If $\vec{F}$ is a path-independent field, then $\int _ { C } \vec { F } \cdot d \vec { r } = 0$ where C has the parameterization $\vec { r } ( t ) = \cos t \vec { i } + \sin t \vec { j } , 0

Question

If $\vec{F}$ is a path-independent field, then $\int _ { C } \vec { F } \cdot d \vec { r } = 0$ where C has the parameterization $\vec { r } ( t ) = \cos t \vec { i } + \sin t \vec { j } , 0 < t < 3 \pi$

Quizplus · Accepted Answer

The integral of a path-independent field over a closed loop is zero. However, the given parameterization describes a path that traces out a circle and a half, not a closed loop, since it starts at $t=0$ and ends at $t=3\pi$, making the path open.

If F⃗\vec{F}F Is a Path-Independent Field, Then ∫CF⃗⋅dr⃗=0\int _ { C } \vec { F } \cdot d \vec { r } = 0∫C​F⋅dr=0

If $\vec{F}$ Is a Path-Independent Field, Then $\int _ { C } \vec { F } \cdot d \vec { r } = 0$