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Found in: Page 761

### Essential Calculus: Early Transcendentals

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

## 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$$\left\langle {1.05,2.95} \right\rangle$$.

See the step by step solution

## 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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