COALITION MATH 151, FALL 1997

        Group II (S. A. Fulling, assisted by Cary Lasher)

                              Day 12.2


(ORIGINALLY PLANNED FOR DAY 12.1)
    
Finding the potential energy once you know that d{F_x}/dy = 
d{F_y}/dx.

    1.  Example:  F_x = y^2,  F_y = 2xy + 1; derivs = 2y.

    2.  Method 0:  Guess the answer "by inspection".  Nice when 
    it works.
GOT THIS FAR.  NOW TWO DAYS BEHIND SCHEDULE!  CONTINUED ON 12.2:

    3.  Method 1:  Solve dV/dx = - F_x,  dV/dy = - F_y.
    The constant of integration in each case may depend on the 
    other variable.  Compare the two answers to find the 
    "constants".

    In the example:  dV/dx = - y^2 => V = - xy^2 + C(y).
                     dV/dy = - 2xy -1 => V = - xy^2 - y + D(x).
    Therefore, C(y) = - y + D(x), so C(y) + y = D(x) = constant = K.
    Thus V = - xy^2 - y + K.
    
    4.  Variant of Method 1, preferred by many people:  Find one 
    of the antiderivatives, then differentiate it to find the 
    "constant":

    V = - xy^2 + C(y) => - 2xy - 1 = dV/dy = - 2xy + C'(y).
    Therefore C'(y) = -1, so C(y) = - y + K.  Get same V as 
    before.

    5.  Method 2, more directly related to theory of line 
    integrals but usually harder to implement:  We know

    - V(r) = \int_{r_0}^r F.dr.

    Evaluate the line integral along a convenient path to r = 
    (x,y) from a convenient starting point r_0 (say (0,0)), fixed 
    once and for all.

    6.  Activity:  Implement method 2 on the example.  Let teams 
    choose their paths; there are 3 obvious possibilities.
I DID PART 7 FIRST, THEN ASSIGNED THE 3 PATHS TO TEAMS MOD 3.  
STUDENTS STILL SHAKY ON THIS.  THE FACT THAT THE TOP ENDPOINT IS 
VARIABLE (AND DISTINCT FROM THE VARIABLE OF INTEGRATION) CAUSES A 
NOTATIONAL MUDDLE IF YOU AREN'T CAREFUL.  (THE TA SUGGESTS 
CALLING THE FIXED POINT (a,b), RATHER THAN THE INTEGRATION POINT 
(\tilde x, \tilde y) AS I DID.)  STUDENTS WERE ABLE TO DO THE L-
SHAPED PATHS WITH A FEW HINTS, BUT I HAD TO DO THE DIAGONAL ONE 
MYSELF. 

    7.  DON'T do this:  Treat dx and dy integrals as one-dim 
    integrals with the other variable as a constant.  This is 
    WRONG because y DEPENDS on x and vice versa.  In the example 
    it gives - V = 2xy^2 + y + K (factor 2 is wrong).  This is a 
    particularly vicious mistake because 90% of the time it gives 
    the right answer by accident, except for a factor 2!  
THE POINT IS THAT IN SOLVING EXACT ODEs THE FACTOR 2 IS 
IRRELEVANT.