Q. 45 - Calculus

Question

Determine whether or not each of the vector fields in Exercises 41–48 is conservative. If the vector field is conservative, find a potential function for the field.

$\mathbf{F}(x,y,z)=(y{e}^{2z}+1,x{e}^{2z},2xy{e}^{2z})$

Step-by-Step Solution

Answer

Since, $\frac{\partial F_3}{\partial y}=\frac{\partial F_2}{\partial z}$so the given vector is conservative.

A potential function for the field is $f(x,y,z)=xy{e}^{2z}+x$.

Step 1. Given Information

We have to determine whether or not each of the vector fields in the given exercises is conservative. If the vector field is conservative, find a potential function for the field.

$\mathbf{F}(x,y,z)=(y{e}^{2z}+1,x{e}^{2z},2xy{e}^{2z})$

Step 2. Firstly finding the given vector field is conservative or not.

A vector field $\mathbf{F}(x,y,z)=\left(F_1\right(x,y,z),F_2(x,y,z),F_3(x,y,z\left)\right)$ is conservative if and only if $\frac{\partial F_3}{\partial y}=\frac{\partial F_2}{\partial z},\frac{\partial F_1}{\partial z}=\frac{\partial F_3}{\partial x},\frac{\partial F_2}{\partial x}=\frac{\partial F_1}{\partial y}$

$$\begin{gathered} \frac{\partial F_3}{\partial y}=\frac{\partial }{\partial y}2xy{e}^{2z} \frac{\partial F_2}{\partial z}=\frac{\partial }{\partial z}x{e}^{2z} \\ \frac{\partial F_3}{\partial y}=2x{e}^{2z}\frac{\partial }{\partial y}y \frac{\partial F_2}{\partial z}=x\frac{\partial }{\partial z}{e}^{2z} \\ \frac{\partial F_3}{\partial y}=2x{e}^{2z} \frac{\partial F_2}{\partial z}=2x{e}^{2z} \end{gathered}$$

Hence, $\frac{\partial F_3}{\partial y}=\frac{\partial F_2}{\partial z}$

Step 3. Now finding the potential function for the field.

Since, $\mathbf{F}(x,y,z)=(y{e}^{2z}+1,x{e}^{2z},2xy{e}^{2z})$

$$\begin{gathered} f(x,y,z)=\int (y{e}^{2z}+1)\mathrm{d}x+B+C \\ f(x,y,z)=\int y{e}^{2z}\mathrm{d}x+\int 1\mathrm{d}x+B+C \\ f(x,y,z)=y{e}^{2z}·x+x+B+C \\ f(x,y,z)=xy{e}^{2z}+x+B+C \end{gathered}$$

where $\alpha$ is an arbitrary constant and B is the integral with respect to y of the terms in $F_2(x,y,z)$ in which the factor x does not appear.

Step 4. In this case, that is all of F 2 ( x , y , z ) , so

$$\begin{gathered} B=\int x{e}^{2z}dy \\ B=x{e}^{2z}·y+\beta \\ B=xy{e}^{2z}+\beta \end{gathered}$$

where β is an arbitrary constant.

$$\begin{gathered} C=\int 2xy{e}^{2z}dz \\ C=2xy\int e^{2z}dz \\ C=2xy{e}^{2z}+\gamma \end{gathered}$$

where $\gamma$ is an arbitrary constant.

Step 5. Setting the constants equal to zero since they do not affect the gradient of f ( x , y , z )

We have, 

$f(x,y,z)=xy{e}^{2z}+x$