## Thursday, July 21, 2022

### Line Integrals, Work & Conservation Of Energy Revisited

Consider the work done by a force as it moves a unit particle along  a curved path. We consider this in the context of allowing the direction cosines (l, m, n) to define the orientation of some force vector F.  Where:

l =  cos a   =  x2 -  x1 /     m  =  cos b    =  y2 -  y1 / ℓ     And:

n  =  cos  g   =    z2 -  z1 / ℓ

With:

cos Θ   =   l1 l2+  mm2 +  n1 2

If then Θ  is the angle between the force vector and the path at some specified position of the particle, the component of force acting along the trajectory is F cos Θ . Then the total work done is:

=  òA  B     F cos Θ ds

Where ds is an element of the path. We term an integral of this form a line integral.   More technically we may write:

òA  B     F cos Θ ds =     òB A     F cos Θ (- ds )=  - òB  A cos Θ ds

Then the work done in going from A to B is minus the work done in going from B to A.   For convenience we may also write:

Fl1   =  X,     F m1 =  Y  and:   F n1  =  Z

For the force components parallel to the 3 coordinate axes. And then:

=  òA B     (X dx +  Y dy   + Z  dz)

Assume now there is a function V of the coordinates such that:

X =  -  V / x        Y =  -  V / y        Z =  -  V / z

Then V is a potential function and we can also write:

dV  =  ( V / x) dx  +  ( V / y) dy   +  ( V / z) dz

Or: (using previous coordinate equations):

dV   =  -  (X dx     +  Y dy   +  Z dz)

And the work can be expressed more concisely:

W  =   òA  B     dV   =    - [V]A B    =   V  A   - V  B

And:   WAB  =   - WBA

Thus, the work done depends only on the limits and not the path followed.  So the process is reversible and conservation follows naturally when  forces are derived from a potential.  A more detailed presentation is possible when we incorporate  parametric equations (e.g.  with x= x (t), y = y(t) etc. ) for the curve paths.   In this case we consider the  point of application of a force moving along a curve C, i.e.

F = i X (x,y,z) + Y(x, y, z)  + k Z (x,y,z)

Where X, Y, Z are as previously defined.

Then the work done by a force in the context:

W  =   òC   ·  dR

Where:  R = ix + jy  + kz

is the vector from the origin to the point (x,y,z)  and:

dR = i dx + j dy + kdz

And if one calculates the dot product of the vectors F  and dR then one can write:

W  =   òC   X dx  +  Y dy  + Z dz

Which may be compared with:

=  òA B     (X dx +  Y dy   + Z  dz)

Seen previously.

Example: Consider a force given by:

F = i(2   -   y)  + j (y 2   -   z) +  k (z 2   -   x)

And let the point of application of the force move from the origin 0, to the point (1, 1, 1), along the straight line x= y = z and along the curve:

x = t,  y =  t 2   and z  =   t 3

Over:  (0  <  t   <  1)

Then the integral for evaluation becomes:

W  =   òC  (x 2   -   y) dx + (y 2   -   z) dy  + (z 2   -   x) dz

For the path over straight line: x = y = z:

=  ò0 1     3 (x 2   -   x) dx  =   - 1/2

While along the parameterized curve:

=    ò0 1    2(t4   - 3 ) t dt + 3 (t 6   - t 2 dt  =  - 29/ 60

(Rem:  dy = 2 t dt,      dz =  3 t 2 dt)

So we see the work done is very nearly independent of the curve C connecting the points.

Problems:

1) Let G(x,y) =  3   +   2xy +  2x    and:

x = 2t,  y =  t 2   Over:  (0  <  t   <  1)

Find the work done in going from A= 0  to B = 1 along the curve.

2) Two lines have direction cosines:

(1/2, Ö 6/ 4, - Ö 6/ 4 )  and (Ö 6/ 4, 1/4,  3/4)

Show they are orthogonal.

3) Given the force (vector function):

F = i (2xy) + j (2  + 3y 2 ln z)  + k (y 3/z)

Find the potential function V from which F was derived.

4)Given a force: F = i (y + z) + (z + x)  + k (x + y)

Find the work done as the point of application moves from (0, 0, 0) to (1,1, 1):

(a) Along the straight line: x = y = z

(b) Along the curve: x = t,  y =  t 2   and z  =   t4

(c) Find the direction cosines of the line segment AB (e.g. from (0, 0, 0) to (1,1, 1) ) and the associated  respective angles: a, b,  g.