18.022 Recitation Handout (with solutions)
24 November 2014
1. According to Coulomb’s law, the force between a particle of charge q1 at the origin and a particle
of charge q2 at the point r = (x, y, z) ∈ R3 is given by
F=
q1 q2 r
,
4πε0 |r|3
where ε0 is a physical constant.
(a) Is F a conservative vector field? If so, find a function φ : R3 → R such that ∇φ = F.
(b) If the distance between two charges is tripled, by what factor is the force between them reduced?
(c) How much work is required to move the second particle along the path
√
γ(t) = (1 + (1 − t) cos(t2 ), sin πt, 4t − t2 )
0 ≤ t ≤ 1?
Express your answer in terms of q1 , q2 , and ε0 .
Solution. (a) Writing F as
!
q1 q2
y
x
z
,
,
,
4πε0 (x2 + y2 + z2 )3/2 (x2 + y2 + z2 )3/2 (x2 + y2 + z2 )3/2
we see that F is the gradient of
φ(x, y, z) = −
q1 q2
q1 q2 1
1
= −
.
p
4πε0 x2 + y2 + z2
4πε0 |r|
Therefore, F is conservative.
(b) The magnitude of F is proportional to |r|/|r|3 = |r|−2 , so tripling the distance decreases the force
by a factor of 9 .
R(c) The amount of work required to move the particle from a point r1 to a point r2 along a path γ is
F · ds. Since F = ∇φ is conservative, the value of this integral is φ(r2 ) − φ(r1 ) no matter what path
γ
γ from r1 to r2 is chosen. For the given path, the starting point is γ(0) = (2, 0, 0) and the ending
!
q1 q2 1
1
point is (1, 0, 3). Thus the work is
− √
.
4πε0 2
10
2. (6.2.23 in Colley) Let D be a region to which Green’s theorem applies and suppose that u(x, y)
and v(x, y) are two functions of class C2 whose domains include D. Show that
"
I
∂(u, v)
dA = (u∇v) · ds,
D ∂(x, y)
C
where C = ∂D is oriented as in Green’s theorem.
Solution. Writing out the right-hand side and applying Green’s theorem, we get
I
Z
(u∇v) · ds =
uvx dx + uv y dy
C
"
∂
∂
(uv y ) −
(uvx ) dA
=
∂y
D ∂x
"
=
ux v y − u y vx dA
"D
∂(u, v)
=
dA.
D ∂(x, y)
R
3. (6.1.29 in Colley) Let C be a level set of the function f (x, y). Show that
Solution. Letting a and b be the endpoints of the curve C, we calculate
since f is constant on C.
C
R
C
∇ f · ds = 0.
∇ f · ds = f (b) − f (a) = 0,