20190404

The differential of $\vec F$, written as a linear combination of partial derivatives, is $$ d\vec F = \begin{pmatrix}y^2\\3\\2yz\end{pmatrix}dx + \begin{pmatrix}2xy\\4\\2xz\end{pmatrix}dy + \begin{pmatrix}0\\5\\2xy\end{pmatrix}dz .$$ The above, when we write is as the product of a matrix and a vector, gives $$ d\vec F = \begin{bmatrix} \begin{matrix}y^2\\3\\2yz\end{matrix} &\begin{matrix}2xy\\4\\2xz\end{matrix} &\begin{matrix}0\\5\\2xy\end{matrix} \end{bmatrix} \begin{pmatrix}dx\\dy\\dz\end{pmatrix} .$$ The derivative of $\vec F$ is hence the matrix $$ D\vec F(x,y,z) = \begin{bmatrix} \begin{matrix}y^2\\3\\2yz\end{matrix} &\begin{matrix}2xy\\4\\2xz\end{matrix} &\begin{matrix}0\\5\\2xy\end{matrix} \end{bmatrix} . $$

We need $\vec F(x,y)=(f_x,f_y)$. This means we need a function $f(x,y)$ such that $f_x = 2x+3y$ and $f_y=3x$. Integrating each partial, with respect to the corresponding variable, gives $$\int f_x dx = \int2x+3ydx = x^2+3xy+C_1(y)\quad\text{and} \quad \int f_y dy = \int 3x dy=3xy+C_2(x).$$ Because we assumed $y$ is constant in the first integral, note that $C_1(y)$ could be any function of $y$. Similarly note that $C_2(x)$ could be any function of $x$. One option for the function $f(x,y)$ we seek is $$f(x,y) = x^2+3xy,$$ as this fits both

• $f(x,y)=x^2+3xy+C_1(y)$ with $C_1(y)=0$, and
• $f(x,y)=3xy+C_2(x)$ with $C_2(x) = x^2$.

Since $\vec F = \vec \nabla f$, we know that $\int_C\vec F\cdot d\vec r = \int_C df = f(B)-f(A)$, work done is the difference in potential. We compute

• $f(B) = f(-2,6)=(-2)^2+3(-2)(6) = -32$,
• $f(A) = f(1,2)=(1)^2+3(1)(2) = 7$, and
• $f(B)-f(A) = -32-7 = -39$.

The work done is -39 Nm.