5128_Ch07_pp378-433

Section 7.1 Integral as Net Change 381
(a) Figure 7.2a supports that the position of the particle at t 1 is 163.
(b) Figure 7.2b shows the position of the particle is 44 at t 5. Therefore, the displacement
is 44 9 35.
Now try Exercise 1(b).
T = 1
X = 5.3333333 Y = 1
[–10, 50] by [–2, 6]
(a)
T = 5
X = 44 Y = 5
[–10, 50] by [–2, 6]
(b)
Figure 7.2 Using TRACE and the
parametrization in Example 2 you can
“see” the left and right motion of the
particle.
The reason for our method in Example 2 was to illustrate the modeling step that will
be used throughout this chapter. We can also solve Example 2 using the techniques of
Chapter 6 as shown in Exploration 1.
EXPLORATION 1
Revisiting Example 2
The velocity of a particle moving along a horizontal s-axis for 0 t 5 is
d s
t
dt
2 .
t
81 2
1. Use the indefinite integral of dsdt to find the solution of the initial value
problem
d s
t
dt
2 , t
81 2 s0 9.
2. Determine the position of the particle at t 1. Compare your answer with the
answer to Example 2a.
3. Determine the position of the particle at t 5. Compare your answer with the
answer to Example 2b.
We know now that the particle in Example 1 was at s0 9 at the beginning of the
motion and at s5 44 at the end. But it did not travel from 9 to 44 directly—it began its
trip by moving to the left (Figure 7.2). How much distance did the particle actually travel?
We find out in Example 3.
EXAMPLE 3
Calculating Total Distance Traveled
Find the total distance traveled by the particle in Example 1.
SOLUTION
Solve Analytically We partition the time interval as in Example 2 but record every
position shift as positive by taking absolute values. The Riemann sum approximating
total distance traveled is
vt k Δt,
and we are led to the integral
Total distance traveled 5
vt dt 5
8
0
0 t2 t 1 2 dt.
Evaluate Numerically We have
NINT ( t 2 t
81 2 , t, 0,5) 42.59. Now try Exercise 1(c).