Question

Find the work done by the force field \(\mathbf{F}\) on a particle moving along the given path. \(\mathbf{F}(x, y)=-y \mathbf{i}-x \mathbf{j}\) \(C:\) counterclockwise along the semicircle \(y=\sqrt{4-x^{2}}\) from (2,0) to (-2,0)

Short answer

The work done by the force field on a particle moving along the given path is \(4\pi\).

Step-by-step solution

Parameterize the Path

First, the path must be parameterized. The semicircle can be parameterized by \(x=2\cos(t)\) and \(y=2\sin(t)\) where \(t\) ranges from 0 to \(\pi\). This will move along the semicircle from (2,0) to (-2,0) counterclockwise.

Substituting Into the Force Field

Next, substitute these parameterizations of \(x\) and \(y\) into the force field \(\mathbf{F}(x, y)\). This provides \( \mathbf{F}(x(t), y(t))= -2\sin(t) \mathbf{i} - 2\cos(t) \mathbf{j}\).

Calculate the Differential of Arc Length

Now, compute the differential of arc length \(ds\) used in the line integral formula. As \(dx= -2\sin(t) dt\) and \( dy= 2\cos(t) dt \), the differential \(ds\) can be calculated through the Pythagorean theorem as \(ds = \sqrt{(dx/dt)^2+(dy/dt)^2} dt = dt\).

Compute the Line Integral

Now, compute the line integral \(\int \mathbf{F} \cdot d\mathbf{r}\) over the closed curve C by substituting and simplifying. This results in \(\int_0^\pi [-2\sin(t) \mathbf{i} - 2\cos(t) \mathbf{j}] \cdot [-2\sin(t), 2\cos(t)] dt = \int_0^\pi 4\sin^2(t) + 4\cos^2(t) dt. \) The simplified integral is \(\int_0^\pi 4 dt = 4t |_0^\pi = 4\pi.\)

Key concepts

Work Done by Force Field

When a particle moves through a force field, the concept of work done becomes very important. Work can be seen as the measure of energy transfer when a force moves an object. In a force field represented by a vector function \( \mathbf{F}(x,y)\), the work done by the force on a particle as it travels along a path 'C' is given by the line integral \( \int_C \mathbf{F} \cdot d\mathbf{r}\).

The dot product \( \mathbf{F} \cdot d\mathbf{r}\) indicates that we are considering the component of the force that acts along the direction of motion. The total work done is the sum of these products over the entire path. From the solution of our exercise, the work done by the force field as the particle moves in a counterclockwise direction along the semicircle is calculated to be \( 4\pi \), which implies the particle has been acted upon by the cumulative effect of the force field as it moved along the path.

Parameterization of a Path

Describing a path through which the particle moves involves parameterization. It's the process where we use one or more parameters to express the coordinates of a curve. By parameterizing the path, we create a set of equations that describe the curve in terms of a single variable, often 't'.

In our case, the semicircle is parameterized using trigonometric functions. The parameterization \( x=2\cos(t), y=2\sin(t)\) with \( t\) ranging from 0 to \( \pi \) ensures that we can describe the movement of a particle along the curved path piece by piece. Parameterization is fundamental in calculus as it allows us to convert problems in multiple dimensions into single-variable calculations that are often much easier to solve.

Differential of Arc Length

The differential of arc length, denoted as \(ds\), is a vital concept in understanding the distance a particle travels along a curved path. It's an infinitesimally small segment of the curve that can be used to approximate the curve's length when added together.

To calculate \(ds\), we usually use the derivatives of the parameterized functions \(dx/dt\) and \(dy/dt\) and combine them using the Pythagorean theorem. In the step-by-step solution provided, \(ds\) is obtained as \( \sqrt{(dx/dt)^2+(dy/dt)^2} dt\), which simply equals \(dt\). This relation is very handy; it transforms our integral computation into a straightforward process for finding the total arc length of the path the particle has traveled, thus enabling us to evaluate the line integral that represents the work done.