$$
f(x, y)=\left(\frac{1}{2 \sqrt{x}}, \frac{1}{2 \sqrt{y}}\right)
$$

Find $F$ such that $f=\nabla F$.
$$
F(x, y)=\square+\bar{*}+C
$$

Question

$$
f(x, y)=\left(\frac{1}{2 \sqrt{x}}, \frac{1}{2 \sqrt{y}}\right)
$$

Find $F$ such that $f=\nabla F$.
$$
F(x, y)=\square+\bar{*}+C
$$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find a function $F(x, y)$ such that $f = abla F$, we need to integrate the components of $f(x, y)$ with respect to their respective variables. ‖ ‖ step 2 ⋮ Integrate the first component of $f(x, y)$ with respect to $x$: $\int \frac{1}{2\sqrt{x}} \, dx$. ‖ ‖ step 3 ⋮ The integral of $\frac{1}{2\sqrt{x}}$ with respect to $x$ is $\sqrt{x}$. ‖ ‖ step 4 ⋮ Integrate the second component of $f(x, y)$ with respect to $y$: $\int \frac{1}{2\sqrt{y}} \, dy$. ‖ ‖ step 5 ⋮ The integral of $\frac{1}{2\sqrt{y}}$ with respect to $y$ is $\sqrt{y}$. ‖ ‖ step 6 ⋮ Combine the results of the integrals and add the constant of integration $C$: $F(x, y) = \sqrt{x} + \sqrt{y} + C$. ‖ ∻[question number] Answer∻ ⚹ $F(x, y) = \sqrt{x} + \sqrt{y} + C$ ⚹ ∻Key Concept∻ ⚹Finding a potential function for a vector field⚹ ∻Explanation∻ ⚹To find a potential function $F(x, y)$ for a vector field $f(x, y)$, we integrate the components of the vector field with respect to their respective variables and combine them, including a constant of integration $C$.⚹◊What is the relationship between the vector field $f(x, y)$ and the function $F(x, y)$ in terms of their gradients?⍭ Generate me a similar question◊

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