Areas, Volumes, and Arc Lengths 307
(c)a(t)=v′(t)
Enterd(x∗cos(x∧ 2 +1), x)|x=2 and
obtain 7.95506.
Thus, the velocity of the particle is
increasing att=2, sincea(2)>0.
- (See Figure 12.10-6)
[−π,π] by [−1,2]
Figure 12.10-6
(a)Point of Intersection: Use the
[Intersection] function of the calculator
and obtain (0.517757, 0.868931).
Area =
∫ 0. 51775
0
(cosx−xex)dx
Enter
∫
(cos(x)−x∗e∧x, x,
0, 0. 51775 )and obtain 0.304261.
The area of the region is approximately
0.304.
(b) Step 1. Using the Washer Method:
Outer radius=cosx;
Inner radius=xex.
V=π
∫ 0. 51775
0
[
(cosx)^2 −(xex)^2
]
dx
Step 2. Enter
∫ (
π((cos(x)∧ 2 )−
(x∗e∧(x))∧ 2
)
, x,0. 51775
)
and obtain 1.16678.
The volume of the solid is
approximately 1.167.
- Convert to a parametric representation
withx=rcosθ=5 cosθcos 2θand
y=rsinθ=5 cos 2θsinθ. Differentiate
with respect toθ.
dx
dθ
=−5 cos 2θsinθ−10 sin 2θsinθand
dy
dθ
=5 cos 2θcosθ−10 sin 2θsinθ.
Divide to find
dy
dx
=
5 cos 2θsinθ−10 sin 2θsinθ
−5 cos 2θsinθ−10 sin 2θsinθ
=
−cos 2θcosθ+2 sin 2θsinθ
cos 2θsinθ+2 sin 2θsinθ
. Evaluated
atθ=
3 π
2
,
dy
dx
=0.
The slope of the tangent line is zero,
including a horizontal tangent.
32.
∫
2
x^2 − 4 x
dx=
∫
2
x(x−4)
dxcan be
integrated with a partial fraction
decomposition. Since
A
x
+
B
x− 4
=
2
x(x−4)
,
A=
− 1
2
andB=
1
2
. Therefore,