If c(t) = (x(t),y(t)) is a path in the plane and f(x,y)
is a height function defined over c, then the path
integral of f along c is given by ∫c
f(x,y) ds = ∫abf(x(t),y(t))s'(t)dt. where s'(t)
= √(x'(t)2 + y'(t)2).
If f(x,y) = 1 then the path integral is simply the arc length of c.
1. While path integrals
can be carried out in two-space and three-space, they could still be
considered to be part of single variable calculus. Why is this?
2. Find the arc length of the curve (2cos(t), sin(t)),
0 ≤ t ≤ 2π.
