Question

Math

Posted 3 months ago

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

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Answer from Sia

Posted 3 months ago

Solution by Steps

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.

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