Download Math 601 Solutions to Homework 10

Math 601 Solutions to Homework 10
1. Consider the vector field F(x, y) = (2x + 2y)i + (2x − y)j.
(a) Find the parametric equations for the flow line of F beginning at
the point (0, 5).
(b) Find div(F).
(c) Find rot(F).
Answer:
(a) We need to solve the system of differential equations:
dx
= 2x + 2y
dt
dy
= 2x − y
dt
This is a system of linear differential equations, which we can write
as:
· ¸ ·
¸· ¸
d x
2
2
x
=
2 −1
y
dt y
To solve this system, we need to compute the eigenvalues and
eigenvectors of the matrix.
We can quickly find the eigenvalues if we remember that the product of the eigenvalues must equal the determinant of the matrix,
and the sum of the eigenvalues must equal the trace of the matrix.
So, λ1 λ2 = −6 and λ1 + λ2 = 1. By inspection we see that the
eigenvalues are λ1 = 3 and λ2 = −2. (This method only works for
2 × 2 matrices.)
Now, we find the eigenvectors associated with each eigenvalue.
For λ1 = 3, we have:
·
¸
·
¸
λ1 − 2
−2
1 −2
nullspace
= nullspace
−2
λ1 + 1
−2
4
·
¸
1 −2
= nullspace
0
0
µ· ¸¶
2
= Span
1
· ¸
2
Thus, the vector
is an eigenvector associated with the eigen1
value λ1 = 3.
For λ2 = −2, we have:
·
λ2 − 2
nullspace
−2
·
Thus, the vector
1
−2
−2
λ2 + 1
¸
·
−4 −2
= nullspace
−2 −1
·
¸
2 1
= nullspace
0 0
µ·
¸¶
1
= Span
−2
¸
¸
is an eigenvector associated with the
eigenvalue λ2 = −2.
Thus, the general solution to the system of linear differential equations is:
· ¸
· ¸
·
¸
x
2
1
3t
−2t
= c1 e
+ c2 e
y
1
−2
We would like to find the solution which begins at the point (0, 5),
so we want x = 0, y = 5 when t = 0. We can plug those numbers into the above equation and solve for c1 and c2 . We get the
equations:
0 = 2c1 + c2
5 = c1 − 2c2
Solving for c1 and c2 , we get c1 = 1 and c2 = −2. Thus, the
parametric equation for the flow line is:
x = 2e3t − 2e−2t
y = e3t + 4e−2t
(b) div(F) =
∂
∂
(2x + 2y) +
(2x − y) = 2 − 1 = 1
∂x
∂y
¯
¯ ∂
¯
(c) rot(F) = ¯¯ ∂x
¯ 2x + 2y
¯
∂ ¯
¯
∂y ¯¯ = 2 − 2 = 0
2x − y ¯
2. Find a vector field F whose flow lines are the parametric curves
y = C(t2 + 1)
x=t
dx
Answer: The components of the vector field F must be Fx =
=1
dt
dy
and Fy =
= 2Ct. We just need to write these in terms of x and y.
dt
Using the equations for the flow lines, we know that t = x. Also, the
y
y
equation y = C(t2 + 1) gives us C = 2
= 2
. Thus, the vector
t +1
x +1
field F is:
F(x, y) = i +
2xy
j
x2 + 1
x2 y 2
3. (a) Let C be the portion of the ellipse
+
= 1 in the first quad4
9
rant. Evaluate the following integral:
Z
xy ds
C
(b) Let C be the curve in R3 described by the equations ρ = θ
π
and φ = . with endpoints (x, y, z) = (0, 0, 0) and (x, y, z) =
4√
√
(π 2, 0, π 2). Evaluate the following integral:
Z
x
y
dx + dy + z 3 dz
z
C z
Answer:
(a) First, we parameterize the portion of the ellipse:
x = 2 cos t
y = 3 sin t
0≤t≤
π
2
Thus:
dx = −2 sin t dt
dy = 3 cos t dt
Thus:
ds =
p
dx2 + dy 2 =
p
4 sin2 t + 9 cos2 t dt =
√
4 + 5 cos2 t dt
Thus, the line integral is:
Z
Z π/2
√
xy ds =
6 cos t sin t 4 + 5 cos2 t dt
C
0
We can evaluate this integral using the substitution u = 4 + 5 cos2 t,
du = −10 cos t sin t. Thus:
Z
Z π/2
√
xy ds =
6 cos t sin t 4 + 5 cos2 t dt
C
Z0 4
3√
=
u du
−
5
9
·
¸4
38
2 3/2
= − u
=
5
5
9
(b) First we parameterize the curve C using t = θ. Then:
t
x = ρ sin φ cos θ = √ cos t
2
t
y = ρ sin φ sin θ = √ sin t
2
t
z = ρ cos φ = √
2
0 ≤ t ≤ 2π
The bounds for t are chosen√so that√at t = 0, (x, y, z) = (0, 0, 0)
and at t = 2π, (x, y, z) = (π 2, 0, π 2). Thus:
µ
¶
1
t
√ cos t − √ sin t dt
dx =
2
2
µ
¶
1
t
√ sin t + √ cos t dt
dy =
2
2
1
dz = √ dt
2
Thus, the integral becomes:
Z
C
x
y
dx + dy + z 3 dz =
z
z
Ã
√ µ
¶
t cos t/ 2
1
t
√
√ cos t − √ sin t
t/ 2
2
2
0
√ µ
¶
t sin t/ 2
1
t
√
√ sin t + √ cos t
+
t/ 2
2
2
µ
¶!
3
t
1
√
+ √
dt
2 2
2
Z
2π
This simplifies to:
¶
Z
Z 2π µ
√
x
y
1
t3
3
√ +
dx + dy + z dz =
dt = π 2 + π 4
z
4
2
C z
0
4. Use geometric reasoning to evaluate the following line integrals:
(a) The integral
Z
¡
x2 + y 2
¢2
ds
C
where C is the circle of radius 3 centered at the origin.
(b) The integral
Z
−ry dx + rx dy
C
where C is the circle of radius 2 centered at the origin.
(c) The integral
Z
x dx + y dy + z dz
C
where C is any curve on the sphere x2 + y 2 + z 2 = 9.
Answer:
(a) On the circle C, the value of (x2 + y 2 )2 is 34 = 81. If we integrate
this function over the circle, we get 81 times the circumference of
the circle, which is 81(3)(2π) = 486π
(b) The problem should have specified the orientation of C. We will
assume that C is oriented counterclockwise.
Z
We want to write the integral as a scalar line integral
F · T ds
C
where T is the unit tangent vector to the curve. Since C is a
circle centered at the origin, the vector −yi + xj is tangent to C
and points in the counterclockwise direction. So:
−yi + xj
−yi + xj
T= p
=
r
x2 + y 2
µ
¶
−y x
,
Thus, F · T = (−ry, rx) ·
= y 2 + x2 , so on the circle C,
r r
F · T = r2 = 4. If we integrate this function over the circle, we get
4 times the circumference of the circle, which is 4(2)(2π) = 16π
Note: If we orient C clockwise, we obtain −16π.
(c) The vector field F(x, y, z) = x i+y j+z k points radially outwards,
so the tangent vectors
Z to the curve C are perpendicular to F.
F · T ds = 0
Thus, F · T = 0, so
C
5. Consider the vector field F = (2x + y) i + (x + 3y 2 ) j.
(a) Show that F is conservative.
(b) Find a function f such that ∇f = F.
Z
(c) Evaluate the integral
F · ds where C is the curve y = sin(x2 )
C
√
from (0, 0) to ( π, 0).
Answer:
(a) We can show that F is conservative by showing that rot(F) = 0:
¯
¯ ∂
¯
rot(F) = ¯¯ ∂x
¯ 2x + y
¯
¯
¯
¯=1−1=0
¯
x + 3y 2 ¯
∂
∂y
Since rot(F) = 0, F is conservative.
(b) We want a function f such that
∂f
= 2x + y
∂x
and
∂f
= x + 3y 2
∂y
From the first equation, we see that
f (x, y) = x2 + xy + g(y)
where g(y) is some function of y. If we differentiate this equation
for f with respect to y, we obtain
∂f
dg
=x+
∂y
dy
Comparing with the previous expression for
Thus:
∂f
dg
, we see that
= 3y 2 .
∂y
dy
g(y) = y 3 + C
Thus:
f (x, y) = x2 + xy + y 3 + C
(c) It is now easy to evaluate the line integral:
Z
h
i(√π,0)
2
3
= π
F · ds = x + xy + y
C
(0,0)
6. Let C be the circle of radius 1 centered at the origin and oriented counterclockwise. Use Green’s Theorem to evaluate the following integral:
I
(ey − y 3 ) dx + (xey + x3 ) dy
C
Anwser: First, we compute the rotation of the vector field:
¯
¯
∂
¯
¯ ∂
¯
¯
¡
¢
¯ = ey + 3x2 − ey − 3y 2 = 3x2 + 3y 2
¯ ∂x
∂y
¯
¯
¯ ey − y 3 xey + x3 ¯
Thus, by Green’s Theorem:
I
ZZ
¡ 2
¢
y
3
y
3
(e − y ) dx + (xe + x ) dy =
3x + 3y 2 dA
C
D
where D is the region inside the circle of radius 1 centered at the origin.
This integral will be easiest to compute using polar coordinates:
ZZ
¡
D
2
3x + 3y
2
¢
Z
2π
Z
1
dA =
0
0
3r3 dr dθ =
3π
2