To determine if the vector field $F(x, y) = (16x - y) \, \mathbf{i} + (-x + 2y) \, \mathbf{j}$ is conservative, we can examine the partial derivatives to see if they meet the criteria for a conservative vector field in two dimensions.

A vector field $F(x, y) = P(x, y) \, \mathbf{i} + Q(x, y) \, \mathbf{j}$ is conservative if: $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$

Step 1: Define $P$ and $Q$

From $F(x, y) = (16x - y) \, \mathbf{i} + (-x + 2y) \, \mathbf{j}$:

• $P(x, y) = 16x - y$
• $Q(x, y) = -x + 2y$

Step 2: Compute the Partial Derivatives

1. $\frac{\partial P}{\partial y} = \frac{\partial}{\partial y} (16x - y) = -1$
2. $\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x} (-x + 2y) = -1$

Since $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$, the field is conservative.

Step 3: Find the Potential Function $f(x, y)$

To find $f(x, y)$ such that $\nabla f = F$, we integrate $P(x, y)$ with respect to $x$ and $Q(x, y)$ with respect to $y$.

1. Integrate $P(x, y) = 16x - y$ with respect to $x$: $f(x, y) = \int (16x - y) \, dx = 8x^2 - yx + g(y)$ where $g(y)$ is an arbitrary function of $y$.

2. Differentiate $f(x, y) = 8x^2 - yx + g(y)$ with respect to $y$: $\frac{\partial f}{\partial y} = -x + g'(y)$

3. Since $\frac{\partial f}{\partial y}$ should equal $Q(x, y) = -x + 2y$, we set $g'(y) = 2y$, so $g(y) = y^2$.

Thus, the potential function $f(x, y)$ is: $f(x, y) = 8x^2 - yx + y^2$

Final Answer

$f(x, y) = 8x^2 - yx + y^2$

Would you like further explanation on any part of this solution?

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1. How do you verify if a vector field is conservative?
2. What is the role of partial derivatives in determining conservativeness?
3. Can conservative vector fields exist in three dimensions, and how does the test change?
4. What is a potential function in the context of vector fields?
5. How is the potential function related to the line integral of a vector field?

Tip: For a vector field to be conservative, it must have a curl of zero in two dimensions (i.e., partial derivatives must match).