determine whether or not the vector field is conservative. if it is conservative, find a function f such…

determine whether or not the vector field is conservative. if it is conservative, find a function f such that f = ∇f. (if the vector field is not conservative, enter dne.) f(x, y, z) = ⟨ln(8y), x/y + ln(5z), y/z⟩ f(x, y, z) =

determine whether or not the vector field is conservative. if it is conservative, find a function f such that f = ∇f. (if the vector field is not conservative, enter dne.) f(x, y, z) = ⟨ln(8y), x/y + ln(5z), y/z⟩ f(x, y, z) =

Answer

Explanation:

Step1: Recall the curl - test for conservativity

Un campo vectorial $\mathbf{F}(x,y,z)=\langle P(x,y,z),Q(x,y,z),R(x,y,z)\rangle$ es conservativo si $\nabla\times\mathbf{F}=\mathbf{0}$, donde $P = \ln(8y)$, $Q=\frac{x}{y}+\ln(5z)$ y $R=\frac{y}{z}$. El rotacional $\nabla\times\mathbf{F}=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\ln(8y)&\frac{x}{y}+\ln(5z)&\frac{y}{z}\end{vmatrix}$

Step2: Calcular las componentes del curl

La componente $i$ es: $\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}=\frac{1}{z}-\frac{1}{z}=0$ La componente $j$ es: $\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}=0 - 0=0$ La componente $k$ es: $\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=\frac{1}{y}-\frac{1}{y}=0$ Como $\nabla\times\mathbf{F}=\mathbf{0}$, el campo vectorial es conservativo.

Step3: Encontrar la función potencial $f$

Sabemos que $\frac{\partial f}{\partial x}=P=\ln(8y)$. Integrando con respecto a $x$: $f(x,y,z)=x\ln(8y)+g(y,z)$ Ahora, diferenciando $f$ con respecto a $y$: $\frac{\partial f}{\partial y}=\frac{x}{y}+\frac{\partial g}{\partial y}$ Como $\frac{\partial f}{\partial y}=Q=\frac{x}{y}+\ln(5z)$, entonces $\frac{\partial g}{\partial y}=\ln(5z)$ Integrando con respecto a $y$: $g(y,z)=y\ln(5z)+h(z)$ Así, $f(x,y,z)=x\ln(8y)+y\ln(5z)+h(z)$ Diferenciando $f$ con respecto a $z$: $\frac{\partial f}{\partial z}=\frac{y}{z}+h^{\prime}(z)$ Como $\frac{\partial f}{\partial z}=R = \frac{y}{z}$, entonces $h^{\prime}(z)=0$, y podemos tomar $h(z)=C$ (una constante). Tomando $C = 0$, la función potencial es $f(x,y,z)=x\ln(8y)+y\ln(5z)$

Answer:

$x\ln(8y)+y\ln(5z)$