Thomas Calculus 13th [Solutions]

Section 7.4 Relative Rates of Growth 529
2
85. To find the length of the curve: 1
b
b
y cosh ax y sinh ax L 1 (sinh ax) dx L cosh ax dx
a
0 0
b
1 sinh 1
b
b
ax sinh ab . The area under the curve is A 1 cosh ax dx 1 sinh ax 1 sinh ab
a 0 a
0 a 2 2
a 0 a
1 1 sinh ab which is the area of the rectangle of height 1 and length L as claimed, and which is illustrated
a a
a
below.
86. (a) Let the point located at (cosh u ,0) be called T. Then A( u ) area of the triangle OTP minus the area
2
cosh 2
under the curve y x 1 from A to T
1
u
A( u) cosh u sinh u x 1 dx.
2 1
cosh 2 2 2 2
(b) A ( u ) 1 1
2 cosh u sinh u u x
1
1 dx A ( u ) 2
cosh u sinh u cosh u 1 sinh u
1 2 1 2 2 1 2 2
cosh u sinh u sinh u cosh u sinh u 1
(1)
1
2 2 2 2 2
( ) ( )
u
,
A(0) 0 C 0 A( u) u u 2A
2 2
2
(c) A u 1 A u C and from part (a) we have
7.4 RELATIVE RATES OF GROWTH
1. (a) slower, lim
3
lim
1
0
(b) slower,
x
x
x
e x
x
e
3
x
2
sin x
2
3x 2sin x cos x 6x 2cos 2x 6 4sin 2x
x
x
e x
x
e x
x
e x
x
e
2
x
for all reals, and lim
2
0 lim
10
x x x
x
x
e e e
x e x e
1 1/2
1/2
x
x
2
lim lim
x
lim lim
1
x x x x
x e x e x e x 2 xe
0
x
x
lim lim since 4 x
x e x
e
e
1
lim lim lim lim 0
Theorem because
6 4sin 2 10
(c) slower,
(d) faster,
4 4
(e) slower,
(f ) slower,
(g) same,
(h) slower,
x
3
x
2 3
x
x
2e
lim lim 0 since 3 2
x
x
e
x/2
x
e
e
x
1
e
lim lim 0
lim
x
x/2
e
x
2
lim
1 1
x
e x
2 2
1
log10
x
ln x
x
1
x x x x
lim lim lim lim 0
x e x (ln10) e x (ln10) e x (ln10) xe
e
1
by the Sandwich
2. (a) slower,
(b) slower,
4 3 2
10x 30x 1 40x 30 120x 240x
240
x x x x x
x e x e x e x e x e
lim lim lim lim lim 0
ln x 1 x
1 1
x
x
x x x x x x x
x ln x x x(ln x 1) ln x 1 1 ln x
1
x e x e x e x e x e x e x xe
lim lim lim lim lim lim lim 0
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