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James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

A five star textbook for college calculus

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546 Chapter 8 Further Applications of Integration

As a check on our answer to Example 1,

notice from Figure 5 that the arc length

ought to be slightly larger than the

distance from s1, 1d to s4, 8d, which is

s58 < 7.615773

According to our calculation in

Example 1, we have

L − 1 27 (80 s10 2 13s13 )

< 7.633705

Sure enough, this is a bit greater than

the length of the line segment.

Therefore

L − 4 9 y 10

10

13y4 s u du − 4 9 ? 2 3 u 3y2 g 13y4

− 8

27 f10 3y2 2 ( 13

4 ) 3y2 g − 1

27 (80s10 2 13s13 ) n

If a curve has the equation x − tsyd, c < y < d, and t9syd is continuous, then by

interchanging the roles of x and y in Formula 2 or Equation 3, we obtain the following

formula for its length:

4

L − y d

s1 1 d

ft9sydg2 dy −

c

y Î1 1S dx dy

dyD2

c

Example 2 Find the length of the arc of the parabola y 2 − x from s0, 0d to s1, 1d.

SOLUtion Since x − y 2 , we have dxydy − 2y, and Formula 4 gives

1

L − y Î1 1S dx 2

dy −

0 dyD y 1

s1 1 0 4y2 dy

We make the trigonometric substitution y − 1 2 tan , which gives dy − 1 2 sec2 d and

s1 1 4y 2 − s1 1 tan 2 − sec . When y − 0, tan − 0, so − 0; when y − 1,

tan − 2, so − tan 21 2 − , say. Thus

L − y

sec ? 1 2

0 sec2 d − 1 2 y

0 sec3 d

− 1 2 ? 1 2 fsec tan 1 ln | sec 1 tan

|g 0

− 1 4 ssec tan 1 ln | sec 1 tan |d

(from Example 7.2.8)

Figure 6 shows the arc of the parabola

whose length is computed in Example 2,

together with polygonal approximations

having n − 1 and n − 2 line segments,

respectively. For n − 1 the approximate

length is L 1 − s2 , the diagonal of a

square. The table shows the approximations

L n that we get by dividing f0, 1g

into n equal subintervals. Notice that

each time we double the number of sides

of the polygon, we get closer to the

exact length, which is

L − s5 ln( 2 1 s5 1 2)

< 1.478943

4

FIGURE 6

(We could have used Formula 21 in the Table of Integrals.) Since tan − 2, we have

sec 2 − 1 1 tan 2 − 5, so sec − s5 and

y

1

x=¥

L − s5

2 1 lnss5 1 2d

4

0

1

x

n

L n

1 1.414

2 1.445

4 1.464

8 1.472

16 1.476

32 1.478

64 1.479

n

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