17.1 SOLUTIONS 1197 CHAPTER SEVENTEEN - …fd.valenciacollege.edu/file/tsmith143/ch17.pdf17.1...

17.1 SOLUTIONS 1197

CHAPTER SEVENTEENSolutions for Section 17.1

Exercises

1. One possible parameterization isx = 3 + t, y = 2t, z = −4− t.

2. One possible parameterization isx = 1 + 3t, y = 2− 3t, z = 3 + t.

3. One possible parameterization is

x = −3 + 2t, y = 4 + 2t, z = −2− 3t.

4. One possible parameterization isx = 5, y = −1 + 5t, z = 1 + 2t.

5. One possible parameterization isx = t, y = 1, z = −t.

6. One possible parameterization isx = 1, y = 0, z = t.

7. The displacement vector from the first point to the second is ~v = 4~i − 5~j − 3~k . The line through point (1, 5, 2) andwith direction vector ~v = 4~i − 5~j − 3~k is given by parametric equations

x = 1 + 4t,

y = 5− 5t,

z = 2− 3t.

Other parameterizations of the same line are also possible.

8. The vector connecting the two points is 3~i −~j + ~k . So a possible parameterization is

x = 2 + 3t, y = 3− t, z = −1 + t.

9. The displacement vector from the first point to the second is ~v = (−1 − (−3))~i + (−3 − (−2))~j + (−1 − 1)~k =

2~i − ~j − 2~k . The line through point (−3,−2, 1) and with direction vector ~v = 2~i − ~j − 2~k is given by parametricequations

x = −3 + 2t,

y = −2− t,z = 1− 2t.


10. The line passes through (3,−2, 2) and (0, 2, 0). The displacement vector from the first of these points to the second is~v = (0 − 3)~i + (2 − (−2))~j + (0 − 2)~k = −3~i + 4~j − 2~k . The line through point (3,−2, 2) and with directionvector ~v = −3~i + 4~j − 2~k is given by parametric equations

x = 3− 3t,

y = −2 + 4t,

z = 2− 2t.


1198 Chapter Seventeen /SOLUTIONS

11. The line passes through (3, 0, 0) and (0, 0,−5). The displacement vector from the first of these points to the second is~v = −3~i − 5~k . The line through point (3, 0, 0) and with direction vector ~v = −3~i − 5~k is given by parametricequations

x = 3− 3t,

y = 0,

z = −5t.


12. If the two lines intersect, there must be times t1, t2 such that each of the following three equations are satisfied:

x = 2 + 3t1 = 1 + 3t2,

y = 3− t1 = 2− 3t2,

z = −1 + t1 = 3 + t2.

(Note that the lines need not to go through the intersection point at the same time, so t1 and t2 may be different. Addingthe last two equations gives −2 = 5 − 2t2 and so t2 = 3

2, while the last equation gives −1 + t1 = 3 + 3

2, so t1 = 11

2.

But these don’t satisfy the first equation, therefore the two lines don’t intersect.)

13. The xy-plane is where z = 0, so one possible answer is

x = 3 cos t, y = 3 sin t, z = 0.

This goes in the counterclockwise direction because it starts at (3, 0, 0) and heads in the positive y-direction.

14. The xy-plane is where z = 0, and to make the particle go in the clockwise direction we start at (2, 0, 0) and head in thenegative y-direction. Thus one possible answer is

x = 2 cos t, y = −2 sin t, z = 0.

15. The xz-plane is y = 0, so one possible answer is

x = 2 cos t, y = 0, z = 2 sin t.

16. The circle lies in the plane z = 2, so one possible answer is

x = 3 cos t, y = 3 sin t, z = 2.

17. The yz-plane is x = 0, so the circle of radius 3 in the yz-plane centered at the origin would have equations

x = 0, y = 3 cos t, z = 3 sin t.

To move the center to (0, 0, 2) we add 2 to the equation for z, so one possible answer is

x = 0, y = 3 cos t, z = 2 + 3 sin t.

18. The circle of radius 5 in the yz-plane centered at the origin has equations

x = 0, y = 5 cos t, z = 5 sin t.

To move the center to (−1, 0,−2), we add −1 to the equation for x and −2 to the equation for z, so one possible answeris

x = −1, y = 5 cos t, z = −2 + 5 sin t.

19. The xy-plane is z = 0, so a possible answer is

x = t, y = t3, z = 0.

17.1 SOLUTIONS 1199

20. The xy-plane is z = 0, so a possible answer is

x = t2, y = t, z = 0.

21. The xz-plane is y = 0, so a possible answer is

x = −3t2, y = 0, z = t.

22. Since the curve is parallel to the xy-plane, z is constant, and since it passes through (0, 4, 4), we have z = 4. One possibleanswer is

x = t, y = 4− 5t4, z = 4.

23. Since its diameters lie along the x and y-axes and its center is the origin, the ellipse must lie in the xy-plane, hence atz = 0. The x-coordinate ranges between −3 and 3 and the y-coordinate between −2 and 2. One possible answer is

x = 3 cos t, y = 2 sin t, z = 0.

24. Since its diameters are parallel to the y and z-axes and its center is in the yz-plane, the ellipse must lie in the yz-plane,x = 0. The ellipse with the same diameters centered at the origin would have its y-coordinate range between −5/2 and5/2 and its z-coordinate range between −1 and 1. Thus this ellipse has equation

x = 0, y =5

2cos t, z = sin t.

To move the center to (0, 1,−2), we add 1 to the equation for y and −2 to the equation for z, so one possible answer forour ellipse is

x = 0, y = 1 +5

2cos t, z = −2 + sin t.

25. Since its diameters are parallel to the x and z-axes, the ellipse must be parallel to the xz-plane. The ellipse with the samediameters, but centered at the origin, would have its x-coordinate range between−3/2 and 3/2 and its z-coordinate rangebetween −1 and 1. Thus this ellipse has equation

x =3

2cos t, y = 0, z = sin t.

Since our ellipse has center (0, 1,−2), it must be in the plane y = 1. To move the center to (0, 1,−2), we add 1 to theequation for y and −2 to the equation for z, so one possible answer for our ellipse is

x =3

2cos t, y = 1, z = −2 + sin t.

26. The vector from P0 to P1 is ~v = (4 − 1)~i + (1 + 3)~j + (−3 − 2)~k = 3~i + 4~j − 5~k . Since P0 has position vector~i − 3~j + 2~k , the line is

~r (t) =~i − 3~j + 2~k + t(3~i + 4~j − 5~k ) for 0 ≤ t ≤ 1.

In coordinate form the equations are x = 1 + 3t, y = −3 + 4t, z = 2− 5t.

27. The vector from P0 to P1 is ~v = (5 + 1)~i + (2 + 3)~j = 6~i + 5~j . Since P0 = −~i − 3~j , the line is

~r (t) = −~i − 3~j + t(6~i + 5~j ) for 0 ≤ t ≤ 1.

In coordinate form, the equations are x = −1 + 6t, y = −3 + 5t, 0 ≤ t ≤ 1

28. Since the semicircle is in the xy-plane we have z = 0. A circle of radius 1 in the xy-plane, centered at the originand parameterized in the counterclockwise direction, goes from (1, 0, 0) to (−1, 0, 0). It has equations x = cos t andy = sin t. The semicircle where y ≥ 0 is the obtained by restricting t to 0 ≤ t ≤ π. Thus a possible answer is

x = cos t, y = sin t, z = 0, 0 ≤ t ≤ π.


29. Since the semicircle is in the yz-plane we have x = 0. A circle of radius 5 in the yz-plane, centered at the origin andparameterized in the clockwise direction (from the positive z-axis toward the positive y-axis), goes from (0, 0, 5) to(0, 0,−5). It has equations y = 5 cos t and z = −5 sin t. The semicircle where y ≥ 0 is the obtained by restricting t to−π/2 ≤ t ≤ π/2. Thus a possible answer is

x = 0, y = 5 cos t, z = −5 sin t, −π/2 ≤ t ≤ π/2.

30. Since the direction vectors for these lines (~v 1 = 2~i − ~j + 3~k and ~v 2 = −3~i + 5~j ) are not multiples of each other,these lines are not parallel. To determine if the lines intersect, we see if there are values of t for which the x, y, and zvalues are equal. Since we are not requiring that the particles be at the intersection point at the same time, the value of tfor line l1 might be different than the value of t for line l2. We use t = t1 for l1 and t = t2 for l2, and try to solve thesystem of equations:

5 + 2t1 = 4− 3t2,

−t1 = 1 + 5t2,

−2 + 3t1 = 4.

From the third equation, we see that t1 = 2. Substituting this into the second equation shows that t2 = − 35

. Substitutingthese values of t1 and t2 into the first equation yields the false statement 9 = 29

5. Thus, the two lines do not intersect.

31. The direction vectors for these two lines are ~v 1 = −2~i + ~j + 2~k and ~v 2 = 6~i − 3~j − 6~k . Since ~v 2 = −3~v 1, thelines have the same direction. To see that the lines are not the same, check that a point on the first line (say (3, 5,−4)) isnot on the second line: if 3 = 7 + 6t, then t = −2/3, so y = 1− 3(−2/3) = 3 6= 5. So the lines are parallel and do notintersect.

32. Since the direction vectors for these lines (~v 1 = 4~i − 3~j + 2~k and ~v 2 = −10~i + 2~k ) are not multiples of each other,these lines are not parallel. To determine if the lines intersect, we see if there are values of t for which the x, y, and zvalues are equal. Since we are not requiring that the particles be at the intersection point at the same time, the value of tfor line l1 might be different than the value of t for line l2. We use t = t1 for l1 and t = t2 for l2, and try to solve thesystem of equations:

1 + 4t1 = 3− 10t2,

1− 3t1 = 7,

2t1 = −6 + 2t2.

From the second equation, we see that t1 = −2. Substituting this into the third equation shows that t2 = 1. Substitutingthese values of t1 and t2 into the first equation yields the true statement −7 = −7. The lines intersect at the point wheret1 = −2 and t2 = 1, namely the point (−7, 7,−4).

33. Since the direction vectors for these lines (~v 1 = 2~i − 2~j + ~k and ~v 2 = −~i + 4~j − 2~k ) are not multiples of eachother, these lines are not parallel. To determine if the lines intersect, we see if there are values of t for which the x, y, andz values are equal. Since we are not requiring that the particles be at the intersection point at the same time, the value oft for line l1 might be different than the value of t for line l2. We use t = t1 for l1 and t = t2 for l2, and try to solve thesystem of equations:

3 + 2t1 = −1− t2,5− 2t1 = 4t2,

1 + t1 = 4− 2t2.

Adding the first two equations produces the equation 8 = −1+3t2 and so t2 = 3. Substituting this into the third equationshows that t1 = −3. Substituting these values of t1 and t2 into the first equation yields the false statement −3 = −4.Thus, the two lines do not intersect.

Problems

34. The graph is parameterized by x = t, y =√t. To obtain the segment, we restrict t to 1 ≤ t ≤ 16. Thus one possible

answer isx = t, y =

√t, 1 ≤ t ≤ 16.

17.1 SOLUTIONS 1201

35. The equation for z is z = 3. The x-coordinate goes from 4 to 0 and the y-coordinate from 0 to −3, so possible equationsfor x and y are x = 4 cos t and y = −3 sin t, with t from 0 to π/2. Thus one possible answer is

x = 4 cos t, y = −3 sin t, z = 3 0 ≤ t ≤ π/2.

36. The line segment PQ has length 10, so it must be a diameter of the circle. The center of the circle is therefore the midpointof PQ, which is the point (5, 0). The upper arc of the circle between P and Q can be parameterized as follows:

~r (t) = 5~i + 5(− cos t~i + sin t~j ), 0 ≤ t ≤ π.

The lower arc can be parameterized as follows:

~r (t) = 5~i + 5(cos t~i + sin t~j ), π ≤ t ≤ 2π.

37. Since the center of the circle must lie on the perpendicular bisector of the line segment PQ, it must be of the form (5, c).Since the center must be at a distance of 10 (the radius) from P = (0, 0), we have the equation 10 =

√52 + c2. Thus

c2 = 75, and so c = ±5√

3. There are two circles of radius 10 through the points P and Q, which divide each of the twocircles into two arcs.

The circle of radius 10 with center C = (5, 5√

3) can be parameterized by

~r (t) = 5~i + 5√

3~j + 10(± cos t~i + sin t~j ).

With the plus sign, the circle is traced out counterclockwise, whereas with the minus sign, the circle is traced out clock-wise. We must determine intervals of t-values that correspond to the arcs between P and Q. The fourth quadrant angle θbetween~i and −−→CQ is determined by the equation

10 cos θ = ‖~i ‖‖−−→CQ‖ cos θ =~i · −−→CQ = 5,

so θ = −π/3 or 5π/3. The third quadrant angle φ between~i and −−→CP is determined by the equation

10 cosφ = ‖~i ‖‖−−→CP‖ cosφ =~i · −−→CP = −5,

so φ = 4π/3. The top arc from P to Q can therefore be parameterized by

~r (t) = 5~i + 5√

3~j + 10(− cos t~i + sin t~j ), −π/3 ≤ t ≤ 4π/3

and the bottom arc by~r (t) = 5~i + 5

√3~j + 10(cos t~i + sin t~j ), 4π/3 ≤ t ≤ 5π/3.

The two arcs of the circle of radius 10 with center (5,−5√

3) can be obtained by reflecting the two previous param-eterizations across the x-axis (changing the sign of the y-coordinate), and so can be parameterized by

~r (t) = 5~i − 5√

3~j + 10(− cos t~i − sin t~j ), −π/3 ≤ t ≤ 4π/3

and~r (t) = 5~i − 5

√3~j + 10(cos t~i − sin t~j ), 4π/3 ≤ t ≤ 5π/3.

38. We find the parameterization in terms of the displacement vector −−→OP = 2~i + 5~j from the origin to the point P and thedisplacement vector −−→PQ = 10~i + 4~j from P to Q.

~r (t) =−−→OP + t

−−→PQ or, expressed in coordinates, ~r (t) = (2 + 10t)~i + (5 + 4t)~j . To see that this is correct, note

that the equation parameterizes a line because it is linear, that t = 0 corresponds to −−→OP + 0−−→PQ =

−−→OP , the vector from

the origin to P , and that t = 1 corresponds to −−→OP + 1−−→PQ =

−−→OQ, the vector from the origin to Q.


~r (t) =−−→OP + (t/5)

−−→PQ or ~r (t) = (2 + (t/5)10)~i + (5 + (t/5)4)~j


~r (t) =−−→OP + (t− 10)

−−→PQ or ~r (t) = (2 + (t− 10)10)~i + (5 + (t− 10)4)~j



~r (t) =−−→OP +

(t− 20

10

)−−→PQ =

(2 +

(t− 20

10

)10)~i +

(5 +

(t− 20

10

)4)~j .


~r (t) =−−→OP − t−−→PQ or ~r (t) = (2− 10t)~i + (5− 4t)~j

43. (a) The line segment starting at P0 and ending at P1 is parametrized by

~r (t) =−−→OP0 + t

−−−→P0P1, 0 ≤ t ≤ 1.

We write this in coordinates: Let P0 = (x0, y0, z0) and P1 = (x1, y1, z1). Then a vector between the points is−−−→P0P1 = (x1 − x0)~i + (y1 − y0)~j + (z1 − z0)~k , so

~r (t) = (x0~i + y0

~j + z0~k ) + t((x1 − x0)~i + (y1 − y0)~j + (z1 − z0)~k ),

or

x(t) = x0 + t(x1 − x0) = (1− t)x0 + tx1

y(t) = y0 + t(y1 − y0) = (1− t)y0 + ty1

z(t) = z0 + t(z1 − z0) = (1− t)z0 + tz1

Thus we have~r (t) = (1− t)−−→OP0 + t

−−→OP1.

(b) The parametric equation ~r (t) = t−−→OP0 + (1 − t)−−→OP1, 0 ≤ t ≤ 1 is the line segment from P1 to P0, the same line

segment as in part (a), but traversed in the opposite direction.

44. (a) A vector on the line will lie in both planes and will therefore be orthogonal to both normal vectors. To produce avector orthogonal to two given vectors, you can take their cross product.

(b) The vector (~i + 2~j − 3~k )× (3~i −~j + ~k ) = −~i − 10~j − 7~k is parallel to the line.(c) We need a point on the line and a vector parallel to the line. We found a vector in part (b). To find a point, we

set z = 0 and solve for x and y in the equations for the planes. We have x + 2y = 7 and 3x − y = 0 fromwhich x = 1 and y = 3. Hence, the point (1, 3, 0) is on the line. Finally, a parametric equation for the line is~r = (1− t)~i + (3− 10t)~j − 7t~k . Other answers are possible.

45. (a) Normal vectors to the two planes are

~n1 = 2~i −~j − 3~k and ~n2 =~i +~j + ~k .

The vector ~n1 × ~n2 is perpendicular to both planes and parallel to the line of intersection:

~n1 × ~n2 =

∣∣∣∣∣∣

~i ~j ~k

2 −1 −3

1 1 1

∣∣∣∣∣∣= 2~i − 5~j + 3~k .

(b) To check that the point (1,−1, 1) lies on the planes, substitute into each equation.

2x− y − 3z = 2 · 1− (−1)− 3 · 1 = 0

x+ y + z = 1− 1 + 1 = 1.

Thus, the point lies on both planes.(c) Parametric equations of the line are

x = 1 + 2t, y = −1− 5t, z = 1 + 3t.

46. Add the two equations to get 2x+3z = 5, or x = − 32z+ 5

2. Subtract the two equations to get 2y−z = 1, or y = 1

2z+ 1

2.

So a possible parameterization is

x = −3

2t+

5

2, y =

1

2t+

1

2, z = t.

17.1 SOLUTIONS 1203

47. Add the two equations to get 3x = 8, or x = 83

. Then we have

−y + z =1

3.

So a possible parameterization is

x =8

3, y = t, z =

1

3+ t.

48. Let f(x, y, z) = x2 + y2 − z. Then the surface z = x2 + y2 is a level surface of f at the value 0. The gradient of f isperpendicular to the level surface.

grad f = 2x~i + 2y~j − ~k = 2~i + 4~j − ~k .

So a possible parameterization isx = 1 + 2t, y = 2 + 4t, z = 5− t.

49. Parametric equations for a line in 3-space are in the form

x = x0 + at

y = y0 + bt

z = z0 + ct

where (x0, y0, z0) is a point on the line and the direction vector is ~v = a~i +b~j +c~k . We are given the point as (−4, 2, 3).The vector ~j + ~k is parallel to the yz-plane and at an angle of 45◦ to both positive y- and z-axes. Thus, the directionvector for this line is ~v = ~j + ~k . Parametric equations for this line are

x = −4,

y = 2 + t,

z = 3 + t.

50. The lines intersect if

c+ t = s

1 + t = 1− s5 + t = 3 + s.

Solving the last two equations gives t = −1 and s = 1. Substituting into the first equation gives c = 2.

51. These equations parameterize a line. Since (3 + t) + (2t) + 3(1 − t) = 6, we have x + y + 3z = 6. Similarly,x − y − z = (3 + t) − 2t − (1 − t) = 2. That is, the curve lies entirely in the plane x + y + 3z = 6 and in the planex− y− z = 2. Since the normals to the two planes, ~n1 =~i +~j + 3~k and ~n2 =~i −~j −~k are not parallel, the line isthe intersection of two nonparallel planes, which is a straight line in 3-dimensional space.

52. (a) Both paths are straight lines, the first passes through the point (−1, 4,−1) in the direction of the vector~i −~j + 2~k

and the second passes through (−7,−6,−1) in the direction of the vector 2~i + 2~j + ~k . The two paths are notparallel.

(b) Is there a time t when the two particles are at the same place at the same time? If so, then their coordinates will bethe same, so equating coordinates we get

−1 + t = −7 + 2t

4− t = −6 + 2t

−1 + 2t = −1 + t.

Since the first equation is solved by t = 6, the second by t = 10/3, and the third by t = 0, no value of t solves allthree equations. The two particles never arrive at the same place at the same time, and so they do not collide.


(c) Are there any times t1 and t2 such that the position of the first particle at time t1 is the same as the position of thesecond particle at time t2? If so then

−1 + t1 = −7 + 2t2

4− t1 = −6 + 2t2

−1 + 2t1 = −1 + t2.

We solve the first two equations and get t1 = 2 and t2 = 4. This is a solution for the third equation as well, so thethree equations are satisfied by t1 = 2 and t2 = 4. At time t = 2 the first particle is at the point (1, 2, 3), and at timet = 4 the second is at the same point. The paths cross at the point (1, 2, 3), and the first particle gets there first.

53. It is a straight line through the point (3, 5, 7) parallel to the vector~i −~j + 2~k . A linear parameterization of the same lineis x = 3 + t, y = 5− t, z = 7 + 2t.

54. The x and y coordinates describe circular motion around a circle of radius 5, while simultaneously the z coordinate isincreasing steadily. The motion is upward on the graph of the helix shown in Figure 17.1.

xy

z

Figure 17.1

55. The x and z coordinates describe circular motion around a circle of radius 1, while the y coordinate shows the particlemoving in along the positive y-axis and then moving back out again along the positive y-axis. See Figure 17.2.

xy

z

Figure 17.2

56. (a) The particle moves clockwise around a circle with center (a, a) and radius b, starting at (a, a + b). The motion hasperiod 2π/k.

(b) (i) Increasing b increases the radius.(ii) Increasing a moves the center away from the origin along the line y = x.

(iii) Increasing k makes the particle move faster and reduces the period.(iv) If a = b, the circle touches both the x- and y-axes at the points (a, 0) and (0, a), respectively.

57. The question is equivalent to asking if the line through (−3,−4, 2) and (4, 5, 0) enters the sphere x2 + y2 + z2 = 1. Aparameterization for this line is given by

x = −3 + 7t, y = −4 + 9t, z = 2− 2t.

17.1 SOLUTIONS 1205

We want to see whether the line intersects the sphere x2 + y2 + z2 = 1. Substituting we have

(−3 + 7t)2 + (−4 + 9t)2 + (2− 2t)2 = 1

29− 122t+ 134t2 = 0

Since (122)2 − 4(29)134 < 0, this equation has no real solutions. Thus, the line does not enter the sphere and the pointis visible.

58. The three shadows appear as a circle, a cosine wave and a sine wave, respectively.

−1 1

−1

1

x

yx = cos ty = sin t

−1 0 1

5

10

x

z

x = cos tz = t

−1 0 1

5

10

y

z

y = sin tz = t

Figure 17.3

59. (a) Parametric equations arex = 2 + at, y = 1 + bt, z = 3 + ct.

(b) The line goes through the origin if the position vector 2~i + ~j + 3~k is parallel to the vector a~i + b~j + c~k . Thisoccurs if a, b, c are in the ratio 2 : 1 : 3; that is if

a

2=b

1=c

3.

60. (a) The vector −2~i + 7~j + 4~k is parallel to the line. A normal to the plane is a~i + b~j + c~k . We want the normal tothe plane to be parallel to the line, so we take a = −2, b = 7, c = 4. Any value of d will do, for example d = 0.

(b) The same values of a, b, c as in part (a) work, though now we need to choose d so that the point (5, 3, 0) lies on theplane. So a = −2, b = 7, c = 4 and

d = −2(5) + 7(3) + 4(0) = 11.

(c) The normal a~i + b~j + c~k must be perpendicular to the vector −2~i + 7~j + 4~k , so

−2a+ 7b+ 4c = 0

We can choose any values of a, b, c which satisfy this equation, so a = 7, b = 2, c = 0 work. To ensure that the point(5, 3, 0), which lies on the line, also lies on the plane, substitute the coordinates of the point into the plane, giving

d = 7x+ 2y + 0z = 7(5) + 2(3) = 41.

61. Since the origin is beneath Denver and 1650 meters = 1.65 km, Denver’s coordinates, in kilometers, are (0, 0, 1.65). FromFigure 17.4, we see the x and y coordinates of Bismark are given by

x = 850 cos 60◦ = 425 km and y = 850 sin 60◦ = 736 km.

Since 550 meters = 0.55 km, the coordinates of Bismark in kilometers are (425, 736, 0.55).


60◦

850 km

Bismark

Denver xx

y

y

Figure 17.4

The velocity vector, ~v , of the plane is parallel to the vector −−→DB joining Denver to Bismark, where −−→DB = 425~i +736~j + (0.55− 1.65)~k = 425~i + 736~j − 1.1~k .

Since ||−−→DB|| =√

4252 + 7362 + 1.12 ≈ 850 km and the plane is moving at 650 km/hr, the velocity vector is givenby

~v =650

850(425~i + 736~j − 1.1~k ) = 325~i + 563~j − 0.84~k .

Since the plane is 8000 m = 8 km above Denver, it passes through the point (0, 0, 9.65). Therefore the parametricequation is

~r = 9.65~k + t(325~i + 563~j − 0.84~k ).

62. The line ~r = ~a + t~b is parallel to the vector~b and through the point with position vector ~a .

(a) is (vii). The equation~b · ~r = 0 is a plane perpendicular to~b and satisfied by (0, 0, 0).(b) is (ii). For any constant k, the equation ~b · ~r = k is a plane perpendicular to ~b . If k = ||~a || 6= 0, the plane does not

contain the origin.(c) is (iv). The equation (~a ×~b ) · (~r −~a ) = 0 is the equation of a plane which is satisfied by ~r = ~a , so the point with

position vector ~a lies on the plane. Since ~a ×~b is perpendicular to~b , the plane is parallel to the line, and thereforeit contains the line.

63. (a) We look along the line that passes through P = (1,−2,−1) and is parallel to ~v = ~i + 2~j + ~k . The question iswhich plane, the blue or the yellow, this line first meets.

Parametric equations for the line are

x = 1 + t, y = −2 + 2t, z = −1 + t.

We substitute these into the equations of the respective planes and solve for t in each case:

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

5t− 3 = 6 5t− 1 = 3

t = 95

t = 45

From this we see that the line first intersects the yellow plane 2x+ y + z = 3, when t = 4/5. So you see the yellowplane. (Note that we did not need to find the points of intersection of the line with the planes.)

(b) A vector from P to a point on the green line gives a direction looking directly at the line. If we get a parametricequation for the green line then we can write down a vector from P to any variable point on the line.

To get a parametric equation we need a vector parallel to the green line and a point that lies on the green line.We take the cross product of the normal of the blue plane, ~nb =~i + 3~j − 2~k , and the normal of the yellow plane,~ny = 2~i +~j + ~k . This gives a vector 5~i − 5~j − 5~k , so we take ~u =~i −~j − ~k as a vector parallel to the greenline.

We also need one point on the line. For that, we can choose a value of z, and find the corresponding values of xand y on both the blue and yellow planes. Taking z = 0, say, gives the equations x+ 3y = 6 and 2x+ y = 3, whichhave x = 3/5 and y = 9/5 as solutions. So a point on the green line is Q = (3/5, 9/5, 0). Therefore a parametricequation for the green line is

x =3

5+ t, y =

9

5− t, z = −t.

17.2 SOLUTIONS 1207

A vector from P = (1,−2,−1) to a variable point on the line is then, for −∞ < t <∞,

~v =(

1−(

3

5+ t))~i +

(−2−

(9

5− t))

~j + (−1− (−t))~k =(−2

5+ t)~i +

(−19

5+ t)~j + (−1 + t)~k .

Thus, if we look in the direction of ~v , for any value of t, we look at the line.(c) Consider the plane that contains the point P and the green line; let’s call it the green plane. The green plane divides

3-space into two half-spaces. From P , if we look in a direction pointing into one of the half-spaces we see the yellowplane (as in part (a)) and if we look in a direction pointing into the other half-space we see the blue plane. We haveto figure out which half-space is which.

We need a normal vector to the green plane. We know the point P = (1,−2,−1) on the plane and the equationof the green line. We find that a normal vector to the green plane is ~n = 2~i +~j + ~k .

From part (a) we know that the vector ~v = ~i + 2~j + ~k points from P into the half-space where we see theyellow plane. The dot product of ~n and ~v is

~n · ~v = 2 · 1 + 1 · 2 + 1 · 1 = 5 > 0.

This means that any vector pointing into this half-space has a positive dot product with ~n . Thus the condition on ageneral vector ~w = a~i + b~j + c~k to point into this half-space is

2a+ b+ c > 0;

Similarly, ~w points into the half-space where we see the blue plane if

2a+ b+ c < 0.

64. (a) Parametric equations arex = 1 + 2t, y = 5 + 3t, z = 2− t.

(b) We want to minimize D, the square of the distance of a point to the origin, where

D = (x− 0)2 + (y − 0)2 + (z − 0)2 = (1 + 2t)2 + (5 + 3t)2 + (2− t)2.

Differentiating to find the critical points gives

dD

dt= 2(1 + 2t)2 + 2(5 + 3t)3 + 2(2− t)(−1) = 0

2 + 4t+ 15 + 9t− 2 + t = 0

t =−15

14.

Thus

x = 1 + 2(−15

14

)=−8

7

y = 5 + 3(−15

14

)=

25

14

z = 2−(−15

14

)=

43

14.

Since the distance of the point on the line from the origin increases without bound as the magnitude of x, y, z increase,the only critical point of D must be a global minimum. Therefore, the point (−8/7, 25/14, 43/14) is the point onthe line closest to the origin.

Solutions for Section 17.2

Exercises

1. The velocity vector ~v is given by:

~v =d(t)

dt~i +

(d

dt(t3 − t)

)~j =~i + (3t2 − 1)~j .

The acceleration vector ~a is given by:

~a =d~v

dt=d(1)

dt~i +

(d

dt(3t2 − 1)

)~j = 6t~j .



~v =d

dt(2 + 3t)~i +

d

dt(4 + t)~j +

d

dt(1− t)~k = 3~i +~j − ~k .


~a =d~v

dt=d(3)

dt~i +

d(1)

dt~j − d(1)

dt~k = ~0


~v =d

dt(3 cos t)~i +

d

dt(4 sin t)~j = −3 sin t~i + 4 cos t~j .


~a =d~v

dt=d

dt(−3 sin t)~i +

d

dt(4 cos t)~j = −3 cos t~i − 4 sin t~j .

4. Since ~r (t) = 3 cos(t2)~i + 3 sin(t2)~j + t2~k , we have

~v (t) = −6t sin (t2)~i + 6t cos (t2)~j + 2t~k ,

~a (t) = (−6 sin (t2)− 12t2 cos (t2))~i + (6 cos (t2)− 12t2 sin (t2))~j + 2~k .

5. To find ~v (t) we first find dx/dt = 6t and dy/dt = 3t2. Therefore, the velocity vector is ~v = 6t~i + 3t2~j . The speed ofthe particle is given by the magnitude of the vector,

‖~v ‖ =

√(dx

dt

)2

+(dy

dt

)2

=√

(6t)2 + (3t2)2 = 3|t| ·√

4 + t2.

The particle stops when ~v = ~0 , so when 6t = 3t2 = 0. Therefore, the particle stops when t = 0.


~v =d

dt((t− 1)2)~i +

d

dt(2)~j +

d

dt(2t3 − 3t2)~k

= 2(t− 1)~i + (6t2 − 6t)~k .

The speed is given by:‖~v ‖ =

√(2(t− 1))2 + (6t2 − 6t)2 = 2|t− 1|

√1 + 9t2.

The particle stops when ~v = ~0 , so when 2(t − 1) = (6t2 − 6t) = 0. Since these are all satisfied only by t = 1, this isthe only time that the particle stops.

7. To find ~v (t) we first find dx/dt = 6t cos(t2) and dy/dt = −6t sin(t2). Therefore, the velocity is ~v = 6t cos(t2)~i −6t sin(t2)~j . The speed of the particle is given by

‖~v ‖ =√

(6t cos(t2))2 + (−6t sin(t2))2

=√

36t2(cos(t2))2 + 36t2(sin(t2))2

= 6|t|√

cos2(t2) + sin2(t2)

= 6|t|.The particle comes to a complete stop when speed is 0, that is, if 6|t| = 0, and so when t = 0 .


~v =d

dt(3 sin2 t)~i +

d

dt(cos t− 1)~j +

d

dt(t2)~k = 6 sin t cos t~i − sin t~j + 2t~k .

The speed is given by:‖~v ‖ =

√36 sin2 t cos2 t+ sin2 t+ 4t2.

The particle comes to a stop when ~v = ~0 , so when when 6 sin t cos t = − sin t = 2t = 0, and so the particle stops whent = 0.

17.2 SOLUTIONS 1209

9. We have

Length =

∫ 2

1

√(x′(t))2 + (y′(t))2 + (z′(t))2 dt =

∫ 2

1

√52 + 42 + (−1)2 dt =

√42.

This is the length of a straight line from the point (8, 5, 2) to (13, 9, 1).

10. We have

Length =

∫ 1

0

√(−et sin(et))2 + (et cos(et))2 dt

=

∫ 1

0

√e2t dt =

∫ 1

0

et dt

= e− 1.

This is the length of the arc of a unit circle from the point (cos 1, sin 1) to (cos e, sin e)—in other words between theangles θ = 1 and θ = e. The length of this arc is (e− 1).

11. We have

Length =

∫ 2π

0

√(−3 sin 3t)2 + (5 cos 5t)2 dt.

We cannot find this integral symbolically, but numerical methods show Length ≈ 24.6.

12. The velocity vector ~v is

~v =dx

dt~i +

dy

dt~j +

dz

dt~k = 3(2π)(− sin(2πt))~i + 3(2π) cos(2πt)~j + 0~k

= −6π sin(2πt)~i + 6π cos(2πt)~j .

The acceleration vector ~a is

~a =d2x

dt2~i +

d2y

dt2~j +

d2z

dt2~k = −6π(2π) cos(2πt)~i + 6π(2π)(− sin(2πt))~j

= −12π2 cos(2πt)~i − 12π2 sin(2πt)~j .

To check that ~v and ~a are perpendicular, we check that the dot product is zero:

~v · ~a = (−6π sin(2πt)~i + 6π cos(2πt)~j ) · (−12π2 cos(2πt)~i − 12π2 sin(2πt)~j )

= 72π3 sin(2πt) cos(2πt)− 72π3 cos(2πt) sin(2πt) = 0

The speed is‖~v ‖ = ‖ − 6π sin(2πt)~i + 6π cos(2πt)~j ‖ = 6π

√sin2(2πt) + cos2(2πt) = 6π,

and so is constant. The magnitude of the acceleration is

‖~a ‖ = ‖ − 12π2 cos(2πt)~i − 12π2 sin(2πt)~j ‖ = 12π2√

cos2(2πt) + sin2(2πt) = 12π2,

which is also constant.

13. The velocity vector ~v is

~v =dx

dt~i +

dy

dt~j +

dz

dt~k = 0~i + 2(3) cos(3t)~j + 2(3)(− sin(3t))~k

= 6 cos(3t)~j − 6 sin(3t)~k .


~a =d2x

dt2~i +

d2y

dt2~j +

d2z

dt2~k = 6(3)(− sin(3t))~j − 6(3) cos(3t)~k

= −18 sin(3t)~j − 18 cos(3t)~k .

To check that ~v and ~a are perpendicular, we check that the dot product is zero:

~v · ~a = (6 cos(3t)~j − 6 sin(3t)~k ) · (−18 sin(3t)~j − 18 cos(3t)~k )

= −108 cos(3t) sin(3t) + 108 sin(3t) cos(3t) = 0.


The speed is‖~v ‖ = ‖6 cos(3t)~j − 6 sin(3t)~k ‖ = 6

√sin2(3t) + cos2(3t) = 6,

and so is constant. The magnitude of the acceleration is

‖~a ‖ = ‖ − 18 sin(3t)~j − 18 cos(3t)~k ‖ = 18√

sin2(3t) + cos2(3t) = 18,

which is also constant.

14. In vector form the parameterization is

~r = 2~i + 3~j + 5~k + t2(~i − 2~j − ~k ).

Thus the motion is along the straight line through (2, 3, 5) in the direction of~i − 2~j − ~k . The velocity vector ~v is

~v =dx

dt~i +

dy

dt~j +

dz

dt~k = 2t(~i − 2~j − ~k )


~a =d2x

dt2~i +

d2y

dt2~j +

d2z

dt2~k = 2(~i − 2~j − ~k ).

The speed is‖~v ‖ = 2|t|‖~i − 2~j − ~k ‖ = 2

√6|t|.

The acceleration vector is constant and points in the direction of ~i − 2~j − ~k . When t < 0 the absolute value |t| isdecreasing, hence the speed is decreasing. Also, when t < 0 the velocity vector 2t(~i − 2~j − ~k ) points in the directionopposite to~i − 2~j − ~k . When t > 0 the absolute value |t| is increasing and hence the speed is increasing. Also, whent > 0 the velocity vector points in the same direction as~i − 2~j − ~k .

15. In vector form the parameterization is

~r =~i +−5~j − 2~k + (2t3 + 3t)(−~i + 2~j + 3~k ).

Thus the motion is along the straight line through (1,−5,−2) in the direction of−~i + 2~j + 3~k . The velocity vector ~v is

~v =dx

dt~i +

dy

dt~j +

dz

dt~k = (6t2 + 3)(−~i + 2~j + 3~k )


~a =d2x

dt2~i +

d2y

dt2~j +

d2z

dt2~k = 12t(−~i + 2~j + 3~k ).

The speed is‖~v ‖ = |6t2 + 3|‖ −~i + 2~j + 3~k ‖ = 3

√14|2t2 + 1| = 3

√14(2t2 + 1).

The graph of the speed is a parabola opening upward with vertex at t = 0. Thus the speed is decreasing when t < 0and increasing when t > 0. The velocity vector always points in the same direction −~i + 2~j + 3~k , since 6t2 + 3 isalways positive. The acceleration vector points in the opposite direction to −~i + 2~j + 3~k when t < 0 and in the samedirection when t > 0. Thus the acceleration vector points in the opposite direction to the speed when t < 0 and in thesame direction when t > 0.

16. At t = 2, the position and velocity vectors are

~r (2) = (2− 1)2~i + 2~j + (2 · 23 − 3 · 22)~k =~i + 2~j + 4~k ,

~v (2) = 2 · (2− 1)~i + (6 · 22 − 6 · 2)~k = 2~i + 12~k .

So we want the line going through the point (1, 2, 4) at the time t = 2, in the direction 2~i + 12~k :

x = 1 + 2(t− 2), y = 2 z = 4 + 12(t− 2).

17.2 SOLUTIONS 1211

Problems

17. The motion is circular, in a counterclockwise direction, around a circle centered at the origin of radius 3. At time t = 0,the particle is at point (3, 0). The velocity vector is

~v = (−6 sin 2t)~i + (6 cos 2t)~j .

At t = 0, the velocity is ~v = 6~j . The acceleration vector is

~a = (−12 cos 2t)~i + (−12 sin 2t)~j .

At t = 0, the acceleration is ~a = −12~i . The vectors ~v = 6~j and ~a = −12~i are shown in Figure 17.5.

−9 3 9

−6

−3

3

6

9

~v

~ax

y

Figure 17.5

18. Table 17.1 shows values near t = 1 with t changing by increments of 0.01.

Table 17.1 Values for the position vector ~r (t) = cos t~i + sin t~j near t = 1

t ~r

0.98 0.5570~i + 0.8305~j

0.99 0.5487~i + 0.8360~j

1.00 0.5403~i + 0.8415~j

1.01 0.5319~i + 0.8468~j

1.02 0.5234~i + 0.8521~j

As we go down the table, the x-values are decreasing by about 0.0084 and the y-values are increasing by about0.0054. Thus, a change in time of ∆t = 0.01 produces the change in position vector ∆x~i + ∆y~j ≈ −0.0084~i +0.0054~j . So, the velocity vector is approximately

~v ≈ 1

∆t

(∆x~i + ∆y~j

)≈ −0.84~i + 0.54~j .

Note that this velocity vector is perpendicular to the radius vector from the origin (0, 0) to the position (0.54, 0.84) attime t = 1. Finally, the x and y-values in the table are almost indistinguishable from those of linear motion given byx = 0.5403 − 0.84(t − 1), y = 0.8415 + 0.54(t− 1), which are the parametric equations for the tangent line throughthe point (0.5403, 0.8415).

19. (a) The curve is a spiral as shown in Figure 17.7.(b) We have:

~v (2) ≈ 2.001 cos 2.001− 2 cos 2

0.001~i +

2.001 sin 2.001− 2 sin 2

0.001~j

= −2.24~i + 0.08~j ,

~v (4) ≈ 4.001 cos 4.001− 4 cos 4

0.001~i +

4.001 sin 4.001− 4 sin 4

0.001~j

= 2.38~i − 3.37~j ,

~v (6) ≈ 6.001 cos 6.001− 6 cos 6

0.001+

6.001 sin 6.001− 6 sin 6

0.001~j

= 2.63~i + 5.48~j .


(c) Evaluating the exact formula ~v (t) = (cos t− t sin t)~i + (sin t+ t cos t)~j gives :

~v (2) = −2.235~i + 0.077~j ,

~v (4) = 2.374~i − 3.371~j ,

~v (6) = 2.637~i + 5.482~j .

See Figure 17.6.

x

y

t = 2

t = 4

t = 6

Figure 17.6: The spiralx = t cos t, y = t sin t and three velocity

vectors

x

y

Figure 17.7: The spiralx = t cos t, y = t sin t for 0 ≤ t ≤ 4π

20. Plotting the positions on the xy plane and noting their times gives the graph shown in Figure 17.8.

2 4 6 80

2

4

6

8

10

x

y

t = 0

t = .5

t = 1

t = 1.5

t = 2

t = 2.5, 3.5

t = 3

t = 4

Figure 17.8

(a) We approximate dx/dt by ∆x/∆t calculated between t = 1.5 and t = 2.5:

dx

dt≈ ∆x

∆t=

3− 7

2.5− 1.5=−4

1= −4.

Similarly,dy

dt≈ ∆y

∆t=

10− 5

2.5− 1.5=

5

1= 5.

So,~v (2) ≈ −4~i + 5~j and Speed = ‖~v ‖ =

√41.

(b) The particle is moving vertically at about time t = 1.5.(c) The particle stops at about time t = 3 and reverses course.

17.2 SOLUTIONS 1213

21. The velocity vector for this motion is~v = (2t− 6)~i + (3t2 − 3)~j .

The motion is vertical when the component in the~i direction is 0 and motion in~j direction is not 0. Motion in~i directionis 0 when

2t− 6 = 0,

t = 3.

At that time, motion in ~j direction is not 0. The motion is horizontal when the component in the ~j direction is 0 andmotion in~i direction is not 0. Motion in ~j direction is 0 when

3t2 − 3 = 0,

t = 1,−1.

At these times, motion in~i direction is not 0. To determine the end behavior, recall that a polynomial is approximated byits highest powered term for large values (positive or negative) of the independent variable. Thus, as t → ±∞, we havex ≈ t2 and y ≈ t3. The end behavior, and the x and y coordinates when the motion is vertical or horizontal, are shown inTable 17.2. The graph is shown in Figure 17.9.

Table 17.2

t x y

−∞ +∞ −∞−1 7 2

1 −5 −2

3 −9 18

+∞ +∞ +∞

−10 10 20−10

40

90

140

190

x

y

Figure 17.9

22. The velocity vector for this motion is

~v =dx

dt~i +

dy

dt~j = (3t2 − 12)~i + (2t+ 10)~j .

The motion is vertical when the component in the~i direction is 0 and motion in~j direction is not 0. Motion in~i directionis 0 when

3t2 − 12 = 0,

t = 2,−2.

At these times, motion in ~j direction is not 0. The motion is horizontal when the component in the ~j direction is 0 andmotion in~i direction is not 0. Motion in ~j direction is 0 when

2t+ 10 = 0,

t = −5.

At this time, the motion in~i direction is not 0. To determine the end behavior, recall that a polynomial is approximated byits highest powered term for large values (positive or negative) of the independent variable. Thus, as t → ±∞, we havex ≈ t3 and y ≈ t2. The end behavior, and the x and y coordinates when the motion is vertical or horizontal, are shown inTable 17.3. The graph is shown in Figure 17.10.


Table 17.3

t x y

−∞ −∞ +∞−5 −65 −25

−2 16 −16

2 −16 24

+∞ +∞ +∞

−800 −400

−40

40

80

x

y

Figure 17.10

23. A parameterization is~r (t) = 5~i + 4~j − 2~k + (t− 4)(2~i − 3~j + ~k )

or equivalentlyx = 5 + 2(t− 4), y = 4− 3(t− 4), z = −2 + (t− 4).

24. (a) The vector −−→PQ between the points is given by−−→PQ = 2~i + 5~j + 3~k .

Since ||−−→PQ|| =√

22 + 52 + 32 =√

38, the velocity vector of the motion is

~v =5√38

(2~i + 5~j + 3~k ).

(b) The motion is along a line starting at the point (3, 2,−5) and with the velocity vector from part (a). The equation ofthe line is

~r = 3~i + 2~j − 5~k + t~v = 3~i + 2~j − 5~k +5√38

(2~i + 5~j + 3~k )t,

sox = 3 +

10√38t, y = 2 +

25√38t, z = −5 +

15√38t.

25. (a) The particle starts at (2,−1, 5) so ~r 0 = 2~i −~j +5~k . In 5 seconds, the particle moves through a displacement givenby −−→PQ = 3~i + 4~j − 6~k . Its velocity, ~v , is given by

~v =3

5~i +

4

5~j − 6

5~k = 0.6~i + 0.8~j − 1.2~k .

Thus, the equation of the line is

~r = 2~i −~j + 5~k + t(0.6~i + 0.8~j − 1.2~k )

orx = 2 + 0.6t, y = −1 + 0.8t, z = 5− 1.2t.

(b) The velocity vector in part (a), ~v = 0.6~i + 0.8~j − 1.2~k , means that the particle is moving with

Speed = ||~v || =√

(0.6)2 + (0.8)2 + (1.2)2 = 1.562.

To make the speed = 5, take a new velocity vector given by

~v =5

1.562(0.6~i + 0.8~j − 1.2~k ) = 1.92~i + 2.56~j − 3.84~k .

Thus the equation of the line is

~r = 2~i −~j + 5~k + t(1.92~i + 2.56~j − 3.84~k )

orx = 2 + 1.92t, y = −1 + 2.56 t, t = 5− 3.84t.

17.2 SOLUTIONS 1215

26. (a) At t = 0, we have ~r (0) = 0~i + 0~j + 6.4~k , so the stone’s initial position is (0, 0, 6.4). Thus the rooftop is 6.4meters above the ground.

(b) The stone hits the ground when the height above the ground is 0; that is, when its z coordinate is 0:

6.4− 4.9t2 = 0

t = ±√

6.4

4.9= ±1.14.

Since t must be positive, the stone hits the ground about 1.14 seconds after it is thrown.(c) The velocity of the stone at time t is given by

~v (t) = ~r ′(t) = 10~i − 5~j − 9.8t~k ,

so when the stone hits the ground at t = 1.14 seconds,

~v (1.14) = 10~i − 5~j − 9.8(1.14)~k = 10~i − 5~j − 11.172~k .

The stone’s speed is given by ||~v (1.14)|| =√

102 + 52 + 11.1722 = 15.81 meters/sec.(d) The stone hits the ground at the point with position vector

~r (1.14) = 10(1.14)~i − 5(1.14)~j + (6.4− 4.9(1.14)2)~k ,

which is the point (11.4,−5.7, 0).(e) The acceleration of the stone at time t is given by

~a (t) = ~v ′(t) = −9.8~k .

Thus, the acceleration is constant; the stone hits the ground at an acceleration of−9.8 meters/sec2; that is 9.8 meters/sec2

downward.

27. (a) The top of the tower is at the point (0, 0, 20), so we want ~r (0) = 20~k . This is (I) and (IV). Only (IV) is goingdownward.

Projectile (IV) hits the ground when z = 0, which occurs when 20 − t2 = 0, so t =√

20 = 4.5. (We take thepositive root since the projectile is launched when t = 0.) At this time, ~r (

√20) = 8.9~j , so the projectile hits the

ground at the point (0, 8.9, 0), which is 8.9 meters from the base of the tower in the direction of the tree.(b) To hit the top of the tree, the projectile must go through the point (0, 20, 20). This is (II).

The projectile reaches the top of the tree when 2t2 = 20, so (taking the positive root) t =√

10 = 3.2 sec. Theprojectile is launched from ~r (0) = ~0 , the base of the tower.

(c) Projectiles launched from somewhere on the tower have x(0) = y(0) = 0 and 0 ≤ z(0) ≤ 20. Only (III) and (V)have nonzero x(0) and y(0).

To hit the tree, there must be a time for which the projectile is at a point (0, 20, z) for some 0 ≤ z ≤ 20.Since (III) has x(t) = 20 for all t, it does not hit the tree. So (V) is the answer.For (V), we have 2t = 20, when t = 10 sec. Then ~r (10) = 20~j + 10~k , so the projectile hits the tree at

(0, 20, 10), which is half way up.

28. (a) For any positive constant k, the parameterization

x = −5 sin(kt) y = 5 cos(kt)

moves counterclockwise on a circle of radius 5 starting at the point (0, 5). We choose k to make the period 8 seconds.If k · 8 = 2π, then k = π/4 and the parameterization is

x = −5 sin(πt

4

)y = 5 cos

(πt

4

).

(b) Since it takes 8 seconds for the particle to go around the circle

Speed =Circumference of circle

8=

2π(5)

8=

5π

4cm/sec.


29. Since the acceleration due to gravity is −9.8 m/sec2, we have ~r ′′(t) = −9.8~k . Integrating gives

~r ′(t) = C1~i + C2

~j + (−9.8t+ C3)~k ,

~r (t) = (C1t+ C4)~i + (C2t+ C5)~j + (−4.9t2 + C3t+ C6)~k .

The initial condition, ~r (0) = ~0 , implies that C4 = C5 = C6 = 0, thus

~r (t) = C1t~i + C2t~j + (−4.9t2 + C3t)~k .

To find the position vector, we need to find the values of C1, C2, and C3. This we do using the coordinates of thehighest point. When the rocket reaches its peak, the vertical component of the velocity is zero, so−9.8t+C3 = 0. Thus,at the highest point, t = C3/9.8. At that time

~r (t) = 1000~i + 3000~j + 10000~k ,

so, for the same value of t:C1t = 1000,

C2t = 3000,

−4.9t2 + C3t = 10, 000,

Substituting t = C3/9.8 into the third equation gives

−4.9(C3

9.8

)2

+C2

3

9.8= 10,000

C23 = 2(9.8)10,000

C3 = 442.7

Then C1 = 1000C3/9.8

= 22.1 and C2 = 3000C3/9.8

= 66.4. Thus,

~r (t) = 22.1t~i + 66.4t~j + (442.7t− 4.9t2)~k .

30. (a) No. The height of the particle is given by 2t; the vertical velocity is the derivative d(2t)/dt = 2. Because this is apositive constant, the vertical component of the velocity vector is upward at a constant speed of 2.

(b) When 2t = 10, so t = 5.(c) The velocity vector is given by

~v (t) =d~r

dt=dx

dt~i +

dy

dt~j +

dz

dt~k

= −(sin t)~i + (cos t)~j + 2~k .

From (b), the particle is at 10 units above the ground when t = 5, so at t = 5,

~v (5) = 0.959~i + 0.284~j + 2~k .

Therefore, ~v (5) = − sin(5)~i + cos(5)~j + 2~k .(d) At this point, t = 5, the particle is located at

~r (5) = (cos(5), sin(5), 10) = (0.284,−0.959, 10).

The tangent vector to the helix at this point is given by the velocity vector found in part (c), that is, ~v (5) = 0.959~i +

0.284~j + 2~k . So, the equation of the tangent line is

~r (t) = 0.284~i − 0.959~j + 10~k + (t− 5)(0.959~i + 0.284~j + 2~k ).

17.2 SOLUTIONS 1217

31. (a) The parametric equation describing Emily’s motion is

x = 10 cos(

2π

20t)

= 10 cos(π

10t), y = 10 sin

(2π

20t)

= 10 sin(π

10t)

z = constant.

Her velocity vector is

~v =dx

dt~i +

dy

dt~j +

dz

dt~k = −π sin

(π

10t)~i + π cos

(π

10t)~j .

Her speed is given by:

‖~v ‖ =

√(−π sin

(π

10t))2

+(π cos

(π

10t))2

+ 02

= π

√sin2

(π

10t)

+ cos2

(π

10t)

= π√

1 = π m/sec,

which is independent of time (as we expected). This is certainly the long way to solve this problem though, since wecould have simply divided the circumference of the circle (20π) by the time taken for a single rotation (20 seconds)to arrive at the same answer.

(b) When Emily drops the ball, it initially has Emily’s velocity vector, but it immediately begins accelerating in the z-direction due to the force of gravity. The motion of the ball will then be tangential to the merry-go-round, curvingdown to the ground. In order to find the tangential component of the ball’s motion, we must know Emily’s velocityat the moment she dropped the ball. Then we can integrate the velocity and obtain the position of the ball. AssumingEmily drops the ball at time t = 0, her position and velocity vector are

~r (0) = 10~i + 3~k and ~v (0) = π~j .

Thus, the ball has velocity only in the y-direction when it is dropped. In the z-direction, we have

Acceleration =d2z

dt2= −9.8 m/sec2.

Since the initial velocity 0 and initial height 3, we have

z = 3− 4.9t2.

The ball touches the ground when z = 0, that is, when t = 0.78 sec. In that time, the ball also travels π(0.78) = 2.45meters in the y-direction. So, the final position is (10, 2.45, 0). The distance between this point and P = (10, 0, 0)is 2.45 meters.

(c) The distance of the ball from Emily when it hits the ground is found by finding Emily’s position at t = 0.78 sec andusing the distance formula. Emily’s position when the ball hits the ground is (10 cos(0.078π), 10 sin(0.078π), 3) =(9.70, 2.43, 3). The distance between this point and the point where the ball struck the ground is:

d ≈√

(10− 9.70)2 + (2.45− 2.43)2 + (0− 3)2 = 3.01 meters.

Note that the merry-go-round does not rotate very much in the 0.78 sec needed for the ball to reach the ground, soour answer makes sense.

32. (a) The center of the wheel moves horizontally, so its y-coordinate will never change; it will equal 1 at all times. In onesecond, the wheel rotates 1 radian, which corresponds to 1 meter on the rim of a wheel of radius 1 meter, and so therolling wheel advances at a rate of 1 meter/sec. Thus the x-coordinate of the center, which equals 0 at t = 0, willequal t at time t. At time t the center will be at the point (x, y) = (t, 1).

(b) By time t the spot on the rim will have rotated t radians clockwise, putting it at angle −t as in Figure 17.11. Thecoordinates of the spot with respect to the center of the wheel are (cos(−t), sin(−t)). Adding these to the coordinates(t, 1) of the center gives the location of the spot as (x, y) = (t+ cos t, 1− sin t). See Figure 17.12.


(t, 1)

−t

x

y

Figure 17.11

x

y

Figure 17.12

33. (a) Let the ant begin the trip at time t = 0, and let’s place the origin of our coordinate system at the center of the disk.We align the axes so that at time t = 0 the radius along which the ant crawls falls on the positive x-axis. At time tseconds, the ant is at a distance of r = t cm from the origin and at angle θ = 2πt radians from the positive x-axis.The Cartesian coordinates of this point are (x, y) = (r cos θ, r sin θ) = (t cos(2πt), t sin(2πt)). We can write theparametric equations of the ant’s motion in vector form as

~r (t) = t cos(2πt)~i + t sin(2πt)~j , 0 ≤ t ≤ 100.

(b) The velocity vector of the ant is the derivative

~v (t) = ~r ′(t) = (cos(2πt)− 2πt sin(2πt))~i + (sin(2πt) + 2πt cos(2πt))~j .

The speed is the magnitude of the velocity vector

‖~v ‖ = ((cos(2πt)− 2πt sin(2πt))2 + (sin(2πt) + 2πt cos(2πt))2)1/2

= (1 + 4π2t2)1/2 cm/sec.

Observe that the speed of the ant is increasing. Even though the ant is crawling at constant rate on the disk, the turningof the disk moves the ant faster and faster as it gets closer to the edge.

(c) The acceleration vector is

~a = ~v ′(t) = (−4π sin(2πt)− 4π2t cos(2πt))~i + (4π cos(2πt)− 4π2t sin(2πt))~j .

The magnitude of the acceleration is

‖~a ‖ = ((−4π sin(2πt)− 4π2t cos(2πt))2 + (4π cos(2πt)− 4π2t sin(2πt))2)1/2

= 4π(1 + π2t2)1/2 cm/sec2.

34. At time t object B is at the point with position vector ~r B(t) = ~r A(2t), which is exactly where object A is at time 2t.Thus B visits the same points as A, but does so at different times; A gets there later. While B covers the same path as A,it moves twice as fast. To see this, note for example that between t = 1 and t = 3, object B moves along the path from~r B(1) = ~r A(2) to ~r B(3) = ~r A(6) which is traversed by object A during the time interval from t = 2 to t = 6. It takesA twice as long to cover the same ground.

In the case where ~r A(t) = t~i + t2~j , both objects move on the parabola y = x2. Both A and B are at the origin attime t = 0, but B arrives at the point (2, 4) at time t = 1, whereas A does does not get there until t = 2.

35. In uniform circular motion the velocity vector is tangent to the circle of motion and the acceleration vector is directedtoward the center of the circle. At all times the velocity ~v and acceleration ~a are perpendicular. Since ~v ·~a = (2~i +~j ) ·(~i +~j ) = 3 6= 0, ~v and ~a are not perpendicular, and so the object can not be in uniform circular motion.

36. The acceleration vector points from the object to the center of the orbit, and the velocity vector points from the objecttangent to the circle in the direction of motion. From Figure 17.13 we see that the movement is counterclockwise.

17.3 SOLUTIONS 1219

~a

~v

Center of orbit

Figure 17.13

37. (a) If ∆t = ti+1 − ti is small enough so that Ci is approximately a straight line, then we can make the linear approxi-mations

x(ti+1) ≈ x(ti) + x′(ti)∆t,

y(ti+1) ≈ y(ti) + y′(ti)∆t,

z(ti+1) ≈ z(ti) + z′(ti)∆t,

and so

Length of Ci ≈√

(x(ti+1)− x(ti))2 + (y(ti+1)− y(ti))2 + (z(ti+1)− z(ti))2

≈√x′(ti)2(∆t)2 + y′(ti)2(∆t)2 + z′(ti)2(∆t)2

=√x′(ti)2 + y′(ti)2 + z′(ti)2∆t.

(b) From point (a) we obtain the approximation

Length of C =∑

length of Ci

≈∑√

x′(ti)2 + y′(ti)2 + z′(ti)2∆t.

The approximation gets better and better as ∆t approaches zero, and in the limit the sum becomes a definite integral:

Length of C = lim∆t→0

∑√x′(ti)2 + y′(ti)2 + z′(ti)2 ∆t

=

∫ b

a

√x′(t)2 + y′(t)2 + z′(t)2 dt.


Exercises

1. ~V = x~i

2. ~V = −y~i3. ~V = x~i + y~j = ~r

4. ~V = −x~i − y~j = −~r5. ~V = −y~i + x~j

6. ~V =~r

‖~r ‖ : vectors are of unit length and point outward.


7. See Figure 17.14.

x

y

Figure 17.14: ~F (x, y) = 2~i + 3~j

x

y

Figure 17.15: ~F (x, y) = y~i



x

y

Figure 17.16: ~F (x, y) = −y~j

x

y

Figure 17.17: ~F (~r ) = 2~r



x

y

Figure 17.18: ~F (~r ) =~r

‖~r ‖ Figure 17.19: ~F (~r ) = −~r /‖~r ‖3


17.3 SOLUTIONS 1221


x

y

Figure 17.20: ~F (x, y) = −y~i + x~j

x

y

Figure 17.21: ~F (x, y) = 2x~i + x~j


15.

x

y

Figure 17.22: ~F (x, y) = (x+y)~i +(x−y)~j

16. Notice that for a repulsive force, the vectors point outward, away from the particle at the origin, for an attractive force, thevectors point toward the particle. So we can match up the vector field with the description as follows:

(a) IV(b) III(c) I(d) II

Problems

17. Vector fields (B) and (C) both appear to be constant, and therefore correspond to the equally spaced level curves in (I) and(II). Since the gradient points toward increasing values of the function, (B) corresponds to (II) and (C) corresponds to (I).

Vector field (A) points away from the center, so it corresponds to (IV), which has a minimum in the center.Vector field (D) points toward the center, so it corresponds to (III) which has a maximum at the center.

18. The sketches show that the vector fields point in different directions on the y-axis, so we examine the formulas for thevector fields on the y-axis. On the y-axis, where x = 0, we have:

~F (0, y) = y~j , a vector pointing up if y > 0 and down if y < 0, as in I~G (0, y) = −y~i , a vector pointing left if y > 0 and right if y < 0, as in II~H (0, y) = −y~j , a vector pointing down if y > 0 and up if y < 0, as in IIISo ~F is I, ~G is II, and ~H is III.


19. The sketches show that the vector fields can be distinguished by the directions they point on the coordinate axes, so weexamine the formulas for the vector fields on the axes.

(a) ~F (0, y) + ~G (0, y) = −y~i + y~j = y(−~i + ~j ), a vector pointing up to the left if y > 0 and down to the right ify < 0, as in II.

(b) ~F (0, y) + ~H (0, y) = ~0 , the zero vector as in III.(c) ~G (0, y) + ~H (0, y) = −y~i − y~j = −y(~i + ~j ), a vector pointing down to the left if y > 0 and up to the right if

y < 0, as in I and IV. ~G (x, 0) + ~H (x, 0) = x~i + x~j = x(~i +~j ), a vector pointing up to the right if x > 0 anddown to the left if x < 0, as in II and IV, so ~G + ~H is IV.

(d) − ~F (x, 0) + ~G (x, 0) = −x~i + x~j = x(−~i +~j ), a vector pointing up to the left if x > 0 and down to the right ifx < 0, as in I.

20. One possible solution is ~F (x, y) = x~i . See Figure 17.23.

−3 −2 −1 1 2 3

−3

−2

−1

1

2

3

x

y

Figure 17.23

−3 −2 −1 1 2 3

−3

−2

−1

1

2

x

y

Figure 17.24

21. If we let ~F (x, y) =−x~i − y~j√x2 + y2

, then all vectors will be of unit length and will point toward the origin. See Figure 17.24.

22. The position vector at each point is ~r = x~i +y~j . We want to find ~F (x, y) = A~i +B~j such that ~F ·~r = Ax+By = 0.One possible answer is let A = y and B = −x. So ~F (x, y) = y~i − x~j . Since the vectors are of unit length, we get

~F (x, y) =y~i − x~j√x2 + y2

.

−3 −2 −1 1 2 3

−3

−2

−1

1

2

3

x

y

23. (a) The gradient is perpendicular to the level curves. See Figures 17.26 and 17.25. A function always increases in thedirection of its gradient; this is why the values on the level curves of f and g increase as we approach the origin.

17.3 SOLUTIONS 1223

5

4

32

10

x

y

Figure 17.25: Level curves z = f(x, y)

543

21

0

x

y

Figure 17.26: Level curves z = g(x, y)

(b) f climbs faster at outside, slower at center; g climbs slower at outside, faster at center:

x

z

f(x, y)

g(x, y)

Figure 17.27

This can be understood if we notice that the magnitude of the gradient of f decreases as one approaches theorigin whereas the magnitude of the gradient of g increases (at least for a while - what happens very close to theorigin depends on the behavior of grad g in the region. One possibility for g is shown in Figure 17.27; the graph of gcould also have a sharp peak at 0 or even blow up.)

24. (a) The line l is parallel to the vector ~v =~i − 2~j − 3~k . The vector field ~F is parallel to the line when ~F is a multipleof ~v . Taking the multiple to be 1 and solving for x, y, z we find a point at which this occurs:

x = 1

x+ y = −2

x− y + z = −3

gives x = 1, y = −3, z = −7, so a point is (1,−3,−7). Other answers are possible.(b) The line and vector field are perpendicular if ~F · ~v = 0, that is

(x~i + (x+ y)~j + (x− y + z)~k ) · (~i − 2~j − 3~k ) = 0

x− 2x− 2y − 3x+ 3y − 3z = 0

−4x+ y − 3z = 0.

One point which satisfies this equation is (0, 0, 0). There are many others.(c) The equation for this set of points is −4x+ y − 3z = 0. This is a plane through the origin.

25. (a) Since the velocity of the water is the sum of the velocities of the individual fields, then the total field should be

~v = ~v stream + ~v fountain.


It is reasonable to represent ~v stream by the vector field ~v stream = A~i , since A~i is a constant vector field flowing inthe i-direction (provided A > 0). It is reasonable to represent ~v fountain by

~v fountain = K~r r/r2 = K(x2 + y2)−1(x~i + y~j ),

since this is a vector field flowing radially outward (provided K > 0), with decreasing velocity as r gets larger.We would expect the velocity to decrease as the water from the fountain spreads out. Adding the two vector fieldstogether, we get

~v = A~i +K(x2 + y2)−1(x~i + y~j ), A > 0,K > 0.

(b) The constants A and K signify the strength of the individual components of the field. A is the strength of the flow ofthe stream alone (in fact it is the speed of the stream), and K is the strength of the fountain acting alone.

(c)

x

y

Figure 17.28: A = 1,K = 1

x

y

Figure 17.29: A = 2,K = 1

x

y

Figure 17.30: A = 0.2,K = 2

26. (a) The vector field ~L = 0 ~F + ~G = −y~i + x~j is shown in Figure 17.31.(b) The vector field ~L = a~F + ~G = (ax− y)~i + (ay + x)~j where a > 0 is shown in Figure 17.32.(c) The vector field ~L = a~F + ~G = (ax− y)~i + (ay + x)~j where a < 0 is shown in Figure 17.33.

x

y

Figure 17.31

x

y

Figure 17.32

x

y

Figure 17.33

17.4 SOLUTIONS 1225

27. (a) The vector field ~L = ~F + 0 ~G = x~i + y~j is shown in Figure 17.34.(b) The vector field ~L = ~F + b ~G = (x− by)~i + (y + bx)~j where b > 0 is shown in Figure 17.35.(c) The vector field ~L = ~F + b ~G = (x− by)~i + (y + bx)~j where b < 0 is shown in Figure 17.36.

x

y

Figure 17.34

x

y

Figure 17.35

x

y

Figure 17.36

28. (a) Dividing a vector ~F by its magnitude always produces the unit vector in the same direction as ~F .(b) Since

‖ ~N ‖ = ‖(1/F )(−v~i + u~j )‖ = (1/F )√v2 + u2 = (1/F )F = 1,

then ~N is a unit vector. We check that ~N is perpendicular to ~F using the dot product of ~N and ~F :

~N · ~F = (1/F )(−v~i + u~j ) · (u~i + v~j ) = 0.

Which side of ~F does ~N point? The vector ~k is pointing out of the diagram. Since the cross product ~k × ~F isperpendicular to both ~k and ~F , then ~N lies in the xy-plane and points at a right angle to the direction of ~F . By theright-hand rule, ~N points to the left as shown in the figure.


Exercises

1. Since x′(t) = 3 and y′(t) = 0, we have x = 3t+ x0 and y = y0. Thus, the solution curves are y = constant.

−18 18

−18

18

x

y

Figure 17.37: The field ~v = 3~i

−18 18

−18

18

x

y

Figure 17.38: The flow y =constant


2. Since x′(t) = 0 and y′(t) = 2, we have x = x0 and y = 2t+ y0. Thus, the solution curves are x = constant.

−12 12

−12

12

x

y

Figure 17.39: The field ~v = 2~j

−12 12

−12

12

x

y

Figure 17.40: The flow x = constant

3. Since x′(t) = 3 and y′(t) = −2, we have x = 3t+ x0 and y = −2t+ y0. Thus the flow lines are straight lines parallelto the vector 3~i − 2~j . Alternatively, we have dy

dx= − 2

3. Thus, y = − 2

3x+ c, where c is a constant.

−9 9

−9

9

x

y

Figure 17.41: The field ~v = 3~i − 2~j

−9 9

−9

9

x

y

Figure 17.42: The flow y = − 23x+ c

4. y

x

Figure 17.43: ~v (t) = x~i + y~j

x

y

Figure 17.44: The flow x = aet,y = bet.

As~v (t) =

dx

dt~i +

dy

dt~j ,

17.4 SOLUTIONS 1227

the system of differential equations is {dxdt

= xdydt

= y.

Sincedx(t)

dt=

d

dt[aet] = aet = x(t)

anddy(t)

dt=

d

dt[bet] = bet = y(t),

the given flow satisfies the system. By eliminating the parameter t in x(t) and y(t), the solution curves obtained arey = b

ax.

5.

x

y

Figure 17.45: ~v (t) = x~i

−18 18

−18

18

x

y

Figure 17.46: The flow x(t) = aet, y(t) = b

As~v (t) =

dx

dt~i +

dy

dt~j ,


= xdydt

= 0.

Sinced

dt(x(t)) =

d

dt(aet) = x

andd

dt(y(t)) =

d

dt(b) = 0,

the given flow satisfies the system. The solution curves are the horizontal lines y = b..

6.

x

y

Figure 17.47: ~v (t) = x~j

−12 12

−12

12

x

y

Figure 17.48: The flowx(t) = a, y(t) = at+ b


As~v (t) =

dx

dt~i +

dy

dt~j ,


= 0dydt

= x.

Sinced(x(t))

dt=

d

dt(a) = 0

andd(y(t))

dt=

d

dt(at+ b) = a = x,

the given flow satisfies the system. The solution curves are the vertical lines x = a.

7.

x

y

Figure 17.49: ~v (t) = x~i − y~j

x

y

Figure 17.50: The flow x = aet,y = be−t

As~v (t) =

dx

dt~i +

dy

dt~j ,


= xdydt

= −y.Since

dx(t)

dt=

d

dt[aet] = aet = x(t)

anddy(t)

dt=

d

dt[be−t] = −be−t = y(t),

the given flow satisfies the system. By eliminating the parameter t in x(t) and y(t), the solution curves obtained arexy = ab.

8. y

x

Figure 17.51: ~v (t) = y~i − x~j

y

x

Figure 17.52: The flow x = a sin t,y = a cos t

As~v (t) =

dx

dt~i +

dy

dt~j ,

17.4 SOLUTIONS 1229


= ydydt

= −x.Since

dx(t)

dt=

d

dt[a sin t] = a cos t = y(t)

anddy(t)

dt=

d

dt[a cos t] = −a sin t = −x(t),

the given flow satisfies the system. By eliminating the parameter t in x(t) and y(t), the solution curves obtained arex2 + y2 = a2.

9.

x

y

Figure 17.53: ~v (t) = y~i + x~j

x

y

Figure 17.54: The flowx(t) = a(et + e−t), y(t) = a(et − e−t)

As~v (t) =

dx

dt~i +

dy

dt~j ,


= ydydt

= x.

Sincedx(t)

dt=

d

dt[a(et + e−t)] = a(et − e−t) = y(t)

anddy(t)

dt=

d

dt[a(et − e−t)] = a(et + e−t) = x(t),

the given flow satisfies the system. By eliminating the parameter t in x(t) and y(t), the solution curves obtained arex2 − y2 = 4a2.

10. The vector field is given by ~v = y2~i + 2x2~j , that is, the flow line (x(t), (y(t)) satisfies

x′(t) = y2

y′(t) = 2x2

We’ll use Euler’s method with ∆t = 0.1 to find the parameterized curve (x(t), y(t)) through (1, 2). So

xn+1 = xn + 0.1y2n

yn+1 = yn + (0.1)2x2n

Initially, that is when t = 0, we have (x0, y0) = (1, 2). Then

x1 = x0 + 0.1y20 = 1 + 0.1 · 22 = 1.4

y1 = y0 + 0.1 · 2x20 = 2 + 0.1 · 2 · 12 = 2.2


Thus, we see that after one step, x1 = 1.4 and y1 = 2.2. Further values are given in the Table 17.4.

Table 17.4x 1.4 1.884 2.556 3.646 5.770y 2.2 2.592 3.302 4.609 7.268

Problems

11. This corresponds to area A in Figure 17.55.

0 100 200 300 400 500 6000

100

200

300

400

500

600

km

km

-A

D

� C

-B

Figure 17.55

12. This corresponds to area B in Figure 17.55 in Problem 11.

13. This corresponds to area C in Figure 17.55 in Problem 11.

14. This corresponds to area D in Figure 17.55 in Problem 11.

15. (a) At every point (x, y) in the plane, the vector ~G (x, y) has the same direction as ~F (x, y) but ~G (x, y) is twice as long.For the case where ~F (x, y) = −y~i + x~j see Figures 17.56 and 17.57.

x

y

Figure 17.56: ~F = −y~i + x~j

x

y

Figure 17.57: G(x, y) = 2(−y~i + x~j )

17.4 SOLUTIONS 1231

(b) At every point in the plane the two vector fields ~F and ~G have the same direction. Therefore the flow lines of the twovector fields have the same slopes at every point. By the uniqueness of solutions of differential equations with initialconditions, the flow lines of the two vector fields through any given point must be the same. This means that if twoobjects are placed at the same point, one into the flow of ~F and the other into the flow of ~G , they will move on exactlythe same paths. However, they will move at different speeds. The two flows will have different parameterizations.

For the case where ~F (t) = −y~i + x~j both flows are circular about the origin, but the flow of ~G is twice asfast as the flow of ~F .

16. The object’s motion is described by a function ~r (t) = x(t)~i + y(t)~j , whered~r

dt(t) = ~F (x(t), y(t)) = u(x(t),

y(t))~i + v(x(t), y(t))~j . Using the chain rule to differentiate, we have

~a (t) =d2~r

dt2

=du

dt~i +

dv

dt~j

= (uxdx

dt+ uy

dy

dt)~i + (vx

dx

dt+ vy

dy

dt)~j

= (uxu+ uyv)~i + (vxu+ vyv)~j

17. (a) Perpendicularity is indicated by zero dot product. We have ~v · gradH = (−Hy~i +Hx

~j ) · (Hx~i +Hy

~j ) = 0.(b) If ~r (t) = x(t)~i + y(t)~j is a flow line we have, using the chain rule,

d

dtH(x(t), y(t)) = Hx

dx

dt+Hy

dy

dt= Hx(−Hy) +Hy(Hx) = 0.

Thus H(x(t), y(t)) is constant which shows that a flow line stays on a single level curve of H .For a different solution, use geometric reasoning. The vector field ~v is tangent to the level curves of H because,

by part (a), ~v and the level curves are both perpendicular to the same vector field gradH . Thus the level curves ofH and the flowlines of ~v run in the same direction.

18. The directions of the flow lines are as shown.

(a) III(b) I(c) II(d) V(e) VI(f) IV

−2 −1 1 2

−1

1

x

y(I)

−2 −1 1 2

−1

1

x

y(II)

−2 −1 1 2

−1

1

x

y(III)

−2 −1 1 2

−1

1

x

y(IV)

−5 5

−5

5

x

y(V)

−55

−5

5

x

y(VI)

19. (a) Each vector in the vector field ~v is horizontal, tangent to a circle whose center is on the z-axis, and pointing coun-terclockwise when viewed from above. Thus, ~v is parallel to −y~i + x~j . The point (x, y, z) is moving on a circle ofradius r =

√x2 + y2 and has

Speed =2πr

24=πr

12.


Since the vector at the point (x, y, z) has magnitude πr/12 and is parallel to the unit vector(−y~i + x~j )/

√x2 + y2, we have

~v =πr

12

(−y~i + x~j√x2 + y2

)=

π

12(−y~i + x~j ) meters/hr.

(b) A point moves in a horizontal circle, centered on the z-axis, and oriented counter-clockwise when viewed from above.These circles are the flow lines.


Exercises

1. A horizontal disk of radius 5 in the plane z = 7.

2. A circle of radius 5 in the plane z = 7.

3. A cylinder of radius 5 centered around the z-axis and stretching around from z = 0 to z = 7.

4. A helix (curve) of radius 5 which makes one turn about the z-axis, starting at the point (5, 0, 0) and ending at the point(5, 0, 10π).

5. Since z = r =√x2 + y2, we have a cone around the z-axis. Since 0 ≤ r ≤ 5, we have 0 ≤ z ≤ 5, so the cone has

height and maximum radius of 5.

6. Since x2 + y2 = 4z2, we have z = 12

√x2 + y2. Thus we have a cone of height 7 and maximum radius 14, centered

around the z-axis.

7. Since (x3)2 + ( y

2)2 = 1, so x2

9+ y2

4= 1, which is the equation for an ellipse, we have a cylinder with an elliptical

cross-section. The ellipse is centered around the z-axis from z = 0 to z = 7.

8. This is a parabolic cylinder y = x2, between x = −5 and x = 5, with its axis along the z-axis, stretching from z = 0 toz = 7.

9. The top half of the sphere (z ≥ 0).

10. The half of the sphere with y ≤ 0.

11. A vertical segment lying between two longitudinal lines (θ = π4

and θ = π3

) and stretching between the poles.

12. Half the horizontal ring around the sphere between two latitude lines (φ = π4

and φ = π3

) in the northern hemisphere.

Problems

13. We take two unit vectors, ~u and ~v , parallel to the plane, and perpendicular to each other, say ~u = ~j and ~v =~i√

2− ~k√

2.

Then,

~r = ~j + 2 cos θ~u + 2 sin θ~v

= ~j + 2 cos θ~j + 2 sin θ(~i√

2−

~k√2

)

=√

2 sin θ~i + (1 + 2 cos θ)~j −√

2 sin θ~k , for 0 ≤ θ ≤ 2π,

or

x =√

2 sin θ

y = (1 + 2 cos θ)

z = −√

2 sin θ 0 ≤ θ ≤ 2π

14. The plane through (1, 3, 4) and orthogonal to ~n = 2~i +~j − ~k is given by 2(x− 1) + (y − 3)− (z − 4) = 0, that is,

2x+ y − z − 1 = 0.

17.5 SOLUTIONS 1233

Thus, thinking of the plane as z = 2x+ y − 1, one possible parameterization is

x = u, y = v, z = 2u+ v − 1.

15. (a) We want to find s and t so that

2 + s = 4

3 + s+ t = 8

4t = 12

Since s = 2 and t = 3 satisfy these equations, the point (4, 8, 12) lies on this plane.(b) Are there values of s and t corresponding to the point (1, 2, 3)? If so, then

1 = 2 + s

2 = 3 + s+ t

3 = 4t

From the first equation we must have s = −1 and from the third we must have t = 3/4. But these values of s and tdo not satisfy the second equation. Therefore, no value of s and t corresponds to the point (1, 2, 3), and so (1, 2, 3)is not on the plane.

16. If the planes are parallel, then their normal vectors will also be parallel. The equation of the first plane can be written

~r = 2~i + 4~j + ~k + s(~i +~j + 2~k ) + t(~i −~j ).

A normal vector to the first plane is ~n 1 = (~i +~j + 2~k )× (~i −~j ) = 2~i + 2~j − 2~k . The second plane can be written

~r = 2~i + s(~i + ~k ) + t(2~i +~j − ~k ).

A normal vector to the second plane is ~n 2 = (~i + ~k ) × (2~i + ~j − ~k ) = −~i + 3~j + ~k . Since ~n 1 and ~n 2 are notparallel, neither are the two planes.

17. Since you walk 5 blocks east and 1 block west, you walk 5 blocks in the direction of ~v 1, and 1 block in the oppositedirection. Thus,

s = 5− 1 = 4,

Similarly,t = 4− 2 = 2.

Hence

x~i + y~j + z~k = (x0~i + y0

~j + z0~k ) + 4~v1 + 2~v2

= (x0~i + y0

~j + z0~k ) + 4(2~i − 3~j + 2~k ) + 2(~i + 4~j + 5~k )

= (x0 + 10)~i + (y0 − 4)~j + (z0 + 18)~k .

Thus the coordinates are:x = x0 + 10, y = y0 − 4, z = z0 + 18.

18. (a) Nearer to the equator.(b) Farther from the north pole.(c) Farther from Greenwich.

19. A horizontal circle in the northern hemisphere at a latitude of 45◦ north of the equator.

20. A vertical half-circle, going from the north to south poles.

21. The cross sections of the cylinder perpendicular to the z-axis are circles which are vertical translates of the circle x2+y2 =a2, which is given parametrically by x = a cos θ, y = a sin θ. The vector a cos θ~i + a sin θ~j traces out the circle, atany height. We get to a point on the surface by adding that vector to the vector z~k . Hence, the parameters are θ, with0 ≤ θ ≤ 2π, and z, with 0 ≤ z ≤ h. The parametric equations for the cylinder are

x~i + y~j + z~k = a cos θ~i + a sin θ~j + z~k ,

which can be written asx = a cos θ, y = a sin θ, z = z.


22. We use spherical coordinates φ and θ as the two parameters. Since the radius is 5, we can take

x = 5 sinφ cos θ, y = 5 sinφ sin θ, z = 5 cosφ.

23. The sphere (x−a)2 +(y−b)2 +(z−c)2 = d2 has center at the point (a, b, c) and radius d. We use spherical coordinatesθ and φ as the two parameters. The parameterization of the sphere with center at the origin and radius d is

x = d sinφ cos θ, y = d sinφ sin θ, z = d cosφ.

Since the given sphere has center at the point (a, b, c) we add the displacement vector a~i + b~j + c~k to the radial vectorcorresponding to a parameterization of the sphere with center at the origin and radius d to give

x = a+ d sinφ cos θ, 0 ≤ φ ≤ π,y = b+ d sinφ sin θ, 0 ≤ θ ≤ 2π,

z = c+ d cosφ.

To check that this is a parameterization for the given, we substitute for x, y, z:

(x− a)2 + (y − b)2 + (z − c)2

= d2 sin2 φ cos2 θ + d2 sin2 φ sin2 θ + d2 cos2 φ

= d2 sin2 φ+ d2 cos2 φ = d2.

24. Let (θ, π/2) be the original coordinates. If θ < π, then the new coordinates will be (θ + π, π/4). If θ ≥ π, then the newcoordinates will be (θ − π, π/4).

25. If we set z = u, x2 + y2 = u2 is the equation of a circle with radius |u|. Hence a parameterization of the cone is:

x = u cos v,

y = u sin v, 0 ≤ v ≤ 2π,

z = u.

26. Since the parameterization in Example 6 on page 863 was r = (1 − zh

)a and since the cone is given by z = r, we havez = (1− r

a)h. The parameterization we want is

x = r cos θ, 0 ≤ r ≤ a,y = r sin θ, 0 ≤ θ ≤ 2π,

z =(

1− r

a

)h.

27. In cylindrical coordinates, the paraboloid has equation z = r2. Thus, a point on the surface is given by

x~i + y~j + z~k = r cos θ~i + r sin θ~j + z~k

=√z cos θ~i +

√z sin θ~j + z~k .

Alternatively, we could writex~i + y~j + z~k = r cos θ~i + r sin θ~j + r2~k .

28. The vase obtained by rotating the curve z = 10√x− 1, 1 ≤ x ≤ 2, around the z-axis is shown in Figure 17.58.

17.5 SOLUTIONS 1235

2−2x

y

z10

z = 10√x− 1

Figure 17.58

At height z, the cross-section is a horizontal circle of radius a. Thus, a point on this horizontal circle is given by

~r = a cos θ~i + a sin θ~j + z~k .

However, the radius a varies, so we need to express it in terms of the other parameters θ and z. If you look at the xz-plane,the radius of this circle is given by x, so solving for x in z = 10

√x− 1 gives

a = x =(z

10

)2

+ 1.

Thus, a parameterization is

~r =

((z

10

)2

+ 1

)cos θ~i +

((z

10

)2

+ 1

)sin θ~j + z~k

so

x =

((z

10

)2

+ 1

)cos θ, y =

((z

10

)2

+ 1

)sin θ, z = z,

where 0 ≤ θ ≤ 2π, 0 ≤ z ≤ 10.

29. The plane in which the circle lies is parameterized by

~r (p, q) = x0~i + y0

~j + z0~k + p~u + q~v .

Because ~u and ~v are perpendicular unit vectors, the parameters p and q establish a rectangular coordinate system on thisplane exactly analogous to the usual xy-coordinate system, with (p, q) = (0, 0) corresponding to the point (x0, y0, z0).Thus the circle we want to describe, which is the circle of radius a centered at (p, q) = (0, 0), can be parameterized by

p = a cos t, q = a sin t.

Substituting into the equation of the plane gives the desired parameterization of the circle in 3-space,

~r (t) = x0~i + y0

~j + z0~k + a cos t~u + a sin t~v ,

where 0 ≤ t ≤ 2π.

30. (a) Add second and third equations to get y + z = 1 + 2s. Thus, y + z = 1 + x or −x + y + z = 1, which is theequation of a plane. Now, s = x/2, and t = (y − z + 1)/2, so the conditions 0 ≤ s ≤ 1, 0 ≤ t ≤ 1 are equivalentto 0 ≤ x ≤ 2, 0 ≤ y − z + 1 ≤ 2 or 0 ≤ x ≤ 2, −1 ≤ y − z ≤ 1.

(b) The surface is shown in Figure 17.59.


x y

z

Figure 17.59: The surface x = 2s,y = s+ t, z = 1 + s− t, for

0 ≤ s ≤ 1, 0 ≤ t ≤ 1

31. (a) z2 = 1−s2− t2 = 1−x2−y2. So x2 +y2 +z2 = 1 which is the equation of a sphere. The conditions s2 + t2 ≤ 1,s, t ≥ 0 are equivalent to x2 + y2 ≤ 1 and x, y ≥ 0. But if x2 + y2 + z2 = 1, then x2 + y2 ≤ 1 is satisfiedautomatically, so our surface is defined by:

x2 + y2 + z2 = 1, x, y, z ≥ 0.

(b) The surface x = s, y = t, z =√

1− s2 − t2 for s2 + t2 ≤ 1, s, t ≥ 0 is shown in Figure 17.60.

xy

z

Figure 17.60

x

y

z

Figure 17.61: The surface x = s+ t,y = s− t, z = s2 + t2 for 0 ≤ s ≤ 1,

0 ≤ t ≤ 1

32. (a) From the first two equations we get:

s =x+ y

2, t =

x− y2

.

Hence the equation of our surface is:

z =(x+ y

2

)2

+(x− y

2

)2

=x2

2+y2

2,

which is the equation of a paraboloid.The conditions: 0 ≤ s ≤ 1, 0 ≤ t ≤ 1 are equivalent to: 0 ≤ x + y ≤ 2, 0 ≤ x − y ≤ 2. So our surface is

defined by:

z =x2

2+y2

2, 0 ≤ x+ y ≤ 2 0 ≤ x− y ≤ 2

(b) The surface is shown in Figure 17.61.

17.5 SOLUTIONS 1237

33. (a) As x2 + y2 = 9 and s ∈ [0, π] is equivalent to x ≥ 0, and t ∈ [0, 1] is equivalent to z ∈ [1, 2]. So, x2 + y2 = 9 isthe equation of a cylinder, and our surface is defined by:

x2 + y2 = 9, x ≥ 0, 1 ≤ z ≤ 2.

(b) The surface x = 3 sin s, y = 3 cos s, z = t+ 1 for 0 ≤ s ≤ π, 0 ≤ t ≤ 1 is shown in Figure 17.62.

x

y

z

Figure 17.62

x

y

z

Figure 17.63

34. x = 3t, y = 5t cosu, z = 10t sinu, 0 ≤ u ≤ 2π, 0 ≤ t ≤ 1. See Figure 17.63.

35. x = tu, y = tu2, z = tu− tu2 + 5t, 0 ≤ u ≤ 3, 0 ≤ t ≤ 1. See Figure 17.64.

(0, 0, 5)

(3, 9,−1)

xy

z

Figure 17.64

(2, 3, 4)

(−2, 0,−1)

(2, 0,−1)

x

y

z

Figure 17.65

36. x = 2t cosu+ 2− 2t, y = 2t sinu+ 3− 3t, z = 4− 5t, 0 ≤ u ≤ π, 0 ≤ t ≤ 1. See Figure 17.65.

37. (a) ~R = b cos θ~i + b sin θ~j(b) One vector is always ~k . The other vector we need is in the same direction as ~R in part (a) but has length 1. Therefore

we take the other vector to be ~m = cos θ~i + sin θ~j . See Figure 37. Thus, relative to the center of the small circle,we have

~r = a cosφ~m + a sinφ~k = a cosφ(cos θ~i + sin θ~j ) + a sinφ~k

(c) The parameterization of the torus is given by

x~i + y~j + z~k = ~R + ~r

= (b cos θ + a cosφ cos θ)~i + (b sin θ + a cosφ sin θ)~j + a sinφ~k


Solutions for Chapter 17 Review

Exercises

1. x = t, y = 5.

2. The parameterization x~i + y~j = 2 cos t~i + 2 sin t~j has the right radius but starts at the point (2, 0). To start at (0, 2),we need x~i + y~j = 2 cos(t+ π

2)~i + 2 sin(t+ π

2)~j = 2 sin t~i + 2 cos t~j .

3. The parameterization x~i + y~j = (4 + 4 cos t)~i + (4 + 4 sin t~) j gives the correct circle, but starts at (8, 4). To start onthe x-axis we need

x~i + y~j = (4 + 4 cos(t− π

2))~i + (4 + 4 sin(t− π

2))~j = (4 + 4 sin t)~i + (4− 4 cos t)~j .

4. The parametric equation of a circle isx = cos t, y = sin t.

When t = 0, x = 1, y = 0, and when t = π2

, x = 0, y = 1. This shows a counterclockwise movement, so our originalequation is correct.

5. The vector (~i + 2~j + 5~k )− (2~i −~j + 4~k ) = −~i + 3~j + ~k is parallel to the line, so a possible parameterization is

x = 2− t, y = −1 + 3t, z = 4 + t.

6. A line perpendicular to the xz-plane will have x = constant, z = constant, y = anything: This is given by x = 1, y =t, z = 2.

7. Since the vector ~n = grad(2x − 3y + 5z) = 2~i − 3~j + 5~k is perpendicular to the plane, this vector is parallel to theline. Thus the equation of the line is

x = 1 + 2t, y = 1− 3t, z = 1 + 5t.

8. Since the radius is 2, the circle must be of the form x = 2 cos t, y = 2 sin t, z = 1. But this parameterization traces out thecircle clockwise when viewed from below. Therefore, the parameterization we want is x = 2 cos t, y = −2 sin t, z = 1.

9. Since the circle has radius 3, the equation must be of the form x = 3 cos t, y = 5, z = 3 sin t. But since the circle is beingviewed from farther out on the y-axis, the circle we have now would be seen going clockwise. To correct this, we add anegative to the third component, giving us the equation x = 3 cos t, y = 5, z = −3 sin t.

10. We can find this equation in two ways. First we could find two points on the line of intersection and then proceed as inExample 7 on page 835. To find two points just substitute two different values for z and solve for x and y for each valueof z. Alternatively, assuming the line is not horizontal (which it turns out not to be), we could take z to be the parametert, so z = t. To find x and y as functions of t we solve the two equations for x and y in terms of t. We have

t = 4 + 2x+ 5y

t = 3 + x+ 3y.

Eliminating x we get−t = −2− y and y = −2 + t.

Substituting −2 + t for y in the second equation and solving for x, we get

x = 3− 2t.

Our equations are thereforex = 3− 2t, y = −2 + t, z = t.

or~r = x~i + y~j + z~k = 3~i − 2~j + t(−2~i +~j + ~k ).

SOLUTIONS to Review Problems for Chapter Seventeen 1239

11. No. The first is parallel to the vector 2~i −~j + 3~k and the second is parallel to~i + 2~j + 2~k .

12. The direction vectors of the lines,−~i + 4~j − 2~k and 2~i − 8~j + 4~k , are multiplies of each other (the second is−2 timesthe first). Thus the lines are parallel. To see if they are the same line, we take the point corresponding to t = 0 on the firstline, which has position vector 3~i + 3~j − ~k , and see if it is on the second line. So we solve

(1 + 2t)~i + (11− 8t)~j + (4t− 5)~k = 3~i + 3~j − ~k .This has solution t = 1, so the two lines have a point in common and must be the same line, parameterized in two differentways.

13. (a) We get the part of the line with x < 0 and y < 0 and z < 10.(b) We get the part of the line between the points (0, 0, 10) and (1, 2, 13).

14. The vector field points in a clockwise direction around the origin. Since ‖y~i − x~j ‖ =√y2 + x2, the vectors get longer

as you go away from the origin. See Figure 17.66.

x

y

Figure 17.66

x

y

Figure 17.67

15. At each point, all these vector fields point in the same direction (rotating clockwise around the origin). Since ‖ ~F ‖ =

1x2+y2 ‖y~i − x~j ‖ =

√y2+x2

x2+y2 = 1√x2+y2

, the vectors in the field shrink as you go away from the origin. See Fig-

ure 17.67.

16. The vector field points in a clockwise direction around the origin. Since

‖(

y√x2 + y2

)~i −

(x√

x2 + y2

)~j ‖ =

√x2 + y2

√x2 + y2

= 1,

the length of the vectors is constant everywhere.

x

y

Figure 17.68

Problems

17. (a) A vector field associates a vector to every point in a region of the space. In other words, a vector field is a vector-valuedfunction of position given by ~v = ~f (~r ) = ~f (x, y, z)

(b) (i) Yes, ~r + ~a = (x+ a1)~i + (y + a2)~j + (z + a3)~k is a vector-valued function of position.(ii) No, ~r · ~a is a scalar.

(iii) Yes.(iv) x2 + y2 + z2 is a scalar.


18. (a) To find where the particle is at time equal to 0, we simply substitute 0 in for all t in the equation. Therefore, theparticle is at the point with position vector

~r (0) = [2 + 5(0)]~i + (3 + 0)~j + 2(0)~k

= 2~i + 3~j + 0~k .

Thus, the particle is at the point (2, 3, 0).(b) To find the time at which the particle is at the point (12, 5, 4), we solve for t for each component, and the t should be

the same, if the curve goes through this point. For the x-component, we get

2 + 5t = 12

t = 2.

For the y-component, we get

3 + t = 5

t = 2.

And for the z-component, we get

2t = 4

t = 2.

Therefore, at t = 2 the particle reaches (12, 5, 4).(c) The particle never reaches (12, 4, 4), because the equation

~r = (2 + 5t)~i + (3 + t)~j + 2t~k = 12~i + 4~j + 4~k

has no solution. Thus, the point does not lie on the line.

19. (a) (I) has radius 1 and traces out a complete circle, so I = C4.(II) has radius 2 and traces out the top half of a circle, so II = C1

(III) has radius 1 and traces out a quarter circle, so III = C2.(IV) has radius 2 and traces out the bottom half of a circle, so IV = C6.

(b) C3 has radius 1/2 and traces out a half circle below the x-axis, so

~r = 0.5 cos t~i − 0.5 sin t~j .

C5 has radius 2 and traces out a quarter circle below the x-axis starting at the point (−2, 0). Thus we have

~r = −2 cos(t/2)~i − 2 sin(t/2)~j .

20. Vector fields (A) and (D) both point radially outward, so they correspond to (I) and (II). Since (A) has vectors that are ofconstant length, it corresponds to (II), where the level curves are equally spaced. (D) corresponds to (I).

Vector field (B) corresponds to (III) since the vectors in (B) point away from the origin on the x-axis, and the functionin (III) increases in this direction. To confirm, the vectors in (B) point toward the origin on the y-axis, and the functiondecreases away from the origin on the y-axis.

In vector field (C), vectors point away toward the origin on the x-axis and away from the origin on the y-axis. Thiscorresponds to (IV), in which the function decreases away from the origin on the x-axis and increases on the y-axis.

21. Sketches of the vector fields in Figure 17.69 show that ~E is tangent to (IV), ~F is tangent to (I), ~G is tangent to (II), and~H is tangent to (III).

y

x

~E = x~i + y~j

y

x

~F = x~i − y~j

y

x

~G = y~i − x~j

y

x

~H = y~i + x~j

Figure 17.69


22. At time t the particle is s = t− 7 seconds from P , so the displacement vector from the point P to the particle is ~d = s~v .To find the position vector of the particle at time t, we add this to the position vector ~r 0 = 5~i + 4~j + 3~k for the pointP . Thus a vector equation for the motion is:

~r = ~r 0 + s~v

= (5~i + 4~j + 3~k ) + (t− 7)(3~i +~j + 2~k ),

or equivalently,x = 5 + 3(t− 7), y = 4 + 1(t− 7), z = 3 + 2(t− 7).

Notice that these equations are linear. They describe motion on a straight line through the point (5, 4, 3) that is parallel tothe velocity vector ~v = 3~i +~j + 2~k .

23. The displacement vector from (1, 1, 1) to (2,−1, 3) is ~d = (2~i −~j + 3~k )− (~i +~j +~k ) =~i − 2~j + 2~k meters. Thevelocity vector has the same direction as ~d and is given by

~v =~d

5= 0.2~i − 0.4~j + 0.4~k meters/sec.

Since ~v is constant, the acceleration ~a = ~0 .

24. Parametric equations for a line in 2-space are

x = x0 + at

y = y0 + bt

where (x0, y0) is a point on the line and ~v = a~i + b~j is the direction of motion. Notice that the slope of the line is equalto ∆y/∆x = b/a, so in this case we have

b

a= Slope = −2,

b = −2a.

In addition, the speed is 3, so we have

‖~v ‖ = 3√a2 + b2 = 3

a2 + b2 = 9.

Substituting b = −2a gives

a2 + (−2a)2 = 9

5a2 = 9

a =3√5,− 3√

5.

If we use a = 3/√

5, then b = −2a = −6/√

5. The point (x0, y0) can be any point on the line: we use (0, 5). Theparametric equations are

x =3√5t, y = 5− 6√

5t.

Alternatively, we can use a = −3/√

5 giving b = 6/√

5. An alternative answer, which represents the particle moving inthe opposite direction is

x = − 3√5t, y = 5 +

6√5t.

25. (a) The quantity || grad f || represents the maximum rate of change of temperature with distance at each point. Its unitsare◦

C per cm.(b) The speed of the particle is

√(g′(t))2 + (k′(t))2. Its units are cm/sec.

(c) The rate of change of the particle’s temperature with time is given by the chain rule

dH

dt=∂f

∂x

dx

dt+∂f

∂y

dy

dt= fx · g′(t) + fy · k′(t).


26. (a) With the center at (0, 0, 8) and a point of the circle at (0, 5, 8), we know that the radius is 5. When t = 0, we havex = 0 and y = 5. Since the stone is rotating horizontally, z = 0 for all t. The period is 2π. Thus, the parameterizationis:

x(t) = 5 sin t

y(t) = 5 cos t

z(t) = 8

This parameterization has the correct period (if t is in seconds) and satisfies the initial conditions.(b) From our parameterization with t in seconds, we can see that the stone reaches (5, 0, 8) at time π/2. Thus at t = π/2,

~v = xt(π/2)~i + yt(π/2)~j

= 5 cos(π/2)~i − 5 sin(π/2)~j

= −5~j .

The acceleration of an object is the second derivative of its position. Thus, at t = π/2,

~a = xtt(π/2)~i + ytt(π/2)~j

= −5 sin(π/2)~i − 5 cos(π/2)~j

= −5~i

(c) At the moment in which the stone has left the circle, the only acceleration that acts on the stone is that of gravity.From that, assuming a gravity vector field oriented in the −z direction, we get the differential equations

ztt(t) = −gxtt(t) = ytt(t) = 0.

If we now measure t from the instant the string breaks, then the initial conditions are the velocity and position of thestone at t = 0. Since the velocity at the moment of release is ~v = −5~j , we have

xt(0) = 0, yt(0) = −5, zt(0) = 0.

The initial position at t = 0 is:x(0) = 5, y(0) = 0, z(0) = 8.

27. (a) fx =

[2x(x2 + y2)− 2x(x2 − y2)

]

(x2 + y2)2=

4xy2

(x2 + y2)2.

fy =

[−2y(x2 + y2)− 2y(x2 − y2)

]

(x2 + y2)2=−4yx2

(x2 + y2)2.

∇f(1, 1) =~i −~j , i.e., south-east.(b) We need a vector ~u such that∇f(1, 1) ·~u = 0, i.e., such that (~i −~j ) ·~u = 0. The vector ~u =~i +~j clearly works;

so does ~u = −~i −~j . Dividing by the length to get a unit vector, we have ~u = 1√2~i + 1√

2~j or ~u = − 1√

2~i − 1√

2~j .

(c) f is a function of x and y, which in turn are functions of t. Thus, the chain rule can be used to show how f changedwith t.

df

dt=∂f

∂x· dxdt

+∂f

∂y· dydt

=4xy2

(x2 + y2)2· 2e2t − 4x2y

(x2 + y2)2· (6t2 + 6).

At t = 0, x = 1, y = 1; so,df

dt=

4

4· 2− 4

4· 6 = −4.

28. The displacement from the point (1, 2, 3) to the point (3, 5, 7) is 3~i + 5~j + 7~k − (~i + 2~j + 3~k ) = 2~i + 3~j + 4~k . Sothe equation of the line is

x~i + y~j + z~k = 1~i + 2~j + 3~k + t(2~i + 3~j + 4~k )

orx~i + y~j + z~k = (1 + 2t)~i + (2 + 3t)~j + (3 + 4t)~k .

The square of the distance from a point (x, y, z) on the line to the origin, denoted by D(t) is

D(t) = (x− 0)2 + (y − 0)2 + (z − 0)2

= (1 + 2t)2 + (2 + 3t)2 + (3 + 4t)2


= 1 + 4t+ 4t2 + 4 + 12t+ 9t2 + 9 + 24t+ 16t2

= 14 + 40t+ 29t2

= 29(t2 +

40

29t+

14

29

)

= 29

((t+

20

29

)2

−(

20

29

)2

+14

29

).

Clearly, D(t) is minimum when t = −20/29, and

D(−20/29) = 29

(−(

20

29

)2

+14

29

)=

6

29.

So the shortest distance is√

629

=√

17429

.

29. (a) Separate the ant’s path into three parts: from (0, 0) to (1, 0) along the x-axis; from (1, 0) to (0, 1) via the circle; andfrom (0, 1) to (0, 0) along the y-axis. (See Figure 17.70.) The lengths of the paths are 1, 2π

4= π

2, and 1 respectively.

Thus, the time it takes for the ant to travel the three paths are (using the formula t = dv

) 12

, 13

, and 12

seconds.

(1, 0)

(0, 1)

x

y

Figure 17.70

From t = 0 to t = 12

, the ant is heading toward (1, 0) so its coordinate is (2t, 0). From t = 12

to t = 12

+ 13

= 56

,the ant is veering to the left and heading toward (0, 1). At t = 1

2, it is at (1, 0) and at t = 5

6, it is at (0, 1). Thus its

position is (cos[ 3π2

(t − 12)], sin[ 3π

2(t − 1

2)]). Finally, from t = 5

6to t = 5

6+ 1

2= 4

3, the ant is headed home. Its

coordinates are (0,−2(t− 43)).

In summary, the function expressing the ant’s coordinates is

(x(t), y(t)) =

(2t, 0) when 0 ≤ t ≤ 12(

cos( 3π2

(t− 12)), sin( 3π

2(t− 1

2)))

when 12< t ≤ 5

6

(0,−2(t− 43)) when 5

6≤ t ≤ 4

3.

(b) To do the reverse path, observe that we can reverse the ant’s path by interchanging the x and y coordinates (flippingit with respect to the line y = x), so the function is

(x(t), y(t)) =

(0, 2t) when 0 ≤ t ≤ 12(

sin( 3π2

(t− 12)), cos( 3π

2(t− 1

2)))

when 12< t ≤ 5

6

(−2(t− 43), 0) when 5

6< t ≤ 4

3.

30. (a) Since x = R cos(ωt) and y = R sin(ωt), and x2 +y2 = R2 cos2(ωt)+R2 sin2(ωt) = R2, we have motion arounda circle of radius R centered at the origin. The particle moves counterclockwise, completing one revolution in time2π/ω. Thus, the period = 2π/ω.

(b) The velocity vector is

~v =d~r

dt= −ωR sin(ωt)~i + ωR cos(ωt)~j .

We expect the velocity, ~v , to be tangent to the circle. To verify that this, we compute

~v · ~r = (−ωR sin(ωt)~i + ωR cos(ωt)~j ) · (R cos(ωt)~i +R sin(ωt)~j )

= −ωR2 sin(ωt) cos(ωt) + ωR2 cos(ωt) sin(ωt) = 0.


This shows that the velocity vector is perpendicular to the radius from the center of the circle to the particle, whichmoves counterclockwise.

The speed is ‖~v ‖ = ωR, which is constant. Notice that this makes sense, because in time 2π/ω, the particletravels a distance of 2πR, giving a speed of 2πR/(2π/ω) = ωR.

(c) The acceleration vector is

~a =d~v

dt= −ω2R cos(ωt)~i − ω2R sin(ωt)~j = −ω2~r .

The acceleration vector points in the direction opposite to the position vector ~r , and thus points toward the center ofthe circle. It has constant magnitude ‖~a ‖ = ω2R = ‖v‖2/R.

31. All of the points lie on the unit circle. (You can check this since x2 + y2 = 1.) The problem is that there is no value of tthat gives the point x = 0, y = 1. This is because

y =t2 − 1

t2 + 1= 1

has no real solution. Only when t approaches positive or negative infinity does the point get close to (0, 1). Technically, itis not a circle.

32. (a) The equation of the line is y = t(x+ 2). Substitution into the equation of the ellipse gives

2x2 + 3t2(x+ 2)2 = 8

which simplifies to the quadratic equation in x

(2 + 3t2)x2 + 12t2x+ 12t2 − 8 = 0

There are two solutions x = x1 and x = x2 to this equation, namely the x-coordinates of the two points P = (−2, 0)and Q = (x2, y2) that lie on the intersection of the line and the ellipse. The first solution is x = x1 = −2, and thesimplest way to find the second is to remark that x1x2 = (12t2 − 8)/(2 + 3t2). Therefore

x2 =4− 6t2

2 + 3t2

and soy2 = t(x2 + 2) =

8t

2 + 3t2

Thus the desired point is

Q = (x2, y2) =

(4− 6t2

2 + 3t2,

8t

2 + 3t2

)

(b) From (a),

x =4− 6t2

2 + 3t2, y =

8t

2 + 3t2

will parameterize the ellipse. One reason that this parameterization is interesting is that it allows you to find easilypoints on the ellipse whose x- and y-coordinates are both rational numbers, simply by choosing rational values for t.For example, t = 1 corresponds to the point (−2/5, 8/5) and t = 1/2 corresponds to the point (10/11, 16/11).

33. (a) The equation of the sphere is x2 + y2 + z2 = 1. Substituting the parameterization gives

(x(t))2+(y(t))2+(z(t))2 = cos4 t+sin2 t cos2 t+sin2 t = (cos2 t)(cos2 t+sin2 t)+sin2 t = cos2 t+sin2 t = 1.

Therefore the curve lies on this sphere.(b) If t = π/4, we have x(π/4) = 1/2, y(π/4) = 1/2, z(π/4) = 1/

√2. The gradient vector to the sphere at this point

is perpendicular to the sphere and the curve. Since grad f = 2x~i + 2y~j + 2z~k , we have

Normal = grad f

(1

2,

1

2,

1√2

)= 2

(1

2

)~i + 2

(1

2

)~j + 2

(1√2

)~k =~i +~j +

√2~k .

(c) A tangent vector is x′(t)~i + y′(t)~j + z′(t)~k = (−2 cos t sin t)~i + (cos2 t − sin2 t)~j + cos t~k . At t = π/4, wehave

Tangent vector = −~i +1√2~k .


34. (a) Suppose axes are fixed with the origin on the ground below the point at which the baton is thrown. There is nohorizontal acceleration, so if x measures horizontal displacement of the center in meters

d2x

dt2= 0.

Since the initial velocity is 8 m/sec, integrating gives:

dx

dt= 8,

and since x = 0 when t = 0,x = 8t.

The vertical acceleration is due to gravity. So, if y is vertical displacement of the center in meters:

d2y

dt2= −g.

Sody

dt= −gt+ 10,

and

y = −gt2

2+ 10t+ 1.5.

Thus, the parametric equations for the center of the baton are

x = 8t, y = −gt2

2+ 10t+ 1.5.

(b) We put a new origin at the center of the baton. Suppose (h, k) are the coordinates of the end of the baton relative tothe center. Since the radius of the circular motion is 0.2 m and the angular velocity is 2(2π) = 4π radians/sec andsince x = 0.2 and y = 0 when t = 0, we have

h = 0.2 cos(4πt) k = 0.2 sin(4πt).

(c) To find the coordinates of the end of the baton, we add the results from parts (a) and (b), so if x and y represent theposition of the end of the baton relative to the ground, we have

x = 8t+ 0.2 cos(4πt) y = −gt2

2+ 10t+ 1.5 + 0.2 sin(4πt).

(d) To sketch this, use g = 9.8 meters/sec 2.

0.2 5 10 15

1.5

5

x

y

Figure 17.71

35. At time t the particle has polar coordinates r = ‖~r (t)‖ = at and θ = ωt. At time t, the ray from the origin to theparticle is at angle ωt radians from the positive x-axis. The ray is therefore rotating at a rate of ω radians per unit time.The parameter ω is the rate of change of the polar angle θ of the particle measured in radians per unit time. The largerω is, the quicker the particle completes a complete revolution (a 360◦ trip) around the origin. At time t, the particle is atdistance at from the origin. Thus a equals the rate of change of the particle’s distance from the origin. The larger a is, thefaster the particle moves away from the origin.


36. Suppose that the line goes through the point P = (a, b, c) and is parallel to the vector ~u . The position vector of themoving object at time t is then given by the formula ~r (t) = a~i + b~j + c~k + f(t)~u where f(t) is a function so thatf(t)~u is the displacement vector from the point P to the object at time t.

(a) The velocity vector is given by the derivative ~r ′(t) = f ′(t)~u , which is parallel to the line because it is a multiple ofthe vector ~u .

(b) The acceleration vector is given by the second derivative ~r ′′(t) = f ′′(t)~u which is parallel to the line because it isa multiple of the vector ~u .

37. Let’s place the coordinate system so that the origin is at the center of the circle of radius R on which the object is moving.If θ(t) is the polar coordinate angle of the object at time t, the position vector of the object is given by

~r (t) = R cos(θ(t))~i +R sin(θ(t))~j .

(a) The velocity vector is given by the derivative

~r ′ = −Rθ′ sin θ~i +Rθ′ cos θ~j = Rθ′(− sin θ~i + cos θ~j )

The velocity ~r ′ is perpendicular to the radius vector ~r from the center of the circle to the object because

~r ′ · ~r = Rθ′(− sin θ~i + cos θ~j

)·(R cos θ~i +R sin θ~j

)

= R2θ′(− sin θ cos θ + cos θ sin θ) = 0.

So the velocity vector, ~r ′, is tangent to the circle.(b) The acceleration vector is given by the second derivative

~r ′′ = Rθ′′(− sin θ~i + cos θ~j )−R(θ′)2(cos θ~i + sin θ~j )

The acceleration vector ~r ′′ is expressed as the sum of two vector components. Since ~r = R cos θ~i + R sin θ~j , thecomponent −R(θ′)2(cos θ~i + sin θ~j ) is in the direction opposite to ~r , that is, toward the center of the circle. Thecomponent Rθ′′(− sin θ~i + cos θ~j ) is perpendicular to ~r since the dot product is (− sin θ~i + cos θ~j ) · (cos θ~i +sin θ~j ) = 0. This means that Rθ′′

(− sin θ~i + cos θ~j

)is in the direction tangent to the circle. The sum ~r ′′ is not

in general directed toward the center of the circle. The acceleration is only directed toward the center of the circle inthe case where θ′′(t) = 0, as it is for example in the case θ(t) = ωt of uniform circular motion.

38. (a) Since ~F = ~r‖~r ‖3 , the magnitude of ~F is given by

‖~F ‖ =‖~r ‖‖~r ‖3 =

1

‖~r ‖2 .

Now ~r = x~i + y~j + z~k , so the magnitude of ~r is given by

‖~r ‖ =√x2 + y2 + z2.

Thus,

‖~F ‖ =1

‖~r ‖2 =1

x2 + y2 + z2.

(b) ~F · ~r = ~r‖~r ‖3 · ~r = ‖~r ‖2

‖~r ‖3 = 1‖~r ‖ = 1√

x2+y2+z2.

(c) A unit vector parallel to ~F and pointing in the same direction is given by ~U =~F

‖~F ‖ .

~F =~r

‖~r ‖3 , and ‖ ~F ‖ = 1‖~r ‖2 . Putting these into the expression for ~U we have

~U =~F

‖~F ‖=

~r‖~r ‖3

1‖~r ‖2

=~r

‖~r ‖

=x√

x2 + y2 + z2

~i +y√

x2 + y2 + z2

~j +z√

x2 + y2 + z2

~k .


(d) A unit vector parallel to ~F and pointing in the opposite direction is given by:

~V = −~F

‖~F ‖= − ~r

‖~r ‖

=−x√

x2 + y2 + z2

~i +−y√

x2 + y2 + z2

~j +−z√

x2 + y2 + z2

~k .

(e) If ~r = cos t~i + sin t~j + ~k , then ‖~r ‖ =√

cos2 t+ sin2 t+ 1 =√

2.So, ~F = ~r

‖~r ‖3 = cos t√8~i + sin t√

8~j + 1√

8~k = cos t

2√

2~i + sin t

2√

2~j + 1

2√

2~k .

(f) We know that ~F · ~r = 1‖~r ‖ , so if ~r = cos t~i + sin t~j + ~k , ~F · ~r = 1√

2.

39. We use spherical coordinates φ and θ as the two parameters. The parameterization of the sphere center at the origin andradius 5 is:

x = 5 sinφ cos θ, y = 5 sinφ sin θ, z = 5 cosφ.

We have to shift the center of the sphere from the origin to the point (2,−1, 3). This gives

x = 2 + 5 sinφ cos θ, y = −1 + 5 sinφ sin θ, z = 3 + 5 cosφ.

40. (a) The cone of height h, maximum radius a, vertex at the origin and opening upward is shown in Figure 17.72.

y

x

z

h

a

Figure 17.72

By similar triangles, we haver

z=a

h,

soz =

hr

a.

Therefore, one parameterization is

x = r cos θ, 0 ≤ r ≤ a,y = r sin θ, 0 ≤ θ < 2π,

z =hr

a.

(b) Since r = az/h, we can write the parameterization in part (a) as

x =az

hcos θ, 0 ≤ z ≤ h,

y =az

hsin θ, 0 ≤ θ < 2π,

z = z.


41. (a) The surface is the cylinder x2 + y2 = 1 of radius 1 centered on the z-axis.(b) The parameter curves with constant s and varying t are helices that wind clockwise around the cylinder as they

advance up the cylinder with increasing t. See Figure 17.73.The parameter curves with constant t and varying s are helices that wind counterclockwise up the cylinder. See

Figure 17.74.

x y

z

Figure 17.73: Constant s, varying t

x y

z

Figure 17.74: Constant t, varying s

42. (a) The current, and path that the iceberg would travel, would look like Figure 17.75.

Path of iceberg-

Figure 17.75

To determine the location of the iceberg at time t = 7, we must first determine the velocity in the x and ydirection. In this current, Vx = 1, and Vy = 0. To obtain the position, we must integrate the velocity in terms of t.For this current we get

dx

dt= 1

Hence

x(7) = x(0) +

∫ 7

0

1 · dt

= 1 + 7

= 8.

Since Vy = dy/dt = 0, y is a constant. Thus at t = 7, x has moved from x = 1 to x = 8 and y has stayed at y = 3.Therefore the location at t = 7 is (8, 3).

(b) The current and path that the iceberg would travel, would look like Figure 17.76.


*

I

Path of icebergx

y

Figure 17.76

Path of iceberg

x

y

Figure 17.77

Assuming that the iceberg follows the current exactly, we find the position of the iceberg at time t = 7 byrecognizing that the velocity must be equal to the given vector field.

dx

dt= 2x

dy

dt= y

These are separable equations that are solved for x and y as follows:

dx

dt= 2x

∫dx

2x=

∫1 dt

lnx

2= t+ C

x = kxe2t

and for y

dy

dt= y

∫dy

y=

∫1 dt

ln y = t+ C

y = kyet

We can solve for kx and ky , the arbitrary constants, because we know the position of the iceberg at t = 0.

1 = x(0) = kx

3 = y(0) = ky

sox = e2t, y = 3et.

We now substitute t = 7:x = e2·7 = e14 and y = 3e7

(c) The current, and path that the iceberg would travel, would look like Figure 17.77.Since

~v = −y~i + x~j ,

the system of differential equations satisfied by x(t) and y(t) is

dx

dt= −y, dy

dt= x.


We differentiate one of the equations and substitute into the other, giving a second order equation

dx

dt= −y

d2x

dt2= −dy

dt

d2x

dt2= −x

d2x

dt2− x = 0

This differential equation has a solution of this form :

x = A cos t+B sin t

By taking the derivative and using the fact that y = −dx/dt, we get:

y = A sin t−B cos t

We know the initial position of the iceberg, so we can find the constants A and B with the simultaneous equations:

1 = x(0) = A cos 0 +B sin 0

3 = y(0) = A sin 0−B cos 0

Thus, A = 1 and B = −3. Now we evaluate the two expressions for t = 7:

x = cos 7− 3 sin 7 = −1.217

y = sin 7 + 3 cos 7 = 2.919

and find the position of the iceberg, (−1.217, 2.919).

43. The parameterization for a sphere of radius a using spherical coordinates is

x = a sinφ cos θ, y = a sinφ sin θ, z = a cosφ.

Think of the ellipsoid as a sphere whose radius is different along each axis and you get the parameterization:x = a sinφ cos θ, 0 ≤ φ ≤ π,y = b sinφ sin θ, 0 ≤ θ ≤ 2π,

z = c cosφ.To check this parameterization, substitute into the equation for the ellipsoid:

x2

a2+y2

b2+z2

c2=a2 sin2 φ cos2 θ

a2+b2 sin2 φ sin2 θ

b2+c2 cos2 φ

c2

= sin2 φ(cos2 θ + sin2 θ) + cos2 φ = 1.

44. The sphere x2 + y2 + z2 = 1 is shown in Figure 17.78.

x

y

z(x, y, z)

RR

(p, q)

(0, 0, 1)

(0, 0,−1)

Figure 17.78

(a) The origin corresponds to the south pole.(b) The circle x2 + y2 = 1 corresponds to the equator.(c) We get all the points of the sphere by this parameterization except the north pole itself.(d) x2 + y2 > 1 corresponds to the upper hemisphere.(e) x2 + y2 < 1 corresponds to the lower hemisphere.


45. Consider the lines in the directions −−→PQ = −9~i − 15~j + 12~k and −→RS = 57~i − 15~j + 24~k , respectively. If the distancebetween these lines is always greater than 16 then the beads can always pass without touching. If the distance is less than16, and if that occurs at a point along the segment from P to Q and from R to S, then the beads will touch at that point.

Vectors in the directions of the two lines are ~u = −3~i − 5~j + 4~k and ~v = 19~i − 5~j + 8~k . Parametric equationsfor the lines are:

x(t) = 7− 3t, y(t) = 12− 5t, z(t) = −10 + 4t,

andx(s) = −20 + 19s, y(s) = 17− 5s, z(s) = 1 + 8s.

(You’ll see why we used different parameters for the two lines in a moment.)The distance between variable points on the lines is a function of s and t; we want the minimum of this function. It

is easier to work with the square of the distance. Thus we want to find the minimum of

D(s, t) = (−27 + 19s+ 3t)2 + (5− 5s+ 5t)2 + (11 + 8s− 4t)2.

Computing ∂D/∂s and ∂D/∂t and simplifying we find that

∂D

∂s= 900(−1 + s),

∂D

∂t= 100(−2 + t).

The unique critical point of D(s, t) is (s, t) = (1, 2) and the value of D at that point is D(1, 2) = 225. This must be theminimum value of D(s, t), because D(s, t) is a paraboloid opening upward. (We can also check that this is the minimumwith the test for local max and local min.) The distance between the lines is therefore

√225 = 15.

The points on the two lines where s = 1 and t = 2 are A = (−1, 12, 9) and B = (1, 2,−2); these are the pointswhere the lines are closest. The only question now is whetherA andB are along the segments fromR to S and from P toQ. In the parameterizations of the lines, R and P correspond to s = 0 and t = 0, respectively, and S and Q correspondto to s = 3 and t = 3. So A and B do lie on the given segments. If the beads are centered at these points they will hitbecause they each have diameter 8 cm, whereas the lines are only 15 cm apart there.

(The solution also shows why we needed different parameters, s and t, for the two lines. The points where the linesare closest together occur at different values of the two parameters: t = 1 for one line and s = 2 for the other.)

CAS Challenge Problems

46. (a) Since ~e 1 and ~e 2 are perpendicular, we have ~e 1 ·~e 2 = 0. The normal vector to the plane is~i +~j +~k , and since ~e 1

and ~e 2 are parallel to the plane, we have ~e 1 · (~i +~j + ~k ) = 0 and ~e 2 · (~i +~j + ~k ) = 0. Also, since ~e 1 and ~e 2

are unit vectors, we have ~e1 · ~e1 = 1 and ~e2 · ~e2 = 1.(b) We have

ac+ bd = 0, a+ b = 0, c+ d+ e = 0, a2 + b2 = 1, c2 + d2 + e2 = 1

By solving these equations, we can choose, for example,

~e1 =1√2~i − 1√

2~j , ~e2 =

1√6~i +

1√6~j −

√2√3~k .

The equations of the circle will then follow from the given formula, with ~r 0 =~i + 2~j + 3~k and R = 5.

47. (a)

~r · ~F = (x~i + y~j ) · (−y(1− y2)~i + x(1− y2)~j )

= −xy(1− y2) + yx(1− y2) = 0

This means that the tangent line to the flow line at a point is always perpendicular to the vector from the origin to thatpoint. Hence the flow lines are circles centered at the origin.

(b) The circle ~r (t) = cos t~i + sin t~j has velocity vector ~v (t) = − sin t~i + cos t~j = −y~i + x~j = (1− y2)~F . Thusthe velocity vector is a scalar multiple of ~F , and hence parallel to ~F . However, since ~v (t) is not equal to ~F (~r (t)),it is not a flow line.

(c) Using a CAS, we find

~v (t) = − t

(1 + t2)3/2~i +

(− t2

(1 + t2)3/2+

1√1 + t2

)~j = − t

(1 + t2)3/2~i +

1

(1 + t2)3/2~j


and

~F (~r (t)) = −

t(

1− t2

1+t2

)

√1 + t2

~i +

1− t2

1+t2√1 + t2

~j = − t

(1 + t2)3/2~i +

1

(1 + t2)3/2~j = ~v (t).

Although the circle parameterized in part (b) has velocity vectors parallel to ~F at each point of the circle, its speedis not equal to the magnitude of the vector field. The circle in part (c) is parameterized at the correct speed to be theflow line.

48. (a) We have~r ′(t) = (3ae3t − be−t)~i + (6ae3t + 2be−t)~j

and

~F (~r (t)) = ((ae3t + be−t) + (2ae3t − 2be−t))~i + ((4(ae3t + be−t) + (2ae3t − 2be−t))~j

= (3ae3t − be−t)~i + (6ae3t − be−t)~j = ~r ′(t).

(b) We want ~r (0) =~i − 2~j , so we solve

a+ b = 1

2a− 2b = −2

to get a = 0, b = 1. So the flow line is ~r (t) = e−t~i − 2e−t~j , which approaches (0, 0) as t → ∞. For the secondflow line, we solve

a+ b = 1

2a− 2b = −1.99

to get a = 0.0025, b = 0.9975, so the flow line is

~r (t) = (0.0025e3t + 0.9975e−t)~i + (0.005e3t − 1.995e−t)~j ,

which approaches (∞,∞) as t→∞. For the third flow line we solve

a+ b = 1

2a− 2b = −2.01

which gives a = −0.0025, b = 1.0025, so the flow line is

~r (t) = (−0.0025e3t + 1.0025e−t)~i + (−0.005e3t + 2.005e−t)~j ,

which approaches (−∞,−∞) as t→∞.

49. Answers may differ depending on the method and CAS used.

(a) Using a CAS to solve for x and y in terms of z and letting z = t, we get x = 2013− 6t

13, y = −1

13− t

13, z = t.

(b) Using a CAS to solve for y and z in terms of x and letting x = t, we get x = t, y = 16(−2 − 2t + 3t2), z =

16(20− 10t− 3t2).

(c) Using a CAS to solve for x and z in terms of y, we get two solutions

x =√

2− t2, y = t, z = 5 + 5t− 3√

2− t2

andx = −

√2− t2, y = t, z = 5 + 5t+ 3

√2− t2

Each of these is a parameterization of one half of the intersection curve.

CHECK YOUR UNDERSTANDING

1. False. The y coordinate is zero when t = 0, but when t = 0 we have x = 2 so the curve never passes through (0, 0).

2. True. Every point (x, y) on this curve satisfies y = (t2)2 = x2.

3. False. For example, the graph of x = cos t, y = sin t for 0 ≤ t ≤ 2π is a circle. A circle is not the graph of a function,since for some values of x there are two values of y.

4. True. Every y-coordinate is one less than every x-coordinate, so the equation of the line is y = x− 1.

5. False. When t = 0, we have (x, y) = (0,−1). When t = π/2, we have (x, y) = (−1, 0). Thus the circle is being tracedout clockwise.

CHECK YOUR UNDERSTANDING 1253

6. True. The functions et and ln t are inverses, so ln et = t. Thus if x = et, y = t, we have y = t = ln et = lnx.

7. True. Taking two values for t, say t = 0 and t = 1 give the points (1, 0) and (0, 2), which lie on a line with equationy = −2x+2. The second parameterization describes the same set of points, since y = −4s+2 = −2(2s)+2 = −2x+2.

8. True. Adding the equations z = x+ y and z = 1− x− y gives 2z = 1 or z = 12. Thus the line of intersection is parallel

to the xy-plane at height z = 12. Letting x be the parameter t and z = 1

2in the first plane’s equation gives 1

2= t + y or

y = 12− t. The same result is obtained by setting x = t and z = 1

2in the second plane’s equation.

9. True. To find an intersection point, we look for values of s and t that make the coordinates in the first line the same as thecoordinates in the second. Setting x = t and x = 2s equal, we see that t = 2s. Setting y = 2 + t equal to y = 1− s, wesee that t = −1−s. Solving both t = 2s and t = −1−s yields t = − 2

3, s = − 1

3. These values of s and t will give equal

x and y coordinates on both lines. We need to check if the z coordinates are equal also. In the first line, setting t = − 23

gives z = 73. In the second line, setting s = − 1

3gives z = − 1

3. As these are not the same, the lines do not intersect.

10. False. All points on this line lie in the plane x = 1, so the line is parallel to the yz-plane.

11. True. The ~j component of ~r is always one more than twice the~i component, so the line is y = 2x+ 1.

12. False. The line ~r 1(t) is in the direction of the vector~i − 2~j , while the line ~r 2(t) is in the direction of the vector ~2 i−~j .Since these vectors are not parallel (they are not scalar multiples of one another) the lines are not parallel.

13. False. The velocity vector is ~v (t) = ~r ′(t) = 2t~i − ~j . Then ~v (−1) = −2~i − ~j and ~v (1) = 2~i − ~j , which are notequal.

14. True. The velocity vector is ~v (t) = ~r ′(t) = 2t~i −~j , so the speed is s(t) =√

4t2 + 1. Then s(−1) = s(1) =√

5.

15. False. While this is true for motion in a circle with constant speed, it is not true in general. For a counterexample, considermotion along a parabola ~r (t) = t~i + t2~j . Then ~v (t) = ~i + 2t~j and ~a (t) = 2~j . Taking the dot product gives~v (t) · ~a (t) = 4t, which is not zero for all t. Thus the velocity and acceleration vectors are not always perpendicular.

16. False. If a particle is moving along a line with nonconstant speed, then the acceleration and velocity vectors are parallel.For a counterexample, consider motion along the line ~r (t) = t2~i + t2~j . Then ~v (t) = 2t~i + 2t~j and ~a (t) = 2~i + 2~j ,so ~v (t) = t~a (t). Thus in this case the velocity vector and acceleration vectors are parallel at all points.

17. False. As a counterexample, consider the curve ~r (t) = t2~i + t2~j for 0 ≤ t ≤ 1. In this case, when t is replaced by −t,the parameterization is the same, and is not reversed.

18. True. The length of the curve C is given by∫ ba||~v (t)|| dt =

∫ ba

1 dt = b− a.19. False. The velocity of the particle is given by ~v (t) = ~r ′(t) = 3~i + 2~j + ~k , so speed is constant: ||~v (t)|| =√

32 + 22 + 12 =√

14. So the particle never stops.

20. False. As a counterexample, consider motion along the helix ~r (t) = cos t~i + sin t~j + t~k . In this case the speed is||~v (t)|| =

√(− sin t)2 + (cos t)2 + 12 =

√2. Thus the particle has constant speed, but is traveling along a helix, not a

line.

21. True, since the vectors x~j are parallel to the y-axis.

22. False. The flow lines are circles centered at the origin.

23. False. The flow lines are lines parallel to the x-axis.

24. False. Each flow line stays in the quadrant in which it originates.

25. True. Any flow line which stays in the first quadrant has x, y →∞.

26. False. The flow lines for ~F are perpendicular to the contours for f , since the flow lines follow∇f , which is perpendicularto the contours of f .

27. True. Since a vector tangent to a circle centered at the origin is perpendicular to the radius vector ~r and ~F (~r ) is tangentto its flow lines, ~F (~r ) must be perpendicular to ~r . Thus ~F (~r ) · ~r = 0 for all ~r .

28. False. If the flow lines are all straight lines parallel to ~v , we need ~F (x, y) to be parallel to ~v for all x and y. Thatdoes not mean that ~F (x, y) must be equal to ~v ; it only needs to be a scalar multiple ~v . For example, the vector field~F (x, y) = 6~i + 10~j has all its flow lines parallel to ~v . Another example is ~F (x, y) = 3ex−y~i + 5ex−y~j = ex−y~v ,where the scalar multiplied times ~v varies as x and y vary.

29. True. If (x, y) were a point where the y-coordinate along a flow line reached a relative maximum, then the tangent vectorto the flow line, namely ~F (x, y), there would have to be horizontal (or ~0 ), that is its ~j component would have to be 0.But the ~j component of ~F is always 2.

30. False. At all points on the x-axis, y = 0, so the vector field is a horizontal vector, ~F (x, 0) = ex~i , pointing to the right,since it is a positive multiple of~i . Thus the x-axis itself is a flow line for ~F . Since there can be only one flow line throughany point, no flow line can cross the x-axis.


31. False. Suppose that there were a function f such that ~F = ∇f = fx~i + fy~j . Then fx = y and fy = 1. But thenfxy = 1 and fyx = 0, which is impossible since we know fxy = fyx.

32. True

33. False. There is only one parameter, s. The equations parameterize a line.

34. True. The position vector of a point on the lower hemisphere is the negative of the position vector of the opposite pointon the upper hemisphere. As ~r ranges over all points in the upper hemisphere, −~r ranges over all points in the lowerhemisphere.

35. False

36. True. Adding a constant vector shifts the plane by a corresponding displacement, keeping it parallel to the original plane.

37. True. If the surface is parameterized by ~r (s, t) and the point has parameters (s0, t0) then the parameter curves ~r (s0, t)and ~r (s, t0).

38. False. For example, the lines of longitude on a sphere correspond to different values of the parameter θ, but all passthrough the north and south poles.

39. True. We find the tangent vector to a parametrically defined curve by differentiating ~r (t).

40. False. Suppose ~r (t) = (cos t)~i + (sin t)~j , so ~r (t) traces out the unit circle. Then ~r (t) lies along the radius of the circleand

~r ′(t) = (− sin t)~i + (cos t)~j

and ~r ′(t) is tangent to the circle. Thus ~r (t) and ~r ′(t) are perpendicular, so their cross product is not zero.

41. False. Suppose ~r (t) = t~i + t~j . Then ~r ′(t) =~i +~j and

~r ′(t) · ~r (t) = (t~i + t~j ) · (~i +~j ) = 2t.

So ~r ′(t) · ~r (t) 6= 0 for t 6= 0.

42. False. This result is true for uniform circular motion, but is not true in general.

PROJECTS FOR CHAPTER SEVENTEEN

1. Set up the coordinate system as shown in Figure 17.79.

(−15, 0)

basket(0, 10)

6

?6 player

y

x

Figure 17.79

(a) We separate the initial velocity vector into its x and y components.

Vx = V cosA

Vy = V sinA.

Since there is no force acting in the x direction, the x-coordinate of the basketball is just

x = (V cosA)t− 15.

For the y-coordinate, we know thaty′′(t) = −32,

PROJECTS FOR CHAPTER SEVENTEEN 1255

soy′(t) = −32t+ C1

andy(t) = −16t2 + C1t+ C2.

We also know that y′(0) = V sinA and y(0) = 6. Substituting these values in, we get C1 = V sinA,C2 = 6 and thus

y = −16t2 + (V sinA)t+ 6.

(b) Use a graphing calculator or computer to plot the path of the basketball for various values of V and A.Many pairs of V and A put the shot in the basket. For example, V = 26, A = 60◦, V = 32, A = 30◦.

(c) Now that we have the equations, we need to find a relationship between V and A that ensures that thebasketball goes through the hoop (i.e., the curve passes through (0, 10)). So we set

x = (V cosA)t− 15 = 0

y = −16t2 + (V sinA)t+ 6 = 10.

From the first equation, we get t = 15V cosA . Then we substitute that into the second equation:

−16

(15

V cosA

)2

+ (V sinA)(15

V cosA) = 4

− 3600

V 2 cos2A+ 15 tanA = 4

V 2 =3600

cos2A(15 tanA− 4)

Keeping in mind that tan2 θ + 1 = 1cos2 θ , we have:

V 2 =3600(1 + tan2A)

15 tanA− 4.

We can minimize V by minimizing V 2 (since V > 0).

d(V 2)

dA=

2 tanA(15 tanA− 4)− 15(tan2A+ 1)

(15 tanA− 4)2· 3600

cos2A= 0

3600

cos2A

[15 tan2A− 8 tanA− 15

(15 tanA− 4)2

]= 0

15 tan2A− 8 tanA− 15 = 0

tanA =8 +√

964

30≈ 1.30

A ≈ 52◦.

2. (a) The product rule gives

d~L

dt=

d

dt(~r × ~v ) =

d~r

dt× ~v + ~r × d~v

dt= ~v × ~v + ~r × ~a .

But the cross product of any vector with itself is ~0 . So ~v × ~v = ~0 . Hence d~Ldt = ~r × ~a .


(b) The area swept out by the planet is approximately a triangle, with sides ~r , ~r + ∆~r , and ∆~r . Since‖∆~r × ~r ‖ is the area of the parallelogram formed by ∆~r and ~r , and since the triangle is half the size ofthe parallelogram, we have ∆A ≈ 1

2‖∆~r × ~r ‖.(c) Dividing by ∆t gives

∆A

∆t≈ 1

2

∥∥∥∥∆~r

∆t× ~r

∥∥∥∥ .

Taking the limit as ∆t→ 0 and recalling that ~L = ~r × ~v , we get

dA

dt=

1

2‖~L ‖.

(d) Since ~a is directed from the earth to the sun, and ~r from the sun to the earth, we see that ~r and ~a areparallel. So ~r × ~a = ~0 , as the cross product of parallel vectors is ~0 . By part (a), this means d~L /dt = ~0 .

So ~L must be a constant.(e) Since ‖~L ‖ is a constant, part (c) implies that

area swept out betweent = t0 and t = t1

=

∫ t1

t0

dA

dtdt =

1

2‖~L ‖

∫ t1

t0

dt =1

2‖~L ‖(t1 − t0).

So the area swept out over a time interval t1 − t0 only depends on t1 − t0, not t0 and t1 individually.(f) Let’s compare the triangles of are swept out by the planet when it is closest to and furthest from the sun,

for a given size time interval. Since the ~r and ~r + ∆~r sides are shorter when the planet is closest to thesun, the central angle and the third side must be larger then. So ∆~r , and hence ~v = ∆~r /∆t, are largerwhen the planet is closest to the sun, compared to when the planet is furthest from the sun (for a fixed ∆t).

3.

x

y(a)

Figure 17.80

x

y(b)

Figure 17.81

x

y(c)

Figure 17.82

x

y(d)

Figure 17.83

17.1 SOLUTIONS 1197 CHAPTER SEVENTEEN - …fd.valenciacollege.edu/file/tsmith143/ch17.pdf17.1...

Documents

Transcript of 17.1 SOLUTIONS 1197 CHAPTER SEVENTEEN - …fd.valenciacollege.edu/file/tsmith143/ch17.pdf17.1...