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Do an internet search to find a reliable definition of a piecewise-smooth curve

ID: 3372272 • Letter: D

Question

Do an internet search to find a reliable definition of a piecewise-smooth curve and write out the "best" definition you find. Figure out on your own, or read through the pages you find to learn, why for a contour integral in a conservative vector field, a piecewise-smooth curve is "just as good" as a smooth curve in that you can evaluate the contour integrals in the same way. Write a description of why this is the case. Even more advanced treatments allow an even broader class of curves called rectifiable curves. Every smooth and piecewise-smooth curve is rectifiable but not every rectifiable curve is smooth or even piecewise-smooth. Do an internet search to find a reliable definition of a rectifiable curve and write out the "best" definition you find Read through the pages you find to learn why allowing rectifiable curves that are not piecewise- smooth does not introduce any new problems when considering contour integrals. Write out a description of why this is the case.

Explanation / Answer

1. A piecewise smooth curve is any curve that can be written as the union of a finite number of smooth curves, i.e. it is continuously differentiable except at a finite number of points, which are differentiable from the right or left (for a rigorous definition, refer to http://planetmath.org/piecewisesmooth).

2. A contour integral in a conservative vector field is independent of path, hence it follows from the fundamental theorem for line integrals that if a curve is piecewise-smooth, it is "just as good" as a smooth curve, since the integral holds for each of the pieces of the piecewise-smooth curve.

3. A rectifiable curve is a curve of finite length (for a rigorous definition, refer to http://www.encyclopediaofmath.org/index.php/Rectifiable_curve).

4. Every rectifiable curve can be approximated by a piecewise-smooth function. Moreover, a rectifiable curve must be continuous, and is differentiable almost everywhere, hence it does not introduce any new problems when considering contour integrals.

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