Answer
(a)True
(b)True
(c)True
Work Step by Step
Determine whether the following statements about the vector field
\[
\mathbf{F}(x, y) = x^2 \mathbf{i} - y \mathbf{j}
\]
are true or false. If false, explain why.
Solution
(a) \(\|\mathbf{F}(x, y)\| \to 0\) as \((x, y) \to (0, 0)\)
Compute the magnitude:
\[
\|\mathbf{F}(x, y)\| = \sqrt{(x^2)^2 + (-y)^2} = \sqrt{x^4 + y^2}.
\]
As \((x, y) \to (0, 0)\), we have \(x^4 \to 0\) and \(y^2 \to 0\), hence
\[
\|\mathbf{F}(x, y)\| \to 0.
\]
\[
\boxed{\text{True.}}
\]
---
(b) On the positive \(y\)-axis, the vector points in the negative \(y\)-direction.}
For points on the positive \(y\)-axis, \(x = 0\) and \(y > 0\). Then
\[
\mathbf{F}(0, y) = 0^2 \mathbf{i} - y \mathbf{j} = -y \mathbf{j}.
\]
Since \(y > 0\), the vector \(-y \mathbf{j}\) points in the negative \(y\)-direction.
\[
\boxed{\text{True.}}
\]
---
(c) In the first quadrant, the vector points down and to the right.
In the first quadrant, \(x > 0\) and \(y > 0\). Then
\[
\mathbf{F}(x, y) = x^2 \mathbf{i} - y \mathbf{j}.
\]
The \(x\)-component (\(x^2\)) is positive, so the vector points to the right.
The \(y\)-component (\(-y\)) is negative, so the vector points downward.
\[
\boxed{\text{True.}}
\]
---
Summary
\[
\begin{array}{c|c|c|l}
\text{Part} & \text{Statement} & \text{True/False} & \text{Explanation} \\ \hline
(a) & \|\mathbf{F}\| \to 0 \text{ as } (x, y) \to (0, 0) & \text{True} & \sqrt{x^4 + y^2} \to 0 \\[4pt]
(b) & \text{On positive $y$-axis, vector points in negative $y$-direction} & \text{True} & \mathbf{F}(0, y) = -y\mathbf{j} \\[4pt]
(c) & \text{In first quadrant, vector points down and right} & \text{True} & x^2 > 0, -y < 0
\end{array}
\]
