Download PARAMETRIZED CURVES AND LINE INTEGRAL Let`s first recall

PARAMETRIZED CURVES AND LINE INTEGRAL
Let’s first recall the definitions of two important types of vector-valued functions:
−
Definition 0.1. A parametrized curve C is a function →
r : I → Rn , where
I = [a, b] ⊆ R is an interval.
→
−
Definition 0.2. A vector field is a function F : Rn → Rn .
Notice that the domain and the target have the same dimension. You can regard
this as assignment to each point in Rn an arrow based at that point.
1. Parametrized Curves
→
−
→
−
n
Let r : I → Rn be
a parametrized curve C in R . In coordinates, r (t) =
y1 (t), y2 (t), . . . , yn (t) , where yi (t)’s are functions of t ∈ I.
Definition 1.1. We define the velocity vector to be
→
−
−
r (t + ∆t) − →
r (t)
d−
→
−
r (t) = lim
r 0 (t) = →
∆t→0
dt
∆t
−
−
and it is tangent to the curve. We define the speed of →
r to be k→
r 0 (t)k.
→
−
→
−
→
−
−
d2 →
0
00
If v = r (t) is also differentiable, then we call r (t) = dt2 r the acceleration of
→
−
r (t).
−
Remark 1.2. In coordinates, →
r 0 (t) = y10 (t), y20 (t), . . . , yn0 (t) .
−
−
Proposition 1.3. Let →
r1 and →
r2 be two paths in Rn , then
(1)
d →
d−
d−
−
(−
r1 ± →
r2 ) = →
r1 ± →
r2 .
dt
dt
dt
(2)
−
−
d →
d→
r1 →
d→
r2
−
−
(−
r1 • →
r2 ) =
•−
r2 + →
r1 •
.
dt
dt
dt
(3)
−
−
d→
r1 →
d→
r2
d →
−
−
(−
r1 × →
r2 ) =
×−
r2 + →
r1 ×
.
dt
dt
dt
−
−
Proposition 1.4. If →
r (t) has constant length (i.e., k→
r (t)k is constant for all t),
→
−
→
−
0
then r is perpendicular to its derivative r (t).
Definition 1.5. Let h(s) be a function form [c, d] to [a, b] that is 1-1 and onto.
−
−
Then we call the composed function →
r (h(s)) a re-parametrization of →
r (t).
That is, we change the parameter from t to s via t = h(s).
−
−
Remark 1.6. For re-parametrization →
r (h(s)) of →
r (t), we have:
−
(1) If h(s) is 1-1 and onto, then the image set of →
r (h(s)) is the same with the
→
−
→
−
−
image set of r (t). That is, Im r (h(s))) = Im →
r.
1
2
PARAMETRIZED CURVES AND LINE INTEGRAL
(2) Re-parametrization changes the speed of the curve. By chain rule,
d→
d→
−
−
r (t)
ds r (h(s)) = dt
· |h0 (s)| .
t=h(s)
So the Speed is changed by the factor of |h0 (s)| .
−
Definition 1.7. The arc length L of a path →
r : [a, b] → Rn is found by integrating
its speed:
Z b
−
k→
r 0 (t)k dt
L=
a
For that parametrization, we can also define the arc length parameter as a function of t
Z t
−
s(t) =
k→
r 0 (τ )k dτ.
a
This function tells us how far we travel along the curve from time a to time t.
Notice that, by Fundamental Theorem of Calculus, we have
d
d
s(t) =
dt
dt
t
Z
−
−
k→
r 0 (τ )k dτ = k→
r 0 (t)k = Speed.
a
This function s(t) is important when studying the geometry of curves. We shall
revisit it in the near future.
−
−
Definition 1.8. A curve →
r : I → Rn is parametrized by arc length if k→
r 0 (t)k = 1
for any t ∈ I.
Rt −0
Rt
In this case, s(t) = a k→
r (τ )k dτ = a 1 dτ = t − a.
2. Line Integral
2.1. Scalar Version. Given a scalar-valued function f : Rn → R and a path
→
−
r : [a, b] → Rn , we define:
Definition 2.1. The line integral of f along the path is
Z
b
−0
−
f →
r (t) k→
r (t)k dt.
a
Remark 2.2. We denote this integral as
−
d
that dt
s(t) = k→
r 0 (t)k.
R
→
−
r
−
f ds, where ds = k→
r 0 (t)k dt. See above
Example 2.3. Consider function f (x, y) = x − 1. Integrate it along the unit circle
in R2 .
PARAMETRIZED CURVES AND LINE INTEGRAL
3
−
To compute it, we can first parametrize the unit circle as →
r (θ) = (cos θ, sin θ),
where θ ∈ [0, 2π]. Then the integral is given by
Z
Z θ=2π
−
f (cos θ, sin θ) · k→
f ds =
r 0 (θ)k dθ
→
−
|
{z
} |
{z
}
θ=0
r
f
Z
ds
θ=2π
(cos θ − 1) ·
=
p
sin2 θ + cos2 θ dθ
θ=0
θ=2π
Z
(cos θ − 1) · 1 dθ
=
θ=0
h
iθ=2π
= sin θ − θ
θ=0
= 2π
−
Of course we can parametrize the unit circle in another way. Consider →
r2 (θ) =
(cos 2θ, sin 2θ), where θ ∈ [0, π]. Then the integral is:
Z
Z θ=π
−
f ds =
f (cos 2θ, sin 2θ) · k→
r2 0 (θ)k dθ
→
−
|
{z
}
{z
}
|
r2
θ=0
f
Z
ds
θ=π
(cos 2θ − 1) ·
=
p
(−2 sin θ)2 + (2 cos θ)2 dθ
θ=0
Z θ=π
(cos 2θ − 1) · 2 dθ
=
θ=0
iθ=π
h1
= 2 sin 2θ − θ
2
θ=0
= 2π
Remark 2.4. The scalar line integral is independent of the re-parametrization.
A good interpretation to the scalar line integral is the area under the fence. And
the integral in Calculus I is a special case of this, where the curve is just the x-axis.
Figure 1
4
PARAMETRIZED CURVES AND LINE INTEGRAL
→
−
→
−
2.2. Vector Version. Given a vector field F : Rn → Rn and a a path R : [a, b] →
Rn , we can define
→
−
−
Definition 2.5. The line integral of F along →
r is
Z
Z b
−0
→
−
→
− →
−
F • d→
r =
F −
r (t) • →
r (t) dt
→
−
| {z }
r
a
−
d→
r
−
−
Notice that in this formula, d→
r =→
r 0 (t) dt as a vector, and hence dot product
→
−
→
−
between F and d r .
Vector line integral has an important physics interpretation:
→
−
Example 2.6 (Work done by a force). Let E (x, y) = (−y, x) be a electric field
in R2 . If there is a charged particle with electric charge q moving along the path
→
−
r (t) = (et cos t, et sin t), t ∈ [0, 2π], then by Coulomb’s law, the force induced by
→
−
→
−
electric field on the particle is F = q · E . So the work done by the force is given by
Z
→
−
r
→
−
−
F • d→
r =
Z
t=2π
t=0
−
r 0 (t) dt
q · E(et cos t, et sin t) • →
|
{z
} | {z }
→
−
→
−
−
F →
r (t)
Z
dr
t=2π
=
q · (−et sin t, et cos t) • (et cos t − et sin t, et cos t + et sin t) dt
t=0
t=2π
Z
=
q · e2t · 1 dt
t=0
q 2t t=2π
e 2
t=0
q 2π
e −1
=
2
=
Figure 2
As you probably know, given a forcing acting on a moving particle, if we decomposed the force to a component along the direction of the motion and a component
orthogonal to the motion, then only the component along the motion does work.
PARAMETRIZED CURVES AND LINE INTEGRAL
So in the formula
Z
→
−
r
→
−
−
F • d→
r =
Z
b
5
−0
→
− →
F −
r (t) • →
r (t) dt
a
→
−
→
− →
r 0 (t)
−
· k→
r 0 (t)k dt
r (t) • →
F −
−
k r 0 (t)k
a
Z b
→
− −0
→
− →
r (t) • T · k→
r (t)k dt
F −
=
a
Z
→
− →
−
=
F • T ds
Z
b
=
→
−
r
→
− →
−
→
−
→
−
We have F • T , which is the length of F project onto the unit tangent vector T
→
− →
−
(i.e. the direction of the motion). Then we integrate F • T according to the speed
R →
−
−
F • d→
r.
of motion to get the work →
−
r
A last words about vector line integral is that, dislike scalar line integral, vector
line integral depends on the orientation of parametrization.
→
−
Example 2.7. Integrating the vector field F (x, y) = (−y, x) along the unit circle
in R2 .
−
If we parasitized the unit circle clockwise as →
r1 (t) = (cos t, sin t), t ∈ [0, 2π],
then
Z t=2π
Z
→
−
→
−
→
−
−
F •dr =
F (cos t, sin t) • →
r1 0 (t) dt
→
−
r1
t=0
t=2π
Z
(− sin t, cos t) • (− sin t, cos t), dt
=
t=0
Z t=2π
1 · dt
=
t=0
= 2π
−
If we parasitize the unit circle counter-clockwise as →
r2 (t) =
t ∈ [0, 2π], we get
Z
Z t=2π
−0
→
−
→
−
→
−
F •dr =
F cos(−t), sin(−t) • →
r1 (t) dt
→
−
r2
cos(−t), sin(−t) ,
t=0
t=2π
Z
=
− sin(−t), cos(−t) • sin(−t), − cos(−t) , dt
t=0
t=2π
Z
(−1) · dt
=
t=0
= −2π
−
−
Notice that →
r2 has opposite orientation with →
r1 .
−
−
Theorem 2.8. Let →
r1 : [a, b] → Rn and →
r2 : [c, d] → Rn be parametrizations to
→
−
the same path. Given a vector field F : Rn → Rn , we have
R →
R →
−
−
−
−
−
−
(1) If →
r1 has the same direction with →
r2 , then →
F • d→
r = →
F • d→
r.
−
−
r1
r2
R
R
→
−
→
−
→
−
→
−
→
−
−
(2) If r1 has the opposite direction with r2 , then →
F •dr =− →
F • d→
r.
−
−
r1
r2
6
PARAMETRIZED CURVES AND LINE INTEGRAL
Definition 2.9. Given a curve C with a certain orientation, we call the curve with
the opposite orientation −C.
R →
R →
−
−
−
−
So indeed, −C F • d→
r = − C F • d→
r
2.3. Summary.
R
f · ds
→
−
r
R
→
−
r
→
−
−
F • d→
r
f is a scalar-valued function
→
−
F is a vector field
−
ds = k→
r 0 (t)k dt
−
−
d→
r =→
r 0 (t) dt
f · ds usual multiplication
→
−
−
F • d→
r dot product
3. Notation Matters
For the vector line integral, the notation that I prefer is
Z
Z b
→
−
→
−
−
−
•→
F • d→
r =
F →
r 0 (t)dt,
−
→
−
r
r (t)
a
since it makes sense geometrically. But often you see the following notation:
→
−
Given a vector field F (x, y, z) = P (x, y, z), Q(x, y, z), M (x, y, z) where P (x,
y, z),
−
Q(x, y, z), M (x, y, z) are scalar-valued functions. Let →
r (t) = x(t), y(t), z(t) be
the path you integrate along. Then
Z b
Z b
→
−
−
•→
F →
r 0 (t)dt =
(P, Q, M ) • x0 (t), y 0 (t), z 0 (t) dt
−
r (t)
{z } |
{z
}
a
a |
→
−
F
Z
b
=
→
−
r 0 (t)
h
i
P · x0 (t) + Q · y 0 (t) + M · z 0 (t) dt
a
Z
b
=
P · x0 (t) dt + Q · y 0 (t) dt + M · z 0 (t) dt
a
Let x0 (t) dt = dx, y 0 (t) dt = dy and z 0 (t) dt = dz, then
Z
Z b
→
−
→
−
→
−
−
•→
F •dr =
F →
r 0 (t)dt
−
→
−
r
r (t)
a
b
Z
=
P dx + Q dy + M dz
a
Thus we have the following notation:
Z
Z
→
−
−
F • d→
r =
→
−
r
b
P dx + Q dy + M dz.
a
Warning: Even with dx, dy, and dz, this is still a line integral. So to compute
it, you still need a parametrization to the curve.