Part 19 Inner Product Dot Product-Advanced Engineering Mathematics-Solution Manual, Exercises of Engineering Mathematics

Download Part 19 Inner Product Dot Product-Advanced Engineering Mathematics-Solution Manual and more Exercises Engineering Mathematics in PDF only on Docsity! SOLUTION TO PROBLEM SET 9.1, page 370 2. The components are 5, 5, 5. The length is 5
3. Hence the unit vector in the direction of v is [1/
3, 1/
3, 1/
3 ]. 4. v1
2, v2
6, v3
6; v

76. Hence the unit vector in the direction of v is [1/
19, 3/
19, 3/
19]. 6. Components 6, 8, 10; length 10
2. Unit vector [0.6/
2, 0.8/
2, 1/
2]. 8. Q: (0, 0, 0); v

84 10. Q: (3, 2, 6), v
7. Note that the given vector is the position vector of Q. 12. Q: (4, 0, 0); v
3
3 14. [6, 3, 10]. This illustrates (4a). 16. [26, 13, 19]. This illustrates (4b). 18. [2/
5, 1/
5, 0], [0, 0, 1]. These are unit vectors. 20. 5.48 
30 
5 
45  8.94 24. [11, 8, 0] is the resultant,
185
13.6 its magnitude 26. The resultant is 0. The forces are in equilibrium. 28. Resultant [13, 25, 3], magnitude
803
28.3 30. The z-component of the resultant is 0 for c
12. 32. 1  p  q  3, nothing about the direction. If this were the arm of some machine or robot, it could not reach the origin but could reach every point of the annulus (ring) indicated. In the next problem the origin can be reached. 34. vB  vA
[400/
2, 400/
2 ]  [500/
2, 500/
2 ]
[100/
2, 900/
2 ] 36. Choose a coordinate system whose axes contain the mirrors. Let u
[u1, u2] be incident. Then the first reflection gives, say, v
[u1, u2], and the second w
[u1, u2]

u. The reflected ray is parallel to the incoming ray, with the direction reversed. 38. Team Project. (a) The idea is to write the position vector of the point of intersection P in two ways and then to compare them, using that a and b are linearly independent vectors. Thus (a  b)
a  (b  a). 
1   are the coefficients of a and 
 those of b. Together, 

1_2, expressing bisection. (b) The idea is similar to that in part (a). It gives (a  b)
1_2a   1_ 2(b  a). 
1_2  1_ 2 from a and 
1_ 2 from b, resulting in 
1_ 4, thus giving a ratio (3/4)
(1/4)
3:1. (c) Partition the parallelogram into four congruent parallelograms. Part (a) gives 1
1 for a small parallelogram, hence 1
(1  2) for the large parallelogram. (d) v(P)
1_2a  (b  1_ 2a)
1_ 2b  (a  1_ 2b) has the solution 

1_ 3, which gives by substitution v(P)
1_3(a  b) and shows that the third median OQ passes through P and OP equals 2_3 of v(Q)
1_ 2a  b, dividing OQ in the ratio 2
1, too. (e) In the figure in the problem set, a  b  c  d
0; hence c  d
(a  b). Also, AB
1_2(a  b), CD
1_ 2(c  d)
 1_ 2(a  b), and for DC we get  1_ 2(a  b), which shows that one pair of sides is parallel and of the same length. Similarly for the other pair. Instructor’s Manual 175 im09.qxd 9/21/05 12:15 PM Page 175 docsity.com (f) Let a, b, c be edge vectors with a common initial point (see the figure). Then the four (space) diagonals have the midpoints AG: 1_2(a  b  c) BH: a  1_2(b  c  a) EC: c  1_2(a  b  c) DF: b  1_2(a  c  b), and these four position vectors are equal. Section 9.1. Parallelepiped in Team Project 38(f) (g) Let v1, • • • , vn be the vectors. Their angle is 
2 /n. The interior angle at each vertex is
 (2 /n). Put v2 at the terminal point of v1, then v3 at the terminal point of v2, etc. Then the figure thus obtained is an n-sided regular polygon, because the angle between two sides equals  
. Hence v1  v2  • • •  vn
0. (Of course, for even n the truth of the statement is immediately obvious.) SECTION 9.2. Inner Product (Dot Product), page 371 Purpose. We define, explain, and apply a first kind of product of vectors, the dot product a • b, whose value is a scalar. Main Content, Important Concepts Definition (1) Dot product in terms of components Orthogonality Length and angle between vectors in terms of dot products Cauchy–Schwarz and triangle inequalities Comment on Dot Product This product is motivated by work done by a force (Example 2), by the calculation of components of forces (Example 3), and by geometric applications such as those given in Examples 5 and 6. “Inner product” is more modern than “dot product” and is also used in more general settings (see Sec. 7.9). E b c a A H G D F B C 176 Instructor’s Manual im09.qxd 9/21/05 12:15 PM Page 176 docsity.com To prove (15), we note that a • [b  (c  d)] equals (a b [c  d])
(a  b) • (c  d) by the definition of the triple product, as well as (a • c)(b • d)  (a • d)(b • c) by (13) (take the dot product by a). The last formula, (16), follows from familiar rules of interchanging the rows of a determinant. 26. [3, 3, 0]  [0, 0, 5]
[15, 15, 0]; 15
2 28. 0, 0. Q lies on the line of action of p. 30. w  r
[5/
2, 5/
2, 0]  [4, 2, 2]
[5
2, 5
2, 5
2], v

150 32. [3, 4, 7]  [1, 8, 4]
[40, 5, 20]
45 34. 1_2 [0, 3, 0]  [4, 0, 2]
1_ 2 [6, 0, 12]

45 36. [2, 3, 2]  [1, 5, 3]
[19, 4, 13], hence 19x  4y  13z
c
19
2  4
1  13
3
5. 38. From the given points we get three edge vectors whose scalar triple product is k k
8. Hence the answer is 8/6
4/3. SECTION 9.4. Vector and Scalar Functions and Fields. Derivatives, page 384 Purpose. To get started on vector differential calculus, we discuss vector functions and their continuity and differentiability. Main Content, Important Concepts Vector and scalar functions and fields Continuity, derivative of vector functions (9), (10) Differentiation of dot, cross, and triple products, (11)–(13) Partial derivatives Comment on Content This parallels calculus of functions of one variable and can be surveyed quickly. SOLUTIONS TO PROBLEM SET 9.4, page 389 2. Parallel straight lines 4. Circles (x  )2  y2
passing through the origin 6. Hyperbolas x2  (y  4)2
const 1  4c2 1  2c 1 4 7 1 4 5 4 6 4 Instructor’s Manual 179 im09.qxd 9/21/05 12:15 PM Page 179 docsity.com 8. CAS Project. A CAS can graphically handle these more complicated functions, whereas the paper-and-pencil method is relatively limited. This is the point of the project. Note that all these functions occur in connection with Laplace’s equation, so they are real or imaginary parts of complex analytic functions. 10. Elliptic cylinders with vertical generating straight lines 12. Cylinders with cross section z
4y2  c and generating straight lines parallel to the x-axis 14. Elliptic paraboloids z
x2  4y2  c 16. On horizontal lines y
const the i-component of v is constant, and on vertical lines x
const the j-component of v is constant. 18. At each point the vector v equals the corresponding position vector, so that sketching is easy. 20. On y
x the vector v is vertical and on y
x it is horizontal. 22. As a curve this is a helix. The first derivative [4 sin t, 4 cos t, 2] is tangent to this curve, as we shall discuss in the next section, and the second derivative [4 cos t, 4 sin t, 0] is parallel to the xy-plane and perpendicular to that tangent. 24. [cos x cosh y, sin x sinh y], [sin x sinh y, cos x cosh y]; [ex cos y, ex sin y], [ex sin y, ex cos y] SECTION 9.5. Curves. Arc Length. Curvature. Torsion, page 389 Purpose. Discussion of space curves as an application of vector functions of one variable, the use of curves as paths in mechanics (and as paths of integration of line integrals in Chapter 10). Role of parametric representations, interpretation of derivatives in mechanics, completion of the discussion of the foundations of differential–geometric curve theory. Main Content, Important Concepts Parametric representation (1) Orientation of a curve Circle, ellipse, straight line, helix Tangent vector (7), unit tangent vector (8), tangent (9) Length (10), arc length (11) Arc length as parameter [cf. (14)] Velocity, acceleration (16)–(19) Centripetal acceleration, Coriolis acceleration Curvature, torsion, Frenet formulas (Prob. 50) Short Courses. This section can be omitted. Comment on Problems 26–28 These involve only integrals that are simple (which is generally not the case in connection with lengths of curves). 180 Instructor’s Manual im09.qxd 9/21/05 12:15 PM Page 180 docsity.com SOLUTIONS TO PROBLEM SET 9.5, page 398 2. r(t)
[5  3t, 1  t, 2  t] 4. r(t)
[t, 3  2t, 7t] 6. r(t)
[cos t, sin t, sin t] 8. Direction of the line of intersection [1, 1, 1]  [2, 5, 1]
[4, 3, 7]. Point of intersection with the xy-plane z
0 from x  y
2, 2x  5y
3; thus x
13/7, y
1/7. Hence a parametric representation is r(t)
[_137  4t, 1_7  3t, 7t]. 10. Helix [3 cos t, 3 sin t, 4t] 12. Straight line through (4, 0, 3) in the direction of [2, 8, 5] 14. Space curve with projections y
x2, z
x3 into the xy- and xz-coordinate planes, similar to the curve in Fig. 210. 16. Hyperbola x2  y2
1, z
0 18. x
1, (y  5)(z  5)
1; hyperbola 20. No, because the exponential function et  is nonnegative. 22. r
[1, 2t, 0], u
(1  4t2)
1/ 2[1, 2t, 0], q
[2  w, 4  4w, 0] 24. r(t)
[3 sin t, 3 cos t, 4], u
[0.6 sin t, 0.6 cos t, 0.8], q
[3, 3w, 8  4w] 26. t
4 , r
[2 sin t, 2 cos t, 6],
r• r

40,

4
40 28. r
[3a cos2 t sin t, 3a sin2 t cos t]. Taking the dot product and applying trigonometric simplification gives r• r
9a2 cos4 t sin2 t  9a2 sin4 t cos2 t
9a2 cos2 t sin2 t
sin2 2t. From this we obtain as the length in the first quadrant

a  /2 0 sin 2t dt
 (cos  cos 0)
. Answer: 6a 30. We obtain ds2
dx2  dy2
(d cos    sin  d)2  (d sin    cos  d)2
d2  2 d2
(2  2) d2. For the cardioid, 2  2
a2(1  cos )2  a2 sin2 
2a2(1  cos )
4a2 sin2 1_2 so that

2a 2 0 sin 1_2 d
8a. 3a  2 3a  4 3  2 9a2  4 Instructor’s Manual 181 im09.qxd 9/21/05 12:15 PM Page 181 docsity.com