18.01 (Fall 14) Solution to Problem Set 8
Part II
1. solution
a) The integration of sec x is as following.
Z
Z
sec xdx =
cos x
dx.
1 − sin2 x
Let u = sin x, we have du = cos xdx, and therefore
Z
Z
Z
cos x
du
1
1
1
=
+
du
2 dx =
2
1−u
2
1−u 1+u
1 − sin x
i
1h
ln |1 + u| − ln |1 − u| + C
=
2
i
1h
ln(1 + sin x) − ln(1 − sin x) + C
=
2
1 1 + sin x
= ln
+C
2 1 − sin x
r
1 + sin x
= ln
+C
1 − sin x
b) We have
r
ln
1 + sin x
1 + sin x
+ C = ln p
+C
1 − sin x
(1 + sin x)(1 − sin x)
1 + sin x
= ln √
+C
cos2 x
1 + sin x
= ln
+C
| cos x|
= ln | sec x + tan x| + C.
2. solution
For the first hump, solve for y = ex cos x = 0, we have x = π2 . Use the shell method,
the volume is given by
Z π
2
2πxex cos xdx.
0
1
Now, compute the antiderivative,
Z
Z
I = xex cos xdx = xex d sin x
Z
Z
x
x
x
= x sin xe − sin xd(xe ) = x sin xe − sin x(ex + xex )dx
Z
Z
x
x
= x sin xe − sin xe dx − x sin xex dx
Z
Z
x
x
= x sin xe − sin xe dx + xex d cos x
Z
Z
x
x
= x(sin x + cos x)e − sin xe dx − cos xd(xex )
Z
Z
x
x
= x(sin x + cos x)e − sin xe dx − cos x(ex + xex )dx
Z
Z
Z
x
x
x
= x(sin x + cos x)e − sin xe dx − cos xe dx − x cos xex dx
Z
Z
x
x
= x(sin x + cos x)e − sin xe dx − cos xex dx − I.
Therefore,
Z
x
2I = x(sin x + cos x)e −
x
sin xe dx −
Z
cos xex dx,
Z
Z
i
1h
x
x
x
I = x(sin x + cos x)e − sin xe dx − cos xe dx .
2
R
R
Now, we need to compute I1 = sin xex dx and I2 = cos xex dx.
Z
Z
Z
x
x
x
I1 = sin xe dx = − e d cos x = −e cos x + cos xex dx
Z
Z
x
x
x
x
= −e cos x + e d sin x = −e cos x + e sin x − sin xex dx
= −ex cos x + ex sin x + C − I1 .
Therefore, we have
I1 =
1
− ex cos x + ex sin x + C.
2
Also, we have
I2 = I1 + ex cos x =
1 x
e cos x + ex sin x + C.
2
We can get
I=
i 1h
i
1h
x(sin x + cos x)ex − I1 − I2 = x(sin x + cos x)ex − sin xex + C.
2
2
The volume is given by
π
π
π
π
1 π
2π(I( ) − I(0)) = 2π · ( − 1)e 2 = π( − 1)e 2 .
2
2 2
2
2
3. solution
R
After canceling ex sinh xdx from each side, we should get ex cosh x − ex sinh x = C
for some constant C instead of ex cosh x − ex sinh x = 0. Use the original definition
of sinh x and cosh x, we can compute explicitly that C = 1.
4. solution
a) We have
dx
= 1 − cos t,
dt
dy
= sin t.
dt
Therefore,
dy
=
dx
dy
dt
dx
dt
=
sin t
.
1 − cos t
It will be infinite when cos t = 1, i.e. t = 2πk for some integer k. The graph is
as the following:
b) The arclength is given by
Z
0
2π
v
u
dy
u
dy 2
t1 + ( dt )2 dx dt
1 + ( ) dx =
dx
dx
dt
0
dt
r
Z 2π
Z 2π q
dx 2
dy 2
=
( ) + ( ) dt =
(1 − cos t)2 + sin2 tdt
dt
dt
0
0
Z 2π
Z 2π r
√
t
=
2 − 2 cos tdt =
4 sin2 dt
2
0
0
Z 2π
Z 2π
t
t
=
2| sin |dt = 2
sin dt
2
2
0
0
t 2π
= −4 cos = 8.
2 0
r
Z
2π
3
c) The surface area is given by
r
Z 2π
Z 2π
q
dy
(1 − cos t) (1 − cos t)2 + sin2 tdt
2πy 1 + ( )2 dx = 2π
dx
0
0
Z 2π
Z 2π
t
t
t
= 4π
(1 − cos t) sin dt = 4π
2 sin2 sin dt
2
2
2
0
0
Z 2π
Z 2π
t
t
t
sin2 d cos
sin3 dt = −16π
= 8π
2
2
2
0
0
Z 2π
t
t
(1 − cos2 )d cos
= −16π
2
2
0
1
t
64
t
= π
= −16π[cos − cos3 ]2π
2 3
20
3
5. solution
When the curve is vertical, we have
dx
dt
= 0. Since
dx
= (cos t)(ln t),
dt
The first point where
π
2
Z
dx
dt
= 0 is t = π2 . The length is given by
Z
p
x0 (t)2 + y 0 (t)2 dt =
1
π
2
Z
p
(ln t)2 dt =
1
1
π
2
π
ln tdt = [t ln t − t]12 =
π π π
ln − + 1.
2 2 2
6. solution
a) The graph is as below. The curve is traced counterclockwise as t increases. And
2
the algebraic equation is x4 + y 2 = 1.
b) Let (x, y) be a point on the new curve, we know that R−θ (x, y) is a point in the
previous curve. Therefore,
√
√
√
1
[(x + 3) cos θ + y sin θ − 3]2 + [−(x + 3) cos θ + y cos θ]2 = 1.
4
4
We have
√
√
√
4 =[(x + 3) cos θ + y sin θ − 3]2 + 4[−(x + 3) sin θ + y cos θ]2
√
√
√
√
=(x + 3)2 cos2 θ + y 2 sin2 θ + 3 + 2(x + 3)y cos θ sin θ − 2 3(x + 3) cos θ
√
√
√
− 2 3y sin θ + 4[(x + 3)2 sin2 θ + y 2 cos2 θ − 2(x + 3)y sin θ cos θ]
√
√
=(x + 3)2 (cos2 θ + 4 sin2 θ) + y 2 (sin2 θ + 4 cos2 θ) − 6(x + 3)y cos θ sin θ
√
√
√
+ 3 − 2 3(x + 3) cos θ − 2 3y sin θ.
When θ = π4 , we have
√
√
√
√
√
5
5
4 = (x + 3)2 + y 2 − 3(x + 3)y + 3 − 6(x + 3) − 6y.
2
2
When θ = π2 , we have
4 = 4(x +
When θ =
3π
4 ,
√
√
3)2 + y 2 + 3 − 2 3y.
we have
√
√
√
√
√
5
5
4 = (x + 3)2 + y 2 + 3(x + 3)y + 3 + 2 6(x + 3) − 6y.
2
2
The graph is as the following.
5
c) The curve for 0 ≤ t ≤ 2π is the following. The curve does not end at the starting
point.
The curve for 0 ≤ t ≤ 16π is the following.
6