Thomas Calculus 13th [Solutions]

Section 6.3 Arc Length 461
32. (a) From the accompanying figure and definition of the
differential (change along the tangent line) we see
that dy f ( xk
1)
x k length of kth tangent fin is
2 2
2 2
xk dy xk f xk 1 xk
( ) ( ) .
(b) Length of curve
n
n
k 1
n
n
n
2 2
k xk f xk 1 xk
k 1
n
k 1
lim (length of th tangent fin) lim ( )
2 b
2
f xk
1 xk
f x dx
a
lim 1 ( ) 1 ( )
33.
4
2 2 2
2 2
x y 1 y 1 x ; P 0, 1 1 3
4 , 2 , 4
, 1 L x i x i 1 y i y i 1
k 1
2 2 2
2 2 2 2
2
1 15 1 1 3 15 3 1 7 3 3
7
0 1 1 0
4 4 2 4 2 4 4 2 4 2 4 4
1.55225
y y
x2 x1
L x 2
2 2 2 x2
2
y2 y1
1 1 1 2 1 1
x
x x x
x 2 x 1
34. Let ( x1 , y1) and ( x2, y 2),
with x2 x 1 , lie on y mx b , where m 2 1
, then dy
1
2 2
2 1 2 1
2
x2 x1
x x y y
1 2 1
2 2
2 1 2 1
x2 x1
x x y y
2 2
x2 x1 x2 x1 x2 x1 y2 y1 .
dx
m
35.
36.
3/2 dy 1/2
y 2x 3 x ;
dx
x 1/2
2 x
L( x) 1 3t dt 1 9 t dt;
0 0
[ 1 9 9 ; 0 1, 1 9 ] 1 9 3/2
1 9
1
x
2 2 3/2
u t du dt t u t x u x u du u x
2
9 1 27 1 27 (1 9 x ) 27
;
2 3 2 2 2(10 10 1)
L(1) (10)
27 27 27
3
2 1 dy 2 2
y x x x x 2x 1 1 ( x 1) 1 ;
3 4x 4 dx 2 2
4( x 1) 4( x 1)
2 4 2
( ) x 4 2
2 1
4( 1) 1 [4( 1) 1]
0 1 ( 1) x 2
0 1 t
x
2
0
1 t
L x t dt dt dt
4
4( t 1) 4( t 1) 16( t 1)
x 4 8 4 8 4 4 2 4
16( t 1) 16( t 1) 8( t 1) 1 x 16( t 1) 8( t 1) 1 x [4( t 1) 1] x 4( t 1) 1
dt dt dt dt
0
4
0
4
0
4
0
2
16( t 1) 16( t 1) 16( t 1) 4( t 1)
x 2
( t 1) 1 dt;
0
2
4( t 1)
x 1 2 1 2
[ u t 1 du dt; t 0 u 1, t x u x 1] u u du
1 4
3 1
1
1 1 1 3 1 1 1 1 3 1 1
3 u 4 u x
(
3 x 1) ( 1) ;
1
4( x 1) 3 4 3 x L(1) 8 1 1 59
4( x 1) 12 3 8 12 24
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