Transcript
Chapter 2 Problem Solutions
211 .. F1 = mω 2 d sin θ , F2 = mg
cos θ =
9.78
(2π )
2
× 0.5
sin θ g 2π × rpm , F1 = F2 , ∴ cos θ = 2 , ω = = 2π , cos θ 60 ω d
= 0.5 , ∴ θ = 60.3o
d F1
F2
m
mg
21 . .2 W cos 45o , ∴ θ = 813 . o, 200 mi GS + Wsin45o = 200 cos θ , GS = 169.71 hr
W × cos 45o = 200 × sin θ , sin θ =
45° W GS 200 mi/hr
2-1
21 . .3 4 sin θ = 3 , sin θ =
3 , ∴ θ = 48.6 o 4
N 4 mi/hr 3 mi/hr
. .4 21 mv 2 , −W = N cos θ = mg , ∴ tan θ = r 1 hr mi 5280 ft ft v = 60 × × = 88 , ∴ tan θ = 016 . , hr 1mi 3600 s s F = N sin θ =
v2 , rg
∴ θ = 916 . o F
N
—W
N
W
2.3.1 C = A + B, ∴ C 2 = C • C = ( A + B ) • ( A + B ) = A • A + B • B + 2A • B = A2 + B 2 + 2 AB cos α But α = 180 o − θ AB and ∴ cos α = − cos θ AB ∴ C 2 = A2 + B 2 − 2 AB cos θ AB which is the law of cosines.
C AB A
2-2
B ␣
2.4.1 A + B + C = 0 , A × ( A + B + C) = 0 = A ×3 A+A×B+A×C= 0 12
(
0
)
o
A × B = − AB sin α C a n = − AB sin 180 − θ C a n where a n is a unit normal into the
page. Also A × C = AC sin α B a n = AC sin θ B a n . ∴ AB sin θ C = AC sin θ B giving the B C law of sines: = . Similarly for B × ( A + B + C) = 0 and sin θ B sin θ C
C × ( A + B + C) = 0 . B
C C B
B
B
C
A
␣C
C
␣B
A
A
2.4.2 (a) ( A • B) gives a scalar which cannot be “crossed with” a vector. (c) ( A • B) gives a scalar which cannot be “dotted with” a vector. (d) ( B • C) gives a scalar which cannot be added to a vector.
2.5.1 (a) A = (3 − 0)a x + ( −4 − 2)a y + ( 5 − ( −4)) a z = 3a x − 6a y + 9a z (b) A = (c) a A =
(3) 2 + ( −6)2 + (9) 2
= 1122 . m
A = 0.27a x − 0.53a y + 0.8a z A
2.5.2 (a) A + B = 3a x + 4a y − 3a z , (b) B − C = −2a x + 2a y − 3a z , (c)
. , (e) A + 3B - 2C = − a x + 8a y − 9a z , (d) A = 2 2 + 32 + 12 = 374 aB =
(
)
1 B = a x + a y − 2a z =.41a x + 0.41a y − 0.82a z , (f) A • B = 7 , (g) B • A = 7 , B 6
(h) B × C = − a x − 7a y − 4a z , (i) C × B = a x + 7a y + 4a z , (j) A • B × C = −19
2-3
2.5.3 (a) A • B = 2 − 2 − 3 = −3 = A B cos θ . Therefore B cos θ =
A•B 3 = = 0.8 , A 14
−3 A•B = ⇒ θ = 1091 . o , (c) AB 14 6 A × B − a x − 7a y − 5a z unit vector = = . a x − 0.808a y − 0.577a z . = −0115 A×B 1 + 49 + 25
(b) cos θ =
2.5.4 A • B = α + 2 − 3 = 0 ∴α = 1
2.5.5 A × B = ( −18 − 3β )a x + (3α + 9)a y + ( β − 2α )a z = 0 . Hence α = −3 and β = −6 .
2.5.6 A × B = −14a x − 9a y + a z gives a vector that is perpendicular to the planes containing
both A and B and hence is perpendicular to both A and B. The length of this vector is
( −14) 2 + ( −9) 2 + (1) 2 C=
= 16.67 . Hence
(
)
10 −14a x − 9a y + a z = −8.4a x − 5.4a y + 0.6a z . Check that A • C = 0 and 16.67
B•C = 0. 2.5.7
(
)
(
)
B × C = B y Cz − Bz C y a x + ( Bz Cx − Bx Cz )a y + Bx C y − B y Cx a z so that
) ( ( ) + ( Az ( B y Cz − Bz C y ) − Ax ( Bx C y − B y Cx ))a y + ( Ax ( Bz Cx − Bx Cz ) − A y ( B y Cz − Bz C y ))a z
A × ( B × C) = A y Bx C y − B y Cx − Az ( Bz Cx − Bx Cz ) a x
(
)
(
)
(
)
B( A • C) = Bx Ax Cx + A y C y + Az Cz a x + B y Ax Cx + A y C y + Az Cz a y
(
)
+ Bz Ax Cx + A y C y + Az Cz a z
(
)
C( A • B ) = Cx Ax Bx + A y B y + Az Bz a x + C y Ax Bx + A y B y + Az Bz a y
(
)
+ Cz Ax Bx + A y B y + Az Bz a z Matching components we find that A × ( B × C) = B( A • C) − C( A • B) .
2-4
2.5.8
(
)
( ) ( A × B) × C = (( Az Bx − Ax Bz )Cz − ( Ax B y − Ay Bx )C y )a x + (( Ax B y − A y Bx )Cx − ( A y Bz − Az B y )Cz )a y + (( A y Bz − Az B y )C y − ( Az Bx − Ax Bz )Cx )a z
A × B = A y Bz − Az B y a x + ( Az Bx − Ax Bz )a y + Ax B y − A y Bx a z so that
Comparing terms to the result in the previous problem we see that A × ( B × C) ≠ ( A × B ) × C . 2.5.9
The distance is D =
( 3 − ( −1)) 2 + ( −1 − 2)2 + ( 5 − ( −4)) 2 = 10.296 .
By integration we
3 5 integrate D = ∫ PP2 dl . A straight line between the two points is governed by y = − x + 1 4 4 2
and z = D=
9 7 3 9 x − . Hence dl = dx 1 + − + 4 4 4 4
2
= 2.574dx and
x =3
∫ 2.574 dx = 10.296 .
x =−1
2.5.10
The surface is drawn below and lies in the yz plane at x=1. Hence the surface area is z = 2 y = 2 z −1
z =2
z =1
z =1
A= ∫
∫ dydz = ∫ (2 z − 2) dz = 1 . Directly it is the area of a triangle of height 1 and
y =1
base 2 or A =
1 (2)(1) = 1 . 2 z y = 2z – 1
(1, 3, 2)
y (1, 1, 1)
(1, 3, 1)
x
2-5
2.6.1
Drawing the rectangular and cylindrical coordinate system axes as shown below we see that x = r cos φ , y = r sin φ , and z = z . From this we form
(
)
x 2 + y 2 = r 2 cos 2 φ + sin 2 φ and hence r = 1442443 1 y r sin φ = = tan φ . x r cos φ
x 2 + y 2 . Similarly we form
z z
r P(r, , z)
y
r x
2.6.2
Draw the coordinate system and use the right-hand rule. 2.6.3
Drawing the vector in the xy plane as shown below shows that Ax = Ar cos φ − Aφ sin φ and A y = Ar sin φ + Aφ cos φ . Ay y
Ax
Ar
A
x
2-6
2.6.4
At point P, φ =
π 3
= 60o . Hence Ax = 2 cos φ + 3 sin φ =3.598 and
A y = 2 sin φ − 3 cos φ =0.232 and Bx = 4 cos φ − 6 sin φ =-3.196 and B y = 4 sin φ + 6 cos φ =6.464. Directly in cylindrical coordinates A • B = 8 − 18 − 2 = −12 . In rectangular coordinates A • B = (3598 . × −3196 . ) + (0.232 × 6.464) − 2 = −12 . 2.6.5
π From problem 2.6.1 x = r cos φ , y = r sin φ , z=z. At P1 2, ,1 , x1 = 0, y1 = 2, z1 = 1 2 π and at P2 = 3, ,−2 , x 2 = 15 . , y 2 = 2.598, z 2 = −2 . Hence the distance between the two 3
points is D =
( x2 − x1 )2 + ( y2 − y1 )2 + ( z2 − z1 )2
= 3.407 .
2.6.6
The surface is 1/6 of the surface of a cylinder of length 4-1=3 and radius 2. Hence the (2π × 2 × 3) = 2π . By direct integration we have surface area is S = 6 4
π
3
r = 2) dφ dr = 2π . ∫ (1 4243 z =1 φ = 0 dsr
S= ∫
2.6.7
The volume is 1/6 of the volume of a cylinder of radius 2 minus the volume of a cylinder 1 π 2 2 of radius 1 or V = π ( 2) × 1 − π (1) × 1 = . By direct integration, 6 2 1
π
(
)
3 2
π
φ dz = . ∫ rdrd 1 424 3 2 z = 0 φ = 0 r =1
V = ∫
∫
dv
2.7.1
Drawing the rectangular and spherical coordinate system axes as shown below we see that
x = r sin θ cos φ , y = r sin θ sin φ , and z = r cos θ . From this we form
(
)
x 2 + y 2 + z 2 = r 2 sin 2 θ cos 2 φ + sin 2 φ + r 2 cos 2 θ = r 2 and hence 1442443 1
2-7
y r sin θ sin φ r = x 2 + y 2 + z 2 . Similarly we form = = tan φ and x r sin θ cos φ x2 + y2 r sin θ = = tan θ . z r cos θ z z
r sin
r y
x
2.7.2
Draw the coordinate system and use the right-hand rule. 2.7.3
Drawing the vector in the zx or zy plane as shown below shows that Az = Ar cos θ − Aθ sin θ . The components parallel to the xy plane are Ar sin θ + Aθ cos θ
and the φ component, Aφ . Hence the x and y components of these are
Ax = ( Ar sin θ + Aθ cos θ ) cos φ − Aφ sin φ and A y = ( Ar sin θ + Aθ cos θ ) sin φ + Aφ cos φ .
Ay
z Ar sin Ar cos
Ax
Ar
Ar sin + A cos
A
A sin
A
A cos
x x, y plane
2-8
y
2.7.4
2π π = 120 o and φ = = 60o . Hence 3 3
At point P, θ =
Ax = 2 sin θ cos φ + 3 cos θ cos φ − sin φ =-0.75 and A y = 2 sin θ sin φ + 3 cos θ sin φ + cos φ =0.701 and Az = 2 cos θ − 3 sin θ =-3.598. . , B y = 0.634 , and Bz = −3732 . . Directly in spherical coordinates Similarly, Bx = 383 A • B = 8 + 6 − 3 = 11 . In rectangular coordinates A • B = ( −0.75 × 383 . ) + ( 0.701 × 0.634) + ( −3598 . × −3732 . ) = 11 . 2.7.5 π 2π From problem 2.7.1 x = r sin θ cos φ , y = r sin θ sin φ , and z = r cos θ . At P1 2, , , 2 3 π π x1 = −1, y1 = 1732 . , z1 = 0 and at P2 = 3, ,− , x 2 = 2.25, y 2 = −1299 . , z 2 = 15 . . 3 6
Hence the distance between the two points is
( x2 − x1 )2 + ( y2 − y1 )2 + ( z2 − z1 )2
D=
= 4.69 .
2.7.6
The surface is 1/8 of the surface of a sphere of radius 4. Hence the surface area is
(4π × (4) ) S= 2
S=
π
∫
π
∫
θ = π 2φ = π 2
8
= 8π . By direct integration we have
2 r = 4) sin θ dθ dφ = 8π . (1 4442444 3
dsr
2.7.7 π
The volume is S =
3
2π
2
r = 2) sin θ dφ dθ = 5.21 . ∫ (1 4442444 3 π φ = 0 θ= 4 ds ∫
r
2-9
2.8.1 ∫ F • dl =
2
4
1
x = −1
y =3
z =−2
∫ 2 xdx + ∫ 4dy −
∫ ydz . Third integral with respect to z contains y. So we
1 11 z + . Hence the line integral 3 3 2 4 1 11 7 1 becomes ∫ F • dl = ∫ 2 xdx + ∫ 4dy − ∫ z + dz = − . 3 2 x =−1 y =3 z =−2 3 need to determine the equation of the path as y =
2.8.2 P2
1
1
2
P1
x =0
y =0
z =0
W = ∫ F • dl = ∫ 2 xdx + ∫ 3zdy + ∫ 4dz . But along the path z = 2 y which when
substituted into the second integral gives W=1+3+8=12J. 2.8.3 P2
3
2
0
P1
x =0
y =0
z =2
P2
3
2
0
P1
x =0
y =0
z =2
(a) ∫ F • dl = ∫ xdx + ∫ 2 xydy − ∫ ydz . Along this path x =
3 y and y = − z + 2 . 2
substituting these gives ∫ F • dl = ∫ xdx + ∫ 3 y 2 dy − ∫ ( − z + 2) dz =
29 . (b) The 2
integral is the sum of the integrals along the two paths: P2
0 0 0 3 2 0 3 ∫ F • dl = ∫ xdx + ∫ 2 xydy − ∫ ( y = 0) dz + ∫ xdx + ∫ 2 x = y ydy − ∫ ydz =0+0+0+ 2 P x =0 y =0 z =2 x =0 y =0 z =0 1
9/2+8+0=25/2. Along the first segment of this path neither x nor y change and y=0. Hence the integral along this first path is zero. Along the second segment of the path, 3 there is no change in z and we substitute the equation for the path, x = y , into the y 2 integration. 2.8.4 P2
∫ F • dl = ∫ 2rdr + ∫ zrdφ + ∫ 4dz . The two paths are sketched below.
P1
P2
0
0
0
P1 P2
r =0 P3
φ =0
z =3
P4
P2
P1
P1
P3
P4
(a) ∫ F • dl = ∫ 2(r = 0) dr + ∫ z ( r = 0) dφ + ∫ 4dz = 0 + 0 − 12 = −12 (b) ∫ F • dl = ∫ F • dl + ∫ F • dl + ∫ F • dl
2-10
π
P3
8
4
P1
r =0
φ=
P4
8
3
∫ F • dl = ∫ 2rdr + ∫ ( z = 3)rdφ + ∫ 4dz = 8 + 0 + 0 = 8 π
z =3
4
π
∫ F • dl =
∫
(
4
)
(
0
)
2 r = 8 dr + ∫ z r = 8 dφ + ∫ 4dz = 0 + 0 − 12 = −12
P3
r= 8
P2
0
P4
r= 8
φ=
P2
P3
P4
P2
P1
P1
P3
P4
φ=
π
z =3
4
π
∫ F • dl =
0
4
∫ 2rdr + ∫ ( z = 0)rdφ + ∫ 4dz = −8 + 0 + 0 = −8 π
z =0
4
∫ F • dl = ∫ F • dl + ∫ F • dl + ∫ F • dl = 8 − 12 − 8 = −12
z P1(0, 0, 3)
P3(2, 2, 3)
P2(0, 0, 0) y P4(2, 2, 0) x
2.8.5 P2
∫ F • dl = ∫ rdr + ∫ 2rdφ − ∫ zdz . The two paths are sketched below.
P1
P2
P0
P2
P1
P1
P0
(a) ∫ F • dl = ∫ F • dl + ∫ F • dl P0
0
0
0
P1 P2
r =0
φ =0
z =2
3
0
0
P0
r =0
φ =0
z =0
∫ F • dl = ∫ (r = 0) dr + ∫ 2( r = 0) dφ − ∫ zdz = 0 + 0 + 2 = 2 ∫ F • dl = ∫ rdr + ∫ 2rdφ − ∫ ( z = 0) dz =
9 9 +0+0= 2 2
2-11
P2
P0
P2
∫ F • dl = ∫ F • d l + ∫ F • d l = 2 +
P1
P1
P0
9 13 = 2 2
P2
P3
P4
P2
P1
P1
P3
P4
(b) ∫ F • dl = ∫ F • dl + ∫ F • dl + ∫ F • dl π
P3
3
P1
r =0
2
2
∫ F • dl = ∫ rdr + ∫ 2rdφ − ∫ 4dz = φ=
π
z =2
9 9 +0+0= 2 2
2
π
P4
3
P3
r =3
P2
3
P4
r =3
P2
P3
P4
P2
P1
P1
P3
P4
0
2
∫ F • dl = ∫ ( r = 3) dr + ∫ 2( r = 3) dφ − ∫ 4dz = 0 + 0 + 2 = 2 φ=
π
z =2
2
0
0
∫ F • dl = ∫ ( r = 3) dr + ∫ 2( r = 3) dφ − ∫ 4dz = 0 − 3π + 0 = −3π φ=
π
z =0
2
∫ F • dl = ∫ F • dl + ∫ F • dl + ∫ F • dl =
9 13 + 2 − 3π = − 3π 2 2
z P1(0, 0, 2)
P3(0, 3, 2) 2
P4(0, 3, 0) 3
y
P2(3, 0, 0) 3 x
2.8.6 P2
∫ F • dl = ∫ 2rdr + ∫ 3rdθ + ∫ 2r sin θ dφ . The two paths are sketched below.
P1
P2
P0
P2
P1
P1
P0
(a) ∫ F • dl = ∫ F • dl + ∫ F • dl
2-12
P0
0
0
0
P1
r =3
φ =0
P2
3
θ =0 π
P0
r =0
P2
P0
P2
P1
P1
P0
∫ F • dl = ∫ 2rdr + ∫ 3rdθ + ∫ 2r sin θ dφ = −9 + 0 + 0 = −9
∫ F • dl = ∫ 2rdr +
0
2
∫ 3rdθ + ∫ 2r sin θ dφ = −9 + 0 + 0 = 9
θ =π 2
φ =0
∫ F • dl = ∫ F • d l + ∫ F • d l = − 9 + 9 = 0 P2
P3
P2
P1
P1
P3
(b) ∫ F • dl = ∫ F • dl + ∫ F • dl π
π
P3
3
P1
r =3
θ =0
φ =π 2
P2
3
π
0
P3
r =3
P2
P3
P2
P1
P1
P3
2
∫ F • dl = ∫ 2(r = 3) dr + ∫ 3( r = 3) dθ + ∫ F • dl = ∫ 2( r = 3) dr +
2
∫ 3( r = 3) dθ +
θ =π 2
∫ F • dl = ∫ F • d l + ∫ F • d l =
2
∫ 2(r = 3) sin θ dφ = 0 +
π 2( r = 3) sin dφ = 0 + 0 − 3π = −3π 2 φ =π 2 ∫
9π 3π − 3π = 2 2
z P(0, 0, 3)
9π 9π +0= 2 2
3
P3(0, 3, 0)
P0(0, 0, 0)
3 P2(3, 0, 0) 3 x
2-13
y
2.8.7 P2
2 ∫ F • dl = ∫ rdr + ∫ 2r dθ + ∫ 3r sin θ dφ . The two paths are sketched below.
P1
P0
P3
P0
P1
P1
P3
(a) ∫ F • dl = ∫ F • dl + ∫ F • dl π
P3
2
P1
r =2
P0
0
P3
r =2
P0
P3
P0
P1
P1
P3
4
π
2
∫ F • dl = ∫ ( r = 2) dr + ∫ 2( r = 2) dθ + θ =0
π
∫ F • dl = ∫ rdr +
π
4
2
∫ 2r dθ +
θ =π 4
2
∫ 3( r = 2) sin θ dφ = 0 + 2π + 0 = 2π
φ =π 2
2
∫ 3r sin θ dφ = −2 + 0 + 0 = −2
φ =π 2
∫ F • dl = ∫ F • dl + ∫ F • dl = 2π − 2 P0
0
0
0
P1
r =2
θ =0
φ =0
(b) ∫ F • dl = ∫ rdr + ∫ 2r 2 dθ + ∫ 3r sin θ dφ = −2 + 0 + 0 = −2
z P1(0, 0, 2)
2
P3(0, 2 , 2 ) 45°
P0(0, 0, 0)
y
x
2.9.1 A sketch of the surface is given below. ∫ F • ds = ∫∫ xdydz + ∫∫ ydxdz + ∫∫ zdxdy . 1
Top:
1
1
y = −1x =−1
Right: Front:
1
∫ ( z = 1) dxdy = 4 . Bottom: − ∫
∫
1
∫
1
∫
1
1
∫ ( y = 1) dxdz = 4 . Left: − ∫
z = −1x =−1 1 1
∫ ( y = −1) dxdz = 4 .
z = −1x =−1 1 1
∫ ( x = 1) dydz = 4 . Back: − ∫
z = −1 y = −1
∫ ( z = −1) dxdy = 4 .
y =−1x =−1
∫ ( x = −1) dydz = 4 .
z =−1 y =−1
Total=4+4+4+4+4+4=24. 2-14
z
(–1, 1, 1)
(–1, 1, –1) y
x
2.9.2 A sketch of the surface is given below. ∫ F • ds = ∫∫ xydydz + ∫∫ yzdxdz − ∫∫ xzdxdy . 2
Top: − ∫
1
∫ x( z = 3) dxdy = −3 . Bottom:
y = 0x = 0 3
Right: ∫
1
3
∫ ( y = 2) zdxdz = 9 . Left: − ∫
2
∫
1
∫ ( y = 0) zdxdz = 0 .
z =0 x =0 3 2
z =0 y =0
z =0 y =0
∫ ( x = 1) ydydz = 6 . Back: − ∫
∫ x( z = 0) dxdy = 0 .
y = 0x = 0
z =0 x =0 3 2
Front: ∫
1
∫ ( x = 0) ydydz = 0 .
Total=-3+0+9+0+6+0=12. z 3
2 1 x
2-15
y
2.9.3 A sketch of the surface is given below. ∫ F • ds = ∫∫ 2r 2 dφ dz + ∫∫ 3φ drdz − ∫∫ 2rdrdφ . π
π
2 2
2 2
∫ 2rdrdφ = −2π . Bottom: ∫
Top: − ∫
φ = 0r = 0 3
∫ 2rdrdφ = 2π .
φ = 0r = 0
2
3 2 π ∫ 3 φ = drdz = 9π . Left: − ∫ ∫ 3(φ = 0) drdz = 0 . 2 z =0r =0 z =0r =0
Right: ∫
π
3
2
2
∫ 2( r = 2) dφ dz = 12π .
Front: ∫
z = 0φ = 0
Total=-2π+2π+9π+0+12π=21π. z
3 2
y
x
2.9.4 A sketch of the surface is given below. ∫ F • ds = ∫∫ 3r 2 dφ dz + ∫∫ φ drdz − ∫∫ zrdrdφ . π
Top: −
π
2
2
∫ ( z = 4)rdrdφ = 8π . Bottom:
∫
φ =− π r = 0 4
2 2
4
π
2
2
∫ ( z = 0)rdrdφ = 0 .
∫
φ = −π r = 0 2
4
2
π π ∫ φ = drdz = 4π . Left: − ∫ ∫ φ = − drdz = 4π . 2 2 z =0r =0 z =0r =0
Right: ∫ Front: ∫
2
2
∫ z ( r = 2) dφ dz = 48π .
z = 0φ =− π
2
Total=8π+0+4π+4π+48π=64π.
2-16
z 4 2
y
x
2.9.5 A sketch of the surface is given below. 3 ∫ F • ds = ∫∫ r sin θ dφ dθ + ∫∫ 2r sin θ drdφ − ∫∫ φ rdrdθ .
π
π
2
2
π
3
2 2
π
φ = 0r = 0
2
∫ ( r = 2) sin θ dφ dθ = 4π . Bottom: ∫
Front: ∫
θ = 0φ = 0 π
2 2
π
π2
π
∫ 2r sin
2 2
drdφ = 2π .
. Left: ∫ ∫ (φ = 0)rdrdθ = 0 . ∫ φ = rdrdθ = − 2 2 θ = 0r = 0 θ = 0r = 0 2 2 π π + 0 = 6π − Total= 4π + 2π − . 2 2 Right: − ∫
z
2
y
x
2-17
2.9.6 A sketch of the surface is given below. 3 ∫ F • ds = ∫∫ r sin θ dφ dθ − ∫∫ 2r sin θ drdφ + ∫∫ 3φ rdrdθ .
π
2
π
Front: ∫
2
∫ (r = 3) sin θ dφ dθ = 27π . Bottom: −
θ = 0φ =− π π
π
3
3
2
∫ 2r sin
∫
φ =− π r = 0 2
2
π
π 2
drdφ = −9π .
2 3 π 27π 2 π 27π 2 . Left: − ∫ ∫ 3 φ = − rdrdθ = . ∫ 3 φ = rdrdθ = 2 8 2 8 θ = 0r = 0 θ = 0r = 0 2 3
Right: ∫
Total= 27π − 9π +
27π 2 27π 2 27π 2 + = 18π + . 8 8 4
z
3
y
x
2-18
