Mathematical Tools for Physics

13—Vector Calculus 2 400

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=1

F~(~rk).∆~rk=

∫

F~(~r).d~r

r 0

r

r r r r r

1

2 3 4 5 6 (8)

This is the definition of a line integral.
How do you evaluate these integrals? To repeat what I did with Eq. ( 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=~r dt ̇ and the integral is

∫

F~(~r).d~r=

∫

F~

(

~r(t)

)

.~r dt ̇

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

) (9)

The integral of this from an initial point of the motion to a final point is

∫~rf

~ri

F~.d~r=

∫

m

2

d

(

~v.~v

)

=

m

2

[

v^2 f−vi^2

]

(10)

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.