Calculus II
Practice Problems 11: Answers
a cos θ
1. Show that the graph of the polar equation r
the origin. Where is its center?
Answer. Let c
becomes
a2
b2 and θ0
r
b sin θ is a circle of radius
arctan b a , so that a
c cos θ cos θ0
c cos θ 0 and b
sin θ sin θ0 a2
b2 going through
c sin θ0 . Then the equation
c cos θ θ0 which is a circle through the origin with diameter c and center on the ray θ θ 0 . The center is at 1 2 c cos θ 0 c sin θ0 a 2 b 2 . This could also be seen by multiplying the given equation by r and changing to cartesian coordinates, getting
x2
y2
ax
by
and then completing the square.
2. Graph r
3 cos θ
3 sin θ .
Answer. Following problem 1, we have c 32 3 3 the circle r
figure.
6 cos θ π 3 , since arctan 3
2
6 and θ0
arctan 3
π 3. Thus, this is
π 3. The center is at 3 2 3 3 2 . For the graph, see the
10
5
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3. What is the polar equation of an ellipse, with one focus at the origin, vertex at the point (-1,0) and directrix
the line x 3?
Answer. The general equation of such an ellipse is
r
ed
1 e cos θ
where d is the distance between focus F and directrix L , and thus d 3; and e is the eccentricity. Knowing
the vertex tells us the eccentricity: from V F e V L we get 1 e 2 , so e 1 2. Thus the equation is
r
4. Identify the curve: y
2 sin 5θ .
3 2
1 1 2 cos θ Answer. This is the five-petaled rose. The first petal lies between θ
0 and θ
π 5 and has length 2.
5. Graph r 2 cos 2θ . This is called a lemniscate.
Answer. We make this table of the values:
θ
r
π
4
0
1
π
2
3π
4
0
π
1
0
with the star indicating that r is not defined when cos 2θ 0. But on the other hand, when cos 2θ 0,
we should consider the negative square root as well, for r. Finally, as θ goes from π to 2π we get the image
reflected in the x-axis. This gives the graph:
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0
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1
6. Find the length of the spiral r e 2θ from θ 0 to θ 2π .
Answer. Here r e2θ dr 2e2θ d θ . Thus ds2 4e4θ d θ 2 e4θ d θ 2 , and we have
2π
Length 0
e
7. Find the length of the spiral r
Answer. Here r
e
θ
θ
Length
ds
Area
1
2
2π
0
1
2
9. Find the area inside the cardioid r
∞
0
3
1
2
r2 d θ
e
2
0
8. Find the area inside the limaçon r
5e d θ
5 4π
e 1
2
for θ 0.
∞
2θ
0
e θ d θ . Thus ds2
dr
2π ds θ
dθ 2
dθ
e
2θ
dθ 2
2 lim 1 e
A ∞
2θ
2e A
d θ 2 , and we have
2
sin θ .
2π
0
2π
0
e
2θ
9
3
sin θ 2 d θ
1
2
1 cos 2θ dθ
2
2π
0
9
19
π
2 1 sin θ and above the x-axis.
6 sin θ
sin2 θ d θ
Answer. Since the cardioid is symmetric about the y-axis, the desired answer is twice the area inside the first
quadrant:
Area
π 2
2
0
1 2
r dθ
2
π 2
0
10. What is the slope of the spiral r
r eθ at the same points?
π 2
0
1 sin θ 2 d θ
π 2
0
3
cos 2θ 2 sin θ dθ
2
2
θ at the points θ
1 2 sin θ
sin2 θ d θ
3
π 2
4
2π n for n a positive integer? What about the spiral
Answer. We use equation (15) of the notes. Since r θ dr d θ
tangent line) of the first spiral:
θ cos θ sin θ
m
θ sin θ cos θ 1, and we have, for the slope m (of the
so at θ 2π n, we get m 2π n. As for the logarithmic spiral, we saw, in example 26, that the spiral r e aθ
makes a constant angle (whose tangent is a) with the ray from the origin. At the points 2π n, this ray is the
x-axis, and for our curve a 1; thus we have m 1 at all points. A calculation using (15) corroborates this:
m
eθ sin θ cos θ eθ cos θ sin θ tan θ π 4