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 / ℓ    m  =  cos b    =  y2 -  y1 / ℓ     And:

n  =  cos  g   =    z2 -  z1 / ℓ 

With:

cos Θ   =   l₁ l₂+  m₁ m₂ +  n₁ n 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:  

W =  ò_A  ^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:

ò_A  ^B     F cos Θ ds =     ò_B ^A     F cos Θ (- ds )=  - ò_B  ^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:

W =  ò_A ^{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  =   ò_A  ^{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) + j 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  =   ò_C   ^F ·  ^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  =   ò_C   ^{X dx  +  Y dy  + Z dz}

^{Which may be compared with:} 

W =  ò_A ^{B     (X dx +  Y dy   + Z  dz)}

Seen previously.

Example: Consider a force given by:

F = i(x 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  =   ò_C  (x ²   -   y) dx + (y ²   -   z) dy  + (z ²   -   x) dz

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

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

While along the parameterized curve:

W =    ò₀ ¹    2(t⁴   - t ³ ) t dt + 3 (t ⁶   - t ) 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) =  x 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 (x 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) + j (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.