Download Math 2315 - Calculus II

Math 2315 - Calculus II
Homework #9 - 2007.10.21
Due Date - 2007.10.30
Solutions
Part 1: Problems from sections 10.3 - 10.6
Part 2: The fun problems.
1. Let P0 = (d, α) be polar coordinates of the point on the line L closest to the
origin. Show that L has the polar equation r = d sec(θ − α).
We start with the equation y − y0 = m(x − x0). Here notice that y0 = d sin(α)
and x0 = d cos(α). Since the point (d, α) is closest to the origin on L, the slope of
the line connecting the origin to the point (d, α) is given by tan(α) and that line
is perpendicular to the line we are seeking. So in our case, the slope of the line
m = − cot(α). Using the point slope form, we have
y − y0 = − cot(α)(x − x0 ),
where y = r cos(θ) and y = r sin(θ). Plugging everything in we get
r sin(θ) − d sin(α) = − cot(α)(r cos(θ) − d cos(α)).
After some simplification we get
r (sin(θ) + cos(θ) cos(θ)) =
d
sin(α)
and solving for r after using the difference formula for cosine:
r = d sec(θ − α).
2. Identify the curve with polar equation r = 2a cos(θ), where a is constant.
First, we set r =
p
x2 + y 2 and cos(θ) = xr . We get
p
x
x2 + y 2 = 2a p
,
x2 + y 2
and getting rid of denominators:
x2 + y 2 = 2ax.
1
We complete the square and get
(x − a)2 + y 2 = a2 ,
which is a circle of a radius a centered at (a, 0).
3. Find the area of the region between the inner and outer loops of the Limaçon
defined by r = 2 cos(θ) − 1.
First, notice that since the function is symmetric about the x−axis, so we can
compute only half the area and multiply the result by 2. So first, to compute half
the area of the small loop, we have
Z π
3 1
(2 cos(θ) − 1)2 dθ,
0 2
and to compute half the area of the bigger loop,
Z π
1
(2 cos(θ) − 1)2 dθ.
π
2
6
So the total area is is given by
"Z
#
Z π
π
3
1
1
A=2
(2 cos(θ) − 1)2 dθ −
(2 cos(θ) − 1)2 dθ
π
2
0 2
π 3
3
3
π
3
= 2 cos(θ) sin(θ) + θ − 2 sin(θ) π − cos(θ) sin(θ) + θ − 2 sin(θ) 2
2
0
3
√
= 3 3 + π.
4. If (x, y) has polar coordinates (r, θ), find the coordinates of the following:
a) (x, −y)
Polar coordinates are given by (r, −θ).
b) (−x, −y)
Polar coordinates are given by (r, θ + π).
c) (−x, y)
2
Polar coordinates are given by (r, π − θ).
d) (y, x)
Polar coordinates are given by (r, π2 − θ).
e) (2x, 2y)
Polar coordinates are given by (2r, θ).
5. Find the polar equation of the line y = mx + b.
Here we set y = r sin(θ) and x = r cos(θ) and solve for r:
r sin(θ) = m r cos(θ) + b → r =
b
.
sin(θ) − m cos(θ)
6. Prove that the total area of the four petal rose defined by r = sin(2θ) is equal
to half the area of the circle of radius 1.
We set up the integral using only one petal and multiplying by four:
Z π
2 1
A=4
sin2 (2θ)dθ
2
0
1
1 π2
= 4 − sin(2θ) cos(2θ) + θ 8
4
0
π
π
=4· = .
8
2
7. Why does the circle r = sin(θ) lie inside the spiral r = θ? Calculate the area
between the two polar curves in the first quadrant.
Since |sin(θ)| ≤ |θ| for all θ, the circle must lie inside the spiral.
The area is given by
π
2
1 2
θ dθ −
A=
0 2
1
1
= π3 − π
48
8
Z
3
Z
π
2
0
1 2
sin (θ)dθ
2
8. Find the area of the region that lies inside one, but not both of the curves
r = 2 + sin(2θ) and r = 2 + cos(2θ).
First we draw a picture: Notice that all we need to do is calculate the area
Plot of r = 2 + sin(2θ) (blue) and r = 2 + cos(2θ) (green)
3
r = 2 + sin(2θ)
2
1
y
0
−1
−2
r = 2 + cos(2θ)
−3
−3
−2
−1
0
x
1
2
3
between the two red dashed lines and multiply by 4. So to find the angles of those
red lines, we set 2 + sin(2θ) = 2 + cos(2θ). in the first quadrant we have 2θ = π4 or
5π
θ = π8 . In the second quadrant we have 2θ = 5π
4 or θ = 8 . So we now set up the
integral
Z 5π h
i
8 1
2
2
A=4
(2 + sin(2θ)) − (2 + cos(2θ)2) dθ
π
2
5π
8
1
8
= 4 − cos(2θ) − cos(2θ) sin(2θ) − sin(2θ) π
4
8
√
= 8 2.
4