Compute the work done in taking a particle around the circle \(x^{2}+y^{2}=\) \(a^{2}, z=0\) if the force is (a) \(\boldsymbol{F}=y \boldsymbol{i}\), and (b) \(\boldsymbol{F}=x \boldsymbol{i}\). What do you conclude about these forces? (Use the parametrization \(x=a \cos \varphi, y=\) \(a \sin \varphi, z=0 .)\)

In physics, the work done by a force field on an object represents the energy transferred to or from that object when it moves through the force field. The work is calculated as a line integral of the force along the path of motion. Mathematically, for a force field represented by the vector function \boldsymbol{F} and the path \boldsymbol{r}(t)\text{, the work done is expressed as:}

\[ W = \biggl{\int\boldsymbol{F}(\boldsymbol{r}(t)) \biggr{\cdot} d\boldsymbol{r}(t) \biggl{\int\boldsymbol{F}(\boldsymbol{r}(t)) \biggr{\cdot} d\boldsymbol{r}(t)} \]
When a particle moves along a closed path in a conservative force field, the total work done by the force is zero. This principle is evident in the textbook example, where the work done in moving a particle around a circular path under specific force fields resulted in zero work. This concept applies to both non-changing and variable force fields, as long as they are conservative.

A conservative force is a type of force with an important property: the work done by a conservative force on an object moving from one point to another is independent of the path taken. One hallmark is that the work done by a conservative force over any closed loop is always zero. Forces such as gravity and electrostatic forces are classic examples.

This property has profound consequences in physics, especially in energy conservation, as it implies that mechanical energy (kinetic plus potential) is conserved. When the textbook problem showed that the work done by both force fields over a circular loop is zero, it concluded that these forces are conservative. In practical terms, this means if you displaced an object in a gravitational field and returned it to its original position, the total work done by gravity would be zero—no net energy would be lost or gained.

The technique of parametrization involves representing a curve as a vector function of a single parameter. This is essential in various fields in physics and mathematics, particularly when calculating line integrals.

In the example from the textbook, the circular path is parametrized using the angle \(\varphi\) as the parameter, leading to the equations \(x(\varphi) = a\cos(\varphi)\) and \(y(\varphi) = a\sin(\varphi)\) with \(z(\varphi) = 0\). The derivatives of these equations with respect to \(\varphi\) are then used to find the differential elements along the curve, such as \(dx, dy,\) and \(dz\). Parametrization converts complex motions into simple mathematical expressions which can be analyzed or integrated to find quantities such as work done by a force field.

The concept of circular motion plays a pivotal role in classical mechanics, highlighting how forces act on bodies moving in circles. In this kind of motion, any point on the object moves in a circular path in a plane, and at any given time, the entirety of the force is directed towards the center of the circle (the centripetal force).

Circular motion is thoroughly analyzed using parametric equations, where the coordinates of a moving particle are described in terms of a single parameter—usually time or angle. In the exercise, circular motion was key in evaluating the work done by force fields when an object travels along a circular path—a fundamental example of line integrals applied in classical mechanics. Understanding circular motion is fundamental in fields like orbital mechanics, certain aspects of electromagnetism, and various engineering applications.