Find the potential function f for the field F.
-$$\mathbf { F } = \left( \mathrm { e } ^ { 2 } - \sin \mathrm { x } \right) \mathbf { i } + 2 x y \mathrm { e } ^ { 2 } \mathbf { j } + \mathbf { k }$$
A) $f ( x , y , z ) = \cos x + x z y ^ { 2 } + C$
B) $f ( x , y , z ) = \cos x + e ^ { 2 } + z + C$
C) $f ( x , y , z ) = \cos x + x e ^ { 2 } + z + C$
D) $f ( x , y , z ) = \cos x + e ^ { 2 } ( x + z ) + C$

Question

Quizplus · Accepted Answer

We need to find a function whose gradient is the given field. Let's begin by finding the potential function $f$ for the $x$ and $z$ components of $\mathbf{F}$.
\begin{align*}
\frac{\partial f}{\partial x} &= \mathrm{e}^{2}-\sin x \
\frac{\partial f}{\partial y} &= 2xy\mathrm{e}^{2} \
\frac{\partial f}{\partial z} &= 1
\end{align*}
Integrating the first equation with respect to $x$ gives us $f(x,y,z) = -\cos x + x \mathrm{e}^{2} + g(y,z)$, where $g(y,z)$ is a constant of integration that may depend on $y$ and $z$. Differentiating this equation with respect to $y$ and equating to the second component of $\mathbf{F}$, we have
$$\frac{\partial f}{\partial y} = 2xy\mathrm{e}^{2} = e^{2}y + \frac{\partial g}{\partial y}$$
Therefore, $\frac{\partial g}{\partial y} = e^{2}(y-x)$, and integrating with respect to $y$ gives $ g(y,z) = \frac{1}{2}e^{2}(y^{2}-xy) + h(z)$, where $h(z)$ is a constant of integration. Substituting this into the previous equation for $f$, we get
$$f(x,y,z) = \cos x + x\mathrm{e}^{2} + \frac{1}{2}e^{2}(y^{2}-xy)+h(z)$$
Finally, differentiating this expression with respect to $z$ and equating to the third component of $\mathbf{F}$ gives us
$$\frac{\partial f}{\partial z} = 1 = \frac{\partial h}{\partial z}$$
Therefore, $h(z) = z + C$ for some constant $C$, and the potential function for $\mathbf{F}$ is
$$f(x,y,z) = \cos x + x\mathrm{e}^{2} + \frac{1}{2}e^{2}(y^{2}-xy)+z+C$$ 
Thus, the correct answer is $\boxed{\textbf{(C)}}$.

Find the Potential Function F for the Field F F=(e2−sin⁡x)i+2xye2j+k\mathbf { F } = \left( \mathrm { e } ^ { 2 } - \sin \mathrm { x } \right) \mathbf { i } + 2 x y \mathrm { e } ^ { 2 } \mathbf { j } + \mathbf { k }F=(e2−sinx)i+2xye2j+k

Find the Potential Function F for the Field F $\mathbf { F } = \left( \mathrm { e } ^ { 2 } - \sin \mathrm { x } \right) \mathbf { i } + 2 x y \mathrm { e } ^ { 2 } \mathbf { j } + \mathbf { k }$