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Magazine Chapter 6: Problem 17
Which, if either, of the two force fields $$ \mathbf{F}_{3}=-\boldsymbol{y} \mathbf{i}+x \mathbf{j}+z \mathbf{k}, \quad \mathbf{F}_{2}=y \mathbf{i}+x \mathbf{j}+z \mathbf{k} $$ is conservative? Calculate for each field the work done in moving a particle around the circle \(x=\cos t, y=\sin t\) in the \((x, y)\) plane.
Short Answer
Expert verified
The force field \(\textbf{F}_{2}\) is conservative. The work done by \(\textbf{FI_{1}}\) and \(\textbf{FI_{2}}\) around the circle is zero.
Step by step solution
01
Determine if \(\textbf{F}_{1}\) is conservative
A force field \(\textbf{F}\) is conservative if its curl is zero. Compute the curl of \(\textbf{F}_{1} = -y \textbf{i} + x \textbf{j} + z \textbf{k}\). The curl is given by \(abla \times \textbf{F}_{1}\): \[ abla \times \textbf{F}_{1} = \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \ \frac{\text{d}}{\text{d}x} & \frac{\text{d}}{\text{d}y} & \frac{\text{d}}{\text{d}z} \ -y & x & z \ \end{vmatrix} = abla \times [-y \textbf{i}, x \textbf{j}, z \textbf{k}] \ = \bigg( \frac{\text{d}z}{\text{d}y} - \frac{\text{d}x}{\text{d}z} \bigg) \textbf{i} - \bigg( \frac{\text{d}z}{\text{d}x} - \frac{\text{d}(-y)}{\text{d}z} \bigg) \textbf{j} + \bigg( \frac{\text{d}(x)}{\text{d}x} - \frac{\text{d}(-y)}{\text{d}y} \bigg) \textbf{k} = (0 - 0) \textbf{i} - (0 - 0) \textbf{j} + (1 - (-1)) \textbf{k} = 2\textbf{k} \] Since \(abla \times \textbf{F}_{1} eq 0\), the field \(\textbf{F}_{1}\) is not conservative.
02
Determine if \(\textbf{F}_{2}\) is conservative
Now compute the curl for \(\textbf{F}_{2} = y \textbf{i} + x \textbf{j} + z \textbf{k}\). \[ abla \times \textbf{F}_{2} = \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \ \frac{\text{d}}{\text{d}x} & \frac{\text{d}}{\text{d}y} & \frac{\text{d}}{\text{d}z} \ y & x & z \ \end{vmatrix} = abla \times [y \textbf{i}, x \textbf{j}, z \textbf{k}] \ = \bigg( \frac{\text{d}z}{\text{d}y} - \frac{\text{d}x}{\text{d}z} \bigg) \textbf{i} - \bigg( \frac{\text{d}z}{\text{d}x} - \frac{\text{d}y}{\text{d}z} \bigg) \textbf{j} + \bigg( \frac{\text{d}x}{\text{d}x} - \frac{\text{d}y}{\text{d}y} \bigg) \textbf{k} = (0 - 0) \textbf{i} - (0 - 0) \textbf{j} + (1 - 1) \textbf{k} = 0 \] Since \(abla \times \textbf{F}_{2} = 0\), the field \(\textbf{F}_{2}\) is conservative.
03
Calculate the work done by \(\textbf{F}_{1}\)
Using the given path \(x = \text{cos}t\), \(y = \text{sin}t\), parametrize \(x, y, z \) over one period \(t = 0 \) to \(2 \pi\). With \(\text{d}{t}\), the force field \(\textbf{F}_{1}\), we have: \[ \textbf{r}'(t) = \frac{\text{d}}{\text{d}t} (x \textbf{i} + y \textbf{j}) = - \text{sin}t \textbf{i} + \text{cos}t \textbf{j}, \ \textbf{F}_{1}(x(t), y(t), z(t)) = - \text{sin}t \textbf{i} + \text{cos}t \textbf{j}. \ \text{Work} = \text{Line Integral}\bigg[\textbf{F}_{1}\bigg] = \bigg(- \text{sin}t \textbf{i} + \text{cos}t \textbf{j} \bigg) \bigcdot \bigg(- \text{sin}t \textbf{i} + \text{cos}t \textbf{j} \bigg) \text{d}t \ = \bigg[- \text{sin}^2 t + \text{cos}^2 t \bigg] \text{d}t \ = \bigg[ \text{cos}(2t) \bigg] \text{d}t, \ \text{Evaluating:} \ \bigg[ \text{cos}(2t) \bigg] = 0 \] The work done by \(\textbf{F}_{1}\) is zero.
04
Calculate the work done by \(\textbf{F}_{2}\)
Using the same path \(x = \text{cos}t\), \(y = \text{sin}t\), parametrize \(x, y, z\) over one period \(t = 0 \) to \(2 \pi\). With \(\text{d}{t}\), the force field \(\textbf{F}_{2}\), we have: \[ \textbf{r}'(t) = \frac{\text{d}}{\text{d}t} (x \textbf{i} + y \textbf{j}) = - \text{sin}t \textbf{i} + \text{cos}t \textbf{j}, \ \textbf{F}_{2}(x(t), y(t), z(t)) = \text{sin}t \textbf{i} + \text{cos}t \textbf{j}. \ \text{Work} = \text{Line Integral}\bigg[\textbf{F}_{2}\bigg] = \bigg[ \text{sin}t \textbf{i} + \text{cos}t \textbf{j} \bigg] \bigcdot \bigg[- \text{sin}t \textbf{i} + \text{cos}t \textbf{j} \bigg] \text{d}t \ = - \text{sin}^2 t + \text{cos}^2 t \text{d}{t} \ = \bigg[ \text{-cos}(2t) \text{d}t \bigg] = 0 \] The work done by \(\textbf{F}_{2}\) is also zero nodes.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Curl of a Vector Field
The curl of a vector field is an important concept in vector calculus. It measures the rotation of the vector field at a given point. The formula for the curl of a vector field \(\textbf{F} = (F_x, F_y, F_z)\) is given by: \[\abla \times \textbf{F} = \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \ \partial_x & \partial_y & \partial_z \ F_x & F_y & F_z \end{vmatrix}.\] This determinant expands to: \[\abla \times \textbf{F} = (\partial_y F_z - \partial_z F_y) \textbf{i} - (\partial_x F_z - \partial_z F_x) \textbf{j} + (\partial_x F_y - \partial_y F_x) \textbf{k}.\] If the curl of a vector field is zero, the field is conservative. This means the field has no rotation and can be expressed as the gradient of a scalar potential function.
Line Integrals
Line integrals are used to compute the integral of a function along a curve. In the context of force fields and work, they help calculate the total work done by a force along a path. For a vector field \(\textbf{F} = (F_x, F_y, F_z)\) and a parameterized path \(\textbf{r}(t) = (x(t), y(t), z(t))\), the line integral is computed as: \[\int_{a}^{b}\textbf{F} \cdot \textbf{r}'(t) \mathrm{d}t.\] Here, \(\textbf{F}\) is evaluated along the path, and \(\textbf{r}'(t)\) is the derivative of the position vector function with respect to \(t\). This integral gives the total work done by the force field along the curve.
Work in Physics
In physics, work is the energy transferred by a force acting over a distance. When dealing with force fields, the work done by the force as a particle moves is calculated using line integrals. If a force field is conservative, the work done over any closed path is zero. This is because in conservative fields, the potential energy depends only on the positions and not on the path taken between them.
Parametrization
Parametrization is the process of describing a curve using a single parameter, usually denoted as \(t\). For example, a circle in the \(xy\)-plane can be parameterized as \(x(t) = \cos(t), y(t) = \sin(t)\). Parametrization simplifies the calculation of integrals over curves, particularly in line integrals for work and flux calculations. By expressing coordinates as functions of \(t\), we can seamlessly compute derivatives and evaluate the integrals.
Mathematical Methods in Physics
Mathematical methods are fundamental in physics for formulating and solving problems. Techniques like calculating curls and line integrals help us understand the behavior of physical systems. Using methods from vector calculus, we can analyze force fields, determine conservativeness, and compute work done by these fields. Understanding and employing these mathematical tools are essential for solving complex problems in physics and other scientific fields.
