =

22. x and z cannot both be zero, so the plane is

parallel to the y-axis. x -intercept: y = z = 0 ⇒ 3x = 12, x = 4

z -intercept: x = y = 0 ⇒ 4z = 12, z = 3

dt 2 dt

1 ⎝ 4 ⎠ ⎟ ++ 9 16 dt = ∫ 1 2 9 t + 100 dt u = =+ t 9 100

du = 9 dt

18 ∫ 109 u du =

23. This is a sphere with center (0, 0, 0) and radius 3.

⎜ ⎛ 9 2 4 ⎝ 4 ⎟⎜ ⎞⎛ ∫ 1 t ⎠⎝ + 9 4 t ⎟ ⎞ ⎠ + 1 dt = ∫ 2 2 18 t + 4 dt =

u = 18 t + 4 du = 18 dt

36 ∫ 40 u du =

8 L 8 = 2 ∫ 2 0 4 t ++ 4 t 1 dt = ∫ 0 (2 t + 1) dt =

24. This is a sphere with center (2, 0, 0) and radius 2.

1 4 t + 12 t + 9 dt = ∫ 1 (2 t + 3) dt =

4 (2 t + 3) dt = ⎡ t 2 ∫ 4

1 ⎣ + 3 t ⎤ ⎦ = −= 1 28 4 24

31. =− 2sin , t

− π 4sin t + 4 cos 2 t π + 9 dt = ∫ − π 13 dt =

Section 11.1 Instructor’s Resource Manual

© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of

32. =− 2sin , t

= 1 35. =− 2sin , t

0 4sin t + 4 cos 2 t +

L = ∫ 0 4sin t + cos 2 t + 1 dt =

0 40 1601 dt = ⎡ ⎣ 40 1601 t ⎤ =

∫ 0 3sin t + 2 ⎦ dt 0

π 1601 ≈ 25.14 By the Parabolic Rule (n = 10):

11 dt ∫ 1 2 + ⎜ ⎝ 4 t ⎟ ⎠ dt

2 4.3421 By the Parabolic Rule (n = 10):

0 cos t sin t + cos 2 t dt 9 = 5.5 1.4302 4 5.7208

By the Parabolic Rule (n = 10):

L 2 = 14 + t 2 + 9 t dt 4 0 0 1.4142

By the Parabolic Rule (n = 10):

37. The center of the sphere is the midpoint of the

diameter, so it is

2 2 2 2 ⎟ ⎝ The radius is ⎠⎝ ⎠

equation is ( x − 1) 2 + ( –1) y 2 + ⎜ z – = . ⎝

Instructor’s Resource Manual

Section 11.1 671

38. Since the spheres are tangent and have equal

41. a. Plane parallel to and two units above the

1 xy -plane

radii, the radius of each sphere is

of the

b. Plane perpendicular to the xy-plane whose distance between the centers.

trace in the xy-plane is the line x = y.

(–3 – 5) 2 ++ (1 3) 2 + (2 – 6) 2 = 2 6. The

c. Union of the yz-plane (x = 0) and the

spheres are ( x + 3) 2 + ( –1) y 2 + ( – 2) z 2 = 24 and

xz -plane (y = 0) ( – 5) x 2 + ( y + 3) 2 + ( – 6) z 2 = 24. d. Union of the three coordinate planes

e. Cylinder of radius 2, parallel to the z-axis plane. Since it is in the first octant, the center is

39. The center must be 6 units from each coordinate

(6, 6, 6). The equation is

f. Top half of the sphere with center (0, 0, 0) and radius 3

( x − 6) 2 + ( – 6) y 2 + ( – 6) z 2 = 36.

42. The points of the intersection satisfy both

40. x + y = 12 is parallel to the z-axis. The distance ( –1) x 2 + ( y + 2) 2 ++ ( z 1) 2 = 10 and z = 2, so from (1, 1, 4) to the plane x + y = 12 is the same

as the distance in the xy-plane of (1, 1, 0) to the ( –1) x 2 + ( y + 2) 2 ++ (2 1) 2 = 10 . This simplifies line x + y – 12 = 0. That distance is

to ( –1) x 2 + ( y + 2) + 2 = the equation of a circle 1,

= 5 2. The equation of the sphere is

of radius 1. The center is (1, –2, 2).

( –1) x 2 + ( –1) y 2 + ( – 4) z 2 = 50.

43. If P(x, y, z) denotes the moving point, ( –1) x 2 + ( – 2) y 2 ++ ( z 3) 2 = 2 ( –1) x 2 + ( – 2) y 2 + ( – 3) , z 2 which simplifies to ( –1) x 2 + ( – 2) y 2 + ( – 5) z 2 = 16, is a sphere with radius 4 and center (1, 2, 5).

44. If P(x, y, z) denotes the moving point, ( –1) x 2 + ( – 2) y 2 ++ ( z 3) 2 = ( – 2) x 2 + ( – 3) y 2 + ( – 2) , z 2

which simplifies to x + y + 5z = 3/2, a plane.

⎡ ⎛⎞ h ⎤

45. Note that the volume of a segment of height h in a hemisphere of radius r is π h 2 ⎢ r −⎜⎟ ⎥ . ⎣ ⎝⎠ 3 ⎦

The resulting solid is the union of two segments, one for each sphere. Since the two spheres have the same radius, each segment will have the same value for h. h is the radius minus half the distance between the centers of the two spheres.

46. As in Problem 45, the resulting solid is the union of two segments. Since the radii are not the same, the segments will have different heights. Let h 1 be the height of the segment from the first sphere and let h 2 be the height from

the second sphere. r 1 = is the radius of the first sphere and 2 r 2 = is the radius of the second sphere . 3

Solving for the equation of the plane containing the intersection of the spheres ( x − 1) 2 + ( y − 2) 2 +− ( z 1) 2 −= 4 0

and ( x − 2) 2 + ( y − 4) 2 +− ( z 3) 2 − = we get x + 2y + 2z – 9 = 0. 9 0,

The distance from (1, 2, 1) to the plane is , and the distance from (2, 4, 3) to the plane is .

Section 11.1 Instructor’s Resource Manual

47. Plots will vary. We first note that the sign of c will influence the vertical direction an object moves (along the helix) with increasing time; if c is negative the object will spiral downward, whereas if c is positive it will spiral upward. The smaller c is the “tighter” the spiral will be; that is the space between successive “coils” of the helix

decreases as c decreases.

48. Plots will vary. We first note that the sign of a will influence the rotational direction that an object moves (along the helix) with increasing time; if a is negative the object will rotate in a clockwise direction, whereas if a is positive rotation will be counterclockwise. The smaller a is the narrower the spiral will be; that is the circles

traced out will be of smaller radius as a decreases

11.2 Concepts Review

7. w = u cos 60 °+ v cos 60 °=+= 1

1. magnitude; direction

8. w = u cos 45 °+ v cos 45 °= + = 2

2. they have the same magnitude and direction.

9. u +=−+ v 1 3, 0 4 + =

3. the tail of u; the head of v

u −=−− v 1 3, 0 4 − =−− 4, 4

4. 3 u = ( 1) − 2 + (0) 2 = 11 = v = (3) 2 + (4) 2 = 25 = 5

Problem Set 11.2

−− 12 ( 2),12 2 − = 14,10 u = (12) 2 + (12) 2 = 288 12 2 = v = ( 2) − 2 + (2) 2 = 8 = 22

3. 12. u +=− v ( 0.2) ( 2.1), 0.8 1.3 +− + =−

2.3, 2.1 u −=− v ( 0.2) ( 2.1), 0.8 1.3 −− − =

13. u +=−+ v 1 3, 0 4, 0 0 + + =

5. w = 1 ( u + v 1 ) 1 = u + v

2 2 2 u −=−− v 1 3, 0 4, 0 0 − − =−− 4, 4, 0 u = ( 1) − 2 + (0) 2 + (0) 2 = 11 =

Instructor’s Resource Manual

Section 11.2 673

14. u += v +− 0 ( 3), 0 3, 0 1 + +=− 3,3,1 18. Let v be the resulting force. Let θ be the angle of v measured clockwise from south.

u −= v −− 0 ( 3), 0 3, 0 1 − −= −− 3, 3, 1 v cos θ= 60 cos 30 °+ 80 cos 60 °= 30 3 40 +

u = (0) 2 + (0) 2 + (0) 2 = 0 = 0

v = ( 3) − 2 + (3) 2 + (1) 2 = 19 ≈ 4.359

v sin θ= 80sin 60 – 60sin 30 ° °= 40 3 – 30

15. u +=+− v 1 ( 5), 0 0,1 0 + + =− 4, 0,1 = ( 10 4 3 – 3 )

u −=−− v 1 ( 5), 0 0,1 0 − − =

100 3 3 4 ( + ) + 100 4 3 – 3 ( ) v = − 2 + ( 5) 2 (0) + 2 (0) = 25 = 5 = 100(100) = 10,000

The resultant force has magnitude 100 lb in the

direction S 23.13° W.

v = (2.2) 2 + (1.3) 2 +− ( 0.9) 2

19. The force of 300 N parallel to the plane has

17. Let θ be the angle of w measured clockwise from magnitude 300 sin 30° = 150 N. Thus, a force of south.

150 N parallel to the plane will just keep the

w cos θ= u

cos 30 °+ v cos 45 °= 25 3 25 2 +

weight from sliding.

= 25 ( 3 + 2 ) 20. Let a be the magnitude of the rope that makes an angle of 27.34°. Let b be the magnitude of the w sin θ= v sin 45 – ° u sin 30 °= 25 2 – 25 rope that makes an angle of 39.22°.

= 1. a sin 27.34° = b sin 39.22°

25 ( 2 –1 ) 2. a cos 27.34° + b cos 39.22° = 258.5

w = w 2 cos 2 2 θ + w sin 2 θ

Solve 1 for b and substitute in 2.

sin 27.34 °

= 625 ( 3 + 2 ) + 625 ( 2 –1 ) sin 39.22 °

2 2 a cos 27.34 °+ a cos 39.22 °= 258.5

cos 27.34 °+ sin 27.34 cot 39.22 ° °

w = 625 8 2 2 2 6 ( − + ) = 25 8 2 2 2 6 − +

a sin 27.34 °

sin 39.22 °

≈ 79.34 The magnitudes of the forces exerted by the

w sin θ tan θ =

ropes making angles of 27.34° and 39.22° are

w cos θ

3 + 2 178.15 lb and 129.40 lb, respectively.

θ = tan ⎜ ⎜ ⎟ ⎟ = 7.5 ° 21. Let θ be the angle the plane makes from north,

measured clockwise.

w has magnitude 79.34 lb in the direction

425 sin θ = 45 sin 20°

S 7.5° W.

9 sin θ= sin 20 °

85 = –1 ⎛ 9 θ ⎞ sin ⎜ sin 20 °≈ ⎟ 2.08 °

Let x be the speed of airplane with respect to the ground. x = 45 cos 20° + 425 cos θ ≈ 467 The plane flies in the direction N 2.08° E, flying 467 mi/h with respect to the ground.

Section 11.2 Instructor’s Resource Manual

22. Let v be his velocity relative to the surface. Let θ

e. ab ( u ) = ( a bu bu 1 , 2 ) = ( a bu 1 ), ( a bu 2 ) =

be the angle that his velocity relative to the surface makes with south, measured clockwise.

v = 409 ≈ 20.22 au 1 + av au 1 , 2 + av 2 =

≈ 8.53 ° au 1 + bu au 1 , 2 + bu 2 =

20 His velocity has magnitude 20.22 mi/h in the

au au 1 , 2 + bu bu 1 , 2 = a u + b u

direction S 8.53° W.

h. 1 u =⋅ 1 u 1 ,1 ⋅ u 2 = uu 1 , 2 = u

23. Let x be the air speed. x cos 60° = 40

i. a u = au au 1 , 2 = ( au 1 ) 2 + ( au 2 ) 2 =

1 2 1 + u 2 ) = The air speed of the plane is 80 mi/hr

= 80 au 2 2 + au 2 2 = au 2 ( 2 2

cos 60 °

24. Let x be the air speed. Let θ be the angle that the

26. Let u = uuu 1 , 2 , 3 , v = vvv 1 , 2 , 3 , and plane makes with north measured counter-

clockwise.

w = www 1 , 2 , 3

x cos θ = 837 + 63 cos 11.5°

a. u += v u x sin θ = 63 sin 11.5° + 1 vu 1 , 2 + vu 2 , 3 + v 3 =

x 2 = x 2 cos 2 θ + x 2 sin 2 θ v 1 + uv 1 , 2 + uv 2 , 3 + u 3 = v+u

= (837 63cos11.5 ) + °+ 2 (63sin11.5 ) 2 °

x = 704,538 105, 462 cos11.5 + °≈ 898.82

2 + 837 63cos11.5 w 2 ), u 3 + ( v 3 + w 3 ) =

The plane should fly in the direction N 0.80° W at an air speed of 898.82 mi/h.

c. u += 0 u 1 + 0, u 2 + 0, u 3 + 0 = uuu 1 , 2 , 3 = u

u +=+ 0 0u by part a.

25. Let u = uu 1 , 2 , v = vv 1 , 2 , and w = ww 1 , 2

d. u (-u) + = uuu 1 , 2 , 3 +−− u 1 , u 2 , − u 3 =

e. ab ( u ) = ( a bu bu bu 1 , 2 , 3 ) =

u +=+ 0 0u by part a.

au au au 1 , 2 , 3 + bu bu bu 1 , 2 , 3 = a u + b u

d. u (-u) + = uu 1 , 2 +−− u 1 , u 2 =

Instructor’s Resource Manual

Section 11.2 675

27. Given triangle ABC, let D be the midpoint of AB π 2 a regular n-gon is π – . Thus the vectors

and E be the midpoint of BC. u = AB , v = BC ,

w = AC , z = DE (placed head to tail from v 1 to v n ) form a regular n -gon. From Problem 19, the sum of the vectors

is 0.

31. The components of the forces along the lines containing AP, BP, and CP are in equilibrium; that is,

u +v=w

W = W cos α + W cos β W = W cos β + W cos γ z = 1 u + 1 v = 1 ( u + v ) = 1 w W = W cos α + W cos γ

2 2 2 2 Thus, cos α + cos β = 1, cos β + cos γ = 1, and Thus, DE is parallel to AC.

cos α + cos γ = 1. Solving this system of

28. Given quadrilateral ABCD, let E be the midpoint 1 equations results in cos α = cos β = cos γ = . of AB, F the midpoint of BC, G the midpoint of

2 CD , and H the midpoint of AD.

Hence α = β = γ = 60°.

ABC and ACD are triangles. From Problem 17,

Therefore, α + β = α + γ = β + γ = 120°.

EF and HG are parallel to AC. Thus, EF is parallel to HG. By similar reasoning using

be the points where the weights are triangles ABD and BCD, EH is parallel to FG.

32. Let ABC ′′′ ,,

attached. The center of gravity is located

Therefore, EFGH is parallelogram.

AA ′ + BB ′ + CC ′

units below the plane of the

3 triangle. Then, using the hint, the system is in

equilibrium when AA ′ + BB ′ + CC ′ is

maximum. Hence, it is in equilibrium when AP + BP + CP is minimum, because the total length of the string is

29. Let P i

be the tail of v i . Then

AP + AA ′ + BP + BB ′ + CP + CC ′ .

v 1 + v 2 +…+ v n

33. The components of the forces along the lines

⎯⎯→ containing AP, BP, and CP are in equilibrium; = PP 12 + PP 23 ++ PP n 1 that is,

30. Consider the following figure of the circle.

3 cos α + 4 cos γ = 5. Solving this system of equations results in

cos α = , cos β = 0, cos γ = , from which it

4 3 follows that sin α = , sin β = 1, sin γ = . 5 5

n Therefore, cos( αβ + ) = – , cos( αγ + ) = 0,

The vectors have the same length. Consider the

following figure for adding vectors v i and v i + 1 . cos( 3 βγ + ) = –, so

Then β – . Note that the interior angle of

Section 11.2 Instructor’s Resource Manual

This problem can be modeled with three strings

e. aca ⋅= + 4 9[(0)( 2) ( 5)(3)] −+− =− 15 13 going through A, four strings through B, and five

strings through C, with equal weights attached to the twelve strings. Then the quantity to be

f. bb ⋅− b = (2)(2) ( 3)( 3) +−−− 49 +

minimized is 3 AP + 4 BP + 5 CP . = 13 − 13

34. Written response.

35. By symmetry, the tension on each wire will be

b. bc ⋅= (1)(0) (–1)(5) + = –5 the same; denote it by T where T can be the

tension vector along any of the wires. The

c. ( abc + ) ⋅= −⋅ 4, 2 0,5

chandelier exerts a force of 100 lbs. vertically

downward. Each wire exerts a vertical tension of (4)(0) (–2)(5) = –10

T sin 45 upward. Since a state of equilibrium

d. c 2 (3 ⋅ a + 4) b = 2 0, 5 ⋅ ( 9, – 3 + 4, – 4 )

The tension in each wire is approximately 35.36 lbs.

e. bba ⋅= + 1 1[(1)(3) ( 1)( 1)] +−− = 42

36. By symmetry, the tension on each wire will be

f. c −⋅= cc ( + 0 25 ) − [(0)(0) (5)(5)] +

the same; denote it by T where T can be the

tension vector along any of the wires. The

chandelier exerts a force of 100 lbs. vertically downward. Each wire exerts a vertical tension of

T sin 45 upward. Since a state of equilibrium

exists, 3 ⋅ T sin 45 = 100 or T =

The tension in each wire is approximately 47.14 lbs.

11.3 Concepts Review

4. ABC ,, ()

Problem Set 11.3

1. a. 2a – 4b = (–4i + 6j) + (–8i + 12j)

= –12i + 18j

b. ab ⋅= (–2)(2) (3)(–3) + = –13

c. abc ⋅+= ( ) (–2 i + j 3 ) (2 – 8 ) ⋅ i j = (–2)(2) + (3)(–8) = –28

d. (–2 a + 3)5 b ⋅ c = 5[(10 – 15 ) (–5 )] i j ⋅ j = 5[(10)(0) + (–15)(–5)] = 375

Instructor’s Resource Manual

Section 11.3 677

10 ( )( ) ab ( )( 10 2 10 ) 20

5. a. ab i = (1)(0) (2)(1) ( 1)(1) 1 + +− =

2 f. By Theorem A (5), bb i − b = 0

6. a. ac i = ( 2)( 2) ( 2)(2) (0)(1) −+ + = 0

Section 11.3 Instructor’s Resource Manual

2 2 2 9. The basic formulae are

(3)( 2) ( 3)( 2) (0)(0) +− + = 0 a. a =

f. By Theorem A (5), aa − a = 0 a

uv i

b. b = − 1, 1,1

7. The basic formula is cosθ =

cos α b = 1 = 1 ≈ 0.577, α b ≈ 54.74

cos γ = 3 c = , γ c ≈ 70.53

c 3 uv i

8. The basic formula is cos θ uv , =

10. The basic formulae are

cos α = 1 a = 3 ≈ 0.577, α a ≈ 54.74

≈ 0.577, γ a ≈ 54.74 cos −

cos γ

Instructor’s Resource Manual

Section 11.3 679

b = 2 16. If xyz ,, is perpendicular to both 1, -2, -3 and

b 1 2 -3, 2, 0 , then x – 2y – 3z = 0 and –3x + 2y = 0. cos α b =

, α b = 45

b 2 Solving these equations for y and z in terms of x

b 2 2 yields y = xz ,– = x . Thus, all the vectors cos β b =

, β b ≈ 135 2 3

b 0 have the form x , x ,– x where x is a real

cos γ b = 3 =

= 0, γ b = 90 2 3

b 2 number.

c. c =−− 2, 2,1

c = 3 17. A vector equivalent to BA is

c 3 A vector equivalent to BC is

cos β = 2 2 v c = 1 + 4, 0 - 5, 1 - 6 = 5, -5, -5 . =− , β c ≈ 131.81

c = , γ c ≈ 70.53

Therefore the vectors are orthogonal

18. A vector equivalent to BA is

u = 6 - 3, 3 - 1, 3 + 1 = 3, 2, 4 .

12. ab i = (1)(1) (1)( 1) (1)(0) + −+ = 0

A vector equivalent to BC is

ac i = (1)( 1) (1)( 1) (1)(2) −+ −+ = 0 v = -1 - 3, 10 - 1, -2.5 + 1 = -4, 9, -1.5 .

bc i = (1)( 1) ( 1)( 1) (0)(2) −+−−+ = 0

uv ⋅= 3(–4) 2 9 4 (–1.5) +⋅+⋅ = 0 so u is

Therefore the vectors are mutually orthogonal. perpendicular to v. Thus the angle at B is a right

13. ab i = (1)(1) ( 1)(1) (0)(0) +− + = 0 angle.

ac i = (1)(0) ( 1)(0) (0)(2) +− + = 0 19. c ,6

i c ,4 −=⇒ 0 c 2 − 24 =⇒ 0

bc i = (1)(0) (1)(0) (0)(2) + + = 0

c 2 = 24 ⇒=± c 26

Therefore the vectors are mutually orthogonal.

14. ( u +⋅ v )(–) u v =⋅ uu – uv ⋅+⋅ vu – vv ⋅ 20. (2 c i − j 8 ) (3 i i + c j ) =⇒ 0 6 c − 8 c =⇒ 0

−=⇒= 2 c 0 c 0

Thus, u 2 = v 2 or u = v

21. ( c i ++ jk ) (0 i i ++ 2 j d k ) =⇒ 0 0 c ++=⇒ 2 d 0

c is any number, d =− 2

15. If xi + yj + zk is perpendicular to –4i + 5j + k and 4i + j, then –4x + 5y + z = 0 and 4x + y = 0 since

22. a , 0,1 0, 2, i b =⇒= 0 b 0

the dot product of perpendicular vectors is 0.

a , 0,1 1, ,1 i c =⇒+= 0 a 10

Solving these equations yields y = –4x and z = 24x. Hence, for any x, xi – 4xj + 24xk is

0, 2, b i c 1, ,1 =⇒ 0 2 c += b 0

perpendicular to the given vectors.

Thus: a =− 1 and c == b 0

x i − 4 x j + 24 x k = x 2 + 16 x 2 + 576 x 2 For problems 23-34, the formula to use is

= x 593 ⎛

proj a b = ⎜ ba 10 i 2 ⎟ a

This length is 10 when x =±

. The vectors

proj v u =

Section 11.3 Instructor’s Resource Manual

proj i u =

⎛ uu ⋅ ⎞ proj u =

35. a. proj u = ⎜

2 u − + ⎜ ⎟ uu ⋅ proj ( u w − v ) = ( 1)(1) (6)(2)

b. proj − u u =

⎟ () −= u ⎜ ⎟ u =

2, 1 , w 1,5 36. a. proj u ()() −= u ⎜ ⎟

2 () − u ⎜ ⎟

proj j u =

⎟ () −= u u

proj i u = (1)(1) (2)(0)

b. −=

1, 0 proj − u ()()()() u ⎜ 2 ⎟ u

proj v u =

2, 0, 1 , − w = 1,5, 3 − v

proj u v =

+ 1 2 38. uv ⋅= 5–5 () + 5( 5) 2(1) + = 2

2, 0, 1 , − w = 1,5, 3 − v

proj u w =

40. cos (46 ) cos (108 ) cos 2 °+ 2 °+ 2 γ = 1 u =

w += v 3,5, 4 − ⇒ γ ≈ 49.49° or γ ≈ 130.51°

proj ( u w + v ) =

41. There are infinitely many such pairs. Note that

1, 2, 0 is perpendicular to − 4, 2,5 . For

2, 1, c is a candidate. –4, 2, 5 ⋅ –2, 1, c =++ 825 c

proj u

0, 0,1 8 + 2 + 5c = 0 ⇒ c = –2, so one pair is

1, 2, 0 ,v= − 2,1, 2 − .

Instructor’s Resource Manual

Section 11.3 681

42. The midpoint is

52. r = ka + mb ⇒ 6 = k(–4) + m(2) and

–7 = k(3) + m(–1)

vector is

. Solve the system of equations to get

22 k = –4, m = –5.

43. The following do not make sense.

53. a and b cannot both be zero. If a = 0, then the line ax + by = c is horizontal and n = bj is vertical, so

a. vw ⋅ is not a vector. n is perpendicular to the line. Use a similar argument if b = 0. If a ≠ 0 and 0 b ≠ , then

b. uw ⋅ is not a vector.

P 1 ⎜ ,0 ⎟ and P 2 ⎜ 0, ⎟ are points on the line.

44. The following do not make sense.

c. u is not a vector. n ⋅ PP 12 = ( a i + b j ) ⋅ ⎜ – i + j ⎟ = – c += c 0

d. u + v is not a scalar.

2 54. 2 u + v +− u v = ( u +⋅+ v )( u v )

45. a u + b u = au 1 , u 2 + buu 1 , 2 +−⋅− ( u v )( u v )[ =⋅+ uu 2( uv ⋅+⋅ ) vv ] = au 1 , au 2 + bu 1 , bu 2 = au 1 + bu au 1 , 2 + bu 2 +⋅− [ uu 2( uv ⋅+⋅ ) vv ] = 2( uu ⋅+ ) 2( vv ⋅ )

46. uv i = uv 11 + uv 22 = vu 11 + vu 22 = vu i −−⋅− ( u v )( u v )[ =⋅+ uu 2( uv ⋅+⋅ ) vv ]

47. c ( uv ⋅ ) = cu ( 1 , u 2 ⋅ vv , −⋅− [ uu 2( uv ⋅+⋅ ) vv ] = 4( uv ⋅ 1 ) 2 )

cuv ( 11 + uv 22 ) = cuv ( 11 ) + cuv ( 22 ) so uv ⋅= u + v − u − v

4 4 = ( cu v 11 ) + ( cu v 2 ) 2 = cu 1 , cu 2 ⋅ v 1 , v 2

= ( cu 1 , u 2 ) ⋅ vv 1 , 2 = ( c uv ) ⋅ 56. Place the cube so that its corners are at the points (0, 0, 0), (1, 0, 0), (0,1, 0), (1, 1, 0), (0, 0, 1),

(1, 0, 1), (0,1, 1), and (1, 1, 1). The main

48. uv ⋅+ ( w ) = u 1 , u 2 ⋅ ( vv 1 , 2 + ww 1 , 2 ) diagonals are (0, 0, 0) to (1, 1, 1), (1, 0, 0) to

= (0, 1, 1), (0, 1, 0) to (1, 0, 1), and (0, 0, 1) to

uu 1 ,, 2 ⋅ v 1 + wv 1 2 + w 2

(1, 1, 0). The corresponding vectors are

= uv 11 ( + w 1 ) + uv 2 ( 2 + w 2 ) 1, 1, 1 , − 1,1,1 , 1, 1,1 − , and 1,1, 1 −. = ( uv 11 + uv 22 )( + uw 11 + uw 22 ) =⋅+⋅ uv uw Because of symmetry, we need only address one situation; let’s choose the diagonal from

49. 0u i = 0 u 1 + 0 u 2 = 0 (0, 0, 0) to (1,1,1) and the face that lies in the xy − plane. A vector in the direction of the

50. uu i = u 2 2 2 1 2 + u 2 = u 1 + u 2 = u diagonal is d = 1,1,1 and a vector normal to the

chosen face is n =

0, 0,1 . The angle between

51. r = ka + mb ⇒ 7 = k(3) + m(–3) and the diagonal and the face is the complement of –8 = k(–2) + m(4)

the angle between d and n ; that is

Solve the system of equations to get

Section 11.3 Instructor’s Resource Manual

57. Place the box so that its corners are at the points

62. D = 12j

(0, 0, 0), (4, 0, 0), (0, 6, 0), (4, 6, 0), (0, 0, 10), Work = F· D = (–5)(0) + (8)(12) = 96 joules (4, 0, 10), (0, 6, 10), and (4, 6, 10). The main diagonals are (0, 0, 0) to (4, 6, 10), (4, 0, 0) to

63. D = (4 – 0)i + (4 – 0)j + (0– 8)k = 4 +4j – 8 k (0, 6, 10), (0, 6, 0) to (4, 0, 10), and (0, 0, 10) to

Thus, W =⋅= FD 0(4) 0(4) 4( 8) + −−= 32 joules. (4, 6, 0). The corresponding vectors are

4, 6, 10 , − 4, 6,10 , 4, 6,10 − , and

64. D = (9 – 2)i + (4 – 1)j + (6 – 3)k = 7i + 3j + 3k Thus, W − =⋅= FD 3(7) – 6(3) 7(3) + = 24 ft-lbs.

4, 6, 10 . All of these vectors have length

65. 2( x −− 1) 4( y −+ 2) 3( z += 3) 0

+ 16 36 100 + = 152. Thus, the smallest angle

2 x − 4 y + 3 z =− 15

θ between any pair, u and v of the diagonals is found from the largest value of ⋅ uv , since

There are six ways of pairing the four vectors.

67. ( x −+ 1) 4( y −+ 2) 4( z −= 1) 0 The largest value of ⋅ uv is 120 which occurs

x + 4 y + 4 z = 13

with u = 4, 6, 10 and v = − 4, 6,10 . Thus,

69. The planes are 2x – 4y + 3z = –15 and

3x – 2y – z = –4. The normals to the planes are (0, 0, 0), (1, 0, 0), (0,1, 0), (1, 1, 0), (0, 0, 1),

58. Place the box so that its corners are at the points

−− 3, 2, 1 . If θ is the angle (1, 0, 1), (0,1, 1), and (1, 1, 1). The main

− 2, 4,3 and v =

between the planes,

diagonals are (0, 0, 0) to (1, 1, 1), (1, 0, 0) to

, so (1, 1, 0). The corresponding vectors are

All of these vectors have length 1 1 1 ++= 3. 406

Thus, the angle θ between any pair, u and v of the diagonals is found from

70. An equation of the plane has the form

2x + 4y – z = D. And this equation must satisfy cos θ =

uv ⋅

. 2(–1) + 4(2) – (–3) = D, so D = 9. Thus an

3 equation of the plane is 2x + 4y – z = 9. There are six ways of pairing the four vectors,

but due to symmetry, there are only two cases we

71. a. Planes parallel to the xy-plane may be

need to consider. In these cases, uv ⋅= 1 or

expressed as z = D, so z = 2 is an equation of

1 the plane. uv ⋅=− 1 . Thus we get that cos θ = or

3 b. An equation of the plane is

2(x + 4) – 3(y + 1) – 4(z – 2) = 0 or cos

θ = − . Solving for θ gives 1

2x – 3y – 4z = –13.

70.53 or θ ≈ 109.47 .

72. a. Planes parallel to the xy-plane may be expressed as z = D, so z = 0 is an equation of

59. Work =⋅= FD (3 i + j 10 ) (10 ) ⋅ j the plane.

= 0 + 100 = 100 joules

b. An equation of the plane is

60. F = 100 sin 70°i – 100 cos 70°j x ++= y z D ; since the origin is in the

D = 30i

plane, 0 0 0 ++= D . Thus an equation is

Work = ⋅ FD x ++=. y z 0

= (100sin 70 )(30) ( 100 cos 70 )(0) ° +− ° = 3000 sin 70° ≈ 2819 joules

11 Work = F · D = (6)(5) + (8)(8) = 94 ft-lb

Instructor’s Resource Manual

Section 11.3 683

74. The distance is 0 since the point is in the plane.

83. Let x = xyz , , , so

( (–3)(2) 2(6) (3) – 9 + + = 0 ) ( xa −⋅− )( xb ) =

75. (0, 0, 9) is on –3x + 2y + z = 9. The distance from x − a 1 , y − a 2 , z − a 3 ⋅− xb 1 , y − b 2 , z − b 3

(0, 0, 9) to 6x – 4y – 2z = 19 is = x 2 − ( a + bx ) + ab + y 1 2 1 11 − ( a 2 + b 2 ) y + ab 22 6(0) – 4(0) – 2(9) –19

4.9443 is the

Setting this equal to 0 and completing the squares distance between the planes.

yields

a 3 + b ⎞ (1, 0, 0) to –5x + 3y + 2z = 7 is

76. (1, 0, 0) is on 5x – 3y – 2z = 5. The distance from a 2 b ⎛ 2

77. The equation of the sphere in standard form is ⎛ a + b a + b a + b ⎞

2 2 2 A sphere with center 1 1 , 2 2 , 3 ⎜ 3 ⎟ ( x + 1) + ( y + 3) + ( – 4) z = 26, so its center is ⎝ 2 2 2 ⎠

(–1, –3, 4) and radius is 26. The distance from

and radius

the sphere to the plane is the distance from the

1 ( a b 1 2 –) 1 + ( a 2 – b 2 ) 2 + ( a 3 – b ) 2 ⎤ = ab – 2 center to the plane minus the radius of the sphere

be any point on

so the sphere is tangent to the plane. Ax + By + Cz = D, so Ax 0 + By 0 + Cz 0 = D . The

distance between the planes is the distance from

0 to Ax + By + Cz = E, which is (2, 1, 1), is in the plane. Then

78. The line segment between the points is

Px ( 0 , y 0 , z )

perpendicular to the plane and its midpoint,

perpendicular to the plane. The equation of the plane is

85. If a, b, and c are the position vectors of the 8(x – 2) + 0(y – 1) – 6(z – 1) = 0 or 8x – 6z = 10.

vertices labeled A, B, and C, respectively, then the side BC is represented by the vector c – b.

79. uv ⋅= cos θ uv ≤ uv since cos θ ≤ 1. The position vector of the midpoint of BC is

b + (–) cb = ( bc + ). The segment from A to

80. u + v 2 = u 2

+ 2 uv ⋅+ v 2 2 2

≤ + 2 uv ⋅+ v 2 the midpoint of BC is ( bc + )–. a Thus, the

≤ 2 u 2 + 2 uv + v 2 = (

u + v ) position vector of P is

⎤ abc ++

Therefore, u + v ≤ u + v . a +

81. The 3 wires must offset the weight of the object,

If the vertices are (2, 6, 5), (4, –1, 2), and

thus (6, 1, 2), the corresponding position vectors are (3i + 4j + 15k) + (–8i – 2j + 10k) + (ai + bj + ck)

6, 1, 2 . The position = 0i + 0j + 30k

− 4, 1, 2 , and

vector of P is

Thus, 3 – 8 + a = 0, so a = 5;

4, 2, 3 . Thus P is (4, 2, 3).

82. Let v 1 , v 2 , … , v n represent the sides of the polygon connected tail to head in succession

around the polygon.

Then v 1 + v 2 +…+ v n = 0 since the polygon is

closed, so

Section 11.3 Instructor’s Resource Manual

86. Let A, B, C, and D be the vertices of the

Problem Set 11.4

tetrahedron with corresponding position vectors

a , b, c, and d. The vector representing the

segment from A to the centroid of the opposite

1. a. ab ×= –3 2 –2

face, triangle BCD, is ( bcd ++ )– a by

Problem 85. Similarly, the segments from B, C,

j + k and D to the opposite faces are

=(–8 + 4)i – (12 – 2)j + (–6 + 2)k

= –4i – 10j – 4k

( abc ++ )–, d respectively. If these segments

b. b + c = 6i + 5j – 8k, so

i j meet in one point which has a nice formulation as k some fraction of the way from a vertex to the

a ×+= ( bc ) –3 2 –2 centroid of the opposite face, then there is some

6 5 –8 number k, such that

= (–16 + 10)i – (24 + 12)j + (–15 – 12)k

k ⎢ ( abd ++ )– c =+ d k ( abc ++ )– d .

= –6i – 36j – 27k

3 c. abc ⋅+= ( ) –3(6) 2(5) – 2(–8) + = 8

Thus, a (1 – ) k = a , so k = . Hence the

segments joining the vertices to the centroids of

the opposite faces meet in a common point which

d. bc ×= –1 2 –4

of the way from a vertex to the

k corresponding centroid. With k

1 = (–8 + 12)i – (4 + 28)j + (–3 – 14)k above formulas simplify to ( abcd +++ ), the

= 4i – 32j – 17k

position vector of the point.

a ××= ( bc ) –3

87. After reflecting from the xy-plane, the ray has 4 –32 –17 direction ai + bj – ck. After reflecting from the

xz -plane, the ray now has direction ai – bj – ck.

j + k After reflecting from the yz-plane, the ray now

4 –17 4 –32 has direction –ai – bj – ck, the opposite of its

= (–34 – 64)i – (51 + 8)j + (96 – 8)k

original direction. ( a < 0 , b < 0 , c < 0 )

= –98i – 59j + 88k

11.4 Concepts Review

Instructor’s Resource Manual

Section 11.4 685

= (–3 + 4)i – (–3 + 4)j + (–12 + 12)k

=i–j=

1, - 1, 0 = (16 – 9)i – (–8 + 9)j + (–6 + 12)k =

− 7, 1, 6 is perpendicular to the plane.

are the unit vectors perpendicular to the plane. –1

k = (2 + 12)i – (–6 – 10)j + (–36 + 10)k

− 14, 11, 9 − is perpendicular to the plane.

are the unit vectors perpendicular to the plane. 2 3 1 3 = 12 i –

= (–8 – 6)i – (–4 + 6)j + (2 + 4)k = –14i – 2j + 6k

7. ab ×= –1 1 –3

is perpendicular to both a and b. All

perpendicular vectors will have the form

c (–14i – 2j + 6k) where c is a real number.

= (–4 + 6)i – (4 + 12)j + (–2 – 4)k

4 Area of parallelogram = ab ×

= (20 – 4)i – (–8 + 6)j + (4 – 15)k = 16i + 2j – 11k All vectors perpendicular to both a and b will have the form c(16i + 2j – 11k) where c is a real number.

Section 11.4 Instructor’s Resource Manual

1, 1, 1 are in the plane.

Area of parallelogram =× ab =

= (0 + 2)i – (–1 + 2)j + (–1 – 0)k = −− 2, 1, 1 = 146 The plane through (1, 3, 2) with normal

9. a =

2 – 3, 4 – 2, 6 – 1 = –1, 2, 5 and

−− 2, 1, 1 has equation

b = −− 1 3, 2 2,5 1 − − = − 4, 0, 4 are adjacent

2(x – 1) – 1(y – 3) – 1(z – 2) = 0 or 2x – y – z = –3.

sides of the triangle. The area of the triangle is half the area of the parallelogram with a and b as

− 0 1, 0 1,1 2 − − = −−− 1, 1, 1 and adjacent sides.

12. u =

v = −−−− 2 1, 3 1, 0 2 − = −−− 3, 4, 2 are in the

i jk

ab ×= –1 2 5 plane.

–1 –1 = (8 – 0)i – (–4 + 20)j + (0 + 8)k =

Area of triangle =

86 = (2 – 4)i – (2 – 3)j + (4 – 3)k = − 2,1,1

= 46 The plane through (0, 0, 1) with normal − 2,1,1 has equation –2(x – 0) + 1(y – 0) + 1(z – 1) = 0 or

13. u= − 0 7,3 0, 0 0 − − = − 7,3, 0 and sides of the triangle.

3, 3, 3 are adjacent

= (15 – 0)i – (–35 – 0)j + (0 + 21)k = 15,35, 21

Area of triangle =

ab × = 1 + 81 81 = 92 The plane through (7, 0, 0) with normal

15,35, 21 has equation 15(x – 7) +35(y – 0) + 21(z – 0) = 0 or 15x + 35y +21 z = 105.

Instructor’s Resource Manual

Section 11.4 687

14. u= 0 − ab , − 0, 0 0 − = − ab ,,0 and

19. (4i + 3j – k) × (2i – 5j + 6k) = 13i – 26j – 26k = 13(i – 2j – 2k)

v = 0 − a , 0 0, − c − 0 = − a , 0, c are in the

is normal to the plane. An equation of the plane is plane.

0, 0,1 is normal to the xy-plane and

− 3, 2,1 is normal to the plane

k 3 x − 2 y += z 4 . A normal vector to the required

plane must be perpendicular to both the other = bci – (–ac)j + abk =

, bc ac ab , normals; thus one possibility is:

The plane through (a, 0, 0) with normal

, bc ac ab , has equation bc(x – a) + ac(y – 0) +

kv ×= 0 0 1 = 2,3, 0

The plane has an

ab (z – 1) = 0 or bcx +ac y +abz = abc.

15. An equation of the plane is equation of the form: 2 x + 3 y = D . Since the

1(x –2) – 1(y – 5) +2(z – 1) = 0 or point (0, 0, 0) is in the plane, 2(0) 3(0) + = D ; x –y +2z = –1.

thus D = 0 and an equation for the plane is

16. An equation of the plane is 1(x –0) +1(y – 0) +1(z – 2) = 0 or

21. Each vector normal to the plane we seek is x + y + z = 2.

parallel to the line of intersection of the given planes. Also, the cross product of vectors normal

17. The plane’s normals will be perpendicular to the normals of the other two planes. Then a normal is

to the given planes will be parallel to both planes, − hence parallel to the line of intersection. Thus,

− 4, 3, 2 × 3, 2, 1 − = − 1,10,17 is normal to of the plane is 7(x + 1) + 5(y + 2) + 4(z – 3) = 0 or 7x + 5y + 4z = –5.

7, 5, 4 . An equation

the plane we seek. An equation of the plane is –1(x – 6) + 10(y – 2) + 17(z + 1) = 0 or

18. u = 1,1,1 is normal to the plane x ++= y z 2 –x + 10y + 17z = –3.

and v = −− 1, 1, 1 is normal to the plane

22. a × b is perpendicular to the plane containing a

x −−= y z

4 . A normal vector to the required

and b, hence a

b is perpendicular to both a and

b hence it plane must be perpendicular to both the other

b . (a × b) ×

c is perpendicular to a ×

is parallel to the plane containing a and b. normals; thus one possibility is

uv ×= 1 1 1 =

The plane has an

equation of the form: 2 y − 2 z = D . Since the

point (2, 1, 4) − is in the plane, 2( 1) 2(4) −− = D ;

thus D = - 10 and an equation for the plane is

2 y − 2 z =− 10 or y −=− z 5.

23. Volume =

24. Volume = (3 – 4 i j + k 2 ) [(– ⋅ i ++ 2 jk ) (3 – 2 × i j + 5) k = (3 – 4 i j + k 2 ) (12 ⋅ i + 8–4) j k = –4 = 4

25. a. Volume =⋅× uvw ( ) =

b. Area =× uv = 3, − 5, 1 =

Section 11.4 Instructor’s Resource Manual

26. From Theorem C, ab×c ⋅ ( ) = ( a×bc ) ⋅ , which leads to ab×c ⋅ ( ) =⋅ ca×b ( ) . Again from Theorem C, cab ⋅× ( ) = ( cab ×⋅=− ) ( a×cb ) ⋅ , which leads to ca×b ⋅ ( ) =⋅ ba×c ( ) . Therefore, we have that ab×c ⋅ ( ) =⋅ ba×c ( ) =⋅ ca×b ( ).

27. Choice (c) does not make sense because ( ab ⋅ ) is a scalar and can't be crossed with a vector. Choice (d) does not make sense because ( a×b ) is a vector and can't be added to a constant.

28. a × b and c × d will both be perpendicular to the common plane. Hence a ×

b and c × d are parallel so

(a × b ) × (c ×

d) = 0.

29. Let b and c determine the (triangular) base of the tetrahedron. Then the area of the base is

bc × which is half of

the area of the parallelogram determined by b and c. Thus,

(area of base)(height) =⎢ (area of corresponding parallelogram)(height)

(area of corresponding parallelpiped) = abc ⋅× ( )

Volume = abc ⋅× ( ) =

31. Let u = u 1 , u 2 , u 3 and v = v 1 , v 2 , v 3 then uv ×= uv 23 – uv 32 , uv 31 – uv 13 , uv 12 – uv 21

uv × 2 = uv 2 2 − 2 uv + uv 2 2 2 3 2323 3 2 + uv 22 31 − 2 uv 1313 + uv 2 2 2 1 2 3 + uv 1 2 − 2 uv

1212 + uv 2 1

= u 2 1 ( v 2 + v 2 + v 2 )– uv 22 + u 2 ( v 2 + v 2 2 2 2 2 2 2 1 2 2 3 11 2 1 2 + v 3 )– uv 2 2 + u 3 ( v 1 + v 2 + v 3 ) – uv 2 3 2 3 –2 uv 2323 –2 uv 1313 –2 uv 1212

= ( u 2 + u 1 2 2 + u 2 )( v 2 v 2 v 3 2 1 + 2 + 3 )–( uv 22 + uv 2 2 2 11 2 2 2 + uv 3 3 + 2 uv 2323 + 2 uv 1313 + 2 uv 1212 )

33. (v + w) × u = –[u × (v + w)] = –[(u × v) + (u × w)] = –(u × v) – (u × w) = (v × u ) + (w × u)

34. u × v=0 ⇒ u and v are parallel. uv ⋅=⇒ 0 u and v are perpendicular. Thus, either u or v is 0.

Instructor’s Resource Manual

Section 11.4 689

35. PQ = –,,0, ab PR = – , 0, a c , = 39. The area of the triangle is 1 A ab × . Thus, The area of the triangle is half the area of the

parallelogram with PQ and PR as adjacent

A 2 = 1 ab × 2 = 1 2 a 2 b −⋅ ( ab ) 2

sides, so area

36. The area of the triangle is = 1 (2 ab 22 – a 4 + 2 ac 22 – b 4 + 2 bc 22 – c 4 ). 1

x 2 − x 1 , y 2 − y 1 ,0 × x 3 − x 1 , y 3 − y 1 ,0 =

Note that s – a = ( b + c – ), a

ss ( − asbs )( − )( − c )

which is half of the absolute value of the

1 = ( a ++ b cb )( +− c aa )( +− cba )( +− b c ) determinant given. (Expand the determinant

along the third column to see the equality.) = 1 (2 ab 22 – a 4 + 2 ac 22 – b 4 + 2 bc 22 – c 4 )

37. From Problem 35, D 2 = ( bc 22 + ac 22 + ab 22 ) which is the same as was obtained above.

38. Note that the area of the face determined by a

= ( uv 11 )(0) ( + uv 12 )( ) ( k + uv 13 )( ) −+ j

1 ( uv 21 )( −+ k )( uv

Label the tetrahedron so that m = ( ab × ), 22 )(0) ( + uv 23 )( ) i + 2 ( uv 31 )( ) ( j + uv 32 )( ) ( −+ i uv 33 )(0)

n = ( bc × ), and p = ( ca × ) point outward.

= ( uv 23 – uv 32 ) i + ( uv 31 – uv 13 ) j + ( uv 12 – uv 21 ) k

The fourth face is determined by a – c and b – c,

mnpq +++= [( ab ×+×+× )( bc )( ca )

−×−×−× ( ab )( bc )( ca )] = 0

Section 11.4 Instructor’s Resource Manual

7. lim ln( ), ln , 3 + t t 2 tt does not exist because

11.5 Concepts Review

lim ln( ) t =∞ –.

1. a vector-valued function of a real variable

2. f and g are continuous at c; ft ′ () i + gt ′ () j

8. lim − 1/ t 2 − t e ,, t

3. position

4. r ′ ( ); ( ); t r ′′ t tangent; concave = lim e − 1/ t 2 , lim t , lim t = − 0, 1, 0

Problem Set 11.5

9. a. The domain of ft () =

1. lim[2 – t i t j ] lim(2 ) – lim( ) = t i t j = 2– i j t → 1 t → 1 t → 1 ( −∞ , 4) ∪ (4, ) ∞ . The domain of

t –4

gt () = 3– t is (– ∞ , 3]. The domain of

2. lim[2( t − 3) 2 i − 7 t 3 j ]

ht () = ln 4 − t is ( −∞ , 4) ∪ (4, ) ∞ . Thus, the

= lim[2( t − 2 3) ] i − lim(7 ) t 3 j =− 189 j

domain of r is (– ∞ , 3] or { t ∈ : t ≤ 3}.

b. The domain of ft () = t 2 is ( −∞ ∞ ,) . The

t − 1 t +− 2 t 3

3. lim ⎢

domain of gt () =

20 – t is (– ∞ , 20]. The

1 ( t − 1 )( t + 3 ) ⎞

domain of ht () = 3 is ( −∞ ∞ ,) . Thus, the

= lim ⎜

t → 1 ⎜ ⎝ ( t − 1 )( t + 1 ⎟ t → 1 t − 1 domain of r is (– ∞ , 20] or { t ∈ : t ≤ 20}. ) ⎠ ⎝ ⎠

⎟ i − lim ⎜

c. The domain of ft () = cos t is ( −∞ ∞ ,) . The

lim

i − lim( t + 3) j = i − 4 j

⎝ t + ⎟ 1 ⎠ t → 1 2 domain of gt () = sin t is also ( −∞ ∞ ,) . The

domain of ht () = 9 − 2 2 − t 10 t 28 7 t t is [ 3,3 ] . Thus,

4. lim ⎢

the domain of r is [ − 3,3 ] , or

10. a. The domain of f (t) = ln(t – 1) is (1, ∞ ). The

20 – t is (– ∞ , 20]. Thus, = lim (2 t − 14) i −

56 56 domain of gt () =

j =− 18 i −

t →− 2 5 5 the domain of r is (1, 20] or

{ t ∈ :1 <≤ t 20}. ⎡ sin cos t t 7 t 3 t ⎤

t → –1

5. lim ⎢

b. The domain of ft () ln( t ) is (0, ) ).

The domain of gt () = tan t is (– ∞ , ⎞ ∞ ). =

The domain of ht () = t is ( −∞ ∞ ,) . Thus,

the domain of r is (0, ∞ ) or { t ∈ : t > 0}.

⎛ t ⎞ + lim ⎜

c. The domain of gt ( ) 1/ 1 = − t 2 is ( 1,1) − .

⎡ t sin t

7 t 3 sin t ⎤

The domain of ht ( ) 1/ 9 t is ( − 3,3 ) .

6. lim ⎢ 2 i − 3 j −

t →∞ ⎢ ⎣ t

(The function f is fx () = 0 which has

⎛ t sin t ⎞

domain ( −∞ ∞ ,) .) Thus, domain of r is

Instructor’s Resource Manual

Section 11.5 691

11. a.

c. gt ( ) 1/ 1 = − t 2 is continuous on ( 1,1) − .

t –4 −∞

ht ( ) 1/ 9 = ( − (4, ) . gt () 3– t is t 2 , 4) is continuous on ( − 3,3 ) . (The

continuous on (– ∞ , 3]. ht () = ln 4 − t function f is fx () = 0 which is continuous on

is continuous on ( −∞ , 4)

.) Thus, r is continuous on ( 1,1 ) .

(4, ) ∞ . Thus, r is continuous on

(– ∞ , 3] or { t ∈ : t ≤ 3}. 13. a. D t r () t = 9(3 t + 4) 2 i + 2 te t j + 0 k

D t 2 r () t = 54(3 t + 4) i + 2(2 t 2 + 1) e t j

b. ft () = t 2 is continuous on

b. D t r () t = sin 2 t i − 3sin 3 t j + 2 t k (

1, – n )( ∪ n , n + 1 ) where

D t 2 r () t = 2 cos 2 t i − 9 cos 3 t j + 2 k

n is a non-negative integer. gt () =

( −∞ , 20 ) or { t ∈ : t < 20}. ht () = 3 ( )

20 – t is continuous on

2– t 2 – t is continuous on 2 ( −∞ ∞ ,) . Thus, r is r ′′ () t = e + 4 te –2 e i + (ln 2) 2 2 t j

continuous on

( – n + 1, – n )( ∪ k , k + 1 ) where

b. r () t = 2sec 2 2 t i +

n and k are non-negative integers and

k < 400 or

t 2 20, not an integer}. r ′′ () t = 8 tan 2 sec 2 – t 2 t i

(1 + t 22 )

c. ft () = cos t and ft () = sin t are

15. r ′ () t = – e – i – j ; r ''( ) t = e − t i + 2 j ( )

continuous on −∞ ∞ , .

ht () = 9 − t 2 is continuous on

r () t ⋅

r ''( ) t = e − 2 t − 2 ln t 2

[ − . Thus, r is continuous on

D t [() r t ⋅ r ''( )] t =− 2 e − 2 t − ⎢ 2 ⋅ 2 ⋅− 2 t 3 ln () t 2 ⎥

12. a. f(t) = ln(t – 1) is continuous on

20 – t is continuous on (– ∞ ,

20). Thus, r is continuous on (1, 20) or

{ t ∈ :1 << t 20}. 16. r ′ () t = 3cos 3 t i + 3sin 3 t j r () t ⋅ r '( ) t = 0

b. ft () = ln( t –1 ) is continuous on

D t [() r t ⋅ r '( )] 0 t =

( 0, ∞ . gt () = tan –1 ) t is continuous on

17. ht ()() r t =

e –3 ∞ t ∞ e –3 t –1 i + ln(2 ) t 2 j

ht = t

is continuous on

( −∞ ∞ ,) .Thus, r is continuous on (0,

+ e –3 t ⎛ 2 2 ⎜ ⎞ – 3ln(2 ) t j

18. h(t)r(t) = ln(3t – 2)sin 2ti + ln(3t – 2) coshtj

t [ ( ) ( )] r t = ⎢ 2 ln(3 – 2) cos 2 t t +

3sin 2 t

Dht

3–2 t ⎦ ⎥ i

3cosh t ⎤

+ ⎢ ln(3 – 2) sinh t t +

3–2 t ⎥ j ⎦

Section 11.5 Instructor’s Resource Manual

19. v () t = r ′ () t = 4 i + 10 t j + 2 k 25. v () t =− sin t i + cos t jk +

a () t = r ′′ ( ) 10 t = j a () t =− cos t i − sin t j

v (1) = 4i + 10j + 2k; a(1) = 10j; v ( π ) = –j + k; a( π ) = i; s () π= 11 += 2 ≈ 1.414 s (1) =

26. v(t) = 2 cos 2t i – 3 sin 3t j – 4 sin 4t k

(0) = i – 2j + 27k; a(0) = 2j – 18k;

3 e t j – 4sin 4 t k

+ t 2 a () t =

2 26 2sec t tan t i + 3 e t j – 16 cos 4 t k

4 9 s ⎛⎞=+ 49 e π ⎜⎟ /2 ≈ 6.877 1 26 ⎝⎠ 4

a () t = 30 t 4 i + 72(66 t 2 – 5)(6 t 2 – 5) 4 j

v (1) =+ 6 i 72 jka + ; (1) = 30 i + 4392 ; j = e 4 s 2 (2) +

+ (cos t

≈ 4.035 v (2) 2 i j – e ka ; (2) 2 i –2 j e k ;

s (2) = 4 π++ 2 1 e –4 ≈ 6.364

≈ 4.035 a () 1 t = – i 2 2 3 – j – k

v 31. If 2 v C , then =⋅= vv C Differentiate implicitly to get D t ( vv ⋅ ) = 2 vv ⋅= ′ 0. Thus,

vv ⋅=⋅= ′ va 0, so a is perpendicular to v.

Instructor’s Resource Manual

Section 11.5 693

32. If r () t = C , similar to Problem 31,

39. f ′ () u = – sin u i + 3 e 3 u j ; () gt ′ = 6 t

r () t ⋅ r ′ () t = 0. Conversely, if

F ′ () t = f ′ ( ( )) ( ) gtgt ′

r () t ⋅ r ′ () t = 0, then 2() r t ⋅ r ′ () t = 0. But

= –6 sin(3 t t 2 18 t 2 – 4) i + te 9 –12 j

since 2() r t ⋅ r ′ () t = D t [ ( ) ( )], r t ⋅ r t this

means that 2 r ()() t ⋅ r t = r () t is a constant,

40. f ′ () u = 2 u i + sin 2 ; ( ) ugt j ′ = sec 2 t

so r () t is constant.

F ′ () t = f ′ ( ( )) ( ) gtgt ′

2 2 = 2 tan sec t t i + sec t sin(2 tan ) t j

0 cos t + (– sin ) t dt 41. 1 t

2 2 2 2 ∫ 0 ( e i + e − j ) dt = ⎡ ⎣ e i − e − j ⎤ ⎦

= ∫ + 1 cos +

0 t sin tdt = ∫ 0 11 + dt

=− ( e 1) i +− (1 e − 1 ) j

= 2 ∫ dt = 2(2 – 0) =

− 1 [(1 t ) 3/2 i +− (1 t ) 3/2 j ] 2 dt

34. s = ∫ (cos – sin ) t t t 2 0 + (sin t + t cos ) t 2 + 2 dt 1

3 3 = 43. r(t) = 5 cos 6ti + 5 sin 6tj

7 + ln 2 ( + ) 7 – ln 3 ≈ 4.126

2 2 v (t) = –30 sin 6ti + 30 cos 6tj

Use Formula 44 with u = t and a = 3 for

v () t = 900sin 6 2 t + 900 cos 6 2 t = 30

t 2 + 3. dt a ∫ (t) = –180 cos 6ti – 180 sin 6tj

44. a. r ′

() t = cos – sin t i t j + (2 – 3) t k

35. s = ∫ 24 t 2 + 4 t 4 + 36 3 dt 3

2t – 3 < 0 for t < , so the particle moves

= ∫ t + + 9 dt

3 downward for 0 ≤< t .

= ∫ 2( t + 3) dt = 2 ⎢ + 3 t

4 t 2 + 36 t 4 + 324 t dt 4 t ∫ 2 4 –12 t + 10 = has no real-number solutions, so 0

0 the particle never has speed 0, i.e., it never stops

= moving. ∫

135 – 3 –10 t =+ ( t 2)( – 5) t = t = –2, 5. Since t 0, ≥

(91 3/2 – 1) ≈ 6.423 t 2

0, the particle is 12 meters above the ground when

37. s = ∫ 9 t 4 + 36 t 4 + 324 t dt 4

t = 5.

1 1 d. v(5) = cos 5i – sin 5j + 7k

343 t 12 + 98 ∫ 12 t + 1764 t 12 dt

t dt 6 ∫ 1 = ⎡ ⎣ 35 t 7 ⎤ = 35 ⎦ ≈ 6.708

Section 11.5 Instructor’s Resource Manual

45. a. The motion of the planet with respect

10t = π + t or t = .

to the sun can be given by

x = R p cos , t y = R p sin . t Thus, when the radius of the planet’s orbit around

Assume that when t = 0, both the

the sun is ten times the radius of the moon’s orbit

planet and the moon are on the x-axis.

around the planet and t = , the moon is

Since the moon orbits the planet 10

times for every time the planet orbits

motionless with respect to the sun.

the sun, the motion of the moon with respect to the planet can be given by

46. a. Years

x = R m cos10 , t y = R m sin10 . t

Combining these equations, the motion

of the moon with respect to the sun is given by x = R p cos t + R m cos10 , t

y = R p sin t + R m sin10 . t

The moon’s orbit is almost indistinguishable from

a circle.

c. The sun orbits the earth once each year while the moon orbits the earth roughly 13 times each year.

c. xt ′ () = – R p sin t − 10 R m sin10 t d. r (0) = 93.24 i = 93.24 million mi, the sum of

yt ′ () = R p cos t + 10 R m cos10 t the orbital radii.

The moon is motionless with respect to

e. 93 – 0.24 = 92.76 million mi

the sun when ( ) xt ′ and ( ) yt ′ are both

0. f. No; since the moon orbits the earth 13 times for Solve ( ) xt ′ = 0 for sin t and ( ) yt ′ = 0 each time the earth orbits the sun, the moon could not be stationary with respect to the sun unless the

for cos t to get sin t = –

10 R

m sin10 , t 1

radius of its orbit around the earth were th the

radius of the earth’s orbit around the sun.

cos10 t .

g. v(t) = [–186π sin (2 π t) – 6.24 π sin(26 π t)]i + Since sin 2 t + cos 2 t = 1 [186 π cos (2π t) + 6.24 π cos (26 π t)]j

100 R 2 m 2 100 R 2 m

2 a ( ) [ 372 t =− π 2 cos(2 ) 162.24 π− t π 2 cos(26 )] π t 1 i = sin 10 t + cos 10 t

+− [ 372 π 2 sin(2 ) 162.24 π− t π 2 sin(26 )] π t j

100 R 2

m . Thus, R 2 1

v ⎛⎞=+ ⎜⎟ 0 i (–186 – 6.24 ) π π= j –192.24 π j

2 p = 100 R 2 or

s ⎜⎟ ⎛⎞= 192.24 π million mi/yr

R p = 10 R m . Substitute this into

xt ′ () = 0 and ( ) yt ′ = 0 to get

a ⎛⎞= ⎜⎟ (372 π+ 162.24 π ) i + 0 j = 534.24 π 2 i

– R (sin t + sin10 ) t = 0 and

R p (cos t + cos10 ) t = 0. π

If 0 ≤≤ t , then to have sin t + sin

2 10t = 0 and cos t + cos 10t = 0 it must

be that

Instructor’s Resource Manual

Section 11.5 695

47. a. Winding upward around the right circular cylinder x = sin t, y = cos t as t increases.

b. Same as part a, but winding faster/slower by a factor of 3t 2 .

c. With standard orientation of the axes, the motion is winding to the right around the right circular cylinder x = sin t, z = cos t.

d. Spiraling upward, with increasing radius, along the spiral x = t sin t, y = t cos t.

48. For this problem, keep in mind that r, θ , u 1 , and u 2 are all functions of t and that prime indicates differentiation with respect to t .

a. u 1 = cos θ i + sin θ j and u 2 =− sin θ i + cos θ j . Applying the Chain Rule to u 1 gives

u ' 1 =− ( sin θθ ) ' i + ( cos θθ ) ' j

= θ ' ( − sin θ i + cos θ j ) = θ ' u 2

Similarly, applying the Chain Rule to u 2 gives

u ' 2 =− ( cos θθ ) ' i +− ( sin θθ ) ' j

=− θ ' cos ( θ i + sin θ j ) =− θ ' u 1

b. v () t = r '( ) t = Dr

t () u 1 = r u 1 + r ' u 1

a () t = v '( ) t = Dr t ( ' u 1 + r θ ' u 2 )

= r ' u ' 1 + r '' u 1 + r θ ' u ' 2 + r θ '' u 2 + θ '' r u 2

' u 1 ( + ) ( 2'' r θ + r θ '' ) u 2

= 2 r '' − r () θ

c. The only force acting on the planet is the gravitational attraction of the sun, which is a force directed along the line from the sun to the planet. Thus, by Newton’s Second Law,

m a ==− F c u 1 + 0 u 2

From Newton’s Law of Gravitation GMm

so from part (b) GMm

m a =− 2 u 1

( GM

r '' − r ( ') θ 2 ) u 1 + ( 2'' r θ + r θ '' ) u 2 ==− a 2 u 1

r Equating the coefficients of the vectors u 1 and u 2 gives

r '' − r () θ ' =−

2 GM

2'' r θθ r '' 0 =

d. rr × ' is a constant vector by Example 8. Call it D =× rr ' . Thus,

D =×=× rr ' r ( r 11 u + r θ ' u 2 )

= r 1 ru × 1 + r θ ' ru × 2

=+ 0 ( )( ) r θ ' r u 1 × u 2 = r 2 θ ' ( u 1 × u 2 )

Section 11.5 Instructor’s Resource Manual

θ '(0) = v 0 / r 0 . Substituting these into the expression from part (d) gives

r (0) ) θ '(0) k = D

r (0) 20 ) k = D

D = rv 00 k

Since D is a constant vector (i.e., constant for all t), we conclude that

r 2 θ ' k == D rv 00 k

r 2 θ ' = rv 00 for all t.

f. Let q = r ' . From (c)

g. Integrating the result from (f) gives

⎛ rv 22

∫ 3 − ⎟ dr dr ⎜

2 r C r 1 When t = , '(0) 0 r = q (0) = since the rate of change of distance from the origin is 0 at the perihelion. Also, 0

when t =, 0 r (0) =. r 0 Thus

Instructor’s Resource Manual

Section 11.5 697

=− 1 h. Let p 1/ r . Then r 1/ p and r '

2 p ' p . Thus,

= p () '

() r =− ' ' ⎜ = ⎜ ⎟

Dividing both sides of the equation from (g) by () r 2 θ ' and using the result from (e) that r 2 θ ' = rv 00 gives

q 2 2 GM ⎛ 1 1 ⎞

()() r

r 2 θ ' ⎝ r () ⎠

q 2 GM ⎛

00 ) ⎝ r () ⎠

r 2 00 ' r ( rv 2 θ ) ⎝ r 0 ⎠ rv

q 2 2 GM ⎛ 1 1 ⎞ ⎛ 1 1 ⎞

r r ⎟⎜ 2 2 () ⎟

Now substitute the result from above to get

4 () p p ' 2 GM

22 ( p − p 0 ) + ( p

() θ ' 4 vr 00

p vr 2 22

00 ⎛ dp

⎟ ⎞= 2 GM p ( − p 0 ) + vr 00 ( p 0 − p )

() θ ' ⎝ dt ⎠

vr 22 00 ⎛ dp 2 ⎞

⎟ = 2 GM p ( − p

() θ ' ⎝ dt ⎠

i. Continuing with the equation from (h), and using the Chain Rule gives

22 ⎛ dp dt / ⎞

⎟ = 2 GM p ( − p 0 ) + v 0 ⎜ 1 − ⎟

j. Taking the square root of both sides from part (i) gives 2 2 dp 2 ⎛ GMp ⎞ ⎛ GMp 2

0 0 =± ⎞ ⎜ p

00 / r > 0 . Recall that the planet is at its perihelion at time t = 0 , so this is as close as it gets to the sun. Thus, for t near 0, the distance from the sun r must increase with t .

From (e) we have r 2 θ ' = rv 00 , so θ ' = rv

1 dp dp dt / − pr 2 ' Thus,

dr dt / r '

> 0 and from the beginning of (h) r ' =− 2 p ' . Thus,

d θ d θ / dt d θ / dt The minus sign is the correct sign to take.

d θ d θ / dt θ '

Section 11.5 Instructor’s Resource Manual

0 − GMp 0 / v 0 )( − p − GMp 0 / v 0 ) dp = ∫ d θ

⎝ p 0 − GMp 0 / v ⎟ 0 ⎠ The initial condition for this differential equation is p = p 0 when t = 0 . Thus,

The solution is therefore θ = cos − 1 ⎜

0 0 ⎜ p − GMp 2 / v 2

l. Finally (!)

) ( cos θ )

Solving this for p gives p = p

− GMp 0 / v 0

+ GMp 0 / v 0

Recall that p = 1/ r , so that

) ) + GMp 0 / r v 0

p 0 − GMp 2 0 / v 2 0 ( cos θ

GMp 2 2 2 0 2 / v 0 + p 0 − GMp 0 / v 0 ( cos θ ) GM ( ⎛ ) GM ⎞

rv 2 where e = 00 − is the eccentricity. This is the polar equation of an ellipse. 1 GM

Instructor’s Resource Manual

Section 11.5 699

11.6 Concepts Review

9. Set z = . Solving 4 0 x + 3 y = and 1

10 x + 6 y = 10 yields x = 4, y = − . Thus 5

1. 1 + 4t; –3 – 2t; 2 – t P 1 (4, 5, 0) − is on the line. Set y = . Solving 0

x –1 y + 3 z –2

4 x − 7 z = and 10 1 x − 5 z = 10 yields

= 13 3 ⎛ 13 3 x ⎞ , z = . Thus P 2 ⎜ , 0, ⎟ is on the line.

direction vector. Thus,

3 − 27, 50, 6 −=− 10P P 12 is also a direction

Problem Set 11.6

vector. The symmetric equations are thus

1. A parallel vector is

10. With x = 0, y – z = 2 and –2y + z = 3 yield (0, –5, –7).

2. A parallel vector is

With y = 0, x – z = 2 and 3x + z = 3 yield

v = 7 2, 2 + 1, 3 + 5 − − = 5, 1, − 8 ⎛ 5 3 ⎜ ⎞ , 0, – ⎟ .

x = 2 + 5t, y = –1 – t, z = –5 + 8t

A vector parallel to the line is

3. A parallel vector is

11. u = 1, 4, 2 − and v = 2, 1, 2 − − are both v = 5 5, 4 + 3, 2 + 3 − = 0, 7, 5

4. A parallel vector is

perpendicular to the line, so u × v =

x = 5, y = –3 + 7t, z = –3 + 5t − − − , and hence 10, 2,9 is parallel to 10, 2, 9 the line.

5. x = 4 + 3t, y = 5 + 2t, z = 6 + t With y = 0, x – 2z = 13 and 2x – 2z = 5 yield x –4 y –5 z –6 = =

3 2 1 ⎜ − 8, 0, − ⎟ . The symmetric equations are ⎝

Since the second direction number is 0, the line

does not have symmetric equations.

12. u =

1, -3, 1 and v =

6, -5, 4 are both

7. Another parallel vector is 1, 10, 100 . x = 1 + t, y = 1 + 10t, z = 1 + 100t

perpendicular to the line, so u × v = -7, 2, 13 x –1 y –1 z –1

is parallel to the line.

With x = 0, –3y + z = –1 and –5y + 4z = 9 yield

13. 1, -5, 2 is a vector in the direction of the line. x –4 y z –6 = =

Section 11.6 Instructor’s Resource Manual

20. Using t = 0, one point of the plane is (0, 1, 0). direction of the line.

14. 2,1, 3 − × 5,4, 1 − = 11, 13,3 − is in the

0, 1, 1 = −− 2, 2, 2 = –2 1,1, 1 − is x + 5 y –7 z + 2 perpendicular to the normals of both planes,

3 hence parallel to their line of intersection.

3, 1, 2 is parallel to the line in the plane we

15. The point of intersection on the z-axis is (0, 0, 4).

seek, thus

3, 1, 2 × 1,1, 1 − = − 3, 5, 2 is a

A vector in the direction of the line is −−−

normal to the plane. An equation of the plane is

–3(x – 0) + 5(y – 1) + 2(z – 0) = 0 or equations are x = 5t, y = –3t, z = 4.

–3x + 5y + 2z = 5.

21. a. With t = 0 in the first line, x = 2 – 0 = 2, direction of the line since the line is

16. 3,1, 2 − × 2,3, 1 − = 5, 1,7 − is in the

y =+⋅= 340 3, z = ⋅ = so (2, 3, 0) is 20 0, perpendicular to 3,1, 2 − and 2,3, 1 −.

on the first line.

x –2 y + 4 z –5

b. − 1, 4, 2 is parallel to the first line, while

1, 0, 2 is parallel to the second line, so

17. Using t = 0 and t = 1, two points on the first line

8, 4, 4 − =4 2,1, 1 − are (–2, 1, 2) and (0, 5, 1). Using t = 0, a point on

is normal to both. Thus, π has equation the second line is (2, 3, 1). Thus, two nonparallel

vectors in the plane are 2(x – 2) + 1(y – 3) – 1(z – 0) = 0 or + 2x + y – z = 7, and contains the first line.

c. With t = 0 in the second line, x = –1 + 0 = –1, y = 2, z = –1 2 0 +⋅= –1, so

Hence, 2, 4, 1 − × 4,2, 1 − = 2, 2, 12 −−− is a Q (–1, 2, –1) is on the second line.

normal to the plane, and so is

1, 1, 6 . An

d. From Example 10 of Section 11.3, the equation of the plane is

1(x + 2) + 1(y – 1) + 6(z – 2) = 0 or

distance from Q to π is

1 22. With t = 0, (1, –3, –1) is on the first line. simultaneously to get x = 1, y = 2. From the first

perpendicular to both lines, so

11(x – 1) – 2(y + 3) + 14(z + 1) = 0 or first line. This point also satisfies the equations of

11x – 2y + 14z = 3 is parallel to both lines and the second line, so the lines intersect.

contains the first line.

− 4,3, 2 − and 1,1,6 − are parallel to the plane

With t = 0, (4, 1, 0) is on the second line. The distance from (4, 1, 0) to 11x – 2y + 14z = 3 is

determined by the lines, so

1,1, 6 = 20, 26, 1 − is a normal

321 to the plane. An equation of the plane is

20(x – 1) + 26(y – 2) – 1(z – 4) = 0 or

⎛ π 20x + 26y – z = 68.

23. r ⎛⎞=+ ⎜⎟ i 33 j + k , so 1, 3 3, ⎜ ⎟ is on the ⎝⎠ 3 3 ⎝

19. Using t = 0, another point in the plane is

tangent line.

(1, –1, 4) and 2, 3, 1 is parallel to the plane. r ′ () t = –2sin t i + 6 cos t jk + , so Another parallel vector is

r ′ ⎛⎞ ⎜⎟ = –3 i ++ 3 jk is parallel to the tangent

−−+ 1 1, 1 1,5 4 − = 0, 0, 1 . Thus,

2, 3, 1 × 0, 0, 1 = 3, 2, 0 − is a normal to

π line at t = . The symmetric equations of the

the plane. An equation of the plane is

3(x – 1) – 2(y + 1) + 0(z – 5) = 0 or 3x – 2y = 5.

x –1 y –33 z –

line are

Instructor’s Resource Manual

Section 11.6 701

24. The curve is given by r () t = 2 t 2 i + 4 t j + t 3 k . =+ x 1 2 t

r (1) = 2i + 4j + k, so (2, 4, 1) is on the tangent

line.

r ′ () t = 4 t i ++ 4 j 3 t k , so (1) r ′ =++ 4 i 4 j 3 k is

parallel to the tangent line. The parametric

equations of the line are x = 2 + 4t, y = 4 + 4t,

z = 1 + 3t. This line intersects the xz-plane when y = 0 ,

25. The curve is given by r () t = 3 t i + 2 t 2 j + t 5 k . which occurs when 7 0 = + 7 t , that is, r (–1) = –3i + 2j – k, so (–3, 2, –1) is on the

plane.

1 1 when t = − . For this value of t, x =−,

r ′ () t =+ 3 i 4 t j + 5 t 4 k , so (–1) r ′ = 3–4 i j + 5 k is

in the direction of the curve at t = –1, hence

. The normal to the plane. An equation of the plane is

point of intersection is therefore

3x – 4y + 5z = –22.

26. r ⎜⎟ ⎛⎞=+ i j , so

( cos 2 t t + sin ) t i ++ 3 j (2 cos – 2 sin ) t t t k so = (sin cos ) t t + (sin t ) + cos t

= sin 2 t cos 2 t + sin 4 t + cos 2 t

⎛⎞ π r ′ ⎜⎟ =+ i 3– j π k is in the direction of the

= sin 2 t (cos 2 t + sin 2 t ) cos + 2 t

= sin 2 t + cos 2 t = 1

curve at , hence normal to the plane. An

Thus, the curve lies on the sphere

2 equation of the plane is

1 ⎜ x – ⎟ + 3 ⎜ y – ⎟ – ( – 0) π z = 0 or

whose center is at the origin.

r ⎛⎞⎛⎞ π 1 ⎛ 3 ⎞ 1 3 b. ⎜ ⎟ ⎜ ⎟⎜ = ⎜ ⎟ i + j + k

7 t + ⎜ 97 −− t 4 t ⎟

the tangent line.

= 4 t 2 +− 7 t 97 t 4 t 2 = 9 r ′ () t = (cos 2 t – sin 2 t ) i + 2 cos sin – sin t t j t k Thus, the curve lies on the sphere

x + y 2 + z 2 = whose center is at the 9 so r ′ ⎜⎟ = i +

j – k is parallel to the

origin.

line. The line has equations

The line intersects the xy-plane when z = 0,

297 −− t 4 t 2 ⎛ 337 The point is ⎞

r '(1/ 4) = 2 i + 7 j −

The tangent line is therefore

Section 11.6 Instructor’s Resource Manual

29. a. r () t = 2 t i + t 2 j +− (1 t 2 ) k . Notice that

Since x = 2 t , we have x + + = + . , so y z x 1 =

this curve lies on the plane with equation

y +=. z 1

by Lagrange’s Identity. Thus,

b. Since (2) r = 4 i +− 4 j 3 k ,

PQ × n

r '( ) t = 2 i + 2 t j − 2 t k and

r '(2) = 2 i +− 4 j 4 k , the equation of the

tangent line is x =+ 42 t , y =+ 44 t ,

a. P(3, –2, 1) is on the line, so

z = − − . The line intersects the xy-plane 34 t

3 PQ = 1 – 3, 0 2, – 4 –1 + = –2, 2, – 5

when z = , that is, when 0 t = − , which

4 while n = 2, 2,1 − , so

gives x =, y = , and 1 z = . The point of 0 − 2, 2, −× 5 2, − 2, 1

intersection is therefore ⎜ ,1, 0 ⎟ .

30. In Figure 7, d is the magnitude of the scalar

b. P(1, –1, 0) is on the line, so

projection of PQ on n. pr n PQ =

1, 0, 3 while so

The point (0, 0, 1) is on the plane 4x – 4y + 2z = 2. With P(0, 0, 1), Q(4, –2, 3), and

32. d is the distance between the parallel planes

1 containing the lines. Since n is perpendicular to n =

4, – 4, 2 , both n 1 and n

2 , it is normal to the planes. Thus,

d is the magnitude of the scalar projection of PQ

2, – 2, 1 14 on n, which is

From Example 10 of Section 11.3,

a. P(3, –2, 1) is on the first line, Q(–4, –5, 0) is 4(4) – 4(–2) 2(3) – 2 + 28 14 on the second line, n =

be the scalar projection of PQ on n.

Then PQ = PR + d 2

Instructor’s Resource Manual

Section 11.6 703

the second line, n 1 =

11.7 Concepts Review

Problem Set 11.7

1. r t = tt 2 T ′ () , ()

() 1 + t 2 1 + t () 2

14 + t )( 14 + t 2 )

Section 11.7 Instructor’s Resource Manual

3. r () t = t , 2 cos , 2sin t t 5 T ′ () π =

v () t = r ′ () t = − 1, 2sin , 2 cos t t

a () t = v ′ () t = − 0, 2 cos , 2sin t − t

1 2sin t

2sin t

,5cos ,5sin t t

, 5sin ,5cos − t t

, 5cos , 5sin − t − t

20sin t

400 + π 4. 2 r () t = 5cos , 2 ,5sin t t t 400 + π 2

v () t = r ′ () t =− 5sin , 2,5cos t t v () π =

a () t = v ′ () t =− 5cos , 0, 5sin t − t

400 + 2 ( 3/2 t )

a () π =

( t ) cos t t sin t )

T () t =−

5sin t 2 5cos t

( 400 + t )

− ( 20 (400 + t 2 )sin t + t cos t )

( 400 + t )

5cos t

5sin t

Instructor’s Resource Manual Section 11.7 705

4sin t

16 + t 2 16 + t ( 2 400 + π )

− 4 16 + t cos t + t 4 sin t

− ( 20 (400 + π 2 ) sin ππ + cos π )

() t

( 16 + t )

( ) sin t

− t 4 cos t + 4 16 + t 2

) cos π

− π 4 cos π

T ′ () π =

6. r () t = , 2 cos , 2sin t t

v () t = r ′ () t =

, 2sin , 2 cos − t t

, 2 cos , 2sin − t − t

a () π = , 2, 0

Section 11.7 Instructor’s Resource Manual

− 21 () i − 31 () j

T () 1 =

( )( ) ( ) − 2 t = − 6 t = −− 2 − 3 t κ 2 (1)

2 2 2 3/2 ⎞ ⎜ () − 2 t +− 3 t ⎟

9. z ′ () t = –3sin t i + 4 cos t j ⎝

z ′ () t = 9sin 2 t + 16 cos 2 t 12 t 2 − 6 t 2 2

When t = 1 , the curvature is

4 cos =− t

3sin t

9sin 2 t +

16 cos 2 t

9sin 2 t +

⎜⎟ ⎛⎞= – i + j

⎝⎠ 4 5 5 xt ′ () = –3sin t yt ′ () = 4 cos t xt ′′ () = –3cos t yt ′′ () = –4sin t

9sin 2 t

Instructor’s Resource Manual Section 11.7 707

12. r ′ () t = cosh t i + sinh t j 15.

r ′ () t = sinh 2 t + cosh 2 t

r ′ () t

T () t =

cosh t

r ′ () t

sinh 2 t + cosh 2 t

sinh + t

sinh 2 t + cosh 2 t

T () ln 3 =

41 41 y ′ = xy 4, 4 ′′ =

xt ′ () = cosh t yt ′ () = sinh t 4

xt ′′ () = sinh t yt ′′ () = cosh t (1 16 + x 23/2 )

(sinh t + cosh t )

13. r ′ () t = –(cos t + sin ) te – t i + (cos – sin ) t te t j r ′ () t = 2 e – t

xt ′ () = –(cos – sin ) t te − t yt ′ () = (cos – sin ) t te − t

xt ′′ () = (2sin ) te − t yt ′′ () = (–2 cos ) te − t

14. r ′ () t = (cos – sin ) t t t i + (sin t + t cos ) t j ⎡ + x 2 x + 2 ⎣ ⎤ 1 (3 –16 16) ⎦ r ′ () t = t 2 + 1 8 At (4, 0), 1 κ = = and 8 R = .

cos1 – sin1 sin1 cos1 + (1) =

2 2 xt ′ () = cos – sin t t t yt ′ () = sin t + t cos t

xt ′′ () = –2sin – cos t t t yt ′′ () = 2 cos – sin t t t

Section 11.7 Instructor’s Resource Manual

and R = . 1000 25 2

At (1, 0), κ = = and 2 R = . At (2, –6), κ =

Instructor’s Resource Manual Section 11.7 709

y ′ = –2sin 2 , –4 cos 2 xy ′′ = x y ′ = sec 2 xy , 2sec ′′ = 2 x tan x

4 cos 2 κ x =

2sec tan

(1 4sin 2 ) + 2 x 3/2

e 2 x ( 2 + 4 x 23/2 )

Section 11.7 Instructor’s Resource Manual

3 x 2/3

9 x 5/3

2sech 2 x tanh x

6 x 1/ 3

(1 sech + x ) 3/2

Instructor’s Resource Manual Section 11.7 711

28. r(t) = sin 3t, cos 3t, t r ′ () t = 3cos 3 , – 3sin 3 , 1 t t r ′′ () t = –9sin 3 , – 9 cos 3 , 0 t t

⎝⎠ 9 r ′ π ()

r ′ ()() 9 × r ′′ π 2 ,– 2 ,1 × – 2 9 ,–,0 2 1 9 93 1

a N ⎛⎞= ⎜⎟

r ′ () 9 10 10 10

r ′′ π ()()() – a π T T π 1 93 9 3 1

r ′ π ()() 9 × r ′′ π 9

r () 9 () 10 10

29. r () t = 7 sin 3t, 7 cos 3t, 14t r ′ () t = 21cos 3 , t − 21sin 3 , 14 t r ′′ () t = –63sin 3 , – 63cos 3 , 0 t t

π π r ′ ()

T ⎜⎟ ⎛⎞=

⎝⎠ 3 r ′ π ()

r ′ π ()() ⋅ r ′′ π − 21, 0, 14 ⋅ 0, – 63, 0

a T ⎜⎟ ⎛⎞=

⎝⎠ 3 r ′ π ()

π π r ′ ()()

⎝⎠ 3 r ′ π ()

3 – a N T ⎛⎞= 3 T 3 1

N ()

′′ ⎛⎞ r π () r

Section 11.7 Instructor’s Resource Manual

30. r ′ () t = –3cos 2 t sin t i + 3sin 2 t cos t k r ′′ () t = (6 cos sin t 2 t – 3cos ) 3 t i + (6 cos 2 t sin – 3sin ) t 3 t k

π r ′ ⎜⎟ = 0 so the object is motionless at t 1 = .

κ, T, N, and B do not exist.

31. t

r ′ () t = sinh i + j

r () t = cosh i

r ′ (1)

sinh 1 i +

sinh 3 1 j 3 i + j

T (1) =

1 = tanh i + sech j r ′ (1)

sinh 21 + 1 3 cosh 3 3 3

r ′ (1) ⋅ r ′′ (1) ( sinh 3 i +⋅ j

)( 3 cosh 3 i ) 1 1

( sinh 3 i +× j )( 3 cosh 3 i ) −

cosh 1 cosh 3 1 3 3

= 3 ⎢ cosh i – sinh ⎜ tanh i + sech j ⎟ ⎥ = cosh –

i – tanh j

a N (1)

⎝ 3 cosh 1 3 ⎟ 3

sech i – tanh j

r ′ (1) × r ′′ (1) 1 cosh 1 1 1

B (1) = T(1) × N(1) = tanh i sech j

⎟⎜ × sech i – tanh j ⎟ =⎜ –sech – tanh ⎟ k = –k

32. r ′ () t = e 7 t (7 cos 2 – 2sin 2 ) t t i + e 7 t (7 sin 2 t + 2 cos 2 ) t j + 7 e 7 t k r ′′ () t = e 7 t (45cos 2 t − 28sin 2 ) t i + e 7 t (45sin 2 t + 28cos 2 ) t j + 49 e 7 t k

e 7/3 π −− π 7 () ( 2 3 i + e 7/3 π ) 73 − ( 2 1 j + 7 e 7/3 π k

r ′ π ()() 3 ⋅ r ′′ π 714 e π 14 / 3

e 7/3 () π 3 102

e π 14 / 3 ( + 49 14 3 ) i

π ⎛⎞ r π ′ ⎛⎞

⎜⎟ × r ′′ ⎜⎟

⎝⎠ 3 ⎝⎠ 3 + e π 14 / 3 ( − 14 49 3 ) j + 106 e π

r ′′ π ()()() – a T π T π 2–73 723 +

r ′ π r ′′ π ()() 3 × 3 2 5406 e π 14 / 3

⎝⎠ 3 ′ π r 3 ()

Instructor’s Resource Manual Section 11.7 713

33. r ' t =− 2 e − 2 t i + 2 e 2 () t j + 22 k

r ′ (0) ⋅ r ′′ (0) ( –2 i ++ 2 j 22 k ) ⋅ (4 i + 4) j

B (2) = T(2) × N(2)

Section 11.7 Instructor’s Resource Manual

35. y ′ = 1 = ,– 1 y ′′

1 x 2 κ= x =

+ 1 ( x 2 + 1) () 3/2 1

Since 0 < x < ∞ , κ=

κ ′ = when 0 x =

. Since κ ′ > on 0 ⎜ 0,

⎟ and κ ′ < on 1 , , 0 ⎜ ∞ ⎟ so κ is maximum when

. The point of maximum curvature is ⎜

k′ = when 0 x

, . κ ′ is not defined when x =π – , 0. Since κ ′ > on – , 0, 0 ⎜ π− ⎟⎜ ∪ ⎟ and κ ′ < on 0

⎜ − ,0 ⎟⎜ ∪ , π ⎟ , so κ has local maxima when x = – , –1 y = and x = , 1. y =

The points of maximum curvature are () 0, 1 and ⎜ ,1. ⎟

37. y ′ = sinh , cosh xy ′′ = x cosh κ= x

= sech 2

(1 sinh + x )

κ′= –2sech 2 x tanh x

κ ′ = when x = 0. Since 0 κ ′ > on (– ∞ , 0) and 0 κ ′ < on (0, ∞ ), so κ is maximum when x = 0, y = 1. The point 0

of maximum curvature is () 0, 1 .

Instructor’s Resource Manual Section 11.7 715

38. y ′ = cosh , sinh xy ′′ = x sinh κ= x

(1 cosh + 2 x ) 3/2

sinh x cosh (1 cosh x + 2 x ) 3/2 – 3 sinh cosh sinh (1 cosh x x x + 2

x ) 1/ 2

2 sinh coth (2 – cosh 2 x x x )

sinh x

(1 cosh + x ) 3 (1 cosh + 2 x ) 5/2

κ ′ is not defined when x = 0 and κ ′ = when cosh 0 x = 2 or x =± ln ( 21. + ) Since κ ′ > on 0

( – , – ln ∞ ( 21 + ) ) ∪ ( 0, ln ( 21 + ) ) and κ ′ < on 0 ( – ln ( 21,0 + ) ) ∪ ( ln ( 21, + ) ∞ κ has local maxima when ) ,

x = – ln ( + 2 1 , –1 ) y = and x = ln ( + 2 1 , 1. ) y =

– ln ( 21 + ) ) = κ ( ln ( 21 + ) ) =

The points of maximum curvature are ( – ln ( + 2 1 , –1 ) ) and ( ln ( 21,1. + ) )

14 + t 2 14 + t 2 κ ′ = when 1

0 x =− ln 2 . Since κ ′ > on 0 6

κ is maximum when x = – 1 ln 2, y =

. The

42. r ′ () t = 2 t i + j

point of maximum curvature is ⎜ – ln 2, . ⎟ r ′′ () t = 2 i

a N = r () t − a T =− 4 2 = 2 Since – << x , cos . κ = x 4 t + 1 4 t + 1

– sin x a N =

κ ′ = when x = 0. Since 0 κ ′ > on – , 0 0 ⎜ ⎟

At t 1 = 1, a T =

and a N =

and 0 κ ′ < on 0, , ⎜ κ is maximum when

x = 0, y = 0. The point of maximum curvature is (0, 0).

Section 11.7 Instructor’s Resource Manual

2 = ′′ 2 2 4 t 2 4 a N = [() xt ′′ + yt ′′ ()]– a T =; 0 a N = a 0

45. r ′ () t = a sinh t i + a cosh t j

r ′′ () t = a cosh t i + a sinh t j r ′ () t =

a sinh t cosh t

2 cosh sinh

4 a 2 cosh 2 = t 2 2 + 2 sinh 2 t

a (cosh t sinh t )–

1, 2 , 3 t t 2 2 × 2 0, 2, 6 t 6, t 2 − 6,2 t

sinh t + cosh t

a N () t =

2 2 22

2 4 2 a 4 (cosh t – sinh t ) 14 + t + 9 t 14 + t + 9 t

sinh 2

t + cosh 2 t

36 t 4 + 36 t 2 + 4 t 4 + t 2 +

14 + t 2 + 9 t 4 14 + t 2 + 9 t N 4

a (cosh t – sinh t

sinh 2 t + cosh 2 t

a N (2) = 2

181

At t ln 3, cosh t =

1 , sinh t = , and

161

2 40 t a 2 sinh 41 + cosh t = , so a

Instructor’s Resource Manual Section 11.7 717

v(t) = cos , 2 cos 2 t t , a(t) = − sin , 4sin 2 t − t a N (2) = 62

a (t) = 0 if and only if –sin t = 0 and –4 sin 2t = 0, 17 which occurs if and only if t = 0, π , 2 π , so it

occurs only at the origin.

51. r ′ () t = (1 – t 2 ) – (1 i + t 2 ) jk + a (t) points to the origin if and only if a(t) = –kr(t) r ′′ () t = –2 – 2 t i t j for some k and r(t) is not 0. This occurs if and

r ′ () t ⋅ r ′′ () t =

–2 (1 – t t 2 ) 2 (1 + t + t 2 ) = 4t 3 only if 3 t = ,, so it occurs only at (1, 0) and

22 r ′ () t = (1 − t 22 ) ++ (1 t 22 ) += 1 2 t 4 + 3 (–1, 0).

Section 11.7 Instructor’s Resource Manual

54. v(t) = − sin t + t cos t + sin , cos t t + t sin t − cos t 59. It is given that at (–12, 16), st ′ ( ) 10 = ft/s and

= t cos ti + t sin tj

st ′′

1 () = 5 ft/s 2 . From Example 2, κ= .

a (t) = − t sin t + cos , cos tt t + sin t 20

Therefore, a = 5 and a = ⎛ 1 ⎞ (10) 2 = 5, so

a. = v () t = t (cos 2 t + sin 2 t ) 1/ 2 = t T

(since t ≥ 0 )

a = 5T + 5N.

Let r(t) =

20 cos , 20sin t t describe the circle.

a 2 a N 2 = – a T 2 v (t) = − 20sin , 20 cos t t , so v () t = 20. = [ (sin t 2 2 t + cos ) (cos 2 t + 2 t + sin 2 t )] – 1 = t 2 v () Then t T () t = =− sin , cos . t t

Therefore, a N = t . v () t

55. st () a T 0 speed st ′ () = c (a constant)

Thus, at (–12, 16), T =

⎜ ⎟ ⎞==⇒= a N 0 κ 0 or

= 0 ⇒= κ 0 N = ,–

since N is a unit vector

⎝ dt ⎠

perpendicular to T and pointing to the concave

56. r(t) = a cos ωt i + b sin ωt j;

side of the curve.

= ––7. i j

T + 2 2 (0) 32 (4) 23/2 N = 23/2 N . (1 4 + x ) (1 4 +

61. Let μ mg = R . Then v R = μ gR . At the

a sin ω t + b cos ω t

values given,

Then v R = (0.4)(32)(400) = 5120 ≈ 71.55 ft/s

(about 49.79 mi/h).

(from the given

Note that this was done assuming ab > 0; if

ab < 0, drop the negative sign in the numerator.

equations, equating m in each.) Therefore, v R = Rg tan . θ

57. v(5) is tangent to the helix at the point where the particle is 12 meters above the ground. Its path is

b. For the values given,

described by v(5) = cos 5i – sin 5j + 7k. v R = (400)(32)(tan10 ) °≈ 47.51 ft/s.

58. ds

a N = 0 wherever κ = 0 or

63. tan φ y dt ′ =

= 0. κ , the

curvature, is 0 at the inflection points, which (1 + y ′ 23/2 ) =+ (1 tan 2 φ ) 3/2 = sec 3 φ

occur at multiples of . However,

this curve. Therefore, a N = 0 at multiples of . ⎣

Instructor’s Resource Manual Section 11.7 719

() ds

(sin φ + cos 2 φ φ ) 1/ 2 ds

Thus, Px ax 3 φ 4 5 () = 3 + ax 4 + ax 5 5 ,

< N 0, sin , = φ − cos φ , so N points to the ds

Px ′ () = 3 ax 2 + 4 ax 5 3 3 4 + 5 ax 4 5 , and

concave side of the curve in either case.

Px ′′ () = 6 ax + 12 ax 2 + 20 ax 5 3 3 4 5 .

65. B =× TN . Left-multiply by T and use

5 (1) 1, (1) = P ′ 5 = and 0, P 5 (1) = 0 ⇒

Theorem 11.4C

TB ×=× T TN ×

= ( TNT i )( − TTN i )

6 a 3 + 12 a 4 + 20 a 5 = 0

The simultaneous solution to these equations is

=− N

a 10, –15, 6, 4 = a 5 = so

Thus, N =−×=× TB BT .

Px 5 ( ) 10 x –15 x 6. x

To derive a result for T in terms of N and B, begin with N =× BT and left multiply by B:

68. Let

Px () = a + ax + ax 2 3 4 BN 5 ×=× B ( BT × ) 5 0 1 2 + ax 3 + ax 4 + ax 5 . Then = Px (

5 () i must satisfy )( − BBT i )

BTB

5 () 0 =; 0 P 5 () 1 =; 1 P′ 5 (0) =; 0 P′ 5 (1) 1 = =; 0 B − 1 T

=− P′′

5 (0) =; 0 P′′ 5 (1) = 0

As in Problem 67, the three conditions at 0 imply Thus, T =−×=× BN NB .

a 0 = a 1 = a 2 = . The three conditions at 0 lead 0

66. Since lim – y = lim y == 0 y (0), y is

to the system of equations

⎪⎩ 3 x if x > 0 The solution to this system is

is continuous since

a 3 = 6, a 4 =− 8, a 5 = . Thus, the required 3

lim y ′ = lim y ′ == 0 y ′ (0).

polynomial is Px 5 () = 6 x 3 − 8 x 4 + 3 x 5 .

is continuous since

lim

y ′′ = lim + y ′′ == 0 y ′′ (0). Thus,

y ′′ κ =

is continuous also. If x ≠ then 0

(1 + y ′ 23/2 ) y’ and κ are continuous as elementary functions.

Section 11.7 Instructor’s Resource Manual

69. Let the polar coordinate equation of the curve be r = f(θ). Then the curve is parameterized by x = r cos θ and y = r sin θ.

x ′ = – sin r θ + r ′ cos θ

y ′ = r cos θ + r ′ sin θ x ′′ = – cos – 2 sin r θ r ′ θ + r ′′ cos θ y ′′ = – sin r θ + r 2 cos ′ θ + r ′′ sin θ

By Theorem A, the curvature is

[(– sin r θ + r ′ cos ) θ 2 + ( cos r θ + r ′ sin ) ] θ 23/2 r 2 + 2 r ′ 2 – rr ′′

2 23/2 ( . r + r ′ )

70. r ′ = –4sin , –4 cos θ r ′′ = θ 74. r ′ = 3 e 3 θ , r ′′ = 9 e 3 θ

16 cos 2 θ + 32sin 2 θ + 16 cos 2 θ 32 1 At θ = 1, r = e 3 ,3, r ′ = e 3 and r ′′ = 9. e 3 κ =

(16 cos θ + 16sin 2 θ ) 3/2

75. r ′ = 4 cos , 4sin θ r ′′ =− θ

73. r ′ = –4sin , –4 cos θ r ′′ = θ κ =

cos 2 θ (cos 2 ) θ 3/2

2sin 2 cos 2 2 + θ cos 2 2 θ θ + 1 3 cos 2 θ + cos 2 θ κ θ = cos 2

cos 2 θ + sin 2 θ cos 2 θ )

( cos 2 θ )

= 3 ( cos 2 θ ) = 3r

78. r(t) = f(t)i + g(t)j, where x = f(t) and y = g(t); ( ) v t = r ′ () t

x 2 2 2 ′ 23/2 + y ′ x ′ 2 + y ′ 2 ( x ′ + y ′ )

( x ′ 2 + y ′ 23/2 )

T '( ) t = 2 23/2 i +

( x ′ 2 + y ′ 23/2 )

T ′ () t = 2 23/2 x ′ + y ′ =

x ′ 2 + y ′ 2 r ′ () t

( x ′ 2 + y ′ 23/2 )

Instructor’s Resource Manual

Section 11.7 721

84. r ′ () t = a 0 i + b 0 j + c 0 k

r ′′ () t = 0 Thus, ( ) r ′ t × r ′′ () t = 0 and since r ′ () t × r ′′ () t

r ′ 3 () t To show that τ = , note that the curve is 0

confined to a plane. This means that the curve is Maximum curvature ≈ 0.7606,

two-dimensional and thus τ=. 0

minimum curvature ≈ 0.1248

85. r (t) = 6 cos π ti + 6 sin π tj, + 2tk, t > 0

80. BB ⋅= 1 Let

(6 cos π t ) 2 + (6sin π t ) 2 + (2 ) t 2 = 100.

π+ )4 t ds 2 t = 100;

( BB ⋅ ) = 2 B ⋅

= 0 Then 36(cos 2 π+ t sin 2

Thus, is perpendicular to B.

t = 4.

ds r (4) = 6i + 8k, so the fly will hit the sphere at the point (6, 0, 8).

r ( ′ () t = –6 sin π π+π t i 6 cos π+ t j 2, k so the fly will ds ds

have traveled

Since N =

Thus T ⋅ =⋅ T ⎜ T ×

d B is perpendicular to T. ds

86. r () t =

10 cos , 10sin , t t ⎜ ⎟ t

82. N is perpendicular to T, and B = T × N is

d B Using the result of Example 1 with a = 10 and perpendicular to both T and N. Thus, since

c = , the length of one complete turn is is perpendicular to both T and B, it is parallel to

N , and hence there is some number τ(s) such that

2 π (10) 2 +⎜⎟ angstroms

= –(). τ s N ⎝ 2 π ⎠

ds = 10 –8 400 π+ 2 34 2 cm. Therefore, the total

83. Let ax + by + cz + d = 0 be the equation of the

length of the helix is

plane containing the curve. Since T and N lie in

(2.9)(10 )(10 ) 400 8 –8 π+ 2 34 2 ≈ 207.1794 cm. the plane B =±

a i ++ b j c k

. Thus, B is a

constant vector and

= 0 , so τ(s) = 0, since N

will not necessarily be 0 everywhere.

722 Section 11.7

Instructor’s Resource Manual

11.8 Concepts Review

5. ( – 4) x 2 + ( y + 2) 2 = 7

Circular cylinder

1. traces; cross sections

2. cylinders; z-axis

3. ellipsoid

4. elliptic paraboloid

Problem Set 11.8

6. Two planes

Elliptic cylinder

Circular cylinder

1 9 1 Hyperboloid of one sheet

4. Parabolic cylinder

Instructor’s Resource Manual

Section 11.8 723

9. z = +

Elliptic paraboloid

Hyperboloid of two sheets

10. Circular cone

15. y =

4 9 Elliptic Paraboloid

13. Hyperbolic paraboloid

Section 11.8 Instructor’s Resource Manual

18. Cylinder

25. All central hyperboloids of two sheets are symmetric with respect to (a) the origin, (b) the z-axis, and (c) the yz-plane.

26. a. 1, 2, 4, 5, 6, 7, 8, 10, 12, 13, 14, 15, 16, 18

b. 1, 2, 6, 7, 8, 9, 10, 14, 16, 19, 20

27. At y = k, the revolution generates a circle of

radius x =

. Thus, the cross section in

19. Hemisphere k

the plane y = k is the circle x 2 + z 2 = or 2

2 x 2 + 2 z 2 = The equation of the surface is k .

y = 2 x 2 + 2. z 2

28. At z = k, the revolution generates a circle of z radius k

y == . Thus, the cross section in the

2 2 plane

20. One sheet of a hyperboloid of two sheets.

z = k is the circle x 2 + y 2 =

or 4

4 x 2 + 4 y 2 = k 2 . The equation of the surface is z 2 = 4 x 2 + 4 y 2 .

29. At y = k, the revolution generates a circle of 3 3

radius x = 3– y 2 = 3– k 2 . Thus, the 4 4

cross section in the plane y = k is the circle

3– k 2 or

12 – 4 x 2 –4 z 2 = 3 k 2 . The

21. a. Replacing x by x

− results in an equivalent

equation.

equation of the surface is 4 x 2 + 3 y 2 + 4 z 2 = 12.

b. Replacing x by x − and y by − y results in

30. At x = k, the revolution generates a circle of

an equivalent equation.

radius y =

x 2 –4 =

c. Replacing x by x −, y by y , and z with

– 4. Thus, the

− z , results in an equivalent equation.

cross section in the plane x = k is the circle

y 2 + z 2 = 4 k 22. a. Replacing 2 y by − y results in an equivalent –4 or

+ 12 3 y 2 + 3 z 2 = 4 k 2 . The

equation.

equation of the surface is 4 x 2 = + 12 3 y 2 + 3. z 2

b. Replacing x with x − and z with − z results in an equivalent equation.

31. When z = 4 the equation is 4 = + or 4 9

c. Replacing y by − y and z with − z results

in an equivalent equation.

23. All central ellipsoids are symmetric with

c 2 = a 2 – b 2 = 20, hence c =± 2 5. The major

respect to (a) the origin, (b) the x-axis, and

axis of the ellipse is on the y-axis so the foci are

the (c) xy-plane.

at ( 0, ± 2 5, 4 . )

24. All central hyperboloids of one sheet are symmetric with respect to (a) the origin, (b) the y-axis, and (c) the xy-plane.

Instructor’s Resource Manual

Section 11.8 725

16 y 2 36.

32. When x = 4, the equation is z = +

=⋅ 9 = y 9 9( – 4) z 4 ( – 4), z hence p . The

vertex is at (4, 0, 4) so the focus is ⎛

33. When z = h, the equation is h

= or 1 y = x intersects the cylinder when x = y = 2. Thus,

the vertices of the triangle are (0, 0, 0), (2, 2, 1), x 2 y 2 c 2 – h 2 and (2, 2, 4). The area of the triangle with sides

which is equivalent to

represented by

2, 2, 1 and

2, 2, 4 is

= 1, which is

c 2 – h 2 . Thus,

c c 37. ( cos ) t t 2 + ( sin ) t t 2 − t 2

the area is

hence every point on the spiral is on the cone.

c For r = 3t cos ti + t sin tj + tk, every point

34. The equation of the elliptical cross section is satisfies x 2 + 9 y 2 –9 z 2 = so the spiral lies on 0

x 2 y 2 the elliptical cone.

2 + 2 = for each z in [0, h). a 1, (–) h z b (–) h z

38. It is clear that x = y at each point on the curve.

Therefore, Δ≈π V ( ah – z )( bh – z ) Δ z

Thus, the curve lies in the plane x = y. Since =π ab h z z

z = t 2 = y 2 , the curve is the intersection of the (–) Δ , using the area formula

mentioned in Problem 33. plane x = y with the parabolic cylinder z = y 2 .

Let the line y = x in the xy-plane be the u-axis,

Therefore, V = ∫ π 0 (–) ab h z dz =π⎢ ab hz – ⎥ then the curve determined by r is in the uz-plane.

2 ⎥ ⎦ 0 The u-coordinate of a point on the curve, (y, y, z) ⎡

⎛ is the signed distance of the point (y, y, 0) from

h 2 ⎞ ⎤ π abh 2

the origin, i.e., u = 2. y Thus, u 2 = 2 y 2 = 2 z ⎢ ⎣ ⎝

=π ab ⎢ ⎜ ⎜ h – ⎟ ⎟ –0 ⎥ =

, which is the

or u 2 =⎜⎟ 4 z . This is a parabola in the

height times one half the area of the base (z = 0),

⎝⎠ 2 π ( )( ) ah bh =π abh . uz -plane with vertex at u = 0, z = 0 and focus at

. The focus is at ⎜ 0, 0, ⎟ .

35. Equating the expressions for y, 4–x = x + z 2 ⎝

x 2 z 2 or 1 =

which is the equation of an

2 4 ellipse in the xz-plane with major diameter of

24 = and minor diameter 2 2. 4

Section 11.8 Instructor’s Resource Manual

11.9 Concepts Review

4. a. x = 8sin ⎜⎟ cos ⎜⎟ = 22

circular cylinder; sphere

2. plane; cone

3. ρ= 2 r 2 + z 2 z = 8cos ⎜⎟ = 43 ⎝⎠ 6

x 4sin ⎜ ⎟ cos ⎜⎟ = 2

4sin ⎜ ⎟ sin ⎜⎟ = 6

Problem Set 11.9

⎛ 3 π ⎞ z = 4 cos ⎜ ⎟ = –2 2

1. Cylindrical to Spherical: ρ = r 2 + z 2 5. a. ρ= x 2 + y 2 + z 2

Spherical to Cylindrical:

r = ρ sin φ

4th quadrant so θ =

2. a. ( ρθφ ,, ) =⎜ 2, , ⎟ Spherical: ⎜ 4 2, , ⎟

b. Note: ⎜ − 2, ,2 ⎟⎜ = 2, ,2 ⎟

quadrant so θ = . ⎛⎞ 4 π

y = 6sin ⎜⎟ = 3 ⎛ 3 ππ ⎞

Instructor’s Resource Manual

Section 11.9 727

Circular cylinder

12. r = 2 sin 2θ

4-leaved cylinder

ρ cos φ = 1 z =1 Plane

Plane

Section 11.9 Instructor’s Resource Manual

Circular cylinder

29. r 2 cos 2 θ = zr ; (cos 2 2 θ – sin 2 θ ) = z ; ( cos ) – ( sin ) r θ 2 r θ 2 = z ; x 2 – y 2 = z

17. x 2 + y 2 = 9; 9; r 2 = r=3

30. ρ sin φ = 1 (spherical); r = 1 (cylindrical);

18. cos θ – sin θ 25; cos 2 θ 25;

x + y = 1 (Cartesian)

r 2 = 25sec 2 θ ; 5sec r = θ

31. z = 2 x 2 + 2 y 2 = 2( x 2 + y 2 ) (Cartesian); z = 2 r 2

2 2 2 2 2 2 2 32. 2 x 2 + 2 y 2 – z 2 = 2 (Cartesian); 2 r 2 – z 2 = 2 20. ( x + y + z )3 + z = 10; 3 ρ + ρ cos φ = 10;

2 1 1 cos φ = (pole is not lost); cos 2 φ = (or

ρ 2 [sin 2 φ cos 2 θ – sin 2 φ sin 2 θ – 1 sin + 2 φ ] 1; = ρ 2 [sin 2 φ cos 2 θ – 1 sin + 2 φ (1 – sin 2 θ )] 1; = ρ 2 [sin 2 φ cos 2 θ – 1 sin + 2 φ cos 2 θ ] 1; = ρ 2 [2sin 2 φ 1 cos 2 θ – 1] 1; = ρ 2 =

2sin 2 φ cos 2

Instructor’s Resource Manual

Section 11.9 729

33. For St. Paul: ρ = 3960, θ = 360° – 93.1° = 266.9° ≈ 4.6583 rad

rad

π x = 3960sin cos 4.6583 ≈ –151.4 4

y = 3960sin sin 4.6583 ≈ –2796.0

z = 3960 cos ≈ 2800.1

For Oslo: ρ = 3960, θ = 10.5° ≈ 0.1833 rad, φ = 90° – 59.6° = 30.4° ≈ 0.5306 rad x = 3960 sin 0.5306 cos 0.1833 ≈ 1970.4 y = 3960 sin 0.5306 sin 0.1833 ≈ 365.3 z = 3960 cos 0.5306 ≈ 3415.5 As in Example 7,

so γ ≈ 1.0173 and the great-circle distance is d ≈ 3960(1.0173) ≈ 4029 mi

34. For New York: ρ = 3960, θ = 360° – 74° = 286° ≈ 4.9916 rad φ = 90° – 40.4° = 49.6° ≈ 0.8657 rad x = 3960 sin 0.8657 cos 4.9916 ≈ 831.1 y = 3960 sin 0.8657 sin 4.9916 ≈ –2898.9 z = 3960 cos 0.8657 ≈ 2566.5 For Greenwich: ρ = 3960, θ = 0, φ = 90° – 51.3° = 38.7° ≈ 0.6754 rad x = 3960 sin 0.6754 cos 0 ≈ 2475.8 y =0 z = 3960 cos 0.6754 ≈ 3090.6

so γ ≈ 0.8802 and the great-circle distance is d ≈ 3960(0.8802) ≈ 3485 mi

35. From Problem 33, the coordinates of St. Paul are P(–151.4, –2796.0, 2800.1). For Turin:

ρ = 3960, θ = 7.4° ≈ 0.1292 rad, φ =

rad

π x = 3960sin cos 0.1292 ≈ 2776.8

y = 3960sin sin 0.1292 ≈ 360.8

z = 3960 cos ≈ 2800.1

so γ ≈ 1.1497 and the great-circle distance is d ≈ 3960(1.1497) ≈ 4553 mi

Section 11.9 Instructor’s Resource Manual

36. The circle inscribed on the earth at 45° parallel (φ = 45°) has radius 3960 cos . The longitudinal angle between

St. Paul and Turin is 93.1° + 7.4° = 100.5° ≈ 1.7541 rad

Thus, the distance along the 45° parallel is ⎜ 3960 cos ⎟ (1.7541) ≈ 4912 mi

37. Let St. Paul be at P 1 (–151.4, – 2796.0, 2800.1) and Turin be at P 2 (2776.8, 360.8, 2800.1) and O be the center of

the earth. Let β be the angle between the z-axis and the plane determined by O, P 1 , and P 2 . OP 1 × OP 2 is normal

to the plane. The angle between the z-axis and OP 1 × OP 2 is complementary to β. Hence ⎛ ⎛ ⎯⎯→ ⎯⎯→ ⎞ ⎞

⎜ ⎜ OP 1 × OP 2 ⎟ ⋅ k ⎟

–1 ⎛ 7.709 10 × 6 β ⎞ =− cos ⎜ ⎟ ≈ – cos ⎜

⎜ OP 1 × OP 2 k ⎟

The distance between the North Pole and the St. Paul-Turin great-circle is 3960(0.5689) ≈ 2253 mi

= ρ 2 sin 2 φ (cos 2 θ + sin 2 2 2 2 θ 2 ) + ρ 2 2 2 2 2 2 2 1 2 sin φ 1 (cos θ 1 + sin θ 1 ) + ρ 2 cos φ 2 + ρ 1 cos φ 1 –2 ρρ 12 sin φ 1 sin φ 2 (cos θ 1 cos θ 2 + sin θ 1 sin θ 2 )–2 ρρ 12 cos φ 1 cos φ 2 = ρ 2 2 2 2 2 2 2 2 2 sin φ 2 + ρ 1 sin φ 1 + ρ 2 cos φ 2 + ρ 1 cos φ 1 − 2 ρρ 12 sin φ 1 sin φ 2 [cos( θθ 1 − 2 )] − 2 ρρ 12 cos φ 1 cos φ 2

= ρ 2 2 (sin 2 φ 2 + cos 2 φ 2 ) + ρ 2 (sin 2 1 2 φ 1 + cos φ 1 )2 + ρρ 12 [– cos( θθ 1 – 2 ) sin φ 1 sin φ 2 – cos φ 1 cos φ 2 ]

= ρ 2 1 + ρ 2 2 + 2 ρρ 12 [– cos( θθ 1 – 2 ) sin φ 1 sin φ 2 – cos φ 1 cos φ 2 ] = ρ 2 1 –2 ρρ + ρ 2 12 2 + 2 ρρ 12 [1 – cos( θθ 1 – 2 ) sin φ 1 sin φ 2 – cos φ 1 cos φ 2 ] = ( ρ 1 – ρ 2 ) 2 + 2 ρρ 12 [1 – cos( θθ 1 – 2 )sin φ 1 sin φ 2 – cos φ 1 cos φ 2 ]

Hence, d = {( ρ –

1 ρ 2 ) + 2 ρρ 12 [1 – cos( θθ 1 – 2 ) sin φ 1 sin φ 2 – cos φ 1 cos φ 2 ]} 1/ 2 ⎯⎯→

39. Let P 1 be (, a 1 θ φ and 1 , 1 ) P 2 be ( a 2 , θ φ If γ is the angle between 2 , 2 ). OP 1 and OP 2 then the great-circle distance between P 1 and P 2 is aγ. OP 1 = OP 2 = a while the straight-line distance between P 1 and P 2 is (from

Problem 38) d 2 = (–) a a 2 + 2 a 2 [1 – cos( θθ 1 – 2 ) sin φ 1 sin φ 2 – cos φ 1 cos φ 2 ] = 2 a 2 {1 – [cos( θθ 1 – 2 ) sin φ 1 sin φ 2 + cos φ 1 cos φ 2 ]}. Using the Law Of Cosines on the triangle OP P 12 ,

d 2 = a 2 + a 2 –2 a 2 cos γ = 2 a 2 (1 – cos ). γ Thus, γ is the central angle and cos γ = cos( θθ 1 – 2 ) sin φ 1 sin φ 2 + cos φ 1 cos φ 2 .

Instructor’s Resource Manual

Section 11.9 731

40. The longitude/latitude system (α, β) is related to a spherical coordinate system (ρ, θ, φ) by the following relations: ρ = 3960; the trigonometric function values of α and θ are identical but

–π ≤ α ≤ π rather than 0 ≤ θ ≤ 2 π , and β = – φ so sin β = cos φ and cos β = sin φ.

From Problem 39, the great-circle distance between (3960, , ) θ φ and 1 1 (3960, , ) θ φ is 3960γ where 2 2 0 ≤≤ γπ

and cos γ = cos( θθ 1 – 2 ) sin φ 1 sin φ 2 + cos φ 1 cos φ 2 = cos( αα 1 – 2 ) cos β 1 cos β 2 + sin β 1 sin β 2 .

41. a. New York (–74°, 40.4°); Greenwich (0°, 51.3°) cos γ = cos(–74 – 0 ) cos(40.4 ) cos(51.3 ) sin(40.4 ) sin(51.3 ) ° ° ° °+ ° °≈ 0.637

Then γ ≈ 0.880 rad, so d ≈ 3960(0.8801) ≈ 3485 mi.

b. St. Paul (–93.1°, 45°); Turin (7.4°, 45°) cos γ = cos(–93.1° – 7.4°) cos (45°) cos (45°) + sin (45°) sin (45°) ≈ 0.4089 Then γ ≈ 1.495 rad, so d ≈ 3960(1.1495) ≈ 4552 mi.

c. South Pole (7.4°, –90°); Turin (7.4°, 45°) Note that any value of α can be used for the poles.

cos γ = cos 0° cos (–90°) cos (45°) + sin (–90°)sin (45°) = − 1

thus γ = 135 °=

rad, so d = 3960 ⎜ ⎟ ≈ 9331 mi.

d. New York (–74°, 40.4°); Cape

Town (18.4°, –33.9°) cosγ = cos (–74° – 18.4°) cos (40.4°) cos (–33.9°) + sin (40.4°) sin (–33.9°) ≈ –0.3880 Then γ ≈ 1.9693 rad, so d ≈ 3960(1.9693) ≈ 7798 mi.

e. For these points α 1 = 100 ° and α 2 = –80 ° while β 1 = β 2 = 0, hence

cos γ = cos 180° and γ = π rad, so d = 3960 π ≈ 12,441 mi.

42. ρ = 2a sin φ is independent of θ so the cross section in each half-plane, θ = k, is a circle tangent to the origin and with radius 2a. Thus, the graph of ρ = 2a sin φ is the surface of revolution generated by revolving about the z-axis

a circle of radius 2a and tangent to the z-axis at the origin.

11.10 Chapter Review

6. False: It is normal to the plane.

Concepts Test

7. True: Let t = .

1. True: The coordinates are defined in terms of distances from the coordinate planes in

a b c such a way that they are unique.

8. True:

Direction cosines are

2. False: The equation is ( – 2) x 2 + y 2 + z 2 = –5, 9.

True: (2 – 3 ) (6 i j ⋅ i + 4) j = 0 if and only if the so the solution set is the empty set.

vectors are perpendicular.

3. True: See Section 11.2.

10. True: Since u and v are unit vectors,

4. False: See previous problem. It represents a

plane if A and B are not both zero.

5. False: The distance between (0, 0, 3) and (0, 0, –3) (a point from each plane) is 6,

11. False: The dot product for three vectors so the distance between the planes is

( abc ⋅⋅ ) is not defined.

less than or equal to 6 units.

Section 11.10 Instructor’s Resource Manual

12. True:

uv ⋅ = uv cos θ ≤ uv since

26. True: The vectors are both parallel and perpendicular, so one or both must

cos θ ≤ 1. be 0.

13. True: If uv ⋅ = uv , then cos θ = 1 since

27. True:

(2 i × 2)( j ⋅×= ji ) 4 ( ) (– ) k ⋅ k

uv ⋅ = uv cos . θ Thus, u is a scalar

= 4( kk ⋅ ) = 4

multiple of v. If u is a scalar multiple of

v , uv ⋅ =⋅ u k u = k u

2 28. False: Let u = v = i, w = j.

Then (u × v) × w=0 × w = 0; but

= u k u = uv .

u × (v × w) = i × k = –j.

29. True: Since bb 1 , 2 , b 3 is normal to the

14. False: If u = –i and v = i +

j , then

30. False: Each line can be represented by

2 2 parametric equations, but lines with any u = v = u + v = 1. zero direction number cannot be

represented by symmetric equations.

= r ′ ′′ r sin θ = 0. Thus, either ′ r and

16. True: u + v 2 = ( u +⋅+ v )( u v ) r ′′ are parallel or either ′ r or ′′ r is 0, = which implies that the path is a straight

u 2 + v 2 + 2 uv ⋅ line.

17. True: Theorem 11.5A

32. True:

An ellipse bends the sharpest at points on the major axis.

18. True: D t [() F t ⋅ F ( )] t

= F () t ⋅ F ′ () t + F () t ⋅ F ′ () t 33. False: κ depends only on the shape of the

= curve.

2() F t ⋅ F ′ () t

34. True: x′ = 3 y′ = 2

19. 2 True: uu = uu = u x′′ = 0 y′′ = 0 xy ′ ′′ – yx ′ ′′

20. False: The dot product of a scalar and a vector

is not defined.

21. True:

uv × =−×=− vu 1 vu ×

35. False: x ′= –2sin t y ′= 2 cos t

=× vu

x ′′ = –2 cos t y ′′ = –2sin t

22. True: () k v v k ( vv × ) = k () 0 = 0 xy – yx

23. False: Obviously not true if u = v. (More generally, it is only true when u and v are also perpendicular.)

24. False: It multiplies v by a; it multiplies the

length of v by a .

Instructor’s Resource Manual

Section 11.10 733

36. True: T () t =

44. False: Suppose v(t) = cos ti + sin tj, then

r ′ () t

v () t = 1 but a(t) = –sin ti + cos tj

r ′ () t ⋅ r ′′

() t

which is non-zero.

45. True: T depends only upon the shape of the

curve, hence N and B also.

r ′′ ( ); t

r ′ () t

46. True: If v is perpendicular to a, then T is also

T () t ⋅ T ′ () t perpendicular to a, so

⎡ r ′ ⎢ () t ⋅ r ′ ′′ ⋅− () t ′

d ⎛ ds

⎜ ⎟ ⎞==⋅= a T Ta 0. Thus

is a constant.

47. False: If r(t) = a cos ti + a sin tj + ctk then v is

= perpendicular to a, but the path of

2 r motion is a circular helix, not a circle.

() t

r () t

48. False: The circular helix (see Problem 27) has

37. False: Consider uniform circular motion:

constant curvature.

dv dt / = 0 but v = a ω .

49. True:

The curves are identical, although the

y ′′

motion of an object moving along the

38. True: κ =

23/2 = 0 [1 curves would be different. + y ′ ]

50. False: At any time 0 <<, t 1 r 1 () t ≠ r 2 () t

39. False: If y ′′ = k then y ′= kx + C and

The parameterization affects only the κ =

rate at which the curve is traced out.

[1 + y ′ 23/2

[1 ( + kx + C )] 23/2

not constant.

52. True:

If a curve lies in a plane, then T and N will lie in the plane, so T × N = B will

40. False: For example, if u = i and v = j, then

be a unit vector normal to the plane.

uv ⋅= 0.

53. False: For r(t) = cos ti + sin t j, ( ) r t = 1, but

41. False: For example, if r () t = cos t 2 i + sin t 2 j , r ′ () t ≠ 0 . then r ′ () t = –2 sin t t 2 i + t 2 cos t 2 j , so

54. True:

The plane passes through the origin so r () t = 1 but r ′ () t = 2. t its intersection with the sphere is a great circle. The radius of the circle is 1, so is

42. True: If vv ⋅= constant, differentiate both

sides to get vv ⋅+⋅= ′ vv ′ 2 vv ⋅= ′ 0, so

curvature is = 1.

vv ⋅= ′ 0.

55. False: The graph of ρ = 0 is the origin.

43. True If r () t = a cos ω t i + a sin ω t j + ct k , then

r ′ () t =− a ω sin ω t i + a ω cos ω t j + c k so

56. False: It is a parabolic cylinder.

T () t = r ′ ()/ t r ′ () t

57. False: The origin, ρ = 0, has infinitely many

a ω sin ω t i + a ω cos ω t j + c k

spherical coordinates, since any value of

θ and φ can be used.

points directly to the z-axis. Therefore N () t = T ′ ()/ t T ′ () t points directly to the

z -axis.

Section 11.10 Instructor’s Resource Manual

Sample Test Problems

1. The center of the sphere is the midpoint

⎞= (1, 2, 4) of the diameter.

The radius is r = (1 2) + 2 + (2 – 3) 2 + (4 – 3) 2 = 911 ++= 11 . The equation of the sphere is ( –1) x 2 + ( – 2) y 2 + ( – 4) z 2 = 11

2. ( x 2 –6 x ++ 9) ( y 2 + 2 y ++ 1) ( z 2 –8 z + 16) =++ 9 1 16; ( – 3) x 2 + ( y + 1) 2 + ( – 4) z 2 = 26

Center: (3, –1, 4); radius: 26

c. 2, −⋅ 5 ( 1, 1 +− 6, 0 ) = 2, −⋅− 5 5, 1

d. ( 4 2, – 5 + 5 1, 1 ) ⋅ 3 –6, 0 =

≈ 0.3363 b. bc i = (0)(3) (1)( 1) ( 2)(4) + −+− =− 9

abc ⋅×=−++ ( ) ( i j 2 k )( ⋅ 2 i −− 6 j 3 k )

d. bc ⋅ is a scalar, and a crossed with a scalar doesn’t exist.

e. ab − =−++ i 0 j 4 k

f. From part (c), bc ×= 2 i −− 6 j 3 k so

bc ×= 2 2 + 6 2 + 3 2 = 49 = 7 .

Instructor’s Resource Manual

Section 11.10 735

θ = cos − 1 0.121716 83.009 ≈ c. − 2, 1,1 ⋅ − 7,1, 5 − = –20

9. c 3,3, 1 −× −− 1, 2, 4 = c 10, 11, 3 − − for any c θ = cos − 1 0.67200 ≈ 47.608 in R .

7. 10. Two vectors determined by the points are

1, 7, 3 − and 3, 1, 3 −− . Then

6, 3, 5 and

6, 3, 5 are normal to the plane.

are the unit vectors

normal to the plane.

11. a. y = 7, since y must be a constant.

b. x = –5, since it is parallel to the yz-plane.

c. z = –2, since it is parallel to the xy-plane.

a. a = 414 ++= 3;

d. 3x – 4y + z = –45, since it can be expressed

b = ++= 25 1 9 35 as 3x – 4y + z = D and D must satisfy 3(–5) – 4(7) + (–2) = D, so D = –45.

b. = i − j + k , direction cosines

a 3 3 3 12. a.

+ 4 1,1 5,1 7 − + = 5, 4,8 − is along the

line, hence normal to the plane, which has ,–, and . equation x – 2, y + 4, z +⋅ 5 5, – 4, 8 = 0 .

k , direction

Section 11.10 Instructor’s Resource Manual

19. 5, 4, 3 −− is a vector in the direction of the line, also be perpendicular. Thus

13. If the planes are perpendicular, their normals will

1, 5, C ⋅

4, –1, 1 4–5 + C ,

and 2, 2,1 is a position vector to the line.

Then a vector equation of the line is r(t) so C = 1.

= 2, 2,1 − + t 5, 4, 3 −− .

14. Two vectors in the same plane are 3, 2, 3 −− and

20. t x y z normal to the plane. An equation of the plane is

3, 7, 0 . Their cross product, 3 7, 3,9 − , is

–2 –2 2 − 8/3 7(x – 2) – 3(y – 3) + 9(z + 1) = 0 or

7x – 3y + 9z = –4. –1 –1 1/ 2 − 1/ 3

15. A vector in the direction of the line is 8,1, 8 − .

0000 Parametric equations are x = –2 + 8t, y = 1 + t, z = 5 – 8t.

16. In the yz-plane, x = 0. Solve –2y + 4z = 14 and 2y 222 8/3 – 5z = –30, obtaining y = 25 and z = 16.

In the xz-plane, y = 0. Solve x + 4z = 14 and

3 3 9/2 9 –x – 5z = –30, obtaining x = –50 and

z = 16. Therefore, the points are (0, 25, 16) and (–50, 0, 16).

17. (0, 25, 16) and (–50, 0, 16) are on the line, so

50, 25, 0 = 25 2, 1, 0 is in the direction of the line. Parametric equations are x = 0 + 2t,

y = 25 + 1t, z = 16 + 0t or x = 2t, y = 25 + t, z = 16.

18. 3, 5, 2 is normal to the plane, so is in the direction of the line. Symmetric equations of the

x –4 y –5 z –8

line are

. Symmetric equations for the

8 x –2 y –2 z –

tangent line are

3 . Normal 1 2 4 ⎛

8 ⎞ plane is 1( – 2) 2( – 2) 4 x + y + ⎜ z – ⎟ = 0 or

3x + 6y + 12z = 50.

Instructor’s Resource Manual

Section 11.10 737

22. r(t) = t cos t, t sin t, 2t ;

ln 2 c.

e 2 ∫ t , e − t dt

r () t =− t cos t − 2sin , t − t sin t + 2 cos , 0 t

π π r ′ ()

⎝⎠ 2 r ′ π ()

2 π+ 2 20 = e 2 t (2 t + 1), (1 e – t − t )

23. r ′ () t = e t cos t + sin , – sin t t + cos , 1 t e. D t [ (3 r t + 10)] [ (3 = r ′ t + 10)](3) r ′ () t = 3 e t = 32 e t + 6 20 , – e –3 –10 t

Length is

−− t ∫ 3 10

= 3( e 5 −≈ e ) 252.3509.

25. Let the wind vector be

w = 100 cos 30 , 100sin 30 ° ° 27. a. r ′ () t = , –6; () t r ′′ t = – t –2 , –6

b. r ′ () t = cos , – 2sin 2 ; t t Let p = p 1 , p 2 be the plane’s air velocity

r ′′ () t = – sin , – 4 cos 2 t t

vector. We want w + p = 450 j =

450 r () t = 2sec t tan , – 12 t t

Therefore, p = –50 3, 400 . The angle β formed with the vertical satisfies

the heading is N12.22°W. The air speed is

p = 167,500 ≈ 409.27 mi/h.

26. r () t = e 2 t , ; e – t r ′ () t = 2 e 2 t , – e – t () 4

t → 0 t → 0 t → 0 ≈ 0.1934

Section 11.10 Instructor’s Resource Manual

29. v () t =

1, 2 , 3 t t 2 ; () a t =

0, 2, 6 t 33. Circular paraboloid

= = 0 4 18 a 22 T = ≈ 5.880;

30. Circular cylinder

35. Plane

31. Sphere

36. Hyperboloid of one sheet

32. Parabolic cylinder

Instructor’s Resource Manual

Section 11.10 739

ρ 2 sin 2 φ cos 2 θρ + 2 sin 2 φ sin 2 θρ = cos φ ρ 2 sin 2 φ (cos 2 θ + sin 2 θ ) = ρ cos ; φ

ρ sin 2 φ = cos ; φ ρ = cot φ csc φ (Note that when we divided through by ρ in part

c and d we did not lose the pole since it is also a

38. The graph of 3 x 2 + 4 y 2 + 9 z 2 = –36 is the empty

solution of the resulting equations.)

set.

( 2 2, 2 2, 4 3 ) and

42. Cartesian coordinates are

39. a. r = 9; r=3

( 2, 6, – 2 2 ) . Distance

b. ( x 2 + y 2 )3 + y 2 = 16 2 2 ⎡ 1/ 2 ⎤ ⎢ 2 + ( 16 22–6 )( + 4322 2 + 2 2 2 ⎣ ) ⎥ ≈ 9.8659.

r + 3 r sin θ = 16, r =

+ 1 3sin 2 θ

43. (2, 0, 0) is a point of the first plane. The distance between the planes is

c. r 2 = 9 z 2(2) – 3(0) + 3(0) – 9 = 5 =

3, 2, 5 − are normal to the

respective planes. The acute angle between the

b. x 2 + z 2 = 4 two planes is the same as the acute angle θ between the normal vectors.

c. (cos 2

= 1 so θ ≈ 1.3204 rad ≈ 75.65°

41. a. ρ 2 = 4; 2 ρ =

45. If speed =

= a constant, then c ,

T + ⎛ ⎞ κ N = c ⎜ 2 ⎟ κ N since

is in the direction of v, while N is

1 – 2 cos 2

φ= 0; cos 2 φ= ; φ = or

perpendicular to T and hence to v also. Thus,

a = c π 2 κ N is perpendicular to v. φ = .

Any of the following (as well as others) would be acceptable:

Section 11.10 Instructor’s Resource Manual

Review and Preview

Instructor’s Resource Manual

Review & Preview 741

5 x 3 = 15 x 2 b. sin17 t = 17 cos17 t

Review & Preview Instructor’s Resource Manual

7. a. sin 2 a = 2 cos 2 a 14. fx ′ () = 4 x 3 − 54 x 2 + 226 x − 288

= 22 ( x − 9 ) ( x 2 − 9 x + 16 )

f ' () x = 0 when x = or x =

b. sin17 a = 17 cos17 a 9 9 ± 17

(using 2 2

the quadratic formula).

0 when x =

d. sin sa = s cos sa or x ==

f ′′ 4.5. ( ) 12 x = x − 108 x + 226. Since

2 f ′′ (2.438) >

0, a local minimum occurs at

8. a. e 41 t + = 4 e 41 t + = 9 − x 17 . Since (4.5) f ′′ < 0, a local

2 maximum occurs at x = 4.5.

f (4.5) 14.0625. Thus, the minimum value =

c. e = ae

of on [2, 6] is 4 and the maximum value of f −

f on [2, 6] is 14.0625.

Since the volume is to be 8 cubic feet, we have

9. fx () = 1

2 is both continuous and

differentiable at x 2 rh 2 = since rational functions = π 8

are continuous and differentiable at every real

number in their domain.

π r 2 Substituting for h in our surface area equation

10. fx () = tan x is not continuous at x = π 2 since

gives us

f ( π 2 ) is undefined.

11. fx () = − is continuous at x 4 x = since 4 Thus, we can write S as a function of r:

Sr () = 2 π r 2 +

lim x −= 4 f (4) = f is not differentiable at 0. 16

x =. 4

16. The area of two of the sides of the box will be

12. As x approaches 0 from the right, 1/ x approaches lh ⋅ . Two other sides will have area wh ⋅ . The

area of the base is lw ⋅ Thus, the total cost, C, . and lim fx () does not exist. Since the limit at

+∞ so , sin 1/ x () oscillates between − 1 and 1

of the box will be C = 2 lh + 2 wh + 3, lw where C x → 0 27 54 54

x = 0 does not exist, f is not continuous at x = 0 .

is in dollars. Since h =

, C = + + 3. lw

w Consequently, f is not differentiable at l

Geometry in Space CHAPTER

Parts

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