MAT 1332, Assignment #2 Solutions
[5]
1. If you simply integrate, the answer is zero. So that obviously isn’t right. It’s best to sketch the two graphs.
See Figure 1.
1
sin(2x)
A
cos(2x)
0.5
B
0
C
−0.5
−1
0
π /8
π /4
π /2
5π /8
π
3π /4
Figure 1: Area between sin(2x) and cos(2x).
Clearly the area between them isn’t zero. In fact, it’s area A, plus area B, plus area C on the graph.
And before we can find them, we need to find the points of intersection. That is, the values of x where
sin(2x) = cos(2x). If you look at Figure 1 again, you may be able to subtly discern these. But let’s figure
them out:
sin(2x) = cos(2x)
tan(2x) = 1
(dividing both sides by cos(2x))
7π 3π π 5π 9π
2x = ... −
, − , , , ...
4
4 4 4 4
7π 3π π 5π 9π
x = ... −
, − , , , ...
8
8 8 8 8
Actually, since 0 ≤ x ≤ π, we only need values that fall into this range. So the points of intersection within
this range are x = π8 and x = 5π
8 .
Now we’re ready to integrate. The area is the sum of areas A, B and C, so we have
Area = A + B + C
=
Z π/8
0
(cos(2x) − sin(2x)) dx +
Z 5π/8
π/8
(sin(2x) − cos(2x)) dx +
1
Z π
5π/8
(cos(2x) − sin(2x)) dx
·
=
sin(2x) cos(2x)
+
2
2
¸π/8
·µ
·
+ −
0
¶
µ
cos(2x) sin(2x)
−
2
2
¸5π/8
¶¸
·
+
π/8
·µ
sin(2x) cos(2x)
+
2
2
¸π
5π/8
¶
sin(π/4) cos(π/4)
sin(0) cos(0)
cos(5π/4) sin(5π/4)
=
+
−
+
+ −
−
2
2
2
2
2
2
µ
¶¸ ·µ
¶ µ
¶¸
cos(π/4) sin(π/4)
sin(2π) cos(2π)
sin(5π/4) cos(5π/4)
−
+
+
− −
+
−
2
2
2
2
2
2
·µ
¶ µ
¶¸ ·µ
¶ µ
¶¸
1
1
1
1
1
1
1
√ + √
√ + √
− 0+
+
− − √ − √
=
2
2 2 2 2
2 2 2 2
2 2 2 2
·µ
¶ µ
¶¸
1
1
1
+ 0+
− − √ − √
2
2 2 2 2
4
= √ = 2.828 units2 .
2
2. Once again, it’s a good idea to draw a sketch. In this case it’s a cubic, with positive leading term and the
hint has kindly given us the places where it crosses the x-axis, so it’s quite easy to draw. For the absolute
value part, we turn the negative piece upside down, so it looks like Figure 2.
1000
|360t−39t2+t3|
500
0
2
3
360t−39t +t
−500
0
5
10
15
20
25
Figure 2: Graph of |g(x)| (solid curve) and g(x) (dotted curve).
Since we have an absolute value, we need to split it into two pieces. The split occurs at x = 15, so the first
part is the regular (positive) function and the second is the negative of the (negative) function. Thus, the
total energy is
Total =
·Z 15 ³
"
=
0
2
360t − 39t + t
t4
180t − 13t +
4
2
#15
3
3
´
dt +
Z 24 ³
15
"
2
39t − 360t − t
t4
+ 13t − 180t −
4
3
0
2
#24
2
15
3
¸
´
dt
= [9281.25 − 0] + [−6912 − (−9281.25)]
= 11650.5 joules/hour .
3. The derivative is
ρ0 (x) = 4 × 10−8 x(240 − x) − 2 × 10−8 x2
−6
= 9.6 × 10
−6
= 9.6 × 10
−8 2
x − 4 × 10
(using the chain rule)
−8 2
x − 2 × 10
x
−8 2
x − 6 × 10
x .
The turning points occur when ρ0 (x) = 0. Thus x = 0 and x =
9.6×10−6
6×10−8
= 160.
Since ρ(0) = 1 and ρ(160) = 1.04096, it’s clear that the minimum occurs at x = 0 and the maximum occurs
at x = 160. (You could also use the second derivative test here if you like.)
(a) Thus, the maximum is 1.04096.
(b) The minimum is 1.
(c) The maximum occurs at x = 160.
(d) The total mass is
Z 200
Total mass =
0
ρ(x)dx
Z 200 h
=
0
Z 200 h
=
h
0
(remember 2m=200cm)
i
1 + 2 × 10−8 x2 (240 − x) dx
i
1 + 4.8 × 10−6 x2 − 2 × 10−8 x3 ) dx
i200
x + 1.6 × 10−6 x3 − 5 × 10−9 x4 )
=
0
= 200 + 12.8 − 8
= 204.8 g .
(e) The average mass is just
204.8
200
= 1.024.
(f) The graph is shown in Figure 3.
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4. The volume obtained by rotating f (x) = sin
V
¡x¢
4
around the x-axis, between −π and π is given by the formula
= π
= π
Z π
−π
Z π
−π
[f (x)]2 dx
µ ¶
sin
2
x
dx .
4
We can’t integrate sin2 θ directly, so we have to use our long-forgotten-but-recently-looked-up-in-the-textbook
trigonemetric identities. Thus
cos 2θ = cos2 θ − sin2 θ
= (1 − sin2 θ) − sin2 θ
(since cos2 θ + sin2 θ = 1)
= 1 − 2 sin2 θ .
Thus
sin2 θ =
3
1 − cos 2θ
.
2
1.045
density
1.04
1.035
average
1.03
1.025
1.02
1.015
1.01
1.005
1
0
20
40
60
80
100
120
140
160
180
Figure 3: The density (solid curve) and average (dotted curve).
Substituting this into the integral, we have
V
= π
=
=
=
=
=
¡ ¢
Z π
1 − cos 2 x4
−π
Z π
µ
2
¶
dx
π
x
1 − cos
dx
2 −π
2
·
¸
π
x π
x − 2 sin
2
2 −π
·µ
¶ µ
µ
¶¶¸
π
π
π
π − 2 sin
− −π − 2 sin −
2
2
2
π
(2π − 4)
2
π 2 − 2π units3 = 3.586 units3 .
4
200