# Match the vector field F with the correct plot. F(x,y) = (x, -y)

• -A)

Figure 1

• -B)

Figure 2

• -C)

Figure 3

• -D)

Figure 4

This problem aims to familiarize us with the concept of a vector field and vector space. The problem is related to vector calculus and physics, where we will briefly discuss about vector fields and spaces.

When we talk about vector field in vector calculus and physics, it is a selection of a vector to every individual point in a subset of space. For illustration, a vector field in the 2-dimensional plane can be envisioned as a cluster of arrows with an allocated numerical value and direction, each connected to a point in that plane.

Vector fields are universal in engineering and sciences, as they represent things like gravity, fluid flow velocity, heat diffusion, etc.

A vector field on an area $D$ of $R^2$ is a function $F$ that gives to each point $(x,y)$ in $D$ a vector $F(x,y)$  in $R^2$; in different terms, two scalar functions are formed $P(x,y)$ and $Q(x,y)$, forming:

$F(x,y) = P(x,y)\hat{i} + Q(x,y)\hat{j} = < P(x,y), Q(x,y)>$

This vector field might look like a function that inputs a position vector $<x, y>$ and outputs a vector $<P, Q>$, which is indeed an alteration from a subset of $R^2$ to $R^2$. This implies that the graph of this vector field spreads in $4$ dimensions, but there is an alternative way to graph a vector field, which we will graph in a minute.

So in order to figure out the correct option from the given choices, we will take some random points and will plot them against the given equation that is $F(x,y) = <x, -y>$.

Thus, now taking the point $(x,y)$ and computing the $F(x,y) = <x, -y>$:

$(1, 0) = <1, 0>$

$(0, 1) = <0, -1>$

$(-1, 0) = <-1, 0>$

$(0, -1) = <0, 1>$

$(2, 0) = <2, 0>$

$(0, 2) = <0, -2>$

The evaluations of the vector field at the assumed points are $<1, 0>, <0, -1>, <-1, 0>, <0, 1>, <2, 0>, <0, -2>$ respectively. Now plotting the vector field of the above points:

Vector representation of $(x, -y)$

Clearly all points from the $1^{st}$ quadrant map to all points of the $4^{th}$ quadrant and so on. Similarly all points of the $2^{nd}$quadrant map to all points of $3^{rd}$ quadrant and so on.

Hence, the answer is option $D$:

Vector Field of $(x, -y)$

## Example

Plot the vector field $F(x,y) = <1, x>$.

We will take the point $(x,y)$ and compute the $F(x,y) = <1, x>$:

$(-2, -1) = <1, -2>$

$(-2, 1) = <1, -2>$

$(-2, 3) = <1, -2>$

$(0, -2) = <1, 0>$

$(0, 0) = <1, 0>$

$(0, 2) = <1, 0>$

$(2, -3) = <1, 2>$

$(2, -1) = <1, 2>$

$(2, 1) = <1, 2>$

Now plotting the vector field of the above points:

Vector Field of given example