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Q13E

Expert-verifiedFound in: Page 760

Book edition
2nd

Author(s)
James Stewart

Pages
830 pages

ISBN
9781133112280

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

The given vector field is\(F\left( {x,y} \right) = \left\langle {y,y + 2} \right\rangle \).

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\(x,{\rm{ }}y,{\rm{ or }}z\).

\(\begin{aligned}{l}F\left( {x,y} \right) &= \left\langle {y,y + 2} \right\rangle \\ \Rightarrow F\left( {x,y} \right) &= yi + \left( {y + 2} \right)j\end{aligned}\)

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\(\left( {x,y} \right)\).

Let,

\(\begin{aligned}{l}F(x,y) &= Pi + Qj\\ \Rightarrow F(x,y) &= yi + \left( {y + 2} \right)j\end{aligned}\)

Then,

\(\begin{aligned}{l}P &= y\\Q &= y + 2\end{aligned}\)

The slope will be,

\(\frac{Q}{P} = \frac{{y + 2}}{y} = 1 + \frac{2}{y}\)

The magnitude will be,

\(\begin{aligned}{l}\sqrt {{P^2} + {Q^2}} \\ &= \sqrt {{y^2} + {{\left( {y + 2} \right)}^2}} \\ &= \sqrt {{y^2} + {y^2} + 4 + 4y} \\ &= \sqrt {2{y^2} + 4y + 4} \end{aligned}\)

As \(y\) increases from \(0\), the slope is positive and converges to\(1\) because.

\(\begin{aligned}{l}\mathop {\lim }\limits_{y \to \infty } \frac{{y + 2}}{y} &= \mathop {\lim }\limits_{y \to \infty } \frac{{\frac{y}{y} + \frac{2}{y}}}{{\frac{y}{y}}}\\ &= \mathop {\lim }\limits_{y \to \infty } \frac{{1 + \frac{2}{y}}}{1}\\ &= 1\end{aligned}\)

As \(y\) increases from\(0\), the magnitude increases to infinity.

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

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