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We take a path in the P that can be divided into a finite number of pieces
each of which resembles a straight line at small distances, and choose a differentiable
scalar field f defined on and around it. The difference in values of f between
the front and back of a small line like segment of P having unit tangent vector
T and length dl starting at the point (x, y, z) and ending at (x + (Ti)dl,
y + (Tj)dl,
z + (Tk)dl)
will be the directional derivative of f in the direction of T multiplied
by dl; which is dl times the component of the gradient of f in the direction of
T: namely dl(Tf). We have then that f(r + Tdl) - f(r) = dl(Tf). The sum of the change in f over each segment of the path that starts at a and ends at b will be the change in f over the entire path or f(b) - f(a). If we sum the previous statement over the entire path P, we get This means that the integral of the component of the gradient of f along the
path will give the change in f or difference between its values at the front and
back ends of P. Exercise 21.5
Find the integral of over
a path from (0, 0, 0) to (1, 2, 3) that goes from the origin up the x axis to
x = 1, then parallels the y axis to y =2 and then parallel to the z axis to z
= 3. |