   Chapter 16.1, Problem 34E

Chapter
Section
Textbook Problem

At time t = 1, a particle is located at position (1, 3). If it moves in a velocity fieldF(x, y) = ⟨xy – 2, y2 – 10⟩find its approximate location at time t = 1.05.

To determine

To find: The approximate location of particle that moves in a velocity field.

Explanation

Given data:

F(x,y)=xy2,y210 , position (1,3) at t=1

The original time is shifted from 1 to 1.05 .

Find the value of shifted time.

shifted time=1.051=0.05

Consider the velocity field F(x,y) .

F(x,y)=xy2,y210 (1)

Substitute 1 for x and 3 for y ,

F(x,y)=[(1)(3)2],=1,1

Modify equation (1) as follows.

F(x,y)=shiftedtimexy2,y210

Substitute 0

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