If $\mathbf { F } ( x , y ) = \left( \frac { 1 } { x ^ { 2 } } + \frac { 1 } { y ^ { 2 } } \right) \mathbf { i } + \left( \frac { 1 - 2 x } { y ^ { 3 } } \right) \mathbf { j }$ is a conservative field, then its potential function $f ( x , y )$ is

A) $- \frac { 2 x ^ { 2 } - 2 y ^ { 2 } - x } { 2 x y ^ { 2 } } + C$
B) $\frac { 2 x ^ { 2 } + 2 y ^ { 2 } + x } { 2 x y ^ { 2 } } + C$
C) $\frac { 2 x ^ { 2 } - 2 y ^ { 2 } + x } { 2 x y ^ { 2 } } + C$
D) $\frac { 2 x ^ { 2 } + 2 y ^ { 2 } - x } { 2 x y ^ { 2 } } + C$
E) $\frac { 2 x ^ { 2 } - 2 y ^ { 2 } - x } { 2 x y ^ { 2 } } + C$

Question

If

\mathbf { F } ( x , y ) = \left( \frac { 1 } { x ^ { 2 } } + \frac { 1 } { y ^ { 2 } } \right) \mathbf { i } + \left( \frac { 1 - 2 x } { y ^ { 3 } } \right) \mathbf { j }

is a conservative field, then its potential function

f ( x , y )

is

A) $- \frac { 2 x ^ { 2 } - 2 y ^ { 2 } - x } { 2 x y ^ { 2 } } + C$
B) $\frac { 2 x ^ { 2 } + 2 y ^ { 2 } + x } { 2 x y ^ { 2 } } + C$
C) $\frac { 2 x ^ { 2 } - 2 y ^ { 2 } + x } { 2 x y ^ { 2 } } + C$
D) $\frac { 2 x ^ { 2 } + 2 y ^ { 2 } - x } { 2 x y ^ { 2 } } + C$
E) $\frac { 2 x ^ { 2 } - 2 y ^ { 2 } - x } { 2 x y ^ { 2 } } + C$

Quizplus · Accepted Answer

For a vector field $\mathbf{F}(x, y) = P(x, y)\mathbf{i} + Q(x, y)\mathbf{j}$ to be conservative, it must be the gradient of some scalar potential function $f(x, y)$, i.e., $\nabla f = \mathbf{F}$. This means $P = \frac{\partial f}{\partial x}$ and $Q = \frac{\partial f}{\partial y}$. Additionally, for a field to be conservative, a necessary condition is that the mixed partial derivatives of $f$ are equal, i.e., $\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x}$.Given $\mathbf{F}(x, y) = \left( \frac{1}{x^2} + \frac{1}{y^2} \right)\mathbf{i} + \left( \frac{1 - 2x}{y^3} \right)\mathbf{j}$, we integrate $P(x, y) = \frac{1}{x^2} + \frac{1}{y^2}$ with respect to $x$ and $Q(x, y) = \frac{1 - 2x}{y^3}$ with respect to $y$ to find $f(x, y)$.Integrating $P$ with respect to $x$, we get $-\frac{1}{x} + \frac{x}{y^2} + g(y)$, where $g(y)$ is an arbitrary function of $y$. Integrating $Q$ with respect to $y$, we should get the same $f(x, y)$ for the field to be conservative. Integrating $Q$ gives $\frac{1}{2y^2} - \frac{2x}{2y^2} + h(x)$, where $h(x)$ is an arbitrary function of $x$.Comparing both integrations and adjusting for constants and functions of single variables, we find that the potential function that matches the given options is $f(x, y) = \frac{2x^2 - 2y^2 - x}{2xy^2} + C$, which corresponds to option E.

If Is a Conservative Field, Then Its Potential Function f(x,y)f ( x , y )f(x,y)

If Is a Conservative Field, Then Its Potential Function $f ( x , y )$