example: find the conservative vector field for the potential function given by: f(x,y,z)=sqrt(x^2 +…

Explanation:

Step1: Recall the formula

If $f(x,y,z)$ is a potential - function, the conservative vector field $\vec{F}$ is given by $\vec{F}=\nabla f=\left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}\right)$.

Step2: Calculate $\frac{\partial f}{\partial x}$

Let $u = x^{2}+4y^{2}+z^{2}$, then $f=\sqrt{u}=u^{\frac{1}{2}}$. Using the chain - rule $\frac{\partial f}{\partial x}=\frac{\partial f}{\partial u}\cdot\frac{\partial u}{\partial x}$. We have $\frac{\partial f}{\partial u}=\frac{1}{2}u^{-\frac{1}{2}}$ and $\frac{\partial u}{\partial x}=2x$. So, $\frac{\partial f}{\partial x}=\frac{x}{\sqrt{x^{2}+4y^{2}+z^{2}}}$.

Step3: Calculate $\frac{\partial f}{\partial y}$

Using the chain - rule again, $\frac{\partial f}{\partial y}=\frac{\partial f}{\partial u}\cdot\frac{\partial u}{\partial y}$. Since $\frac{\partial u}{\partial y}=8y$, then $\frac{\partial f}{\partial y}=\frac{4y}{\sqrt{x^{2}+4y^{2}+z^{2}}}$.

Step4: Calculate $\frac{\partial f}{\partial z}$

Using the chain - rule, $\frac{\partial f}{\partial z}=\frac{\partial f}{\partial u}\cdot\frac{\partial u}{\partial z}$. Since $\frac{\partial u}{\partial z}=2z$, then $\frac{\partial f}{\partial z}=\frac{z}{\sqrt{x^{2}+4y^{2}+z^{2}}}$.

Answer:

$\vec{F}=\left(\frac{x}{\sqrt{x^{2}+4y^{2}+z^{2}}},\frac{4y}{\sqrt{x^{2}+4y^{2}+z^{2}}},\frac{z}{\sqrt{x^{2}+4y^{2}+z^{2}}}\right)$