Math 227 Problem Set IV Solutions
1. For each of the following vector fields, sketch the vector field and determine its field lines.
(a) v(x, y) = xı̂ı + y̂.
(b) v(x, y) = ∇(x2 − y).
Solution. (a) The field lines obey
dx
x
=
dy
y
=⇒
ln |x| = ln |y| + C
=⇒
y = C ′x
The possible minus signs in the absolute values have been absorbed in the
constant C ′ .
(b)
F = ∇(x2 − y) = 2xı̂ı − ̂. The field lines obey
dx
2x
=
dy
−1
=⇒
1
2
=⇒
y=C−
ln |x| = −y + C
1
2
ln |x|
2. Describe the stream lines of
(a) v(x, y, z) = xı̂ı + y̂ − xk̂.
xı̂ı+y̂

(b) v(x, y, z) = (1+z2 )(x2 +y2 ) .
Solution. (a) The streamlines obey
dx
dt
dy
dt
dz
dt
=x
=⇒
x(t) = cet
=y
=⇒
y(t) = det
= −x = −cet
=⇒
z(t) = −cet + f
for any constants c, d and f . The stream lines are straight lines with x = Ky (or y = 0) and z = −x+K ′
for any constants K and K ′ .
b) Observe that the vector field is parallel to the xy plane and points radially outward from the z–axis.
Hence for each z0 and θ0 , x(t) = s(t) cos θ0 , y(t) = s(t) sin θ0 , z = z0 ought to be a stream line for some
function s(t). Subbing into
x′ =
x
(1+z 2 )(x2 +y 2 )
y′ =
y
(1+z 2 )(x2 +y 2 )
z′ = 0
gives
s′ (t) cos θ0 =
s(t) cos θ0
(1+z02 )s(t)2
d 1 2
1
or dt
( 2 s ) = (1+z
2 . So s(t) =
0)
by a constant. All together
s′ (t) sin θ0 =
√
2t
1+z0
x(t) =
s(t) sin θ0
(1+z02 )s(t)2
z ′ (t) = 0 ⇐⇒ s′ (t) =
1
(1+z0 )2 s(t)
does the job. Putting in an arbitrary constant would just shift t
√
2t
1+z0
cos θ0 , y(t) =
√
2t
1+z0
sin θ0 , z = z0
for all constants θ0 and z0 . These are indeed straight lines with y = Kx (or x = 0), z = K ′ for any
constants K, K ′ .
1
3. Sketch the stream lines for the velocity field v = yı̂ı − sin x̂.
Solution. Observe that
• The x–component of v is positive when y > 0. So the flow is to the right in the upper half plane.
• The x–component of v is negative when y < 0. So the flow is to the left in the lower half plane.
• The y–component of v is negative when 0 < x < π. So the flow is downward in the strip 0 < x < π.
Also in the strip 2π < x < 3π and so on.
• The y–component of v is positive when −π < x < 0. So the flow is upward in the strip −π < x < 0.
Also in the strip π < x < 2π and so on.
y
−π
π
0
2π x
Explicitly
dx
y
=
dy
sin x
=⇒
sin x dx = y dy
=⇒
− cos x = 21 y 2 + C
For each fixed value of C, the equation 21 y 2 = − cos x − C has at most two values of y for each value of
x. That’s why the elliptical looking flow lines in the figure above have to be closed curves rather than
spirals.
4. The vector fields r̂(r, θ) = cos θ ı̂ı + sin θ ̂ and θ̂θ(r, θ) = − sin θ ı̂ı + cos θ ̂ are often used as basis vectors
when working with polar coordinates. (Since r̂(r, θ) and θ̂θ(r, θ) happen to be independent of r, people
usually just write r̂(θ), θ̂θ (θ).) Show that the polar curve r = eθ is a field line of the vector field
F(r, θ) = r̂(r, θ) + θ̂θ (r, θ).
Solution. Let’s parametrize the curve r = eθ by θ. That is parametrize θ(t) = t, r(t) = et , so that
∂r̂
∂r̂
r(t) = r(t) r̂ r(t), θ(t) = et r̂(et , t). Since ∂θ
= θ̂θ and ∂r
= 0, the velocity vector of r(t) = et r̂(et , t) is
∂r̂ t
r′ (t) = et r̂(et , t) + et ∂θ
(e , t) = et r̂(et , t) + θ̂θ(et , t) = et F r(t), θ(t)
As the velocity vector r′ (t) is parallel to the vector field F at r(t), the specified curve is indeed a field
line of the specified vector field.
2
5. Let a particle of mass m move on a path r(t) according to Newton’s law, m ddt2r = F, in a force field
F = −∇
∇V on IR3 , where V is a given function (called the potential energy).
(a) Prove that the energy E(t) = 21 mkr′ (t)k2 + V r(t) is constant in time.
(b) If the particle moves on an equipotential surface (i.e. a surface whose equation is V (x, y, z) = V0 ,
for some constant V0 ), show that its speed is constant.
Solution. (a)
dE
dt
i
· r′ (t) + V r(t) = mr′′ (t) · r′ (t) + ∇ V r(t) · r′ (t)
= −∇
∇V r(t) · r′ (t) + ∇V r(t) · r′ (t) = 0
=
h
′
d 1
dt 2 m r (t)
2
(b) If r(t) remains on the surface V (x) = V0 , then 21 mkr′ (t)k2 = E − V r(t) = E − V0 remains constant
in time. Hence, so does the speed kr′ (t)k.
6. Let f (x, t) be a real–valued function of x ∈ IR3 and t ∈ IR. Define the material derivative of f relative
to a vector field F(x) to be
Df
∂f
Dt (x, t) = ∂t (x, t) + ∇ f (x, t) · F(x)
Here ∇f (x, t) is the spatial gradient. That is,
∇f (x, y, z), t =
∂f
∂f
 + ∂f
∂x ı̂ı + ∂y ̂
∂z k̂
Let r(t) be a flow line of F. Show that
d
dt f
In words,
r(t), t =
Df
Dt
r(t), t
Df
Dt
is the rate of change of f felt by a particle that is moving with the flow F.
Solution. That “r(t) is a flow line of the vector field F” means that r′ (t) = F r(t) . Consequently,
d
dt f
r(t), t = r′ (t) · ∇ f r(t), t + ∂f
∂t (r(t), t)
∂f
= F r(t) · ∇f r(t), t + ∂t (r(t), t)
= Df
Dt r(t), t
7. Determine whether or not each of the following vector fields is conservative. If so, find a function φ so
that F = ∇φ.
xı̂ı−y̂

(a) F(x, y) = x2 +y2
yı̂ı−x̂

(b) F(x, y) = x2 +y2
(c) F(x, y, z) = (2xy + z 2 )ı̂ı + (x2 + 2yz)̂ + (y 2 + 2xz)k̂
2xy
−y
2xy
∂
x
∂
∂y x2 +y 2 = − (x2 +y 2 ) 6= (x2 +y 2 ) = ∂x x2 +y 2 so F is not
∇ tan−1 xy is conservative. Here is a derivation of φ:
Solution. (a)
(b) F(x, y) =
So φ(x, y) = tan−1
∂φ
∂x
=
y
x2 +y 2
=⇒
∂φ
∂y
x
= − x2 +y
2
=⇒
x
y +
2
φ(x, y) =
∂ψ
∂y
Z
=0
y
x2 +y 2
conservative.
dx = tan−1
=⇒
x
y
+ ψ(y)
ψ(y) = C
C.
(c) F(x, y, z) = ∇ x y + xz 2 + y 2 z is conservative. Here is a derivation of φ:
∂φ
∂x
∂φ
∂y
∂φ
∂z
= 2xy + z 2
=⇒
= x2 + 2yz
=⇒
= y 2 + 2xz
=⇒
φ(x, y, z) = x2 y + xz 2 + ψ(y, z)
∂ψ
∂y
dζ
y 2 + dz
x2 +
2xz +
= x2 + 2yz
=⇒
= y 2 + 2xz
=⇒
ψ(y, z) = y 2 z + ζ(z)
ζ(z) = C
So φ(x, y, z) = x2 y + xz 2 + y 2 z + C.
8. Let F be a conservative force field with potential φ. Show that the lines of force are perpendicular to the
equipotential surfaces (i.e. surfaces whose equations are of the form φ(x, y, z) = φ0 with φ0 constant).
3
Solution. The line of force through (x0 , y0 , z0 ) is parallel to F(x0 , y0 , z0 ) = ∇φ(x0 , y0 , z0 ) at (x0 , y0 , z0 ).
On the other hand ∇φ(x0 , y0 , z0 ) is a normal vector to the equipotential surface φ(x, y, z) = φ(x0 , y0 , z0 ).
9. Let ψ(r, θ) be a function in polar ccordinates. In cartesian coordinates, this function becomes
p
φ(x, y) = ψ r(x, y), θ(x, y) where r(x, y) = x2 + y 2 and θ(x, y) = tan−1 xy . By definition, the gradient
(∇
∇ψ)(r, θ) of ψ(r, θ) is defined by the requirement that
∇φ(x, y) = (∇
∇ψ) r(x, y), θ(x, y)
(a) Find ∇ψ(r, θ), expressed in terms of the basis vectors
r̂(θ) = cos θ ı̂ı + sin θ ̂
θ̂θ(θ) = − sin θ ı̂ı + cos θ ̂
∂F2
1
(b) We have seen that ∂F
∂y (x, y) = ∂x (x, y) is a necessary condition for the vector field F(x, y) =
F1 (x, y)ı̂ı + F2 (x, y) ̂ to be conservative. Find the analogous necessary condition for the vector field
G(r, θ) = G1 (r, θ) r̂(θ) + G2 (r, θ) θ̂θ(θ), expressed in polar coordinates, to be conservative, i.e. be of
the form (∇
∇ψ)(r, θ) for some ψ(r, θ).
Solution. (a) Substituting x = r cos θ, y = r sin θ into the definition of φ gives ψ(r, θ) = φ(r cos θ, r sin θ)
and then differentiating gives
∂ψ
∂r (r, θ)
∂ψ
∂θ (r, θ)
Now solve for
∂φ
∂x
and
=
=
∂φ
∂φ
∂x (r cos θ, r sin θ) cos θ + ∂y (r cos θ, r sin θ) sin θ
∂φ
− ∂φ
∂x (r cos θ, r sin θ)r sin θ + ∂y (r cos θ, r sin θ)r cos θ
(1)
(2)
∂φ
∂y .
∂φ
∂x (r cos θ, r sin θ)
∂φ
∂y (r cos θ, r sin θ)
=
=
∂ψ
∂r (r, θ) cos θ −
∂ψ
∂r (r, θ) sin θ +
∂ψ
sin θ
∂θ (r, θ) r
∂ψ
cos θ
∂θ (r, θ) r
cos θ (1) −
sin θ (1) +
1
r sin θ
1
r cos θ
(2)
(2)
By definition
(∇
∇ψ) r, θ) = (∇
∇φ)(r cos θ, sin θ)
∂ψ
sin θ
= ∂r (r, θ) cos θ − ∂ψ
ı̂ı + ∂ψ
∂θ (r, θ) r
∂r (r, θ) sin θ +
=
∂ψ
∂r (r, θ) r̂
+
∂ψ
cos θ
∂θ (r, θ) r
1 ∂ψ
θ
r ∂θ (r, θ) θ̂
̂
(b) If G(r, θ) is conservative then there is a function ψ(r, θ) such that G(r, θ) = (∇
∇ψ)(r, θ) so that, by
part(a),
∂ψ
G2 (r, θ) = 1r ∂ψ
G1 (r, θ) = ∂ψ
∂r (r, θ)
∂θ (r, θ) or rG2 (r, θ) = ∂θ (r, θ)
and so
∂G1
∂θ (r, θ)
The condition is
∂G1
∂θ
=
∂2 ψ
∂θ∂r (r, θ)
=
∂2 ψ
∂r∂θ (r, θ)
2
= r ∂G
∂r + G2 .
4
2
= r ∂G
∂r (r, θ) + G2 (r, θ)