quiz 5 (chapter 18) solutions - kkuniyuk
QUIZ 5 (CHAPTER 18) SOLUTIONS MATH 252 – FALL 2008 – KUNIYUKI 105 POINTS TOTAL, BUT 100 POINTS = 100% Show all work, simplify as appropriate, and use “good form and procedure” (as in class). Box in your final answers! No notes or books allowed. A scientific calculator is allowed.
1) Matching. (9 points total) Fill in each blank below with a true property describing the vector field F. (Assume that we are only evaluating F on its domain.) A. The vectors in the field all have the same direction. B. The non-0 vectors in the field all point away from the origin. C. The vectors in the field are all unit vectors.
( )
I. F x, y = xi + yj . It is true that __B__.
( )
If a point a,b is the initial point for a,b , the position vector to the point, then that vector will point away from the origin.
( )
II. F x, y = 2i + 3j . It is true that __A__. F is a constant vector field.
( )
III. F x, y =
1 x +y 2
xi yj =
2
( xi yj) . It is true that __C__.
( x ) + ( y ) 2
2
= x2 + y2
( )
Thus, F x, y =
xi yj . This represents a normalization process. xi yj
(
)
( )
2) Let F x, y, z = x 2 e2 z , cos 3y , xy 2 z 3 x . (20 points total) a) Find div F. div F = • F , , • x 2 e2 z , cos 3y , xy 2 z 3 x x y z
( )
=
) ( ( )) + z ( xy z x ) = ( 2x ) ( e ) + 3sin ( 3y ) + ( xy ) ( 3z ) =
2 2z xe + cos 3y x y
(
2 3
2z
2
2
( )
= 2xe2 z 3sin 3y + 3xy 2 z 2
b) Find curl F. curl F = F
=
i x x 2 e2 z
j
k
y cos 3y
z 2 3 xy z x
( )
= xy 2 z 3 x cos 3y z y 2 2z cos 3y xe y x
(
( ( )) i
)
( ( ))
( )( )
(
(
= xz 3 2 y 0 i y 2 z 3 =
( 2xyz ) i ( y z 3
2 3
2 2z xy 2 z 3 x x e j + z x
(
)
) k 1) ( x ) ( 2e ) j + 2
)
1 2x 2 e2 z j
or
2xyz 3 , y 2 z 3 + 1 + 2x 2 e2 z , 0
or
2xyz 3 , 2x 2 e2 z y 2 z 3 + 1, 0
2z
(
0 0 k
)
3) C consists of the curves C1 and C2 in xyz-space. That is, C = C1 C2 .
(
) (
)
The curve C1 is the directed line segment from 0, 2, 3 to 2, 8, 4 , and the curve C2 is the portion of the graph of y = x 3 in the plane z = 4 directed from
( 2, 8, 4) to (3, 27, 4) . If the force at ( x, y, z ) is F ( x, y, z ) =
xy, z + 4, 3 , find
the work done by F along C. It is recommended that you write your final answer as a decimal. Hint: If you want, you can analyze C2 first. (32 points) Parameterize C1 :
(
) (
)
The displacement vector from 0, 2, 3 to 2, 8, 4 is given by: v = 2 0, 8 2, 4 3 = 2, 6, 1
Parametric equations for C1 (and dx, dy and dz) are given by:
x = 2t y = 2 + 6t z = 3+ t
dx = 2 dt dy = 6 dt dz = dt
(t : 0 1)
Parameterize C2 :
x = t 3 y = t z = 4
dx = dt
dy = 3t 2 dt dz = 0
(t : 2 3)
Compute the work integral along C1 :
C1
F • dr =
xy, z + 4, 3 • dx, dy, dz
=
xy dx + z + 4 dy + 3dz
C1
C1
(
)
Think:
( 2t )( 2 + 6t )( 2 dt ) + ((3 + t ) + 4)(6 dt ) = (8t + 24t ) dt + ( 42 + 6t ) dt + 3dt = ( 24t + 14t + 45) dt =
1
0
1
2
0
1
2
0
1
t3 t2 = 24 + 14 + 45t 2
3
0
C1
M dx + N dy + P dz
( )
+ 3 dt
= 8t 3 + 7t 2 + 45t
()
()
1 0
()
= 8 1 + 7 1 + 45 1 0 = 8 + 7 + 45 3
2
= 60 Compute the work integral along C2 :
C2
F • dr =
xy, z + 4, 3 • dx, dy, dz
=
xy dx + z + 4 dy + 3dz
C2
C2
(
)
(t )(t )( dt ) + (( 4) + 4)(3t = ( t dt ) + ( 24t dt ) = ( t + 24t ) dt 3
=
3
2
2
3
4
Think:
2
2
3
4
2
2
3
t5 t3 = + 24 3 2 5 3
t5 = + 8t 3 5 2
()
()
25 35 3 3 = +8 3 +8 2
5
5 = 48.6 + 216 6.4 + 64
()
()
= 264.6 70.4 971 = 194.2 or 5
The total work along C is given by:
C
F • dr =
C1
F • dr +
C2
F • dr
= 60 + 194.2 = 254.2 or
(
1271 work units 5
)
()
dt + 3 0
)
C1
M dx + N dy + P dz
4) Find the exact mass of a thin wire C in xyz-space if the density at any point x, y, z where x 0 is given by x, y, z = 5x (i.e., five times the point’s
(
)
(
)
distance from the yz-plane), and if C is parameterized by x = 3cos t , y = 3sin t , and y = 7t , where 0 t . (17 points) 4 mass, m =
( x, y, z ) ds C
2
=
C
4 0
5x
2
2
dx dy dz dt + dt + dt dt
) ( 3sin t ) + (3cost ) + (7 )
( = (15cost ) 9sin t + 9cos t + 49 dt =
2
5 3cost
4 0
= =
2
(15cost )
(15cost )
58 dt
4 0
= 15 58 cost dt 4 0
= 15 58 sin t = 15 58 sin sin 0 4
2 = 15 58 0 2
2 = 15 58 2 1
2 = 15 29 2 2 = 15 29
2
9 sin 2 t + cos 2 t + 49 dt
=1
4 0 4 0
2
( mass units)
Note: 15 29 80.7775
2
dt
5) Use the idea of potential functions and the Fundamental Theorem for Line Integrals to show that the following line integral is independent of path in xyz-space and to evaluate the integral. Show all work, as we have done in class. Use good form. In particular, indicate independent variables for functions; for example, write f x, y, z instead of simply f . Give an exact, simplified
(
)
answer; do not approximate. (3, 4,1) 2 3y 3y 2 (1,0, 2) 4 y + z dx + 8xy 3ze dy + x e + 3z dz
(
)
(
)
(
)
(27 points) (3,4,1)
(
This line integral is of the form
(
)
1,0,2
)
F • dr , where
F x, y, z = 4 y 2 + z, 8xy 3ze3 y , x e3 y + 3z 2 , if F is independent of path in xyzspace. Find a potential function f for F in xyz-space. Compare:
(
)
f x x, y, z = 4 y 2 + z Integrate: Partially integrate both sides with respect to x.
(
)
( )
f x, y, z = 4xy 2 + xz + g y, z
Differentiate: Differentiate both sides with respect to y.
(
) ( )( ) ( ) = 8xy + g ( y, z )
f y x, y, z = 4x 2 y + g y y, z y
Compare:
( ) ( y, z ) = 3ze
We require f y x, y, z = 8xy 3ze3 y Then, g y
3y
Integrate: Partially integrate both sides with respect to y.
1 g y, z = 3z e3 y + h z 3
( ) (
)
()
()
= ze3 y + h z
Rewrite:
(
)
()
f x, y, z = 4xy 2 + xz ze3 y + h z
Differentiate: Differentiate both sides with respect to z.
(
)
()
f z x, y, z = x e3 y + h z Compare:
(
)
We require f z x, y, z = x e3 y + 3z 2
()
Then, h z = 3z
2
Integrate: Integrate both sides with respect to z.
()
h z = z3 + K Rewrite:
(
)
f x, y, z = 4xy 2 + xz ze3 y + z 3 + K We can take K = 0 . The existence of a potential shows that F is independent of path throughout xyz-space. Evaluate the given line integral using the Fundamental Theorem for Line Integrals. (3,4,1) ( 1,0,2) F • dr (3,4,1) = 4 y 2 + z dx + 8xy 3ze3 y dy + x e3 y + 3z 2 dz ( 1,0,2) (3,4,1) = f x, y, z ( 1,0,2) (3,4,1) = 4xy 2 + xz ze3 y + z 3 ( 1,0,2) 2 3 2 3( 4 )
(
(
)
(
)
(
)
)
( )( ) ( )( ) ( )
= 4 3 4 + 3 1 1e
( ) 4 ( 1)(0) + ( 1)( 2) ( 2) e ( ) + ( 2)
+ 1
30
= 192 + 3 e12 + 1 0 2 2 + 8 = 196 e12 4 = 192 e12
Note: 192 e12 162,563 . Forces are really working against us!
3
1) Matching. (9 points total) Fill in each blank below with a true property describing the vector field F. (Assume that we are only evaluating F on its domain.) A. The vectors in the field all have the same direction. B. The non-0 vectors in the field all point away from the origin. C. The vectors in the field are all unit vectors.
( )
I. F x, y = xi + yj . It is true that __B__.
( )
If a point a,b is the initial point for a,b , the position vector to the point, then that vector will point away from the origin.
( )
II. F x, y = 2i + 3j . It is true that __A__. F is a constant vector field.
( )
III. F x, y =
1 x +y 2
xi yj =
2
( xi yj) . It is true that __C__.
( x ) + ( y ) 2
2
= x2 + y2
( )
Thus, F x, y =
xi yj . This represents a normalization process. xi yj
(
)
( )
2) Let F x, y, z = x 2 e2 z , cos 3y , xy 2 z 3 x . (20 points total) a) Find div F. div F = • F , , • x 2 e2 z , cos 3y , xy 2 z 3 x x y z
( )
=
) ( ( )) + z ( xy z x ) = ( 2x ) ( e ) + 3sin ( 3y ) + ( xy ) ( 3z ) =
2 2z xe + cos 3y x y
(
2 3
2z
2
2
( )
= 2xe2 z 3sin 3y + 3xy 2 z 2
b) Find curl F. curl F = F
=
i x x 2 e2 z
j
k
y cos 3y
z 2 3 xy z x
( )
= xy 2 z 3 x cos 3y z y 2 2z cos 3y xe y x
(
( ( )) i
)
( ( ))
( )( )
(
(
= xz 3 2 y 0 i y 2 z 3 =
( 2xyz ) i ( y z 3
2 3
2 2z xy 2 z 3 x x e j + z x
(
)
) k 1) ( x ) ( 2e ) j + 2
)
1 2x 2 e2 z j
or
2xyz 3 , y 2 z 3 + 1 + 2x 2 e2 z , 0
or
2xyz 3 , 2x 2 e2 z y 2 z 3 + 1, 0
2z
(
0 0 k
)
3) C consists of the curves C1 and C2 in xyz-space. That is, C = C1 C2 .
(
) (
)
The curve C1 is the directed line segment from 0, 2, 3 to 2, 8, 4 , and the curve C2 is the portion of the graph of y = x 3 in the plane z = 4 directed from
( 2, 8, 4) to (3, 27, 4) . If the force at ( x, y, z ) is F ( x, y, z ) =
xy, z + 4, 3 , find
the work done by F along C. It is recommended that you write your final answer as a decimal. Hint: If you want, you can analyze C2 first. (32 points) Parameterize C1 :
(
) (
)
The displacement vector from 0, 2, 3 to 2, 8, 4 is given by: v = 2 0, 8 2, 4 3 = 2, 6, 1
Parametric equations for C1 (and dx, dy and dz) are given by:
x = 2t y = 2 + 6t z = 3+ t
dx = 2 dt dy = 6 dt dz = dt
(t : 0 1)
Parameterize C2 :
x = t 3 y = t z = 4
dx = dt
dy = 3t 2 dt dz = 0
(t : 2 3)
Compute the work integral along C1 :
C1
F • dr =
xy, z + 4, 3 • dx, dy, dz
=
xy dx + z + 4 dy + 3dz
C1
C1
(
)
Think:
( 2t )( 2 + 6t )( 2 dt ) + ((3 + t ) + 4)(6 dt ) = (8t + 24t ) dt + ( 42 + 6t ) dt + 3dt = ( 24t + 14t + 45) dt =
1
0
1
2
0
1
2
0
1
t3 t2 = 24 + 14 + 45t 2
3
0
C1
M dx + N dy + P dz
( )
+ 3 dt
= 8t 3 + 7t 2 + 45t
()
()
1 0
()
= 8 1 + 7 1 + 45 1 0 = 8 + 7 + 45 3
2
= 60 Compute the work integral along C2 :
C2
F • dr =
xy, z + 4, 3 • dx, dy, dz
=
xy dx + z + 4 dy + 3dz
C2
C2
(
)
(t )(t )( dt ) + (( 4) + 4)(3t = ( t dt ) + ( 24t dt ) = ( t + 24t ) dt 3
=
3
2
2
3
4
Think:
2
2
3
4
2
2
3
t5 t3 = + 24 3 2 5 3
t5 = + 8t 3 5 2
()
()
25 35 3 3 = +8 3 +8 2
5
5 = 48.6 + 216 6.4 + 64
()
()
= 264.6 70.4 971 = 194.2 or 5
The total work along C is given by:
C
F • dr =
C1
F • dr +
C2
F • dr
= 60 + 194.2 = 254.2 or
(
1271 work units 5
)
()
dt + 3 0
)
C1
M dx + N dy + P dz
4) Find the exact mass of a thin wire C in xyz-space if the density at any point x, y, z where x 0 is given by x, y, z = 5x (i.e., five times the point’s
(
)
(
)
distance from the yz-plane), and if C is parameterized by x = 3cos t , y = 3sin t , and y = 7t , where 0 t . (17 points) 4 mass, m =
( x, y, z ) ds C
2
=
C
4 0
5x
2
2
dx dy dz dt + dt + dt dt
) ( 3sin t ) + (3cost ) + (7 )
( = (15cost ) 9sin t + 9cos t + 49 dt =
2
5 3cost
4 0
= =
2
(15cost )
(15cost )
58 dt
4 0
= 15 58 cost dt 4 0
= 15 58 sin t = 15 58 sin sin 0 4
2 = 15 58 0 2
2 = 15 58 2 1
2 = 15 29 2 2 = 15 29
2
9 sin 2 t + cos 2 t + 49 dt
=1
4 0 4 0
2
( mass units)
Note: 15 29 80.7775
2
dt
5) Use the idea of potential functions and the Fundamental Theorem for Line Integrals to show that the following line integral is independent of path in xyz-space and to evaluate the integral. Show all work, as we have done in class. Use good form. In particular, indicate independent variables for functions; for example, write f x, y, z instead of simply f . Give an exact, simplified
(
)
answer; do not approximate. (3, 4,1) 2 3y 3y 2 (1,0, 2) 4 y + z dx + 8xy 3ze dy + x e + 3z dz
(
)
(
)
(
)
(27 points) (3,4,1)
(
This line integral is of the form
(
)
1,0,2
)
F • dr , where
F x, y, z = 4 y 2 + z, 8xy 3ze3 y , x e3 y + 3z 2 , if F is independent of path in xyzspace. Find a potential function f for F in xyz-space. Compare:
(
)
f x x, y, z = 4 y 2 + z Integrate: Partially integrate both sides with respect to x.
(
)
( )
f x, y, z = 4xy 2 + xz + g y, z
Differentiate: Differentiate both sides with respect to y.
(
) ( )( ) ( ) = 8xy + g ( y, z )
f y x, y, z = 4x 2 y + g y y, z y
Compare:
( ) ( y, z ) = 3ze
We require f y x, y, z = 8xy 3ze3 y Then, g y
3y
Integrate: Partially integrate both sides with respect to y.
1 g y, z = 3z e3 y + h z 3
( ) (
)
()
()
= ze3 y + h z
Rewrite:
(
)
()
f x, y, z = 4xy 2 + xz ze3 y + h z
Differentiate: Differentiate both sides with respect to z.
(
)
()
f z x, y, z = x e3 y + h z Compare:
(
)
We require f z x, y, z = x e3 y + 3z 2
()
Then, h z = 3z
2
Integrate: Integrate both sides with respect to z.
()
h z = z3 + K Rewrite:
(
)
f x, y, z = 4xy 2 + xz ze3 y + z 3 + K We can take K = 0 . The existence of a potential shows that F is independent of path throughout xyz-space. Evaluate the given line integral using the Fundamental Theorem for Line Integrals. (3,4,1) ( 1,0,2) F • dr (3,4,1) = 4 y 2 + z dx + 8xy 3ze3 y dy + x e3 y + 3z 2 dz ( 1,0,2) (3,4,1) = f x, y, z ( 1,0,2) (3,4,1) = 4xy 2 + xz ze3 y + z 3 ( 1,0,2) 2 3 2 3( 4 )
(
(
)
(
)
(
)
)
( )( ) ( )( ) ( )
= 4 3 4 + 3 1 1e
( ) 4 ( 1)(0) + ( 1)( 2) ( 2) e ( ) + ( 2)
+ 1
30
= 192 + 3 e12 + 1 0 2 2 + 8 = 196 e12 4 = 192 e12
Note: 192 e12 162,563 . Forces are really working against us!
3
