Showing a Force Field Is Not a Conservative Vector Field

1. Show that the force field \(\mathbf{F}=( x+y) \mathbf{i} +( y-x) \mathbf{j}\) is not a conservative vector field.
2. Show that the work \(W\) done by \(\mathbf{F}\) in moving an object is dependent on the path by using two paths starting at the origin \((0,0)\) and ending at the point \((1,1)\).

Solution (a) Let \(P(x,y) =x+y\) and \(Q(x,y) =y-x.\) Then \[ \dfrac{\partial P}{\partial y}=\dfrac{\partial }{\partial y}(x+y) =1\qquad\hbox{and}\qquad \dfrac{\partial Q}{\partial x}=\dfrac{\partial }{\partial x}(y-x) =-1 \]

Since \(\dfrac{\partial P}{\partial y}\neq \dfrac{\partial Q}{\partial x}\), the force field \(\mathbf{F}\) is not a conservative vector field.

(b) To show that the work \(W\) done by \(\mathbf{F}\) is dependent on the path, we choose two paths beginning at the origin \((0,0)\) and ending at the point \((1,1)\).

For Path 1, we let \(C\) be defined by \(y=x\), where \(0\leq x\leq 1.\) That is, \(x(t) =t,\) \(y(t) =t,\) \(0\leq t\leq 1.\) Then \(\mathbf{F}( x(t),y(t)) =2t\,\mathbf{i}\), \(\mathbf{r}(t) =t\mathbf{i}+t\mathbf{j}\), and \(d\mathbf{r=\,}\left( \mathbf{i}+\mathbf{j}\right) dt.\) So, \[\begin{equation*} \mathbf{F}\,{\cdot}\, d\mathbf{r}=2t\,dt \end{equation*}\]

The work \(W\) done by \(\mathbf{F}\) along \(C\) is \[ W=\int_{C}\mathbf{F}\,{\cdot}\, d\mathbf{r}=\int_{0}^{1}2t\,dt=\big[ t^{2}\big] _{0}^{1}=1 \]

For Path 2, we let \(C\) be the piecewise-smooth curve made up of:

\(C_{1},\) which is traced out by \(x(t) =t,\) \(y(t) =0,\) \(0\leq t\leq 1,\) and

\(C_{2},\) which is traced out by \(x(t) =1,\) \(y(t) =t,\) \(0\leq t\leq 1\).

For \(C_{1},\) \(\mathbf{F}( x(t),y(t)) =t\,\mathbf{i}-t\mathbf{j}\), \(\mathbf{r}(t) =t\mathbf{i}\), and \(d\mathbf{r}=\mathbf{i}\,dt\). So, \[ \mathbf{F}\,{\cdot}\, d\mathbf{r}=t\,dt \]

For \(C_{2},\) \(\mathbf{F}( x(t),y(t)) =( 1+t) \,\mathbf{i}+( t-1) \mathbf{j}\), \(\mathbf{r}(t) =\mathbf{i}+t \mathbf{j}\), and \(d\mathbf{r}=\mathbf{j}\,dt\). So, \[\begin{equation*} \mathbf{F}\,{\cdot}\, d\mathbf{r}=( t-1)\, dt \end{equation*}\]

The work \(W\) done by \(\mathbf{F}\) along Path 2 is \[\begin{equation*} W=\int_{C_{1}}\mathbf{F}\,{\cdot}\, d\mathbf{r}\,+\int_{C_{2}}\mathbf{F}\,{\cdot}\, d \mathbf{r}=\int_{0}^{1}t\,dt +\int_{0}^{1}( t-1)\, dt=0 \end{equation*}\]

Since the work done along Path 1 and Path 2 are not equal, the work \(W\) done by the force \(\mathbf{F}\) is dependent on the path.