Answer
(a) true only for conservative vector fields
(b) true for all vector fields
(c) true only for conservative vector fields
(d) true for all vector fields
(e) true only for conservative vector fields
(f) true for all vector fields
(g) true only for conservative vector fields
Work Step by Step
(a) By Theorem 1, ${\bf{F}}$ is path-independent if it is conservative vector field. Therefore, the statement is true only for conservative vector fields.
(b) The value of the line integral over an oriented curve C is defined without reference to a parametrization, therefore it does not depend on how C is parametrized. Thus, this statement is true for all vector fields.
(c) By Theorem 1, the statement is true only for conservative vector fields.
(d) By Theorem 3 of Section 17.2, reversing the orientation changes the sign of the line integral. This applies to all vector fields.
(e) By Theorem 1, the statement is true only for conservative vector fields.
(f) By definition of the Vector Line Integral in Section 17.2: the line integral of a vector field ${\bf{F}}$ along an oriented curve $C$ is the integral of the tangential component of ${\bf{F}}$ given by
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_C^{} \left( {{\bf{F}}\cdot{\bf{T}}} \right){\rm{d}}s$
This applies to all vector fields.
(g) Every conservative vector field must satisfy the equality of the cross partials of the components:
$\dfrac{{\partial {F_1}}}{{\partial y}} = \dfrac{{\partial {F_2}}}{{\partial x}}$, $\ \ \ \ $ $\dfrac{{\partial {F_2}}}{{\partial z}} = \dfrac{{\partial {F_3}}}{{\partial y}}$, $\ \ \ \ $ $\dfrac{{\partial {F_3}}}{{\partial x}} = \dfrac{{\partial {F_1}}}{{\partial z}}$
Therefore, the statement is true only for conservative vector fields.
