CHAPTER 1 – 9th Edition 1.1. If π represents a vector two units in length directed due west, π represents a vector three units in length directed due north, and π + π = π − π and 2π − π = π + π, find the magnitudes and directions of π and π. Take north as the positive π¦ direction: With north as positive π¦, west will be -π₯. We may therefore set up: π + π = 2π − π = 6ππ¦ + 2ππ₯ and π − π = π + π = −2ππ₯ + 3ππ¦ Add the equations to find π = 4.5ππ¦ (north), and then π = 2ππ₯ + 1.5ππ¦ (east of northeast). 1.2. Vector π extends from the origin to (1,2,3) and vector π from the origin to (2,3,-2). a) Find the unit vector in the direction of (π − π): First π − π = (ππ₯ + 2ππ¦ + 3ππ§ ) − (2ππ₯ + 3ππ¦ − 2ππ§ ) = (−ππ₯ − ππ¦ + 5ππ§ ) √ [ ]1β2 whose magnitude is |π − π| = (−ππ₯ − ππ¦ + 5ππ§ ) ⋅ (−ππ₯ − ππ¦ + 5ππ§ ) = 1 + 1 + 25 = √ 3 3 = 5.20. The unit vector is therefore ππ΄π΅ = (−ππ₯ − ππ¦ + 5ππ§ )β5.20 b) find the unit vector in the direction of the line extending from the origin to the midpoint of the line joining the ends of π and π: The midpoint is located at πππ = [1 + (2 − 1)β2, 2 + (3 − 2)β2, 3 + (−2 − 3)β2)] = (1.5, 2.5, 0.5) The unit vector is then (1.5ππ₯ + 2.5ππ¦ + 0.5ππ§ ) = (1.5ππ₯ + 2.5ππ¦ + 0.5ππ§ )β2.96 πππ = √ (1.5)2 + (2.5)2 + (0.5)2 1.3. The vector from the origin to the point π΄ is given as (6, −2, −4), and the unit vector directed from the origin toward point π΅ is (2, −2, 1)β3. If points π΄ and π΅ are ten units apart, find the coordinates of point π΅. With π = (6, −2, −4) and π = 13 π΅(2, −2, 1), we use the fact that |π − π| = 10, or |(6 − 23 π΅)ππ₯ − (2 − 32 π΅)ππ¦ − (4 + 13 π΅)ππ§ | = 10 Expanding, obtain 36 − 8π΅ + 94 π΅ 2 + 4 − 38 π΅ + 49 π΅ 2 + 16 + 83 π΅ + 19 π΅ 2 = 100 √ or π΅ 2 − 8π΅ − 44 = 0. Thus π΅ = 8± 64−176 = 11.75 (taking positive option) and so 2 2 2 1 π = (11.75)ππ₯ − (11.75)ππ¦ + (11.75)ππ§ = 7.83ππ₯ − 7.83ππ¦ + 3.92ππ§ 3 3 3 1 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 1.4. A circle, centered at the origin with a radius of 2 units, lies in the π₯π¦ plane. Determine the unit vector √ in rectangular components that lies in the π₯π¦ plane, is tangent to the circle at ( 3, −1, 0), and is in the general direction of increasing values of π₯: A unit vector tangent to this circle in the general increasing π₯ direction is π = +ππ . Its π₯ and √ π¦ components are ππ₯ = ππ ⋅ ππ₯ = − sin π, and ππ¦ = ππ ⋅ ππ¦ = cos π. At the point ( 3, −1), √ π = 330β¦ , and so π = − sin 330β¦ ππ₯ + cos 330β¦ ππ¦ = 0.5(ππ₯ + 3ππ¦ ). 1.5. An equilateral triangle lies in the π₯π¦ plane with its centroid at the origin. One vertex lies on the positive π¦ axis. a) Find unit vectors that are directed from the origin to the three vertices: Referring to the figure,the easy one is π1 = ππ¦ . Then, π2 will have negative π₯ and π¦ components, and can be constructed as π2 = πΊ(−ππ₯ − tan 30β¦ ππ¦ ) where πΊ = (1 + tan 30β¦ )1β2 = 0.87. So finally π2 = −0.87(ππ₯ + 0.58ππ¦ ). Then, π3 is the same as π2 , but with the π₯ component reversed: π3 = 0.87(ππ₯ − 0.58ππ¦ ). y a1 a5 a6 x 30° a a3 2 b) Find unit vectors that are directed from the origin a4 to the three sides, intersecting these at right angles: These will be π4 , π5 , and π6 in the figure, which are in turn just the part π results, oppositely directed: π4 = −π1 = −ππ¦ , π5 = −π3 = −0.87(ππ₯ − 0.58ππ¦ ), and π6 = −π2 = +0.87(ππ₯ + 0.58ππ¦ ). 1.6. Find the acute angle between the two vectors π = 2ππ₯ + ππ¦ + 3ππ§ and π = ππ₯ − 3ππ¦ + 2ππ§ by using the definition of: √ √ 2 + 1 2 + 32 = 2 14, a) the dot product:√ First, π ⋅ π = 2 − 3 + 6 = 5 = π΄π΅ cos π, where π΄ = √ β¦ 2 2 2 and where π΅ = 1 + 3 + 2 = 14. Therefore cos π = 5β14, so that π = 69.1 . b) the cross product: Begin with |π π π | | π₯ π¦ π§| | | π × π = | 2 1 3 | = 11ππ₯ − ππ¦ − 7ππ§ | | | 1 −3 2 | | | √ √ √ 2 2 2 and then |π × π| (√= 11 )+ 1 + 7 = 171. So now, with |π × π| = π΄π΅ sin π = 171, find π = sin−1 171β14 = 69.1β¦ 1.7. Given the field π = π₯ππ₯ + π¦ππ¦ . If π ⋅ π = 2π₯π¦ and π × π = (π₯2 − π¦2 ) ππ§ , find π: Let π = π1 ππ₯ + π2 ππ¦ + π3 ππ§ Then π ⋅ π = π1 π₯ + π2 π¦ = 2π₯π¦, and |π π π | | π₯ π¦ π§| | | π × π = | π₯ π¦ 0 | = π3 π¦ ππ₯ − π3 π₯ ππ¦ + (π2 π₯ − π1 π¦) ππ§ = (π₯2 − π¦2 ) ππ§ | | |π1 π2 π3 | | | From the last equation, it is clear that π3 = 0, and that π1 = π¦ and π2 = π₯. This is confirmed in the π ⋅ π equation. So finally π = π¦ππ₯ + π₯ππ¦ . 2 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 1.8. Demonstrate the ambiguity that results when the cross product is used to find the angle between two vectors by finding the angle between π = 3ππ₯ − 2ππ¦ + 4ππ§ and π = 2ππ₯ + ππ¦ − 2ππ§ . Does this ambiguity exist when the dot product is used? We use the relation π × π = |π||π| sin ππ§. With the given vectors we find [ ] √ 2ππ¦ + ππ§ √ √ π × π = 14ππ¦ + 7ππ§ = 7 5 = 9 + 4 + 16 4 + 1 + 4 sin π π§ √ 5 βββββββββββ ±π§ where π§ is identified as shown; we see that π§ can be positive or negative, as sin π can be positive or negative. This apparent sign ambiguity is not the real problem, however, as we really want √ the magnitude √ √ of the angle anyway. Choosing the positive sign, we are left with sin π = 7 5β( 29 9) = 0.969. Two values of π (75.7β¦ and 104.3β¦ ) satisfy this equation, and hence the real ambiguity. √ In using the dot product, we find π ⋅ π = 6 − 2 − 8 = −4 = |π||π| cos π = 3 29 cos π, or √ cos π = −4β(3 29) = −0.248 ⇒ π = −75.7β¦ . Again, the minus sign is not important, as we care only about the angle magnitude. The main point is that only one π value results when using the dot product, so no ambiguity. 1.9. A field is given as π= 25 (π₯ππ₯ + π¦ππ¦ ) (π₯2 + π¦2 ) Find: a) a unit vector in the direction of π at π (3, 4, −2): Have ππ = 25β(9 + 16) × (3, 4, 0) = 3ππ₯ + 4ππ¦ , and |ππ | = 5. Thus ππΊ = (0.6, 0.8, 0). b) the angle between π and ππ₯ at π : The angle is found through ππΊ ⋅ ππ₯ = cos π. So cos π = (0.6, 0.8, 0) ⋅ (1, 0, 0) = 0.6. Thus π = 53β¦ . c) the value of the following double integral on the plane π¦ = 7: 4 2 ∫0 ∫0 4 ∫0 ∫0 2 π ⋅ ππ¦ ππ§ππ₯ 4 2 4 25 25 350 (π₯ππ₯ + π¦ππ¦ ) ⋅ ππ¦ ππ§ππ₯ = × 7 ππ§ππ₯ = ππ₯ ∫0 ∫0 π₯2 + 49 ∫0 π₯2 + 49 π₯2 + π¦2 [ ( ) ] 4 1 = 350 × tan−1 − 0 = 26 7 7 1.10. By expressing diagonals as vectors and using the definition of the dot product, find the smaller angle between any two diagonals of a cube, where each diagonal connects diametrically opposite corners, and passes through the center of the cube: Assuming a side length, π, two diagonal vectors would be π = π(ππ₯ + ππ¦ + ππ§ ) and π = √ √ π(ππ₯ − ππ¦ + ππ§ ). Now use π ⋅ π = |π||π| cos π, or π2 (1 − 1 + 1) = ( 3π)( 3π) cos π ⇒ cos π = 1β3 ⇒ π = 70.53β¦ . This result (in magnitude) is the same for any two diagonal vectors. 3 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 1.11. Given the points π(0.1, −0.2, −0.1), π(−0.2, 0.1, 0.3), and π (0.4, 0, 0.1), find: a) the vector πππ : πππ = (−0.2, 0.1, 0.3) − (0.1, −0.2, −0.1) = (−0.3, 0.3, 0.4). b) the dot product πππ ⋅ πππ : πππ = (0.4, 0, 0.1) − (0.1, −0.2, −0.1) = (0.3, 0.2, 0.2). πππ ⋅ πππ = (−0.3, 0.3, 0.4) ⋅ (0.3, 0.2, 0.2) = −0.09 + 0.06 + 0.08 = 0.05. c) the scalar projection of πππ on πππ : πππ ⋅ ππ ππ = (−0.3, 0.3, 0.4) ⋅ √ (0.3, 0.2, 0.2) 0.05 = 0.12 =√ 0.09 + 0.04 + 0.04 0.17 d) the angle between πππ and πππ : −1 ππ = cos ( πππ ⋅ πππ |πππ ||πππ | ( ) −1 = cos 0.05 √ √ 0.34 0.17 ) = 78β¦ 1.12. Write an expression in rectangular components for the vector that extends from (π₯1 , π¦1 , π§1 ) to (π₯2 , π¦2 , π§2 ) and determine the magnitude of this vector. The two points can be written as vectors from the origin: π1 = π₯1 ππ₯ + π¦1 ππ¦ + π§1 ππ§ and π2 = π₯2 ππ₯ + π¦2 ππ¦ + π§2 ππ§ The desired vector will now be the difference: π12 = π2 − π1 = (π₯2 − π₯1 )ππ₯ + (π¦2 − π¦1 )ππ¦ + (π§2 − π§1 )ππ§ whose magnitude is |π12 | = 1.13. √ [ ]1β2 π12 ⋅ π12 = (π₯2 − π₯1 )2 + (π¦2 − π¦1 )2 + (π§2 − π§1 )2 a) Find the vector component of π = (10, −6, 5) that is parallel to π = (0.1, 0.2, 0.3): π ||πΊ = (10, −6, 5) ⋅ (0.1, 0.2, 0.3) π ⋅π (0.1, 0.2, 0.3) = (0.93, 1.86, 2.79) π = 0.01 + 0.04 + 0.09 |π|2 b) Find the vector component of π that is perpendicular to π: π ππΊ = π − π ||πΊ = (10, −6, 5) − (0.93, 1.86, 2.79) = (9.07, −7.86, 2.21) c) Find the vector component of π that is perpendicular to π : πππΉ = π − π||πΉ = π − π⋅π 1.3 (10, −6, 5) = (0.02, 0.25, 0.26) π = (0.1, 0.2, 0.3) − 2 100 + 36 + 25 |π | 4 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 1.14. Given that π + π + π = 0, where the three vectors represent line segments and extend from a common origin, a) must the three vectors be coplanar? In terms of the components, the vector sum will be π + π + π = (π΄π₯ + π΅π₯ + πΆπ₯ )ππ₯ + (π΄π¦ + π΅π¦ + πΆπ¦ )ππ¦ + (π΄π§ + π΅π§ + πΆπ§ )ππ§ which we require to be zero. Suppose the coordinate system is configured so that vectors π and π lie in the π₯-π¦ plane; in this case π΄π§ = π΅π§ = 0. Then πΆπ§ has to be zero in order for the three vectors to sum to zero. Therefore, the three vectors must be coplanar. b) If π + π + π + π = 0, are the four vectors coplanar? The vector sum is now π + π + π + π = (π΄π₯ + π΅π₯ + πΆπ₯ + π·π₯ )ππ₯ + (π΄π¦ + π΅π¦ + πΆπ¦ + π·π¦ )ππ¦ + (π΄π§ + π΅π§ + πΆπ§ + π·π§ )ππ§ Now, for example, if π and π lie in the π₯-π¦ plane, π and π need not, as long as πΆπ§ + π·π§ = 0. So the four vectors need not be coplanar to have a zero sum. 1.15. Three vectors extending from the origin are given as π«1 = (7, 3, −2), π«2 = (−2, 7, −3), and π«3 = (0, 2, 3). Find: a) a unit vector perpendicular to both π«1 and π«2 : ππ12 = π«1 × π«2 (5, 25, 55) = = (0.08, 0.41, 0.91) |π«1 × π«2 | 60.6 b) a unit vector perpendicular to the vectors π«1 − π«2 and π«2 − π«3 : π«1 − π«2 = (9, −4, 1) and π«2 − π«3 = (−2, 5, −6). So π«1 − π«2 × π«2 − π«3 = (19, 52, 37). Then ππ = (19, 52, 37) (19, 52, 37) = = (0.29, 0.78, 0.56) |(19, 52, 37)| 66.6 c) the area of the triangle defined by π«1 and π«2 : 1 Area = |π«1 × π«2 | = 30.3 2 d) the area of the triangle defined by the heads of π«1 , π«2 , and π«3 : 1 1 Area = |(π«2 − π«1 ) × (π«2 − π«3 )| = |(−9, 4, −1) × (−2, 5, −6)| = 33.3 2 2 5 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 1.16. In geometrical optics, the path of a light ray is treated as a vector having the usual three components in a rectangular coordinate system. When light reflects from a plane surface, the effect is to reverse the vector component of the ray that is normal to the surface. This yields a reflection angle that is equal to the incidence angle. Explain what happens when light ray reflects from a corner cube reflector, consisting of three mutually orthogonal surfaces that occupy, for example, the π₯π¦, π₯π§, and π¦π§ planes. The ray is incident in an arbitrary direction within the first octant of the coordinate system, and has negative π₯, π¦, and π§ directions of travel: We can model the incident ray as the vector, ππ = π(−ππ₯ ) + π(−ππ¦ ) + π(−ππ§ ), where π, π, and π are arbitrary positive values. With the three orthogonal planes positioned as given, their effect is to reverse all three components of the incident vector, giving ππ = π(ππ₯ ) + π(ππ¦ ) + π(ππ§ ) = −ππ . So the light propagates in precisely the reverse direction after reflection, no matter what direction it arrived from. 1.17. Point π΄(−4, 2, 5) and the two vectors, ππ΄π = (20, 18, −10) and ππ΄π = (−10, 8, 15), define a triangle. a) Find a unit vector perpendicular to the triangle: Use ππ = ππ΄π × ππ΄π (350, −200, 340) = = (0.664, −0.379, 0.645) |ππ΄π × ππ΄π | 527.35 The vector in the opposite direction to this one is also a valid answer. b) Find a unit vector in the plane of the triangle and perpendicular to ππ΄π : ππ΄π = (−10, 8, 15) = (−0.507, 0.406, 0.761) √ 389 Then πππ΄π = ππ × ππ΄π = (0.664, −0.379, 0.645) × (−0.507, 0.406, 0.761) = (−0.550, −0.832, 0.077) The vector in the opposite direction to this one is also a valid answer. c) Find a unit vector in the plane of the triangle that bisects the interior angle at π΄: A non-unit vector in the required direction is (1β2)(ππ΄π + ππ΄π ), where ππ΄π = (20, 18, −10) = (0.697, 0.627, −0.348) |(20, 18, −10)| Now 1 1 (π + ππ΄π ) = [(0.697, 0.627, −0.348) + (−0.507, 0.406, 0.761)] = (0.095, 0.516, 0.207) 2 π΄π 2 Finally, ππππ = (0.095, 0.516, 0.207) = (0.168, 0.915, 0.367) |(0.095, 0.516, 0.207)| 6 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 1.18. Given two vector fields, π = (π΄βπ) sin π ππ and π = (π΅βπ) sin π ππ , where π΄ and π΅ are constants, a) evaluate π = π × π and express the result in rectangular coordinates: Begin with π΄π΅ 2 sin π ππ π2 Now, find the three rectangular components by using: π=π×π= ππ₯ = π ⋅ ππ₯ = π₯(π₯2 + π¦2 ) π΄π΅ 2 sin π π ⋅ π = π΄π΅ π π₯ π2 (π₯2 + π¦2 + π§2 )5β2 where we have used ππ ⋅ ππ₯ = sin π cos π with π = (π₯2 + π¦2 + π§2 )1β2 , sin π = (π₯2 + π¦2 )1β2 β(π₯2 + π¦2 + π§2 )1β2 and cos π = π₯β(π₯2 + π¦2 )1β2 . Then ππ¦ = π ⋅ ππ¦ = π¦(π₯2 + π¦2 ) π΄π΅ 2 sin π π ⋅ π = π΄π΅ π π¦ π2 (π₯2 + π¦2 + π§2 )5β2 where we have used ππ ⋅ ππ¦ = sin π sin π with sin π = π¦β(π₯2 + π¦2 )1β2 . Finally ππ§ = π ⋅ π π§ = π§(π₯2 + π¦2 ) π΄π΅ 2 sin π π ⋅ π = π΄π΅ π π§ π2 (π₯2 + π¦2 + π§2 )5β2 where ππ ⋅ ππ§ = cos π = π§β(π₯2 + π¦2 + π§2 )1β2 . b) Determine π along the π₯, π¦, and π§ axes: Using the part π results, we would have π(π₯, 0, 0) = π΄π΅ ππ₯ , π₯2 π(0, π¦, 0) = π΄π΅ ππ¦ , π¦2 π(0, 0, π§) = 0 1.19. Consider the important inverse-square dependent radial field in spherical coordinates: π = π΄βπ2 ππ where π΄ is a constant. a) Transform the given field into cylindrical coordinates: Using π2 = π2 + π§2 , the given field may be written: ] π΄ [ π = 2 π π + π π + π π π π π π π§ π§ π + π§2 Now π π ππ = ππ ⋅ ππ = sin π = = 2 π (π + π§2 )1β2 ππ = ππ ⋅ ππ = 0 π§ π§ ππ§ = ππ ⋅ ππ§ = cos π = = 2 π (π + π§2 )1β2 Substituting all, we find ] π΄ [ π = 2 ππ + π§π π π§ π + π§2 b) Transform the given field √ into rectangular coordinates: The quickest way is to use√the results of part π by using π = π₯2 + π¦2 and ππ = cos π ππ₯ + sin π ππ¦ , where cos π = π₯β π₯2 + π¦2 and √ sin π = π¦β π₯2 + π¦2 . Incorporating the above leads to: [ ] π΄ π = 2 π₯ππ₯ + π¦ππ¦ + π§ππ§ 2 2 3β2 (π₯ + π¦ + π§ ) 7 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 1.20. If the three sides of a triangle are represented by the vectors π, π, and π, all directed counter-clockwise, show that |π|2 = (π + π) ⋅ (π + π) and expand the product to obtain the law of cosines. With the vectors drawn as described above, we find that π = −(π + π) and so |π|2 = πΆ 2 = π ⋅ π = (π + π) ⋅ (π + π) So far so good. Now if we expand the product, obtain (π + π) ⋅ (π + π) = π΄2 + π΅ 2 + 2π ⋅ π where π ⋅ π = π΄π΅ cos(180β¦ − πΌ) = −π΄π΅ cos πΌ where πΌ is the interior angle at the junction of π and π. Using this, we have πΆ 2 = π΄2 + π΅ 2 − 2π΄π΅ cos πΌ, which is the law of cosines. 1.21. Express in cylindrical components: a) the vector from πΆ(3, 2, −7) to π·(−1, −4, 2): πΆ(3, 2, −7) → πΆ(π = 3.61, π = 33.7β¦ , π§ = −7) and π·(−1, −4, 2) → π·(π = 4.12, π = −104.0β¦ , π§ = 2). Now ππΆπ· = (−4, −6, 9) and π π = ππΆπ· ⋅ ππ = −4 cos(33.7) − 6 sin(33.7) = −6.66. Then π π = ππΆπ· ⋅ ππ = 4 sin(33.7) − 6 cos(33.7) = −2.77. So ππΆπ· = −6.66ππ − 2.77ππ + 9ππ§ b) a unit vector at π· directed toward πΆ: ππΆπ· = (4, 6, −9) and π π = ππ·πΆ ⋅ ππ = 4 cos(−104.0) + 6 sin(−104.0) = −6.79. Then π π = ππ·πΆ ⋅ ππ = 4[− sin(−104.0)] + 6 cos(−104.0) = 2.43. So ππ·πΆ = −6.79ππ + 2.43ππ − 9ππ§ Thus ππ·πΆ = −0.59ππ + 0.21ππ − 0.78ππ§ c) a unit vector at π· directed toward the origin: Start with π«π· = (−1, −4, 2), and so the vector toward the origin will be −π«π· = (1, 4, −2). Thus in cartesian the unit vector is π = (0.22, 0.87, −0.44). Convert to cylindrical: ππ = (0.22, 0.87, −0.44) ⋅ ππ = 0.22 cos(−104.0) + 0.87 sin(−104.0) = −0.90, and ππ = (0.22, 0.87, −0.44) ⋅ ππ = 0.22[− sin(−104.0)] + 0.87 cos(−104.0) = 0, so that finally, π = −0.90ππ − 0.44ππ§ . 1.22. A sphere of radius π, centered at the origin, rotates about the π§ axis at angular velocity Ω rad/s. The rotation direction is clockwise when one is looking in the positive π§ direction. a) Using spherical components, write an expression for the velocity field, π―, which gives the tangential velocity at any point on the sphere surface: The tangential velocity is the product of the angular velocity and the perpendicular distance from the rotation axis. With clockwise rotation, and with sphere radius π, we obtain π―(π) = Ωπ sin π ππ mβs b) Derive an expression for the difference in velocities between two points on the surface having different latitudes, where the latitude difference is Δπ in radians. Assume Δπ is small: Method 1: The velocity difference between the two points can be approximated as . ππ£ | Δπ£ = Δπ | ππ |π0 where π0 is the mean angle between the two points, and where Δπ is small enough so that π£(π) can be approximated as a linear function. Using this along with the part π result, we find: . Δπ£ = πΩΔπ cos π0 mβs 8 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 1.22 b) (continued) Method 2 (harder): Write Δπ£ = πΩ(sin π2 − sin π1 ) where π1 and π2 lie on either side of π0 , and where π2 − π1 = Δπ. Using a trig identity, the above expression becomes ( ) ) . ( 1 1 Δπ£ = 2πΩ cos (π2 + π1 ) sin (π2 − π1 ) = πΩΔπ cos π0 mβs 2 2 βββββββββββββββββββββ βββββββββββββββββββ . cos π0 =Δπβ2 c) Find the difference in velocities at locations ±1.0β¦ on either side of 45β¦ north latitude on Earth. Take the Earth’s radius as 6370 km at 45β¦ : We have π = 6370 km, π0 = 45β¦ (same in spherical and geographic coordinates), Δπ = 2.0β¦ = 0.035 rad, and Ω = 2πβ24 hr = 7.3 × 10−5 rad/s. Using part π, the velocity difference becomes √ . Δπ£ = (6.37 × 106 m)(7.3 × 10−5 radβs)(0.035 rad)( 2β2) = 11.5 mβs with speed increasing southward. 1.23. The surfaces π = 3, π = 5, π = 100β¦ , π = 130β¦ , π§ = 3, and π§ = 4.5 define a closed surface. a) Find the enclosed volume: 130β¦ 4.5 Vol = ∫3 ∫100β¦ ∫3 5 π ππ ππ ππ§ = 6.28 NOTE: The limits on the π integration must be converted to radians (as was done here, but not shown). b) Find the total area of the enclosing surface: 130β¦ Area = 2 4.5 + ∫3 5 ∫100β¦ ∫3 130β¦ ∫100β¦ 130β¦ 4.5 π ππ ππ + 5 ππ ππ§ + 2 ∫3 ∫3 ∫100β¦ 4.5 5 ∫3 3 ππ ππ§ ππ ππ§ = 20.7 c) Find the total length of the twelve edges of the surfaces: [ β¦ ] 30 30β¦ Length = 4 × 1.5 + 4 × 2 + 2 × × 2π × 3 + × 2π × 5 = 22.4 360β¦ 360β¦ d) Find the length of the longest straight line that lies entirely within the volume: This will be between the points A(π = 3, π = 100β¦ , π§ = 3) and B(π = 5, π = 130β¦ , π§ = 4.5). Performing point transformations to cartesian coordinates, these become A(π₯ = −0.52, π¦ = 2.95, π§ = 3) and B(π₯ = −3.21, π¦ = 3.83, π§ = 4.5). Taking A and B as vectors directed from the origin, the requested length is Length = |π − π| = |(−2.69, 0.88, 1.5)| = 3.21 9 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 1.24. Two unit vectors, π1 and π2 lie in the π₯π¦ plane and pass through the origin. They make angles π1 and π2 with the π₯ axis respectively. a) Express each vector in rectangular components; Have π1 = π΄π₯1 ππ₯ +π΄π¦1 ππ¦ , so that π΄π₯1 = π1 ⋅ππ₯ = cos π1 . Then, π΄π¦1 = π1 ⋅ ππ¦ = cos(90 − π1 ) = sin π1 . Therefore, π1 = cos π1 ππ₯ + sin π1 ππ¦ and similarly, π2 = cos π2 ππ₯ + sin π2 ππ¦ b) take the dot product and verify the trigonometric identity, cos(π1 − π2 ) = cos π1 cos π2 + sin π1 sin π2 : From the definition of the dot product, π1 ⋅ π2 = (1)(1) cos(π1 − π2 ) = (cos π1 ππ₯ + sin π1 ππ¦ ) ⋅ (cos π2 ππ₯ + sin π2 ππ¦ ) = cos π1 cos π2 + sin π1 sin π2 c) take the cross product and verify the trigonometric identity sin(π2 − π1 ) = sin π2 cos π1 − cos π2 sin π1 : From the definition of the cross product, and since π1 and π2 both lie in the π₯π¦ plane, | π ππ¦ ππ§ || | π₯ | | π1 × π2 = (1)(1) sin(π1 − π2 ) ππ§ = |cos π1 sin π1 0 | | | |cos π2 sin π2 0 | | | [ ] = sin π2 cos π1 − cos π2 sin π1 ππ§ thus verified. 1.25. Convert the vector field π = π΄(π₯2 + π¦2 )−1 [π₯ ππ¦ − π¦ ππ₯ ] into cylindrical coordinates. π΄ is a constant: The π§ component is obviously zero, so we are left only with the π and π components to find. First, ] π΄ [ π₯ π ⋅ π − π¦ π ⋅ π π¦ π π₯ π π₯2 + π¦2 √ √ where ππ¦ ⋅ ππ = sin π = π¦β π₯2 + π¦2 and ππ₯ ⋅ ππ = cos π = π₯β π₯2 + π¦2 . Thus π»π = π ⋅ ππ = π»π = π΄ [π₯π¦ − π¦π₯] = 0 (π₯2 + π¦2 )3β2 Then ] π΄ [ π₯ π ⋅ π − π¦ π ⋅ π π¦ π π₯ π π₯2 + π¦2 √ √ where ππ¦ ⋅ ππ = cos π = π₯β π₯2 + π¦2 and ππ₯ ⋅ ππ = − sin π = −π¦β π₯2 + π¦2 . Thus π»π = π ⋅ ππ = π»π = [ 2 ] π΄ π΄ π΄ 2 π₯ + π¦ = 2 = π (π₯2 + π¦2 )3β2 (π₯ + π¦2 )1β2 Finally π= π΄ π π π 10 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 1.26. Express the uniform vector field, π = 10 ππ¦ in a) cylindrical components: πΉπ = 10 ππ¦ ⋅ ππ = 10 sin π, πΉπ = 10 ππ¦ ⋅ ππ = 10 cos π, and πΉπ§ = 0. Combining, we obtain π (π, π) = 10(sin π ππ + cos π ππ ). b) spherical components: πΉπ = 10 ππ¦ ⋅ ππ = 10 sin π sin π; πΉπ = 10 ππ¦ ⋅ ππ = 10 cos π sin π; πΉπ = 10 ππ¦ ⋅ ππ = 10 cos π. [ ] Combining, we obtain π (π, π, π) = 10 sin π sin π ππ + cos π sin π ππ + cos π ππ . 1.27. The dipole field is given in spherical coordinates as: π= π΄ (2 cos π ππ + sin π ππ ) π3 where π΄ is a constant and where π > 0. a) Identify the surface on which the field is entirely perpendicular to the π₯π¦ plane, and express the field on that surface in cylindrical coordinates: As there is no ππ component, the cylindrical coordinate field components will be ππ and ππ§ only. We look for the surface on which the ππ component is zero. So we set up: πΈπ = π ⋅ ππ = π΄ (2 cos π ππ ⋅ ππ + sin π ππ ⋅ ππ ) = 0 π3 βββ βββ cos π sin π From which the condition for zero πΈπ is identified: 3 cos π sin π = 0 ⇒ π = 0, 90β¦ The π = 0 option is only a line (the π§ axis – and the answer to part π), so the surface on which the field is perpendicular to the π₯π¦ plane is the π₯π¦ plane itself, on which π = 90β¦ . On that surface: π(π = 90) = π΄ π΄ ππ = − 3 ππ§ π3 π b) Identify the coordinate axis on which the field is entirely perpendicular to the π₯π¦ plane and express the field there in cylindrical coordinates: This will be the π = 0 solution in part π, which is the π§ axis. The field on that axis is 2π΄ π(π = 0) = 3 ππ§ π§ c) Specify the surface on which the field is entirely parallel to the π₯π¦ plane: In this case, we require a zero π§ component, so we construct: πΈπ§ = π ⋅ ππ§ = π΄ (2 cos π ππ ⋅ ππ§ + sin π ππ ⋅ ππ§ ) = 0 π3 βββ βββ cos π − sin π This tells us that for a zero π§ component, we must have 2 cos2 π − sin2 π = 0 ⇒ 3 cos2 π − 1 = 0 ⇒ π = 54.74β¦ The surface is a cone of angle 54.74β¦ to the π§ axis, and extending over all values of π within the limits of validity of the given field. 11 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 1.28. State whether or not π = π and, if not, what conditions are imposed on π and π when a) π ⋅ ππ₯ = π ⋅ ππ₯ : For this to be true, both π and π must be oriented at the same angle, π, from the π₯ axis. But this would allow either vector to lie anywhere along a conical surface of angle π about the π₯ axis. Therefore, π can be equal to π, but not necessarily. b) π × ππ₯ = π × ππ₯ : This is a more restrictive condition because the cross product gives a vector. For both cross products to lie in the same direction, π, π, and ππ₯ must be coplanar. But if π lies at angle π to the π₯ axis, π could lie at π or at 180β¦ − π to give the same cross product. So again, π can be equal to π, but not necessarily. c) π ⋅ ππ₯ = π ⋅ ππ₯ and π × ππ₯ = π × ππ₯ : In this case, we need to satisfy both requirements in parts π and π – that is, π, π, and ππ₯ must be coplanar, and π and π must lie at the same angle, π, to ππ₯ . With coplanar vectors, this latter condition might imply that both +π and −π would therefore work. But the negative angle reverses the direction of the cross product direction. Therefore both vectors must lie in the same plane and lie at the same angle to π₯; i.e., π must be equal to π. d) π⋅π = π⋅π and π×π = π×π where π is any vector except π = 0: This is just the general case of part π. Since we can orient our coordinate system in any manner we choose, we can arrange it so that the π₯ axis coincides with the direction of vector π. Thus all the arguments of part π apply, and again we conclude that π must be equal to π. 1.29. A vector field is expressed as π = 10π§πΜ π§ . Evaluate the components of this field that are π) normal, and π) tangent to a spherical surface of radius π: a) The normal component is the radial component, found by projecting π into the ππ direction. In general, πΉπ = π ⋅ ππ = 10π§ππ§ ⋅ ππ = 10π§ cos π where, on the sphere surface, π§ = π cos π. Therefore πΉπ (on surface) = 10π cos2 π. b) The tangential component is found by projecting π into the ππ or ππ directions. Note that ππ§ ⋅ ππ = 0, so we are left with the tangential component lying in only the direction of ππ : πΉπ = 10π§ππ§ ⋅ ππ |π=π = −10π cos π sin π 12 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 1.30. Consider a problem analogous to the varying wind velocities encountered by transcontinental aircraft. We assume a constant altitude, a plane earth, a flight along the π₯ axis from 0 to 10 units, no vertical velocity component, and no change in wind velocity with time. Assume ππ₯ to be directed to the east and ππ¦ to the north. The wind velocity at the operating altitude is assumed to be: π―(π₯, π¦) = (0.01π₯2 − 0.08π₯ + 0.66)ππ₯ − (0.05π₯ − 0.4)ππ¦ 1 + 0.5π¦2 a) Determine the location and magnitude of the maximum tailwind encountered: Tailwind would be π₯-directed, and so we look at the π₯ component only. Over the flight range, this function maximizes at a value of 0.86β(1 + 0.5π¦2 ) at π₯ = 10 (at the end of the trip). It reaches a local minimum of 0.50β(1 + 0.5π¦2 ) at π₯ = 4, and has another local maximum of 0.66β(1 + 0.5π¦2 ) at the trip start, π₯ = 0. b) Repeat for headwind: The π₯ component is always positive, and so therefore no headwind exists over the travel range. c) Repeat for crosswind: Crosswind will be found from the π¦ component, which is seen to maximize over the flight range at a value of 0.4β(1 + 0.5π¦2 ) at the trip start (π₯ = 0). d) Would more favorable tailwinds be available at some other latitude? If so, where? Minimizing the denominator accomplishes this; in particular, the lattitude associated with π¦ = 0 gives the strongest tailwind. 13 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
0
You can add this document to your study collection(s)
Sign in Available only to authorized usersYou can add this document to your saved list
Sign in Available only to authorized users(For complaints, use another form )