2Curves and Line
IA Vector Calculus
2.3 Gradients and Differentials
Recall that the line integral depends on the actual curve taken, and not just the
end points. However, for some nice functions, the integral does depend on the
end points only.
Theorem. If F = ∇f(r), then
Z
C
F · dr = f(b) − f(a),
where b and a are the end points of the curve.
In particular, the line integral does not depend on the curve, but the end
points only. This is the vector counterpart of the fundamental theorem of
calculus. A special case is when C is a closed curve, then
H
C
F · dr = 0.
Proof.
Let
r
(
u
) be any parametrization of the curve, and suppose
a
=
r
(
α
),
b = r(β). Then
Z
C
F · dr =
Z
C
∇f · dr =
Z
∇f ·
dr
du
du.
So by the chain rule, this is equal to
Z
β
α
d
du
(f(r(u))) du = [f(r(u))]
β
α
= f(b) − f(a).
Definition
(Conservative vector field)
.
If
F
=
∇f
for some
f
, the
F
is called a
conservative vector field.
The name conservative comes from mechanics, where conservative vector
fields represent conservative forces that conserve energy. This is since if the
force is conservative, then the integral (i.e. work done) about a closed curve is 0,
which means that we cannot gain energy after travelling around the loop.
It is convenient to treat differentials
F ·
d
r
=
F
i
d
x
i
as if they were objects
by themselves, which we can integrate along curves if we feel like doing so.
Then we can define
Definition
(Exact differential)
.
A differential
F ·
d
r
is exact if there is an
f
such that F = ∇f. Then
df = ∇f · dr =
∂f
∂x
i
dx
i
.
To test if this holds, we can use the necessary condition
Proposition. If F = ∇f for some f, then
∂F
i
∂x
j
=
∂F
j
∂x
i
.
This is because both are equal to ∂
2
f/∂x
i
∂x
j
.
For an exact differential, the result from the previous section reads
Z
C
F · dr =
Z
C
df = f(b) − f(a).
Differentials can be manipulated using (for constant λ, µ):
Proposition.
d(λf + µg) = λdf + µdg
d(fg) = (df)g + f (dg)
Using these, it may be possible to find f by inspection.
Example. Consider
Z
C
3x
2
y sin z dx + x
3
sin z dy + x
3
y cos z dz.
We see that if we integrate the first term with respect to
x
, we obtain
x
3
y sin z
.
We obtain the same thing if we integrate the second and third term. So this is
equal to
Z
C
d(x
3
y sin z) = [x
3
y sin z]
b
a
.