Question

For Theorem 1, show that$\mathbf{\left(}\mathbf{d}\mathbf{\right)}\mathbf{\Rightarrow }\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{,}\mathbf{}\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{\Rightarrow }\mathbf{\left(}\mathbf{c}\mathbf{\right)}\mathbf{,}\mathbf{}\mathbf{\left(}\mathbf{c}\mathbf{\right)}\mathbf{\Rightarrow }\mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{,}\mathbf{}\mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{\Rightarrow }\mathbf{\left(}\mathbf{c}\mathbf{\right)}$ and $\left(c\right)\mathbf{\Rightarrow }\left(a\right)$

Short answer

1. The statement $\left(a\right)\Rightarrow \left(c\right)$has been shown.
2. The statements $\left(c\right)\Rightarrow \left(b\right)$and $\left(b\right)\Rightarrow \left(c\right)$has been shown.
3. The statement $\left(c\right)\Rightarrow \left(a\right)$ has been shown.

Step-by-step solution

Describe the given information

The theorem (1), describing the line integral of a vector along a closed path, shows that for a closed path $\left(d\right)\Rightarrow \left(a\right),\left(a\right)\Rightarrow \left(c\right),\left(c\right)\Rightarrow \left(b\right),\left(b\right)\Rightarrow \left(c\right)$and $\left(c\right)\Rightarrow \left(a\right)$.

Define the line integral

The line integral of a vector F along a route dl is defined as$\mathbf{\int }\mathit{F}\mathbf{·}\mathit{d}\mathit{l}$ . If the line integral is independent of the path, then the line integral of function F must be zero along that closed path.

Prove (a)⇒(c)

Let F be a gradient of a scalar function as $F=-\nabla v$ . Now find the curl of function F as,

$$\begin{gathered} \nabla \times F=\nabla \times \left[-\left(\nabla v\right)\right] \\ =-\nabla \times \left[\left(\nabla v\right)\right] \\ =0 \end{gathered}$$

The result is zero because the curl of gradient of a vector is always zero. Thus we can say that $d\Rightarrow a$.

According to strokes theorem,

$\oint F·dl=\int \left(\nabla \times F\right)·da$

Since the curl of vector function F is given as zero, that is $\nabla \times F=0$. So, above equation becomes,

$$\begin{gathered} \oint F·dl=\int 0·da \\ =0 \end{gathered}$$

Thus, we can say, $a\Rightarrow c$.

Prove (c)⇒(b) and  (b)⇒(c)

The theorem (1) says that the path dl from a to b is an independent path. As discussed, $F=-\nabla v$, then using strokes theorem , we can write the following line integral,

${\int }_a^{b}F·dl=-{\int }_a^b\left(\nabla v\right)·dl$

Thus, the path from a to b is independent according to the gradient theorem (1). Hence, from$a\Rightarrow b$.

As the line integral ${\int }_a^{b}F·dl$is independent of path from a to b , then for any closed loop, this line integral gives same value as the final and initial points in any closed loop coincide with each other. Thus, we can write, $b\Rightarrow c$.

Prove (c)⇒(a)

As discussed, if the line integral is path independent , then for any closed loop, the value of line integral $\oint F·dl$is always zero, that is,role="math" localid="1657521793564" $\oint F·dl=0$

According to strokes theorem,

$\oint F·dl=\int \left(\nabla \times F\right)·da$

Since the line integral of vector function F is zero, that is, $\oint F·dl=0$. So, above equation becomes,

$$\begin{gathered} 0=\int \left(\nabla \times F\right)·da \\ \nabla \times F=0 \end{gathered}$$

Thus, we can say, $c\Rightarrow a$.