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Essential Calculus: Early Transcendentals
Found in: Page 760
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

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

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Short Answer

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.

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

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.

Most popular questions for Math Textbooks

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