13—Vector Calculus 2 328The basic idea is a combination of Eqs. (13.1) and (13.2). Divide the specified curve into anumber of pieces, at the points{~rk}. Between pointsk− 1 andkyou had the estimate of the arc
length as
√
(∆xk)^2 + (∆yk)^2 , but here you need the whole vector from~rk− 1 to~rkin order to evaluate
the work done as the mass moves from one point to the next. Let∆~rk=~rk−~rk− 1 , then
lim
|∆~rk|→ 0∑N
k=1F~(~rk).∆~rk=
∫
F~(~r).d~r
~r 0 ~r^1
~r 2 ~r (^3) ~r (^4) ~r 5 ~r^6 (13.10)
This is the definition of a line integral.
How do you evaluate these integrals? To repeat what happened with Eq. (13.2), that will depend
on the way that you use to describe the curve itself. Start with the simplest method and assume that
you have a parametric representation of the curve:~r(t), thend~r=~rdt ̇ and the integral is
∫
F~(~r).d~r=
∫
F~(~r(t)).~rdt ̇
This is now an ordinary integral with respect tot. In many specific examples, you may find an easier
way to represent the curve, but this is something that you can always fall back on.
In order to see exactly where this is used, start withF~=m~a, Take the dot product withd~rand
manipulate the expression.
F~=md~v
dt
, so F~.d~r=m
d~v
dt
.d~r=md~v
dt
.d~r
dt
dt=md~v.
d~r
dt
=m~v.d~v
or F~.d~r=
m
2
d
(
~v.~v
) (13.11)
The integral of this from an initial point of the motion to a final point is
∫~rf~riF~.d~r=
∫
m
2
d
(
~v.~v
)
=
m
2
[
vf^2 −vi^2
]
(13.12)
This is a standard form of the work-energy theorem in mechanics. In most cases you have to specify
the whole path, not just the endpoints, so this way of writing the theorem is somewhat misleading. Is
it legitimate to manipulate~v.d~vas in Eq. (13.11)? Yes. Simply write it in rectangular components as
vxdvx+vydvy+vzdvzand you can integrate each term with no problem; then assemble the result as
v^2 / 2.
Example
IfF~=Axyxˆ+B(x^2 +L^2 )yˆ, what is the work done going from point(0,0)to(L,L)along the three
different paths indicated.?
∫
C 1F~.d~r=
∫
[Fxˆx+Fyˆy].[xdxˆ +ydyˆ ]
=
∫ [
Fxdx+Fydy
]
=
∫L
0dx0 +
∫L
0dyB 2 L^2 = 2BL^3
∫
C 2F~.d~r=
∫L
0dxAx^2 +
∫L
0dyB(y^2 +L^2 ) =AL^3 /3 + 4BL^3 / 3
∫
C 3F~.d~r=
∫L
0dyB(0 +L^2 ) +
∫L
0dxAxL=BL^3 +AL^3 / 2
(^21)