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

Section 8.1 Arc Length 545
By applying the Mean Value Theorem to f on the interval fx i21 , x i g, we find that there is
a number x i * between x i21 and x i such that
f sx i d 2 f sx i21 d − f 9sx i *dsx i 2 x i21 d
that is,
Dy i − f 9sx i *d Dx
Thus we have
| P i21 P i | − ssDxd2 1 sDy i d 2 − ssDxd 2 1 f f 9sx i *d Dxg 2
Therefore, by Definition 1,
− s1 1 [ f 9sx i *dg 2 ssDxd 2 − s1 1 f f 9sx i *dg 2 Dx (since Dx . 0)
L − lim | nl ` on P i21 P i | − lim
i−1 nl ` on s1 1 f f 9sx i *dg 2 Dx
i−1
We recognize this expression as being equal to
y b
a s1 1 f f 9sxdg2 dx
by the definition of a definite integral. We know that this integral exists because the function
tsxd − s1 1 f f 9sxdg 2 is continuous. Thus we have proved the following theorem:
2 The Arc Length Formula If f 9 is continuous on fa, bg, then the length of
the curve y − f sxd, a < x < b, is
L − y b
s1 1 f f 9sxdg2 dx
a
If we use Leibniz notation for derivatives, we can write the arc length formula as
follows:
3
b
L − y Î1 1S dy 2
dx
a dxD
y
(4, 8)
¥=˛
(1, 1)
0 x
FIGURE 5
Example 1 Find the length of the arc of the semicubical parabola y 2 − x 3 between the
points s1, 1d and s4, 8d. (See Figure 5.)
SOLUtion For the top half of the curve we have
y − x 3y2
dy
dx − 3 2 x 1y2
and so the arc length formula gives
4
L − y Î1 1S dy 2
dx −
1 dxD y 4
s1 1 9 1 4 x dx
If we substitute u − 1 1 9 4 x, then du − 9 4
13
dx. When x − 1, u − 4 ; when x − 4, u − 10.
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