Thomas Calculus 13th [Solutions]

Section 6.3 Arc Length 459
19. (a)
2
2
2y
2 2 dx dx ( y 1)
dy dy
(b)
L 3 2
1 1 ( y 1) dy
(c) L 9.29
20. (a)
dy
dx
dy
2
2 2
cos x cos x xsin x x sin x
dx
(b)
2 2
L 1 x sin x dx
0
(c) L 4.70
21. (a)
dy
dy
2
tan x
dx
dx
2
tan x
L
/6 2 /6 2 2
1 tan x dx sin x cos x dx
0 0
2
cos x
/6
dx
/6
sec x dx
0 cos x 0
(c) L 0.55
(b)
22. (a)
2
dx 2
sec y 1 dx
dy
dy
2
sec y 1
L
/4 2
/4
1 sec y 1 dy | sec y | dy
/3 /3
/4
sec y dy
/3
(c) L 2.20
(b)
23. (a)
dy
dx
2
corresponds to 1
4x
dy
here, so take
dx as 1 . Then y x C and since (1,1) lies on the curve,
2 x
C 0. So y x from (1,1) to (4, 2).
(b) Only one. We know the derivative of the function and the value of the function at one value of x.
2
24. (a) dx corresponds to 1 here, so take dy
dy
4
y
dx as 1 . Then x 1 C and, since (0, 1) lies on the curve,
2
y
y
C 1. So y 1 .
1 x
(b) Only one. We know the derivative of the function and the value of the function at one value of x.
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