Davis Buenger
Math 1172 11.3
1. Find the slope of the line tangent to the
curve r = 1 + 2 sin(2θ) at the point (3, π/4).
Then find an equation for this line.
Back ground: Recall that to find the slope of
a tangent line to an equation in polar coordinates, we convert the polar equation to a parametric equation. Then we use the techniques
of parametric equations to find the derivative of
the curve. Recall that we change between polar
coordinates and Cartesian coordinates through
the equations
x = r cos θ
November 4, 2015
2. Find the points at which r = 6 + 3 cos θ has
a horizontal or vertical tangent line.
Solution: Recall that for a function polar function f (θ),
dy
f (θ) cos θ + f 0 (θ) sin(θ)
(θ) =
.
dx
−f (θ) sin θ + f 0 (θ) cos(θ)
In this case r = f (θ) and r0 (θ) = −3 sin θ. Thus
(6 + 3 cos θ) cos θ − 3 sin θ sin(θ)
dy
(θ) =
.
dx
−(6 + 3 cos θ) sin θ − 3 sin θ cos(θ)
r(θ) will have a horizontal tangent when
dy
dx (θ) = 0, which will occur only when
y = r sin θ.
Thus if we have a polar function r = f (θ), then
we can derive the parametric equation:
θ 7→ f (θ) cos θ, f (θ) sin θ
(6 + 3 cos θ) cos θ − 3 sin θ sin(θ) = 0.
By replacing sin2 θ with 1−cos2 θ we have a the
following quadratic equation in terms of cos θ.
From here we can find the slope of the tangent
line as a function of θ.
6 cos θ + 3 cos2 θ − 3(1 − cos2 θ) = 0.
dy
dy/dθ
f (θ) cos θ + f 0 (θ) sin(θ)
(θ) =
=
.
dx
dx/dθ
−f (θ) sin θ + f 0 (θ) cos(θ)
Equivalently
−3 + 6 cos θ + 6 cos2 θ = 0,
Solution: Considering the above discussion we
need to find the derivative of r(θ) = 1+2 sin(2θ)
with respect to θ and then plug everything into
the formula we derived. A quick calculation
shows:
r0 (θ) = 4 cos(2θ).
−1 + 2 cos θ + 2 cos2 θ = 0.
The roots of the quadratic 2x2 + 2x − 1 are
p
√
−2 ± 4 − 4(2)(−1)
± 3−1
=
.
2·2
2
Thus
√
− 3−1
2
< −1, only θ such that cos θ =
√
dy
(1 + 2 sin(2θ)) cos θ + (4 cos(2θ)) sin(θ) 3−1 can be solutions. In the interval [0, 2π],
(θ) =
. 2
√
dx
−(1 + 2 sin(2θ)) sin θ + (4 cos(2θ)) cos(θ)
the only such θ are cos−1 ( 3−1
2 ) and 2π −
√
3−1
−1
At the point (3, π/4),
cos ( 2 ).
We will have a vertical tangent line when the
dy π denominator of our derivative is 0. I.e. when
dx 4
(1 + 2 sin(2 π4 )) cos π4 + (4 cos(2 π4 )) sin( π4 )
0 = −(6 + 3 cos θ) sin θ − 3 sin θ cos(θ)
=
(1 + 2 sin(2 π4 )) sin π4 − (4 cos(2 π4 )) cos( π4 )
= −6 sin θ(1 + cos θ).
Since
√
=
(1 + 2 · 1) ·
−(1 + 2 · 1) ·
2
2√
√
+ (4 · 0) ·
2
2
+ (4 · 0) ·
2
2√
2
2
= −1
Thus sin θ = 0 or cos θ = −1, and θ = 0, π.
1
3. Make a sketch of the limacon r = 3 + 6 cos θ
and find the area of the inner region.
Solution: To begin, let us identify the angles
corresponding to the inner region. The graph
transitions between inner loop and outer loop
when 0 = 3 + 6 cos(θ). In other words for
4. Make a sketch of the rose r = cos(3θ)
and find the area of the area within all the leaves.
Solution: As the coefficient in front of theta is
odd, our rose will have 3 leaves.
θ = cos−1 (3/6) = 2π/3, 4π/3.
Furthermore, as our function is of the form
r = cos(mθ) a leaf will be centered on the xaxis. All leaves will be equally spaced, so the
other two leaves will be centered on 2π/3 and
4π/3. Let us find the angle bounds associated
to the leaf on the x-axis.
Hence
1
Area =
2
Z
4π/3
(3 + 6 cos θ)2 dθ.
0 = cos 3θ.
2π/3
3θ =
By symmetry
Z π
A =
(3 + 6 cos θ)2 dθ
2π/3
Z π
= 9
1 + 4 cos θ + 4 cos2 θ dθ
2π/3
Z π
= 9
1 + 4 cos θ + 2(1 + cos(2θ)) dθ
2π/3
Z π
= 9
3 + 4 cos θ + 2 cos(2θ) dθ
for k ∈ N.
π
+ kπ
2
π kπ
+
6
3
for k ∈ N. Thus the leaf on the x-axis ranges
from an angle of −π/6 to π/6. We will find the
area of the top half of this leaf and multiply by
6 to find the area of all leaves. .
Z π/6
1
(cos 3θ)2 dθ
A = 6
2
0
Z π/6
1 + cos 6θ
= 3
dθ
2
0
sin 6θ π/6
3
=
θ+
2
6
0
π
=
.
4
2π/3
h
iπ
= 9 3θ + 4 sin θ + sin(2θ)
2π/3
h
i
= 9 3π + 4 sin π + sin(2π)
h
i
−9 3(2π/3) + 4 sin(2π/3) + sin(4π/3) .
2
θ=
5. Find the area of the region that lies with in
the curve r = 3 sin θ and r = 3 cos θ.
Solution: Recall that each of these graphs is a
circle, one centered on the x-axis and the other
on the y-axis. Here is a quick picture.
Z
π/2
3 cos θ dθ
A2 =
π/4
π/2
= 3 sin θ
π/4
√ !
2
= 3 1−
.
2
Thus the area of the region is
√
A1 + A2 = 3( 2 + 2).
6. (Test Question) Consider the polar curves
r = 3 cos θ and r = sin θ. Find all points of
intersection of the two curves. Give exact
values for the coordinates in (r, θ). Then find
the area of the region that lies inside the graph
of r = 3 cos θ and out side of the graph of
r = sin θ.
Solution: Intesection pts: Set the expressions for r equal to each other, and solve:
√
sin θ = 3 cos θ
√
tan θ = 3,
Let us calculate the intersection points. Note
that each curve intersects the origin, but at different angles. 3 sin θ = 0 at π/2 and 3 cos θ = 0
at 0. By equating the curves and solving for
theta, we can find the second intersection point.
3 sin θ = 3 cos θ
so θ =
tan θ = 1
π
3
+ kπ, for any integer k.
θ = π/4.
Our region can be subdivided into two subregions one from the angle 0 to π/4 and the second from an angle of π/4 to π/2.√The area of
the first region is the area under 3 sin θ. Notice that each has only one top function and no
bottom function. The first has a top function
of r = 3√sin θ and the second has a top function
of r = 3 cos θ.
Z π/4
A1 =
3 sin θ dθ
0
π/4
= −3 cos θ
0
!
√
2
= −3
−1 .
2
Inserting θ =
π
3
we get r = sin
point is
into one of the equations for r,
π
3
√
=
3
2
√
3 , so one intersection
3 π
,
2 3
3
!
.
But were not done, because graphing these two
curves in the same plane shows they intersect at
two points: one in the first quadrant (which we
found above) and the origin. Its acceptable to
label this second intersection point (0, 0), as it
is that point on
√ the graph of r = sin θ. On the
graph of r = 3 cos θ, however, it is actually
(0, π). They are the same point with 2 different
representation in polar coordinates.a
Area: By graphing the two curves, we can divide the region in question into a semicircle,
which well call A (below the x-axis), and a semicircle with a bite taken out of it, which is B
(above the x-axis). Since the diameter of the
semicircle
runs from r = 0 to r = 3, its radius
√
is 3/2, so
1
1
area(A) = πR2 = π
2
2
=
=
1
2
Z
1
2
Z
0
=
π
3
0
Z
=
π
3
1 + cos(2θ) 1 − cos(2θ)
3
dθ
−
2
2
1 1
3 1
+ cos(2θ) − + cos(2θ) dθ
2 2
2 2
π
1 3
1 + 2 cos(2θ) dθ
2 0
π/3 pi √3
1 π/3 1
+ sin(2θ)
=
θ
+
.
2 0
2
6
4
0
Hence
√
√
3π pi
3
13pi
3
T otal area =
+
+
=
+
.
8
6
4
24
4
7. (Test Question) Find the area of the region
bounded by the polar curve r = 2 + sin θ.
Solution: The equation is of the form r =
A + B sin θ with A > B. Hence the graph of
the function is a cardioid and one must integrate from 0 to 2π to find its area.
Z
1 2π
Area =
(2 + sin θ)2 dθ
2 0
Z
1 2π
=
4 + 4 sin θ + sin2 θ dθ
2 0
Z
1 2π
1 − cos(2θ)
=
4 + 4 sin θ +
dθ
2 0
2
Z
1 2π 9
cos(2θ)
=
+ 4 sin θ +
dθ
2 0 2
2
1 9
sin(2θ) 2π
=
x − 4 cos θ −
2 2
4
0
"
9
1
sin(2(2π))
=
(2π) − 4 cos(2π) −
2
2
4
#
sin(2 · 0)
9
· 0 − 4 cos(0) −
−
2
4
√ !2
3
3π b
=
.
2
8
Region B requires an integral. Let f (θ) =
√
3 cos θ and g(θ) = sin θ; then area(B) can be
written as
Z
1 β
f (θ)2 − g(θ)2 dθ.
2 0
where β is the angle where r = f (θ) and
r = g(θ) intersect, which we found above to
be β = π3 . Hence
area(B)
Z π
i
1 3h√
=
( 3 cos θ)2 − (sin θ)2 dθ
2 0
Z π
1 3
=
3 cos2 θ − sin2 θ dθ
2 0
a
If you look carefully, a step in the algebra above
even hints at this: we had to divide both sides by cos θ.
But if r = 3 cos θ = 0, then any value of φ such that
r = sin φ = 0 will put us at the same point, even though
θ 6= φ, since (0, θ) is the same point in the plane as
(0, φ), for any angles θ and φ.
b
You can also find this area by calculating
Z 0
1 √
( 3 cos θ)2 dθ.
−π/2 2
=
4
9π
.
2