Austin Cholley Zhu MAT 267 ONLINE A Fall 2021
Assignment Section 13.5 due 10/05/2021 at 11:59pm MST
Problem 1. (1 point)
Show that the vector field F(x,y,z) = h8ycos(−7x),−7xsin(8y),0i
is not a gradient vector field by computing its curl. How does this
show what you intended?
curl(F) = ∇×F=h, , i.
Solution:
SOLUTION
∇×F=
i j k
∂
∂x
∂
∂y
∂
∂z
8ycos(−7x)−7xsin(8y)0
=0i+0j+ (−7sin(8y)−8cos(−7x))k
=h0,0,−7sin(8y)−8cos(−7x)i
.
Since the curl does not equal the zero vector, the vector field
is not a gradient vector field.
Answer(s) submitted:
•0
•0
•-7sin(8y)-8cos(-7x)
(correct)
Correct Answers:
•0
•0
•-7*sin(8*y) - 8*cos(-7*x)
Problem 2. (1 point)
A)
Consider the vector field F(x,y,z) = h3yz,−4xz,8xyi.
Find the divergence and curl of F.
div(F) = ∇·F=.
curl(F) = ∇×F=h, , i.
B)
Consider the vector field F(x,y,z) = h2x2,3(x+y)2,4(x+y+z)2i.
Find the divergence and curl of F.
div(F) = ∇·F=.
curl(F) = ∇×F=h, , i.
Solution:
SOLUTION
A)
∇·F=∂
∂x(3yz) + ∂
∂y(−4xz) + ∂
∂z(8xy) = 0+0+0=0
∇×F=
i j k
∂
∂x
∂
∂y
∂
∂z
3yz −4xz 8xy
=∂
∂y(8xy)−∂
∂z(−4xz)i−∂
∂x(8xy)−∂
∂z(3yz)j
+∂
∂x(−4xz)−∂
∂y(3yz)k
= (8x+4x)i−(8y−3y)j+ (−4z−3z)k
=h12x,−5y,−7zi
B)
∇·F=∂
∂x(2x2) + ∂
∂y(3(x+y)2) + ∂
∂z(4(x+y+z)2)
=4x+6(x+y) + 8(x+y+z)
=18x+14y+8z
∇×F=
i j k
∂
∂x
∂
∂y
∂
∂z
2x23(x+y)24(x+y+z)2
=∂
∂y(4(x+y+z)2)−∂
∂z(3(x+y)2)i
−∂
∂x(4(x+y+z)2)−∂
∂z(2x2)j
+∂
∂x(3(x+y)2)−∂
∂y(2x2)k
= (8(x+y+z)−0)i−(8(x+y+z)−0)j+ (6(x+y)−0)k
=h8(x+y+z),−8(x+y+z),6(x+y)i
Answer(s) submitted:
•0
•(8+4)x
•(3-8)y
•(-4-3)z
•4x+6(x+y)+8(x+y+z)
•8(x+y+z)
1
•-8(x+y+z)
•6(x+y)
(correct)
Correct Answers:
•0
•(8 - -4)*x
•(3 - 8)*y
•(-4 - 3)*z
•2*2*x + 2*3*(x+y) + 2*4*(x+y+z)
•2*4*(x+y+z)
•-2*4*(x+y+z)
•2*3*(x+y)
Problem 3. (1 point)
I.
Let F=8xi+2yj+10zk. Compute the divergence and the curl.
A. div F=
B. curl F=i+j+k
II.
Let F=h10xy,2y,3zi.
The curl of F=h, , i.
Is there a function fsuch that F=∇f? (yes/no)
Solution:
SOLUTION
I.
A. div F=∂
∂x(8x) + ∂
∂y(2y) + ∂
∂z(10z) = 8+2+10 =20
B.
∇×F=
i j k
∂
∂x
∂
∂y
∂
∂z
8x2y10z
=∂
∂y(10z)−∂
∂z(2y)i−∂
∂x(10z)−∂
∂z(8x)j+∂
∂x(2y)−∂
∂y(8x)k
=0i+0j+0k
II.
∇×F=
i j k
∂
∂x
∂
∂y
∂
∂z
10xy 2y3z
=∂
∂y(3z)−∂
∂z(2y)i−∂
∂x(3z)−∂
∂z(10xy)j+∂
∂x(2y)−∂
∂y(10xy)k
=0i−0j−10xk
Since the curl does not equal the zero vector, the vector field
is not conservative and therefore there is no function fsuch that
F=∇f
Answer(s) submitted:
•20
•0
•0
•0
•0
•0
•-10x
•no
(correct)
Correct Answers:
•20
•0
•0
•0
•0
•0
2
•-10*x
•NO
Problem 4. (1 point)
Consider the vector field F(x,y,z) = hy,x,−6zi. Show that Fis a
gradient vector field F=∇Vby determining the function Vwhich
satisfies V(0,0,0) = 0.
V(x,y,z) =
Solution:
SOLUTION
Vx(x,y,z) = yimplies V(x,y,z) = xy +g(y,z)and Vy(x,y,z) =
x+gy(y,z).
But Vy(x,y,z) = x, so g(y,z) = h(z)and V(x,y,z) = xy +h(z).
Thus Vz(x,y,z) = h0(z), but Vz(x,y,z) = −6zso h(z) = −3z2+K.
Hence a potential function for Fis V(x,y,z) = xy −3z2+K.
Since V(0,0,0) = 0, we have K=0 and V(x,y,z) = xy −3z2
Answer(s) submitted:
•xy-6zzˆ(2)
(incorrect)
Correct Answers:
•1*x*y+-6*z*z/2
Problem 5. (1 point)
Let F=h6xyz +2sin x,3x2z,3x2yi.
Find a function fso that F=∇f, and f(0,0,0) = 0.
Solution:
SOLUTION
fx(x,y,z) = 6xyz +2sin ximplies f(x,y,z) = 3x2yz −2 cos x+
g(y,z)and fy(x,y,z) = 3x2z+gy(y,z).
But fy(x,y,z) = 3x2z, so g(y,z) = h(z)and f(x,y,z) = 3x2yz −
2cosx+h(z).
Thus fz(x,y,z) = 3x2y+h0(z), but fz(x,y,z) = 3x2yso h(z) = K, a
constant.
Hence a potential function for Fis f(x,y,z) = 3x2yz −2cos x+K.
Since f(0,0,0) = 0, we have −2+K=0 and K=2.
Thus f(x,y,z) = 3x2yz −2cosx+2
Answer(s) submitted:
•3xxyz-2cosx+2
(correct)
Correct Answers:
•3*x*x*y*z - 2*cos(x) + 2
3
Problem 6. (1 point)
For each of the following vector fields F, decide whether it is con-
servative or not by computing curl F. Type in a potential function
f(that is, ∇f=F). Assume the potential function has a value of
zero at the origin. If the vector field is not conservative, type N.
A. F(x,y) = (8x−3y)i+ (−3x+4y)j
f(x,y) =
B. F(x,y) = 4yi+5xj
f(x,y) =
C. F(x,y,z) = 4xi+5yj+k
f(x,y,z) =
D. F(x,y) = (4siny)i+ (−6y+4xcosy)j
f(x,y) =
E. F(x,y,z) = 4x2i−3y2j+2z2k
f(x,y,z) =
Note: Your answers should be either expressions of x, y and z (e.g.
“3xy + 2yz”), or the letter “N”
Solution:
SOLUTION
A.
∇×F=
i j k
∂
∂x
∂
∂y
∂
∂z
8x−3y−3x+y0
=∂
∂y(0)−∂
∂z(−3x+4y)i
−∂
∂x(0)−∂
∂z(8x−3y)j
+∂
∂x(−3x+4y)−∂
∂y(8x−3y)k
=0i+0j+ (−3+3)k
=0i+0j+0k
Since the curl equals the zero vector, the vector field is conserva-
tive and it has a potential function f(x,y).
fx(x,y) = 8x−3yimplies f(x,y) = 4x2−3yx+g(y)and fy(x,y) =
−3x+g0(y).
But fy(x,y) = −3x+4y, so g0(y) = 4yand g(y) = 2y2+K.
Thus f(x,y) = 4x2−3yx +2y2+K.
Since f(0,0) = 0, we have K=0 .
Hence f(x,y) = 4x2−3yx +2y2
B.
∇×F=
i j k
∂
∂x
∂
∂y
∂
∂z
4y5x0
=∂
∂y(0)−∂
∂z(5x)i−∂
∂x(0)−∂
∂z(4y)j+∂
∂x(5x)−∂
∂y(4y)k
=0i+0j+ (5−4)k
=0i+0j+1k
Since the curl does not equal the zero vector, the vector field is
not conservative and it does not have a potential function .
C.
∇×F=
i j k
∂
∂x
∂
∂y
∂
∂z
4x5y1
=∂
∂y(1)−∂
∂z(5y)i−∂
∂x(1)−∂
∂z(4x)j+∂
∂x(5y)−∂
∂y(4x)k
=0i+0j+0k
Since the curl equals the zero vector, the vector field is conserva-
tive and it does have a potential function f(x,y,z).
fx(x,y,z) = 4ximplies f(x,y,z) = 2x2+g(y,z)and fy(x,y,z) =
gy(y,z).
But fy(x,y,z) = 5y, so g(y,z) = 5
2y2+h(z)and f(x,y,z) =
2x2+5
2y2+h(z).
Thus fz(x,y,z) = h0(z).
But fz(x,y,z) = 1, so h(y) = z+Kand f(x,y,z) = 2x2+5
2y2+z+
K.
Since f(0,0,0) = 0, we have K=0 .
Hence f(x,y,z) = 2x2+5
2y2+z
D.
∇×F=
i j k
∂
∂x
∂
∂y
∂
∂z
4siny−6y+4xcos y0
=∂
∂y(0)−∂
∂z(−6y+4xcosy)i−∂
∂x(0)−∂
∂z(4siny)j
+∂
∂x(−6y+4xcosy)−∂
∂y(4siny)k
=0i+0j+ (4cosy−4 cos y)k
=0i+0j+0k
Since the curl equals the zero vector, the vector field is conserva-
tive and it has a potential function f(x,y).
fx(x,y) = 4sinyimplies f(x,y) = 4xsin y+g(y)and fy(x,y) =
4xcosy+g0(y).
But fy(x,y) = −6y+4xcosy, so g0(y) = −6yand g(y) =
−3y2+K.
Thus f(x,y) = 4xsiny−3y2+K.
Since f(0,0) = 0, we have K=0 .
Hence f(x,y) = 4xsiny−3y2
E.
4
∇×F=
i j k
∂
∂x
∂
∂y
∂
∂z
4x2−3y22z2
=∂
∂y(2z2)−∂
∂z(−3y2)i−∂
∂x(2z2)−∂
∂z(4x2)j
+∂
∂x(−3y2)−∂
∂y(4x2)k
=0i+0j+0k
Since the curl equals the zero vector, the vector field is conserva-
tive and it does have a potential function f(x,y,z).
fx(x,y,z) = 4x2implies f(x,y,z) = 4
3x3+g(y,z)and fy(x,y,z) =
gy(y,z).
But fy(x,y,z) = −3y2, so g(y,z) = −1y3+h(z)and f(x,y,z) =
4
3x3−1y3+h(z).
Thus fz(x,y,z) = h0(z).
But fz(x,y,z) = 2z2, so h(y) = 2
3z3+Kand f(x,y,z) = 4
3x3−
1y3+2
3z3+K.
Since f(0,0,0) = 0, we have K=0 .
Hence f(x,y,z) = 4
3x3−1y3+2
3z3
Answer(s) submitted:
•4xˆ(2)-3xy+2yˆ(2)
•n
•((4xˆ(2))/2)+(((5yˆ(2)))/2)+z
•4xsiny-3yˆ(2)
•(1/3)(4xˆ(3)-3yˆ(3)+2xˆ(3))
(score 0.8)
Correct Answers:
•4*x**2 + -3*x*y + 2*y**2
•N
•4*x**2/2 + (4+1)*y**2/2 + z
•4*x*sin(y) + -3*y**2
•(1/3)*(4*x**3 + -3*y**3 + 2*z**3)
Problem 7. (1 point)
Let Fbe any nonconstant vector field of the form F=f(x)i+
g(y)j+h(z)kand let Gbe any nonconservative vector field of
the form G=f(y,z)i+g(x,z)j+h(x,y)k. Indicate whether the
following statements are true or false by placing ”T” or ”F” to the
left of the statement.
1. Gis incompressible
2. Fis irrotational
3. Gis irrotational
4. Fis incompressible
Solution:
SOLUTION
1. T.
Explanation:
div(G) = ∂
∂xf(y,z) + ∂
∂yg(x,z) + ∂
∂zh(x,y) = 0. Thus Gis incom-
pressible.
2. T.
Explanation:
curl(F) = 0i+0j+0k. Thus Fis irrotational.
3. F.
Explanation:
curlG=
i j k
∂
∂x
∂
∂y
∂
∂z
f(y,z)g(x,z)h(x,y)
= (hy(x,y)−gz(x,z))i−(hx(x,y)−fz(y,z))j+ (gx(x,z)−fy(y,z))k
6=0i+0j+0k
Since curl(G)does not equal the zero vector, Gis not irrotational.
4. F.
Explanation:
div(F) = ∂
∂xf(x) + ∂
∂yg(y) + ∂
∂zh(z)6=0.
Since div(F)6=0, Fis not incompressible.
Answer(s) submitted:
•t
•t
•f
•f
(correct)
Correct Answers:
•T
•T
•F
•F
5
Problem 8. (1 point)
Determine whether the divergence of each vector field (in green)
at the indicated point P(in blue) is positive, negative, or zero.
? ? ?
(Click on a graph to enlarge it)
Solution:
SOLUTION
If we draw a small circle around the point Pwe can see that the
flow rate out is larger than the flow rate in (the vectors exiting the
circle are longer than the vectors entering the circle). Thus the
divergence is positive.
If we draw a small circle around the point Pwe can see that
the flow rate out equals the flow rate in. Thus the divergence is
zero. The vector field is incompressible at the point P.
If we draw a small circle around the point Pwe can see that
the flow rate out is smaller than the flow rate in (the vectors ex-
iting the circle are shorter than the vectors entering the circle).
Thus the divergence is negative.
Answer(s) submitted:
•Positive
•Zero
•Negative
(correct)
Correct Answers:
•POSITIVE
•ZERO
•NEGATIVE
Problem 9. (1 point)
Determine whether the curl of each vector field ~
Fat the origin (in
red) is~
0 or points in the same direction as ±
~
i,±~
j, or ±
~
k.
? ?
? ?
(Click and drag to rotate)
Solution:
SOLUTION
The vector field is irrotational, so the curl is the zero vector,
h0,0,0i
According to the righ hand rule, if we curl our fingers in the
direction of rotation of the vectors, the thumb points in the direc-
tion of the positive y-axis. Thus the curl points in the direction of
j=h0,1,0i.
According to the righ hand rule, if we curl our fingers in the
direction of rotation of the vectors, the thumb points in the direc-
tion of the negative z-axis. Thus the curl points in the direction of
−k=h0,0,−1i.
The vector field is irrotational, so the curl is the zero vector,
h0,0,0i
Answer(s) submitted:
•<0,0,0>
•j = <0,1,0>
6
•-k = <0,0,-1>
•<0,0,0>
(correct)
Correct Answers:
•<0,0,0>
•J = <0,1,0>
•-K = <0,0,-1>
•<0,0,0>
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