91影视

The force field is$F=y\sin 2xi+{\sin }^{2}xjandF=-鈭�蠄$.

A force is said to be conservative if $\mathbf{鈭�}\mathbf{脳}\mathbf{F}\mathbf{=}\mathbf{0}\mathbf{.}$

The formula for the scalar potential is $\mathbf{w}\mathbf{=}\mathbf{鈭�}\mathbf{F}\mathbf{.}\mathbf{dr}$.

The force is said to be conservative if$鈭�脳F=0$. 鈥︹��. (1)

Put equation (2) in equation (1).

$F=y\sin 2xi+{\sin }^{2}xj$ 鈥︹��. (2)

The equation becomes as follows.

$$\begin{gathered} 鈭�脳F=\left(\begin{array}{ccc}i& j& k\\ \frac{鈭�F}{鈭�x}& \frac{鈭�F}{鈭�y}& \frac{鈭�F}{鈭�z}\\ y\sin 2x& {\sin }^{2}x& 0\end{array}\right) \\ 鈭�脳F=\left(\frac{鈭�}{鈭�y}脳0-\frac{鈭�}{鈭�z}\left({\sin }^{2}x\right)\right)i-\left(\frac{鈭�}{鈭�x}\left(0\right)-\frac{鈭�}{鈭�z}\left(y\sin 2x\right)\right)j+\left(\frac{鈭�}{鈭�x}\left({\sin }^{2}x\right)-\frac{鈭�}{鈭�y}\left(y\sin 2x\right)\right)k \\ 鈭�脳F=0+\left(2\sin x\cos x-\sin 2xk\right) \\ 鈭�脳F=0 \end{gathered}$$

Hence the field is conservative

The formula for the scalar potential is $W=鈭�F.dr$.

$$\begin{gathered} W=鈭�\left(-y\sin 2xi+{\sin }^{2}xj\right).\left(dxi+dyj+dzk\right) \\ W=鈭�\left(y-\sin 2x\right)dx+{\sin }^{2}xdy \end{gathered}$$

$W_{1}Isfrom\left(0,0,0\right)to\left(x,0\right).$

$$\begin{gathered} y=0 \\ dy=0 \end{gathered}$$

Substitute the above value in the equation mentioned below.

$$\begin{gathered} w=鈭�\left(-y\sin 2x\right)dx+{\sin }^{2}xdy \\ {w}_1={鈭�}_0^{x}0dx \\ {w}_1=0 \end{gathered}$$

$w_{2}Isfrom\left(x,0\right)to\left(x,y\right)$

x is constant.

dx=0

Substitute the above value in the equation mentioned below.

$$\begin{gathered} w=鈭�\left(y-\sin 2x\right)dx+{\sin }^{2}xdy \\ {w}_2={鈭�}_0^y{\sin }^{2}xdy \\ {w}_2={\left[y{\sin }^{2}x\right]}_0^y \\ {w}_2=y{\sin }^{2}x \end{gathered}$$

The formula states the equation mentioned below.

$$\begin{gathered} w=w_1+w_2 \\ w=0+y{\sin }^{2}x \\ w=y{\sin }^{2}x \end{gathered}$$

The formula states the equation mentioned below.

$F=鈭�W$

It is given that$F=-鈭�蠁$ .

By both the values of$F,鈭�蠁=鈭�W$.

$蠁=-W$

Put the value of W in above equation.

Hence the scalar potential is $-y{\sin }^{2}x$.