Question

Short answer

The vector field \(F\) matches with the plot III.

Step-by-step solution

Given Information.

The given vector field is\(F\left( {x,y} \right) = \left\langle {y,x - y} \right\rangle \).

Explanation.

Vector fields generally have components in the \(i,{\rm{ }}j,{\rm{ and }}k\)axis in order to present the direction and magnitude of a group of vectors (vector fields).

These components can be either constants or a function of variables\(x,{\rm{ }}y,{\rm{ or }}z\).

\(\begin{aligned}{l}F\left( {x,y} \right) &= \left\langle {x, - y} \right\rangle \\ \Rightarrow F\left( {x,y} \right) &= yi + \left( {x - y} \right)j\end{aligned}\)

The vector field has components as functions of variable either\(x{\rm{ or }}y\). That means, the slope and the magnitude of each of the vector will be depending on the coordinates\(\left( {x,y} \right)\).

Find slope and magnitude.

To find the slope and the magnitude:

Let,

\(\begin{aligned}{l}F(x,y) &= Pi + Qj\\ \Rightarrow F(x,y) &= yi + \left( {x - y} \right)j\end{aligned}\)

Then,

\(\begin{aligned}{l}P &= y\\Q &= x - y\end{aligned}\)

The slope will be,

\(\frac{Q}{P} = \frac{{x - y}}{y} = \frac{x}{y} - 1\)

The magnitude will be,

\(\begin{aligned}{l}\sqrt {{P^2} + {Q^2}} \\ &= \sqrt {{y^2} + {{\left( {x - y} \right)}^2}} \\ &= \sqrt {{y^2} + {x^2} + {y^2} - 2xy} \\ &= \sqrt {{x^2} + 2{y^2} - 2xy} \end{aligned}\)

Analyse slope and magnitude.

Let \(y\)be positive constant, as \(x\) increases from \(x = y\), the slope is positive and increases to infinity.

For the same value of \(y\), as \(x\) increases from \(x = y\), the magnitude increases to infinity.

Let \(x\) be positive constant, as \(y\) increases from \(x = y\), the slope is positive and converges to\( - 1\) because.

\(\begin{aligned}{}\mathop {\lim }\limits_{y \to \infty } \frac{{x - y}}{y} &= \mathop {\lim }\limits_{y \to \infty } \frac{{\frac{x}{y} - \frac{y}{y}}}{{\frac{y}{y}}}\\ &= \mathop {\lim }\limits_{y \to \infty } \frac{{\frac{x}{y} - 1}}{1}\\ &= - 1\end{aligned}\)

For the same value of \(x\), as \(y\) increases from\(x = y\), the magnitude does not change and it remains as the value of \(x\).

Therefore, the vector field \(F\) matches with the plot III.