Q30E At time \(t = 1\), a particle is... [FREE SOLUTION]

Book edition 2nd

Author(s) James Stewart

Pages 830 pages

ISBN 9781133112280

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Short Answer

At time \(t = 1\), a particle is located at position \((1,3)\). If it moves in a velocity field \(V(x,y) = \left\langle {xy - {y^2} - 10} \right\rangle \) find its approximate location at time\(t = 1.05\).

The location of particle is\(\left\langle {1.05,2.95} \right\rangle \).

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given Information.

It is given that,

Velocity field is \(V(x,y) = \left\langle {xy - 2,{y^2} - 10} \right\rangle \)

Position of particle at time \(t = 1\) is \(\left( {1,3} \right)\)

We have to find the location of particle at time \(t = 1.05\)

Therefore, the velocity is \(V(x,y) = \left\langle {\frac{d}{{dt}}x\left( t \right),\frac{d}{{dt}}y\left( t \right)} \right\rangle = \left\langle {xy - 2,{y^2} - 10} \right\rangle \) and \(\left\langle {x\left( t \right),y\left( t \right)} \right\rangle \) is the position vector of the particle.

So, at time \(t = 1\), the position of particle is \(\left\langle {x\left( 1 \right),y\left( 1 \right)} \right\rangle = \left\langle {1,3} \right\rangle \).

Step 2: Find the displacement of the particle.

\(\begin{aligned}{l}V(x,y) & = \left\langle {xy - 2,{y^2} - 10} \right\rangle \\V(1,3) & = \left\langle {(1 \times 3) - 2,{3^2} - 10} \right\rangle \\V(1,3) & = \left\langle {1, - 1} \right\rangle \end{aligned}\)

Now, at time \(t = 1.05\)

The rate of change in time is

\(\left( {\Delta t} \right) = 1.05 - 1 = 0.05\)

And the displacement of particle is

\(\begin{aligned}{l}\Delta r & = \left( {V\left( {1,3} \right)} \right) \cdot \Delta t\\ & = \left( {1, - 1} \right) \cdot \left( {0.05} \right)\\ & = \left( {0.05, - 0.05} \right)\end{aligned}\)

Step 3: Find the location of particle.

Let the position of particle, at time \(t = 1.05\), be \(\left\langle {{x_1}\left( {1.05} \right),{y_1}\left( {1.05} \right)} \right\rangle \).

Then,

\(\begin{aligned}{l}\left\langle {{x_1}\left( {1.05} \right),{y_1}\left( {1.05} \right)} \right\rangle & = \left\langle {x\left( 1 \right),y\left( 1 \right)} \right\rangle + \Delta r\\ &= \left\langle {1,3} \right\rangle + \left\langle {0.05, - 0.05} \right\rangle \\ & = \left\langle {1.05,2.95} \right\rangle \end{aligned}\)

Therefore, the location of particle is\(\left\langle {1.05,2.95} \right\rangle \).

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

At time \(t = 1\), a particle is located at position \((1,3)\). If it moves in a velocity field \(V(x,y) = \left\langle {xy - {y^2} - 10} \right\rangle \) find its approximate location at time\(t = 1.05\).

Step by Step Solution

Step 1: Given Information.

Step 2: Find the displacement of the particle.

Step 3: Find the location of particle.

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

At time \(t = 1\), a particle is located at position \((1,3)\). If it moves in a velocity field \(V(x,y) = \left\langle {xy - {y^2} - 10} \right\rangle \) find its approximate location at time\(t = 1.05\).

Step by Step Solution

Step 1: Given Information.

Step 2: Find the displacement of the particle.

Step 3: Find the location of particle.

Most popular questions for Math Textbooks

Find the gradient vector field of \({\rm{f}}\).

\(f(x,y,z) = \sqrt {{x^2} + {y^2} + {z^2}} \)

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