COALITION MATH 151, FALL 1997

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

                              Day 11.1


STARTED BY REVIEWING THE PREVIOUS DAY STARTING AT ITEM 2D (WORK 
IN DIMENSION 2 WITH CONSTANT FORCE).

Work:  The general case of a curved path and a vector field.

    1.  Coordinate approach:  W = int_C [F_x dx + F_y dy]  

        A.  There are many examples in the handout and the 
        textbooks.

        B.  Example 1:  C = line from (1,1) to (3,2);
        F = <x^2 + y, y^2 + x>

            i.  Evaluate the F_x term (by integration over x).  
            This time the two dimensions are not entirely 
            independent, since y depends on x.  
            Answer = 35/3.  Curve: y = x/2 + 1/2

           ii.  Class exercise:  Evaluate the F_y term (by 
            integration over y).  
            Answer = 13/3.  Curve:  x = 2y - 1.

        C.  Example 2:  C = circle (c-clockwise) centered at 
        origin with radius 1, starting at (1,0); F = <-y,x>.
        Parametrize (justification to emerge later):  
        x = cos t, y = sin t, 0 < t <2\pi;  Answer = 2\pi.

        D.  Remark:  In SOME problems the most convenient 
        parameter is one of the coordinates.  Instead of
        x = 2t - 1, y = t, 1 < t < 2, we write just
        x = 2y - 1, 1 < y < 2.  Use this to revisit Example 1:
   
        \int_{y=1}^{y=2} [(x^2+y)dx + (y^2+x)dy]

        You already did the 2nd term and got 13/3.  Do first term 
        using dx = 2 dy.  Answer = 35/3 as before, so total W = 
        16.

        E.  Example 3:  C = broken line from (0,0) to (2,0) to 
        (0,1) = C_1 + C_2.  It is natural to use x as parameter 
        on C_1 and y as parameter on C_2.  You CAN'T use y on C_1 
        nor x on C_2.  In \int_C [F_x dx + F_y dy], the second 
        term is 0 on C_1 since dy = dy/dx dx = 0 there.  
        Similarly, the first term is 0 on C_2.

GOT THIS FAR.
    2.  Vector approach:  W = \int_C F_{||} ds, where
        F_{||} = component of \vec{F} parallel to C;
        s = arc length measured from endpoint to \vec{r}.
        How do we turn this geometrically obvious definition into 
        something we can calculate?

        A.  Consider a parametrization \vec{r} = <x(t),y(t)>.
        Recall that \vec{v} = dr/dt = <x'(t),y'(t)> is tangent to 
        the curve.

        B.  Therefore, T = v/|v| is a unit tangent vector to C.
        Thus F_{||} = \vec{F}.\vec{T}.

        C.  Also, \Delta x \approx x'(t) \Delta t, etc, so
        \Delta s \approx \sqrt{x'^2 + y'^2} \Delta t.
        Therefore, ds = |v| dt.

        D.  Therefore, F_{||} ds = \vec{F}.\vec{v} dt !

             W = \int_C F(r(t)) . dr(t)/dt dt

        E.  Writing out the components, we reduce this formula to

             \int [F_x dx/dt + F_y dy/dt] dt,

            which is what we got by parametrizing \int F_x dx + 
            F_y dy.  So the definitions/methods are equivalent.
            And either way, the evaluation usually boils down to 
            calculating this parametrized form for some 
            convenient t. 

    3.  The remarks from previous day, comparing and contrasting 
    the two methods, are still applicable.