4.3 Independence of Parametrisation

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Suppose the path is the quarter circle from [1, 0]T to [0, 1]T , the circle being

centred on the origin. One student might write

c : [0, 1] −! R2

c(t) ,

_

cos(_

2 t)

sin(_

2 t)

_

Another might take

c : [0, _/2] −! R2

c(t) ,

_

cos(t)

sin(t)

_

Would these two different parametrisations of the same path give the same

answer? It is easy to see that they would get the same answer. (Check if you

doubt this!) They really ought to, since the original definition of the path

integral was by chopping the path up into little bits and approximating each

by a line segment. We used an actual parametrisation only to make it easier

to evaluate it.

Remark 4.3.1. For any continuous vector field F on Rn and any differentiable

curve c, the value of the integral of F over c does not depend on the

choice of parametrisation of c. I shall prove this soon.

Example 4.3.1. For the vector field F on R2 given by

F

_

x

y

_

,

_

−y

x

_

evaluate the integral of F along the straight line joining

_

1

0

_

to

_

0

1

_

.

4.3. INDEPENDENCE OF PARAMETRISATION 37

Parametrisation 1

c : [0, 1] −! R2

t   (1 − t)

_

1

0

_

+ t

_

0

1

_

=

_

1 − t

t

_

c0 =

_

−1

1

_

Then the integral is

R t=1

t=0

_

−t

1 − t

_

q

_

−1

1

_

dt

=

Z 1

0

t + 1 − tdt =

Z

1dt = t]1

0 = 1

This looks reasonable enough.

Parametisation 2

Now next move along the same curve but at a different speed:

c(t) ,

_

1 − sin t

sin t

_

, t 2 [0, _/2].

notice that x(c) = 1 − y(c) so the ‘curve’ is still along y = 1 − x.

Here however we have c0 =

_

−cos t

cos t

_

t 2

_

0, _

2

_

and

F(c(t)) =

_

−sin t

1 − sin t

_

So the integral is

R t=_/2

t=0

_

−sin t

1 − sin t

_

q

_

−cos t

cos t

_

dt

=

Z _/2

0

sin t cos t + cos t − sin t cos t dt

=

Z _/2

0

cos t dt = sin t]_/2

u = 1

38 CHAPTER 4. FIELDS AND FORMS

So we got the same result although we moved along the line segment at a

different speed. (starting off quite fast and slowing down to zero speed on

arrival)

Does it always happen? Why does it happen? You need to think about this

until it joins the collection of things that are obvious.

In the next proposition, [a, b] , {x 2 R : a _ x _ b}

Proposition 4.3.1. If c : [u, v] ! Rn is differentiable and ' : [a, b] ! [u, v]

is a differentiable monotone function with '(a) = u and '(b) = v and e :

[a, b] ! Rn is defined by e , c _ ', then for any continuous vector field V on

Rn, Z v

u

V(c(t)) q c0(t)dt =

Z b

a

V(e(t)) q e0(t)dt.

Proof By the change of variable formula

Z

V(c(t)) q c0(t)dt =

Z

V(c('(t)) q c0('(t))'0t)dt

= V(e(t)) q e0(t)dt(chain rule)

_

Remark 4.3.2. You should be able to see that this covers the case of the

two different parametrisations of the line segment and extends to any likely

choice of parametrisations you might think of. So if half the class thinks of

one parametrisation of a curve and the other half thinks of a different one,

you will still all get the same result for the path integral provided they do

trace the same path.

Remark 4.3.3. Conversely, if you go by different paths between the same

end points you will generally get a different answer.

Remark 4.3.4. Awful Warning The parametrisation doesn’t matter but

the orientation does. If you go backwards, you get the negative of the result

you get going forwards. After all, you can get the reverse path for any

parametrisation by just swapping the limits of the integral. And you know

what that does.

Example 4.3.2. You travel from

_

1

0

_

to

_

−1

0

_

in the vector field

V

_

x

y

_

,

_

−y

x

_

4.4. CONSERVATIVE FIELDS/EXACT FORMS 39

1. by going around the semi circle (positive half)

2. by going in a straight line

3. by going around the semi circle (negative half)

It is obvious that the answer to these three cases are different. (b) obviously

gives zero, (a) gives a positive answer and (c) the negative of it.

(I don’t need to do any sums but I suggest you do.) (It’s very easy!)

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