Areas, Volumes, and Arc Lengths 305
The acceleration of the object is known to be
a=〈−et,et〉=−eti+etj. Integrate to
find the velocity.v=−eti+etj+C, and
since the initial velocity isv 0 =
〈
3, 1
〉
,
v 0 =−i+j+C= 3 i+jand C= 4 i.
The velocity vector isv=−eti+etj+
4 i=(4−et)i+etj. Integrate again to find
the position vectors=(4t−et)i+etj+C.
The initial position at the origin means
thats 0 =(4. 0 −e^0 )i+e^0 j+C=−i+
j+C=0, and therefore,C=i−j. The
position vectors=(4t−et+1)i+
(et−1)jcan be evaluated att=1 to find
the position as (5〈 −e)i+(e−1)j=
5 −e,e− 1
〉
.
12.10 Solutions to Cumulative Review Problems
- (See Figure 12.10-1.)
[−3,3] by [−1,7]
Figure 12.10-1
∫a
−a
ex^2 dx=
∫ 0
−a
ex^2 dx+
∫a
0
ex^2 dx
Sinceex^2 is an even function, thus
∫ 0
−a
ex^2 dx=
∫a
0
ex^2 dx.
k= 2
∫a
0
ex^2 dxand
∫a
0
ex^2 dx=
k
2
.
- (See Figure 12.10-2.)
9
x
y
θ
Figure 12.10-2
sinθ=
9
x
Differentiate both sides:
cosθ
dθ
dt
=(9)(−x−^2 )
dx
dt
.
Whenx=15, 9^2 +y^2 = 152 ⇒ y= 12.
Thus, cosθ=
12
15
=
4
5
;
dx
dt
=−2 ft/sec.
4
5
dθ
dt
= 9
(
−
1
152
)
(−2)
=
dθ
dt
=
18
152
5
4
=
1
10
radian/sec.
- (See Figure 12.10-3.)
[−2,5] by [−2,6]
Figure 12.10-3