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### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# Let $${\bf{F}}({\bf{x}}) = \left( {{r^2} - 2r} \right){\bf{x}}$$, where $${\bf{x}} = \langle x,y\rangle$$ and $$r = |{\bf{x}}|$$. Use a CAS to plot this vector field in various domains until you can see what is happening. Describe the appearance of the plot and explain it by finding the points where $${\bf{F}}({\bf{x}}) = {\bf{0}}$$.

The graph the magnitude of the vectors are increasing as it goes further away from the origin and also the direction of the vector depends on the value of $$\left( {{r^2} - 2r} \right)$$. When $${r^2} - 2r < 0$$, the vectors point inwards that is, towards the origin and when $${r^2} - 2r > 0$$, the vectors point outwards.

See the step by step solution

## Step 1: Given data

The $${\bf{F}}({\bf{x}}) = \left( {{r^2} - 2r} \right){\bf{x}}$$, where $${\bf{x}} = \langle x,y\rangle$$ and $$r = |{\bf{x}}|$$.

## Step 2: Concept of vector field

Let $$D$$ be a set in $${\mathbb{R}^2}$$ (a plane region). A vector field on $${\mathbb{R}^2}$$ is a function $${\bf{F}}$$ that assigns to each point $$(x,y)$$ in $$D$$ a two-dimensional vector $${\bf{F}}(x,y)$$.

## Step 3: Calculate the value of $$F(x,y)$$

Since, $${\bf{x}} = \langle x,y\rangle$$.

Then,

$$\begin{array}{c}r = |{\bf{x}}|\\r = \sqrt {{x^2} + {y^2}} \\{r^2} = {x^2} + {y^2}\end{array}$$

Put these values into the given vector function and obtain the equation for given function as follows:

$$\begin{array}{c}{\bf{F}}({\bf{x}}) = \left( {\left( {{x^2} + {y^2}} \right) - 2\sqrt {{x^2} + {y^2}} } \right)\langle x,y\rangle \\{\bf{F}}({\bf{x}}) = \left( {\left( {{x^2} + {y^2}} \right) - 2\sqrt {{x^2} + {y^2}} } \right)(x{\bf{i}} + y{\bf{j}})\end{array}$$

Now, for different values of $$(x,y)$$, calculate the value of $$F(x,y)$$ in table 1 as follows:

Table 1

## Step 4: Sketch the given value of $$F(x,y)$$

Plot for the points $$\begin{array}{l}\left( {1,0} \right),\left( {0,1} \right),\left( { - 1,0} \right),\left( {0, - 1} \right),\left( {1,1} \right),\left( { - 1,1} \right),\left( {1, - 1} \right),{\rm{ }}\left( { - 1, - 1} \right),\\\left( {1,2} \right),\left( { - 1,2} \right),\left( {1, - 2} \right),\left( { - 1, - 2} \right),\left( {2,1} \right),\left( { - 2,1} \right),\left( {2, - 1} \right),\left( { - 2, - 1} \right),{\rm{ }}\\\left( {2,2} \right),\left( { - 2,2} \right),\left( {2, - 2} \right),\left( { - 2, - 2} \right),\left( {0,3} \right),\left( {3,0} \right),\left( { - 3,0} \right),\left( {0, - 3} \right)\end{array}$$ in figure 1 as follows:

Figure 1

Since,

$$\begin{array}{c}{\bf{F}}({\bf{x}}) = 0\\\left( {{r^2} - 2r} \right){\bf{x}} = 0\\{r^2} - 2r = 0\quad ({\rm{ suppose }}{\bf{x}} \ne 0)\end{array}$$

Simplify further as follows:

$$\begin{array}{c}r(r - 2) = 0\\r = 0,\;\;\;\;2\end{array}$$

Thus, from the graph the magnitude of the vectors increases as it goes further away from the origin. Also, the direction of the vector depends on the value of $$\left( {{r^2} - 2r} \right)$$.

When $${r^2} - 2r < 0$$, the vectors point inwards that is, towards the origin and when $${r^2} - 2r > 0$$, the vectors point outwards.