# Solucionario hayt

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CHAPTER 1 1.1. Given the vectors M = −10ax + 4ay − 8az and N = 8ax + 7ay − 2az , ﬁnd: a) a unit vector in the direction of −M + 2N. −M + 2N = 10ax − 4ay + 8az + 16ax + 14ay − 4az = (26, 10, 4) Thus a= b) the magnitude of 5ax + N − 3M: (5, 0, 0) + (8, 7, −2) − (−30, 12, −24) = (43, −5, 22), and |(43, −5, 22)| = 48.6. c) |M||2N|(M + N): |(−10, 4, −8)||(16, 14, −4)|(−2, 11, −10) = (13.4)(21.6)(−2,11, −10) = (−580.5, 3193, −2902) 1.2. Given three points, A(4, 3, 2), B(−2, 0, 5), and C(7, −2, 1): a) Specify the vector A extending from the origin to the point A. A = (4, 3, 2) = 4ax + 3ay + 2az b) Give a unit vector extending from the origin to the midpoint of line AB. The vector from the origin to the midpoint is given by M = (1/2)(A + B) = (1/2)(4 − 2, 3 + 0, 2 + 5) = (1, 1.5, 3.5) The unitvector will be m= (1, 1.5, 3.5) = (0.25, 0.38, 0.89) |(1, 1.5, 3.5)| (26, 10, 4) = (0.92, 0.36, 0.14) |(26, 10, 4)|

c) Calculate the length of the perimeter of triangle ABC: Begin with AB = (−6, −3, 3), BC = (9, −2, −4), CA = (3, −5, −1). Then |AB| + |BC| + |CA| = 7.35 + 10.05 + 5.91 = 23.32 1.3. The vector from the origin to the point A is given as (6, −2, −4), and the unit vector directed fromthe origin toward point B is (2, −2, 1)/3. If points A and B are ten units apart, ﬁnd the coordinates of point B. With A = (6, −2, −4) and B = 1 B(2, −2, 1), we use the fact that |B − A| = 10, or 3 |(6 − 2 B)ax − (2 − 2 B)ay − (4 + 1 B)az | = 10 3 3 3 Expanding, obtain 36 − 8B + 4 B 2 + 4 − 8 B + 4 B 2 + 16 + 8 B + 1 B 2 = 100 9 3 9 3 9 or B 2 − 8B − 44 = 0. Thus B = B=
√ 8± 64−176 2

= 11.75(taking positive option) and so

2 1 2 (11.75)ax − (11.75)ay + (11.75)az = 7.83ax − 7.83ay + 3.92az 3 3 3 1

1.4. given points A(8, −5, 4) and B(−2, 3, 2), ﬁnd: a) the distance from A to B. |B − A| = |(−10, 8, −2)| = 12.96 b) a unit vector directed from A towards B. This is found through aAB = B−A = (−0.77, 0.62, −0.15) |B − A|

c) a unit vector directed from the origin to the midpoint ofthe line AB. a0M = (3, −1, 3) (A + B)/2 = √ = (0.69, −0.23, 0.69) |(A + B)/2| 19

d) the coordinates of the point on the line connecting A to B at which the line intersects the plane z = 3. Note that the midpoint, (3, −1, 3), as determined from part c happens to have z coordinate of 3. This is the point we are looking for. 1.5. A vector ﬁeld is speciﬁed as G = 24xyax + 12(x 2 + 2)ay + 18z2 az .Given two points, P (1, 2, −1) and Q(−2, 1, 3), ﬁnd: a) G at P : G(1, 2, −1) = (48, 36, 18) b) a unit vector in the direction of G at Q: G(−2, 1, 3) = (−48, 72, 162), so aG = (−48, 72, 162) = (−0.26, 0.39, 0.88) |(−48, 72, 162)|

c) a unit vector directed from Q toward P : aQP = (3, −1, 4) P−Q = √ = (0.59, 0.20, −0.78) |P − Q| 26

d) the equation of the surface on which |G| = 60: We write 60 =|(24xy, 12(x 2 + 2), 18z2 )|, or 10 = |(4xy, 2x 2 + 4, 3z2 )|, so the equation is 100 = 16x 2 y 2 + 4x 4 + 16x 2 + 16 + 9z4

2

1.6. For the G ﬁeld in Problem 1.5, make sketches of Gx , Gy , Gz and |G| along the line y = 1, z = 1, for 0 ≤ x ≤ 2. We ﬁnd√ G(x, 1, 1) = (24x, 12x 2 + 24, 18), from which Gx = 24x, Gy = 12x 2 + 24, Gz = 18, and |G| = 6 4x 4 + 32x 2 + 25. Plots are shown below.1.7. Given the vector ﬁeld E = 4zy 2 cos 2xax + 2zy sin 2xay + y 2 sin 2xaz for the region |x|, |y|, and |z| less than 2, ﬁnd: a) the surfaces on which Ey = 0. With Ey = 2zy sin 2x = 0, the surfaces are 1) the plane z = 0, with |x| < 2, |y| < 2; 2) the plane y = 0, with |x| < 2, |z| < 2; 3) the plane x = 0, with |y| < 2, |z| < 2; 4) the plane x = π/2, with |y| < 2, |z| < 2. b) the region in whichEy = Ez : This occurs when 2zy sin 2x = y 2 sin 2x, or on the plane 2z = y, with |x| < 2, |y| < 2, |z| < 1. c) the region in which E = 0: We would have Ex = Ey = Ez = 0, or zy 2 cos 2x = zy sin 2x = y 2 sin 2x = 0. This condition is met on the plane y = 0, with |x| < 2, |z| < 2. 1.8. Two vector ﬁelds are F = −10ax + 20x(y − 1)ay and G = 2x 2 yax − 4ay + zaz . For the point P (2, 3, −4), ﬁnd: a)...