$$
f(x, y)=P(x, y) \hat{\imath}+Q(x, y) \hat{\jmath} .
$$

Let $P(x, y)=4 y^{2}(x+1)$ and $Q(x, y)=4 x^{2}(y+1)$.
$$
\begin{array}{l}
P_{y}=\square \quad \underset{\sim}{\underline{\underline{x}}} \
Q_{x}=\square \
\end{array}
$$

Is $f$ a conservative vector field?
Choose 1 answer:
(A) Yes
(B) No

Question

$$
f(x, y)=P(x, y) \hat{\imath}+Q(x, y) \hat{\jmath} .
$$

Let $P(x, y)=4 y^{2}(x+1)$ and $Q(x, y)=4 x^{2}(y+1)$.
$$
\begin{array}{l}
P_{y}=\square \quad \underset{\sim}{\underline{\underline{x}}} \
Q_{x}=\square \
\end{array}
$$

Is $f$ a conservative vector field?
Choose 1 answer:
(A) Yes
(B) No

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find the partial derivative of $P$ with respect to $y$, we differentiate $4y^2(x+1)$ with respect to $y$. ‖ ‖ step 2 ⋮ Using the power rule for differentiation, the derivative of $4y^2$ with respect to $y$ is $8y$. ‖ ‖ step 3 ⋮ Therefore, the partial derivative of $P$ with respect to $y$ is $8y(x+1)$. ‖ ‖ step 4 ⋮ To find the partial derivative of $Q$ with respect to $x$, we differentiate $4x^2(y+1)$ with respect to $x$. ‖ ‖ step 5 ⋮ Using the power rule for differentiation, the derivative of $4x^2$ with respect to $x$ is $8x$. ‖ ‖ step 6 ⋮ Therefore, the partial derivative of $Q$ with respect to $x$ is $8x(y+1)$. ‖ ‖ step 7 ⋮ To determine if $f$ is a conservative vector field, we need to check if the mixed partial derivatives of $P$ and $Q$ are equal, that is, if $P_y = Q_x$. ‖ ‖ step 8 ⋮ Since $P_y = 8y(x+1)$ and $Q_x = 8x(y+1)$, they are not equal for all $(x, y)$, hence $f$ is not a conservative vector field. ‖ ∻[question 1] Answer∻ ***B*** ∻Key Concept∻ ⚹Conservative Vector Field⚹ ∻Explanation∻ ⚹A vector field is conservative if its curl is zero, which implies that the mixed partial derivatives of its components must be equal. In this case, since $P_y$ is not equal to $Q_x$, the vector field $f$ is not conservative.⚹◊What is the expression for the divergence of the gradient of a function in vector calculus?⍭ Generate me a similar question◊

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