Answer
See proof
Work Step by Step
From the given figure, the vector field is constant. Consider \[ \vec{F} = a\vec{i} + b\vec{j} \] Then, \[ \frac{\partial \vec{r}}{\partial x} = 0 = \frac{\partial \vec{r}}{\partial y} \quad \frac{\partial f}{\partial x} = 0 = \frac{\partial g}{\partial y} \] Hence, $\vec{F}$ is conservative, and \[ \frac{\partial \vec{r}}{\partial x} = a, \quad \frac{\partial \vec{r}}{\partial y} = b \] Then, \[ \vec{r} = a\vec{i} + by\vec{j} \] and \[ \frac{\partial \vec{r}}{\partial x} = a, \quad \frac{\partial \vec{r}}{\partial y} = b \] Hence, \[ \vec{F} = a\vec{i} + b\vec{j} = \nabla \phi = \frac{\partial \phi}{\partial x}\vec{i} + \frac{\partial \phi}{\partial y}\vec{j} \] Then, \[ \phi = ax + by + C \] Result $\vec{F}$ is conservative, and \[ \vec{F} = a\vec{i} + b\vec{j} = \nabla \phi = \frac{\partial \phi}{\partial x}\vec{i} + \frac{\partial \phi}{\partial y}\vec{j} \] Then, \[ \phi = ax + by + C \]
