Thomas Calculus 13th [Solutions]

10. Converting to pounds and feet,
2
1/2
0
Chapter 6 Additional and Advanced Exercises 503
2 lb 12 in
2 lb/in 24 lb/ft. Thus,
1 ln 1ft
1 2 1 2
2 2
1/2
F 24x W 24x dx
0
1 1 1
10 32 ft/sec 320
12x 3 ft lb. Since W mv0 mv 1 , where W 3 ft lb, m lb
slugs,
2
1 1
2 320
2 2
and v 1 0 ft/sec, we have 3 v0 v 0 3 640. For the projectile height,
2
s 16t v t (since
s 0 at t 0 )
ds
v0
v 32 t v 0.
At the top of the balls path, v 0 t and the height is
dt
32
2
2
0 0 0
s 16 v v
v v 3 640
32 0
30 ft.
32 64 64
0
n
11. From the symmetry of y 1 x , n even, about the y -axis for 1 x 1, we have x 0. To find y ,
we use the vertical strips technique. The typical strip has center of mass: ( x, y) x ,
1 x
, length: 1 x ,
n
n
width: dx, area: dA 1 x dx , mass: dm 1 dA 1 x dx . The moment of the strip about the x -axis is
n
2
n
2
1 x
1 1 x
1
1 n 2n n
y dm dx M 2 1 2
2x x
1
2 1
2 x dx x x dx x
1 2 0 2 n 1 2n 1
0
n 1 2n
1
( n 1)(2n 1) 2(2n 1) ( n 1) 2 2
2n 3n 1 4n 2 n 1 2n
( n 1)(2n 1) ( n 1)(2n 1) ( n 1)(2n
1) .
n 1 1
Also,
2
1
1 2
0 1 n
x dx 2 x x
1 2 1 n
n
0
n 1 n 1
. Therefore, y
2
n
1 2n
1 1
1 1
1 n
M dA x dx
1 1
M x
M
2
x 2n ( n 1) n
0,
n
M ( n 1)(2n 1) 2n 2n 1 2n
1
is the location of the centroid. As n , y 1
so the limiting position of the centroid is 0,
1
.
2
2
n
M
12. Align the telephone pole along the x -axis as shown
in the accompanying figure. The slope of the top
14.5 9
8 8
length of pole is
1 1
(14.5 9)
40 8 40
5.5 11
8 40 8 80 . Thus, y 9 11 x
8 8 80
1
9
11 x is an equation of the line
8 80
representing the top of the pole. Then
b 2 40 2 40
2
M 1 11 1 11
y x y dx x
0 8 x dx x
80 64 x dx
0 80
a
b 2 40 2 40 2
M y dx 1 11 1 11
0 8 x dx
80 64 x dx
0 80
Thus,
a
calculator to compute the integrals). By symmetry about the x-axis,
0
from the top of the pole.
M y
M
129,700
5623.3
x 23.06 (using a
y so the center of mass is about 23 ft
13. (a) Consider a single vertical strip with center of mass ( x, y ). If the plate lies to the right of the line, then the
moment of this strip about the line x b is ( x b) dm ( x b)
dA the plates first moment about
x b is the integral ( x b) dA x dA b dA M y b A.
(b) If the plate lies to the left of the line, the moment of a vertical strip about the line x b is ( b x)
dm
( b x)
dA the plates first moment about x b is ( b x) dA b dA x dA b A M y .
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