6.3 Conservative vector fields  (Page 9/16)

We show that F does positive work on the particle by showing that F is conservative and then by using the Fundamental Theorem for Line Integrals.

To show that F is conservative, suppose $f\left(x,y\right)$ were a potential function for F . Then, $\text{∇}f=\text{F}=⟨2x{y}^2,2x^{2}y⟩$ and therefore $f_x=2x{y}^2$ and $f_y=2x^{2}y.$ Equation $f_x=2x{y}^2$ implies that $f\left(x,y\right)=x^2{y}^2+h\left(y\right).$ Deriving both sides with respect to y yields $f_y=2x^{2}y+h^{\prime }\left(y\right).$ Therefore, $h^{\prime }(y)=0$ and we can take $h\left(y\right)=0.$

If $f(x,y)=x^2{y}^2,$ then note that $\text{∇}f=⟨2x{y}^2,2x^{2}y⟩=\text{F},$ and therefore $f$ is a potential function for F .

Let $(a,b)$ be the point at which the particle stops is motion, and let C denote the curve that models the particle’s motion. The work done by F on the particle is ${\displaystyle {\int }_C\text{F}·d\text{r}}.$ By the Fundamental Theorem for Line Integrals,

Since $a\ne 0$ and $b\ne 0,$ by assumption, $a^2{b}^2>0.$ Therefore, ${\displaystyle {\int }_C\text{F}·d\text{r}}>0,$ and F does positive work on the particle.