📝 Graphical Representations of Vector Fields (14 MCQs)

📖 Explanation: A vector-field plot represents a vector at each point through an arrow whose direction indicates orientation and whose length usually indicates magnitude. Comparing arrows at different locations reveals how the field varies spatially. Color or arrow count may provide additional information, but neither alone communicates the vector field as completely as direction and length.

📖 Explanation: Arrow direction represents the direction of the vector, while arrow length commonly represents its magnitude. If corresponding arrows point in the same directions but one set is longer, the fields differ in magnitude. This conclusion does not require calculating components, divergence, or curl, although a plotting scale should be considered.

📖 Explanation: For $\mathbf{F}(x,y)=(x,y)$, the vector at each point points directly away from the origin, and its magnitude is $\sqrt{x^2+y^2}$, which increases with distance from the origin. The other choices describe rotational or inward-pointing behavior rather than outward radial growth.

📖 Explanation: A vector can have any nonzero magnitude while pointing tangentially to a circular path. For example, $\mathbf{F}(x,y)=(-y,x)$ is perpendicular to the radial direction and produces counterclockwise circulation. Its magnitude is $\sqrt{x^2+y^2}$, so it is generally nonzero away from the origin.

📖 Explanation: In a vector-field visualization of fluid velocity, arrow direction represents flow direction and arrow length commonly represents speed. If nearby arrows become longer toward a region while maintaining a broadly similar direction, the graphical model suggests increasing flow speed there. Pressure cannot be determined from the velocity arrows alone without additional physical information.

📖 Explanation: Along the positive $x$-axis, $y=0$, so the field becomes $\mathbf{F}(x,0)=(2x,0)$. Therefore every arrow points in the positive $x$-direction, and its magnitude is $2x$. As $x$ increases, the arrows should become progressively longer while keeping the same horizontal direction.

📖 Explanation: For $\mathbf{F}(x,y)=(-y,x)$, the vector is tangent to circles centered at the origin because its dot product with the radial vector $(x,y)$ is zero. Its magnitude is $\sqrt{x^2+y^2}$, so on any fixed circle the arrow lengths are constant while their directions rotate continuously.

📖 Explanation: For $\mathbf{F}(x,y)=(x,-y)$, the horizontal component has the same sign as $x$, while the vertical component has the opposite sign of $y$. Thus on the positive $x$-axis arrows point right, and on the positive $y$-axis arrows point downward. The field is therefore not uniformly outward.

📖 Explanation: If velocity equals $\mathbf{F}(x,y)$, an arrow pointing northeast means both velocity components are positive at that instant. This determines the particle's instantaneous direction of motion, but not its entire future path, acceleration, or constant speed. Those require information about how the field changes along the trajectory.

📖 Explanation: Arrow length in a vector-field plot is often scaled for visualization rather than displayed in physical units. If two plots use different scale factors, visual arrow lengths cannot be compared directly. A valid magnitude comparison requires the same plotting scale or numerical information about the vectors.

📖 Explanation: A reliable graphical verification requires testing representative points from different regions. For a proposed formula $\mathbf{F}(x,y)=(P(x,y),Q(x,y))$, compare the signs and relative sizes of $P$ and $Q$ with the plotted horizontal and vertical behavior. Checking only one point can hide systematic errors.

📖 Explanation: For $\mathbf{F}_1=(-x,-y)$, the vector points from each point toward the origin, and its magnitude $\sqrt{x^2+y^2}$ increases with distance. For $\mathbf{F}_2=(x,y)$, the vector points away from the origin with the same distance-dependent magnitude. Thus the two plots show opposite radial behavior.

📖 Explanation: The position vector is $\mathbf{r}=(x,y)$, while the velocity is \mathbf{r}'=(-y,x). Their dot product is $x(-y)+y(x)=0$, so the instantaneous velocity has no radial component. Consequently, \frac{d}{dt}|\mathbf{r}|^2=2\mathbf{r}\cdot\mathbf{r}'=0, meaning the distance from the origin remains constant along the motion.

📖 Explanation: For $\mathbf{F}(x,y)=(x,-y)$, points on the $x$-axis have $y=0$, giving horizontal vectors $(x,0)$. Points on the $y$-axis have $x=0$, giving vertical vectors $(0,-y)$. The magnitude is $\sqrt{x^2+y^2}$, so it increases with distance from the origin. This simultaneously satisfies all three visual requirements.