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Mathematics I
Chapter 11
Dr. Devendra Kumar
Department of Mathematics
Birla Institute of Technology & Science, Pilani
2015–2016
Devendra Kumar BITS, Pilani Mathematics I
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CHAPTER 11
Conic Sections and Polar
Coordinates
Devendra Kumar BITS, Pilani Mathematics I
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In this Chapter, we will be discussing the following
topics:
1 Polar Coordinates
Devendra Kumar BITS, Pilani Mathematics I
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In this Chapter, we will be discussing the following
topics:
1 Polar Coordinates
2 Curve Tracing
Devendra Kumar BITS, Pilani Mathematics I
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In this Chapter, we will be discussing the following
topics:
1 Polar Coordinates
2 Curve Tracing
3 Faster Graphing
Devendra Kumar BITS, Pilani Mathematics I
![Page 6: Maths Lecture 6-7](https://reader030.fdocuments.net/reader030/viewer/2022020200/563dbb1e550346aa9aaa685c/html5/thumbnails/6.jpg)
In this Chapter, we will be discussing the following
topics:
1 Polar Coordinates
2 Curve Tracing
3 Faster Graphing
4 Areas and Lengths of the Curves
Devendra Kumar BITS, Pilani Mathematics I
![Page 7: Maths Lecture 6-7](https://reader030.fdocuments.net/reader030/viewer/2022020200/563dbb1e550346aa9aaa685c/html5/thumbnails/7.jpg)
In this Chapter, we will be discussing the following
topics:
1 Polar Coordinates
2 Curve Tracing
3 Faster Graphing
4 Areas and Lengths of the Curves
5 Conic Sections
Devendra Kumar BITS, Pilani Mathematics I
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Section 11.3
Polar Coordinates
Devendra Kumar BITS, Pilani Mathematics I
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What are cartesian coordinates of a point in the
plane?
Devendra Kumar BITS, Pilani Mathematics I
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What are cartesian coordinates of a point in the
plane?
What are polar coordinates of a point in the plane?
Devendra Kumar BITS, Pilani Mathematics I
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Cartesian Coordinates
To define cartesian coordinates in a plane, we start
with two perpendicular lines (called x-axis and
y-axis respectively).
Devendra Kumar BITS, Pilani Mathematics I
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Polar Coordinates
To define polar coordinates in a plane, we start with
an origin (called the pole) and an initial ray.
Devendra Kumar BITS, Pilani Mathematics I
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Figure: θ is the angle made by the ray OP with initial ray and
r is the distance of P from O along ray OP.
Devendra Kumar BITS, Pilani Mathematics I
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Polar Coordinates Can Have Negative r Values
We extend the meaning of polar coordinates (r,θ) in
the case in which r is negative by agreeing that the
points (−r,θ) and (r,θ) lie on the same line through
the pole O and at the distance |r| form O, but on
opposite sides of O.
Devendra Kumar BITS, Pilani Mathematics I
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Polar Coordinates Can Have Negative r Values
We extend the meaning of polar coordinates (r,θ) in
the case in which r is negative by agreeing that the
points (−r,θ) and (r,θ) lie on the same line through
the pole O and at the distance |r| form O, but on
opposite sides of O.
If r > 0, the point (r,θ) is in the same quadrant as
θ.
Devendra Kumar BITS, Pilani Mathematics I
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Polar Coordinates Can Have Negative r Values
We extend the meaning of polar coordinates (r,θ) in
the case in which r is negative by agreeing that the
points (−r,θ) and (r,θ) lie on the same line through
the pole O and at the distance |r| form O, but on
opposite sides of O.
If r > 0, the point (r,θ) is in the same quadrant as
θ.
If r < 0, the point (r,θ) is in the quadrant opposite
of the angle θ, i.e., opposite of pole.
Devendra Kumar BITS, Pilani Mathematics I
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Remarks.
We have the convention that an angle is positive if
it is measured in counterclockwise (anti clock
wise) direction and negative in clockwise
direction.
Devendra Kumar BITS, Pilani Mathematics I
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Remarks.
We have the convention that an angle is positive if
it is measured in counterclockwise (anti clock
wise) direction and negative in clockwise
direction.
If P ≡O, then r = 0 and we agree that (0,θ)
represents pole for any value of θ.
Devendra Kumar BITS, Pilani Mathematics I
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The Point (r,θ) Has Infinitely Many Polar
Coordinate Pairs
The different polar coordinates of a point (r,θ) are:
Devendra Kumar BITS, Pilani Mathematics I
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The Point (r,θ) Has Infinitely Many Polar
Coordinate Pairs
The different polar coordinates of a point (r,θ) are:
(r,θ+2nπ), n= 0,±1,±2, . . . . . .
Devendra Kumar BITS, Pilani Mathematics I
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The Point (r,θ) Has Infinitely Many Polar
Coordinate Pairs
The different polar coordinates of a point (r,θ) are:
(r,θ+2nπ), n= 0,±1,±2, . . . . . .
(−r,θ+ (2n+1)π), n= 0,±1,±2, . . . . . ..
Devendra Kumar BITS, Pilani Mathematics I
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The Point (r,θ) Has Infinitely Many Polar
Coordinate Pairs
The different polar coordinates of a point (r,θ) are:
(r,θ+2nπ), n= 0,±1,±2, . . . . . .
(−r,θ+ (2n+1)π), n= 0,±1,±2, . . . . . ..
(0,0)= (0,θ), for all θ.
Devendra Kumar BITS, Pilani Mathematics I
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Figure: The point(
2, π6
)
has infinitely many polar coordinate
pairs
Devendra Kumar BITS, Pilani Mathematics I
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Remark.
Thus we conclude that there is a unique
representation of a point in cartesian coordinate
system (in the x− y plane) while the same point can
have infinitely many representation in polar
coordinate system (in θ− r plane).
Devendra Kumar BITS, Pilani Mathematics I
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The Polar Equation of a Circle
Figure: r = a is the equation of a circle of radius |a| centered at
O.
Devendra Kumar BITS, Pilani Mathematics I
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The Polar Equation of a Line
The equation θ = θ0 is the equation of a line through
O and making an angle θ0 with the initial ray.
Figure: The line θ = θ0.
Devendra Kumar BITS, Pilani Mathematics I
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The Graph of Some Inequalities
Devendra Kumar BITS, Pilani Mathematics I
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Homework� Graph the set of points whose polar
coordinates satisfy given conditions:
r = 1
0É r É 1
r Ê 1
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Q:14� Graph the set of points whose polar
coordinates satisfy the inequality 1É r É 2.
Devendra Kumar BITS, Pilani Mathematics I
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Q:14� Graph the set of points whose polar
coordinates satisfy the inequality 1É r É 2.
Sol.
Figure: A finite region: An annulus
Devendra Kumar BITS, Pilani Mathematics I
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Homework� Graph the set of points whose polar
coordinates satisfy the inequality:
−2É r É−1.
−1É r É 2.
r É 1.
r Ê−1.
Devendra Kumar BITS, Pilani Mathematics I
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Q:15� 0É θ É π
6, r Ê 0.
Devendra Kumar BITS, Pilani Mathematics I
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Q:15� 0É θ É π
6, r Ê 0.
Sol.
Figure: An infinite region
Devendra Kumar BITS, Pilani Mathematics I
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Q:17� θ = π
3, −1É r É 3.
Devendra Kumar BITS, Pilani Mathematics I
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Q:17� θ = π
3, −1É r É 3.
Sol.
Figure: A line segment
Devendra Kumar BITS, Pilani Mathematics I
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Equation Relating Polar and Cartesian Coordinates
Figure: The usual way to relate polar and cartesian
coordinates
Devendra Kumar BITS, Pilani Mathematics I
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From figure, we have
x= r cosθ, y= rsinθ,
on squaring and adding:
x2+ y2 = r2,
on dividing:
θ = tan−1 y
x.
Devendra Kumar BITS, Pilani Mathematics I
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Q:37� Replace r = 5sinθ−2cosθ
by an equivalent
cartesian equation.
Devendra Kumar BITS, Pilani Mathematics I
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Q:37� Replace r = 5sinθ−2cosθ
by an equivalent
cartesian equation.
Sol.
r =5
sinθ−2cosθ⇒ rsinθ−2r cosθ = 5
⇒ y−2x= 5.
Devendra Kumar BITS, Pilani Mathematics I
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Example
Replace r2(1+sin2θ)= 1 by an equivalent cartesian
equation.
Devendra Kumar BITS, Pilani Mathematics I
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Example
Replace r2(1+sin2θ)= 1 by an equivalent cartesian
equation.
Sol.
r2+2r2 sinθcosθ = 1
⇒ r2+2(rsinθ)(r cosθ)= 1
⇒ x2+ y2+2xy= 1
⇒ x+ y=±1.
Devendra Kumar BITS, Pilani Mathematics I
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Q:64� Replace (x−5)2+ y2 = 25 by an equivalent
polar equation.
Devendra Kumar BITS, Pilani Mathematics I
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Q:64� Replace (x−5)2+ y2 = 25 by an equivalent
polar equation.
Sol.
(x−5)2+ y2 = 25
⇒ x2−10x+25+ y2 = 25
⇒ r2−10r cosθ = 0
⇒ r = 10cosθ.
Devendra Kumar BITS, Pilani Mathematics I
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Section 11.4
Graphing in Polar
Coordinates
Devendra Kumar BITS, Pilani Mathematics I
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Curves in Polar Coordinates
For polar coordinates (r,θ), the equation f (r,θ) = 0
(implicit form) or r = f (θ) (explicit form) defines a
curve C in the plane.
Devendra Kumar BITS, Pilani Mathematics I
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Curves in Polar Coordinates
For polar coordinates (r,θ), the equation f (r,θ) = 0
(implicit form) or r = f (θ) (explicit form) defines a
curve C in the plane.
A point P lies on C if and only if for at least one
polar coordinate (r0,θ0) of P, f (r0,θ0)= 0.
Devendra Kumar BITS, Pilani Mathematics I
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Symmetry Test
Symmetry about the x-axis: If the point (r,θ)
lies on the graph, the point (r,−θ) or (−r,π−θ)
also lies on the graph.
Devendra Kumar BITS, Pilani Mathematics I
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Symmetry Test
Symmetry about the x-axis: If the point (r,θ)
lies on the graph, the point (r,−θ) or (−r,π−θ)
also lies on the graph.
Symmetry about the y-axis: If the point (r,θ)
lies on the graph, the point (−r,−θ) or (r,π−θ)
also lies on the graph.
Devendra Kumar BITS, Pilani Mathematics I
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Symmetry about the origin: If the point (r,θ)
lies on the graph, the point (−r,θ) or (r,π+θ) also
lies on the graph.
Devendra Kumar BITS, Pilani Mathematics I
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Symmetry about the origin: If the point (r,θ)
lies on the graph, the point (−r,θ) or (r,π+θ) also
lies on the graph.
Symmetry about the line y= x: If the point
(r,θ) lies on the graph, the point(
r, π2−θ
)
or(
−r,−π
2−θ
)
also lies on the graph.
Devendra Kumar BITS, Pilani Mathematics I
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Figure: Symmetry about x-axis and y-axis
Devendra Kumar BITS, Pilani Mathematics I
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Figure: Symmetry about the pole and the line y= x
Devendra Kumar BITS, Pilani Mathematics I
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Slope of a Polar Curve
The parametric equations of r = f (θ) are
x= r cosθ, y= rsinθ.
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Slope of a Polar Curve
The parametric equations of r = f (θ) are
x= r cosθ, y= rsinθ.
The slope of the curve r = f (θ) at any point (r,θ) is
given by
dy
dx
∣
∣
∣
∣
(r,θ)
=dydθ
dxdθ
∣
∣
∣
∣
∣
∣
(r,θ)
=f ′(θ)sinθ+ f (θ)cosθ
f ′(θ)cosθ− f (θ)sinθ
provided dxdθ, 0 at any point (r,θ).
Devendra Kumar BITS, Pilani Mathematics I
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Example
Find the slope of the curve r = 1+ cosθ at θ = π
2.
Devendra Kumar BITS, Pilani Mathematics I
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Example
Find the slope of the curve r = 1+ cosθ at θ = π
2.
Sol.
dy
dx
∣
∣
∣
∣
(r,θ)
=f ′(θ)sinθ+ f (θ)cosθ
f ′(θ)cosθ− f (θ)sinθ
=−sinθsinθ+ (1+ cosθ)cosθ
−sinθcosθ− (1+ cosθ)sinθ.
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Example
Find the slope of the curve r = 1+ cosθ at θ = π
2.
Sol.
dy
dx
∣
∣
∣
∣
(r,θ)
=f ′(θ)sinθ+ f (θ)cosθ
f ′(θ)cosθ− f (θ)sinθ
=−sinθsinθ+ (1+ cosθ)cosθ
−sinθcosθ− (1+ cosθ)sinθ.
Thus
dy
dx
∣
∣
∣
∣
θ=π/2
=−sin(π/2)sin(π/2)+ (1+ cos(π/2))cos(π/2)
−sin(π/2)cos(π/2)− (1+ cos(π/2))sin(π/2)
= 1.
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Q:1� Trace the curve r = 1+ cosθ.
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Q:1� Trace the curve r = 1+ cosθ.
Sol.
Step 1. Since (r,−θ) lies on the curve, so symmetric
about x-axis. So enough to consider the steps
for 0É θ Éπ. (Note that it is not symmetrical
about others).
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Q:1� Trace the curve r = 1+ cosθ.
Sol.
Step 1. Since (r,−θ) lies on the curve, so symmetric
about x-axis. So enough to consider the steps
for 0É θ Éπ. (Note that it is not symmetrical
about others).
Step 2. r = 0 gives cosθ =−1 so that θ =π. Thus
θ =π is tangent to the curve at pole.
Devendra Kumar BITS, Pilani Mathematics I
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Step 3. drdθ
=−sinθ.
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Step 3. drdθ
=−sinθ.
�drdθ
> 0⇒ sinθ < 0, thus no value of θ.
Devendra Kumar BITS, Pilani Mathematics I
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Step 3. drdθ
=−sinθ.
�drdθ
> 0⇒ sinθ < 0, thus no value of θ.
�drdθ
< 0⇒ sinθ > 0⇒ 0< θ <π, thus r
decreases in the interval [0,π].
Devendra Kumar BITS, Pilani Mathematics I
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Step 3. drdθ
=−sinθ.
�drdθ
> 0⇒ sinθ < 0, thus no value of θ.
�drdθ
< 0⇒ sinθ > 0⇒ 0< θ <π, thus r
decreases in the interval [0,π].
Clearly max r = 2 at θ = 0 and min r = 0 at
θ =π.
Devendra Kumar BITS, Pilani Mathematics I
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Step 3. drdθ
=−sinθ.
�drdθ
> 0⇒ sinθ < 0, thus no value of θ.
�drdθ
< 0⇒ sinθ > 0⇒ 0< θ <π, thus r
decreases in the interval [0,π].
Clearly max r = 2 at θ = 0 and min r = 0 at
θ =π.
Step 4.dy
dx
∣
∣
∣
∣
θ=0
=∞,dy
dx
∣
∣
∣
∣
θ=π
2
= 1.
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Step 5. Table θ vs r:
θ 0 π
4π
3π
22π3
3π4
π
r 2 1+ 1p2
32
1 12
1− 1p2
0
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Figure: r = 1+ cosθ: The cardioid
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Q:22� Trace the curve r = 1− cosθ.
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Q:22� Trace the curve r = 1− cosθ.
Sol. We can write r = 1− cosθ = 1+ cos(θ− (−π)).
Thus the curve r = 1− cosθ is obtained replacing θ by
θ−π in r = 1+ cosθ. Therefore just rotate the curve
of r = 1+ cosθ by an angle −π.
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Figure: r = 1− cosθ: The cardioid
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Q:3� Trace the curve r = 1−sinθ.
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Q:3� Trace the curve r = 1−sinθ.
Sol. We can write r = 1−sinθ = 1+ cos(
θ−(
−π
2
))
.
Thus the curve r = 1−sinθ is obtained rotating
r = 1+ cosθ by an angle −π
2.
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Q:3� Trace the curve r = 1−sinθ.
Sol. We can write r = 1−sinθ = 1+ cos(
θ−(
−π
2
))
.
Thus the curve r = 1−sinθ is obtained rotating
r = 1+ cosθ by an angle −π
2.
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Q:20� Trace the curve r = cos2θ.
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Q:20� Trace the curve r = cos2θ.
Sol.
Step 1. 1 Since (r,−θ) lies on the curve, so
symmetric about x-axis.
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Q:20� Trace the curve r = cos2θ.
Sol.
Step 1. 1 Since (r,−θ) lies on the curve, so
symmetric about x-axis.2 Since (r,π−θ) lies on the curve, so
symmetric about y-axis.
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Q:20� Trace the curve r = cos2θ.
Sol.
Step 1. 1 Since (r,−θ) lies on the curve, so
symmetric about x-axis.2 Since (r,π−θ) lies on the curve, so
symmetric about y-axis.3 Since (−r,−π
2−θ) lies on the curve, so
symmetric about the line y= x.
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Q:20� Trace the curve r = cos2θ.
Sol.
Step 1. 1 Since (r,−θ) lies on the curve, so
symmetric about x-axis.2 Since (r,π−θ) lies on the curve, so
symmetric about y-axis.3 Since (−r,−π
2−θ) lies on the curve, so
symmetric about the line y= x.
So enough to consider the steps in the region
0É θ É π
4.
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Step 2. r = 0 gives cos2θ = 0 so that θ = π
4. Thus
θ = π
4is tangent to the curve at pole.
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Step 2. r = 0 gives cos2θ = 0 so that θ = π
4. Thus
θ = π
4is tangent to the curve at pole.
Step 3. drdθ
=−2sin2θ.
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Step 2. r = 0 gives cos2θ = 0 so that θ = π
4. Thus
θ = π
4is tangent to the curve at pole.
Step 3. drdθ
=−2sin2θ.
�drdθ
> 0⇒ sin2θ < 0, thus no value of θ.
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Step 2. r = 0 gives cos2θ = 0 so that θ = π
4. Thus
θ = π
4is tangent to the curve at pole.
Step 3. drdθ
=−2sin2θ.
�drdθ
> 0⇒ sin2θ < 0, thus no value of θ.
�drdθ
< 0⇒ sin2θ > 0⇒ 0< θ < π
2, thus r
decreases in the interval[
0, π4
]
.
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Step 2. r = 0 gives cos2θ = 0 so that θ = π
4. Thus
θ = π
4is tangent to the curve at pole.
Step 3. drdθ
=−2sin2θ.
�drdθ
> 0⇒ sin2θ < 0, thus no value of θ.
�drdθ
< 0⇒ sin2θ > 0⇒ 0< θ < π
2, thus r
decreases in the interval[
0, π4
]
.
Clearly Max r = 1 at θ = 0 and min r = 0 at
θ = π
4.
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Step 4.dy
dx
∣
∣
∣
∣
θ=0
=∞.
Thus the tangent to the curve at θ = 0 is
perpendicular to x-axis.
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Step 4.dy
dx
∣
∣
∣
∣
θ=0
=∞.
Thus the tangent to the curve at θ = 0 is
perpendicular to x-axis.
Step 5. Table
θ 0 π
12π
6π
4
r 1 0.86 0.5 0
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Figure: r = cos2θ: Four leaved rose
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