Engenharia

Disponível somente no TrabalhosFeitos
  • Páginas : 364 (90830 palavras )
  • Download(s) : 0
  • Publicado : 23 de outubro de 2012
Ler documento completo
Amostra do texto
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=

(26, 10, 4)
= (0.92, 0.36, 0.14)
|(26, 10, 4)|

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 unit vector will be
m=

(1, 1.5, 3.5)
= (0.25, 0.38, 0.89)
|(1, 1.5, 3.5)|

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 from the
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 of the line AB .
a0 M =

(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 arelooking for.
1.5. A vector field is specified as G = 24xy ax + 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√ (x, 1, 1) = (24x, 12x 2 + 24, 18),from which Gx = 24x , Gy = 12x 2 + 24,
G
Gz = 18, and |G| = 6 4x 4 + 32x 2 + 25. Plots are shown below.

1.7. Given the vector field E = 4zy 2 cos 2x ax + 2zy sin 2x ay + y 2 sin 2x az 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 which Ey = 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...
tracking img