Question

Which of the following forces is conservative?
A. $\overrightarrow F = y\widehat i - x\widehat j$
B. $\overrightarrow F = xy\widehat i - xy\widehat j$
C. $\overrightarrow F = y\widehat i + x\widehat j$
D. $\overrightarrow F = {x^2}y\widehat i - {y^2}x\widehat j$

Answer 1

In order to find a conservative force we have to find the partial derivative with respect to x and y. Then we have to make a matrix, where the partial derivative will form a column, Force’s x and y part another column and dimension the other column. The determinant of the option which comes out to be $0$ is the conservative force.

Formula used:
\left( {\begin{array}{*{20}{c}} {\widehat i}&{\widehat j}&{\widehat k} \\\ {\dfrac{{\partial F}}{{\partial x}}}&{\dfrac{{\partial F}}{{\partial y}}}&{\dfrac{{\partial F}}{{\partial z}}} \\\ {{F_x}}&{{F_y}}&{{F_z}} \end{array}} \right) is the matrix whose determinant is to be found. $\dfrac{{\partial F}}{{\partial x}}$ is the partial derivative of force $\overrightarrow F$ with respect to $x$,$\dfrac{{\partial F}}{{\partial y}}$ is the partial derivative of force $\overrightarrow F$ with respect to $y$and $\dfrac{{\partial F}}{{\partial z}}$ is the partial derivative of force $\overrightarrow F$ with respect to $z$. ${F_x}$ is the force in $x$-axis, ${F_y}$ is the force in $y$- axis and ${F_z}$ is the force in $z$- axis.

Complete step by step answer:
(A) $\overrightarrow F = y\widehat i - x\widehat j$ we get,
${F_x} = y$ and ${F_y} = - x$,
Now the partial derivative is,
$\dfrac{{\partial F}}{{\partial x}} = 0$ and $\dfrac{{\partial F}}{{\partial y}} = 0$
Now, by putting them in matrix we get,
\left( {\begin{array}{*{20}{c}} {\widehat i}&{\widehat j}&{\widehat k} \\\ {\dfrac{{\partial F}}{{\partial x}}}&{\dfrac{{\partial F}}{{\partial y}}}&{\dfrac{{\partial F}}{{\partial z}}} \\\ {{F_x}}&{{F_y}}&{{F_z}} \end{array}} \right)
Substituting the values we get,
\left( {\begin{array}{*{20}{c}} {\widehat i}&{\widehat j}&{\widehat k} \\\ 0&0&0 \\\ y&{ - x}&0 \end{array}} \right)
The determinant of the given matrix is $0$. Hence $\overrightarrow F = y\widehat i - x\widehat j$ is a conservative force.

(B) $\overrightarrow F = xy\widehat i - xy\widehat j$,
${F_x} = xy$ and ${F_y} = - xy$,
Now the partial derivative is,
$\dfrac{{\partial F}}{{\partial x}} = y$ and $\dfrac{{\partial F}}{{\partial y}} = - x$
Now, by putting them in matrix we get,
\left( {\begin{array}{*{20}{c}} {\widehat i}&{\widehat j}&{\widehat k} \\\ {\dfrac{{\partial F}}{{\partial x}}}&{\dfrac{{\partial F}}{{\partial y}}}&{\dfrac{{\partial F}}{{\partial z}}} \\\ {{F_x}}&{{F_y}}&{{F_z}} \end{array}} \right)
Substituting the values we get,
\left( {\begin{array}{*{20}{c}} {\widehat i}&{\widehat j}&{\widehat k} \\\ y&{ - x}&0 \\\ {xy}&{ - xy}&0 \end{array}} \right)
The determinant of the given matrix is $( - {y^2}x + {x^2}y)\widehat k$. Hence $\overrightarrow F = xy\widehat i - xy\widehat j$ is not a conservative force.

(C) $\overrightarrow F = y\widehat i + x\widehat j$ we get,
${F_x} = y$ and ${F_y} = x$,
Now the partial derivative is,
$\dfrac{{\partial F}}{{\partial x}} = 0$ and $\dfrac{{\partial F}}{{\partial y}} = 0$
Now, by putting them in matrix we get,
\left( {\begin{array}{*{20}{c}} {\widehat i}&{\widehat j}&{\widehat k} \\\ {\dfrac{{\partial F}}{{\partial x}}}&{\dfrac{{\partial F}}{{\partial y}}}&{\dfrac{{\partial F}}{{\partial z}}} \\\ {{F_x}}&{{F_y}}&{{F_z}} \end{array}} \right)
Substituting the values we get,
\left( {\begin{array}{*{20}{c}} {\widehat i}&{\widehat j}&{\widehat k} \\\ 0&0&0 \\\ y&x;&0 \end{array}} \right)
The determinant of the given matrix is $0$. Hence $\overrightarrow F = y\widehat i + x\widehat j$ is a conservative force.

(D) $\overrightarrow F = {x^2}y\widehat i - {y^2}x\widehat j$ we get,
${F_x} = {x^2}y$ and ${F_y} = - x{y^2}$,
Now the partial derivative is,
$\dfrac{{\partial F}}{{\partial x}} = 2xy$ and $\dfrac{{\partial F}}{{\partial y}} = - 2xy$
Now, by putting them in matrix we get,
\left( {\begin{array}{*{20}{c}} {\widehat i}&{\widehat j}&{\widehat k} \\\ {\dfrac{{\partial F}}{{\partial x}}}&{\dfrac{{\partial F}}{{\partial y}}}&{\dfrac{{\partial F}}{{\partial z}}} \\\ {{F_x}}&{{F_y}}&{{F_z}} \end{array}} \right)
Substituting the values we get,
\left( {\begin{array}{*{20}{c}} {\widehat i}&{\widehat j}&{\widehat k} \\\ {2xy}&{ - 2xy}&0 \\\ {{x^2}y}&{ - x{y^2}}&0 \end{array}} \right)
The determinant of the given matrix is $( - 2{x^2}{y^3} + 2{x^3}{y^2})\widehat k$. Hence $\overrightarrow F = {x^2}y\widehat i - {y^2}x\widehat j$ is not a conservative force.

Note: Alternative method: It is for only two-dimensional vector if $\dfrac{{d{F_x}}}{{dy}} = \dfrac{{d{F_y}}}{{dx}}$ then it is also conservative, where is the force in $x$-axis and ${F_y}$ in $y$-axis. It is necessary to partially derive the terms. It is very important to separate the whole Force into the terms of $x$,$y$ and $z$. Till the determinant of any vector is not found out to be $0$ then it is not conservative.