Maths Lecture 6-7

Mathematics I

Chapter 11

Dr. Devendra Kumar

Department of Mathematics

Birla Institute of Technology & Science, Pilani

2015–2016

Devendra Kumar BITS, Pilani Mathematics I

CHAPTER 11

Conic Sections and Polar

Coordinates


In this Chapter, we will be discussing the following

topics:

1 Polar Coordinates



topics:

1 Polar Coordinates

2 Curve Tracing



topics:

1 Polar Coordinates

2 Curve Tracing

3 Faster Graphing



topics:

1 Polar Coordinates

2 Curve Tracing

3 Faster Graphing

4 Areas and Lengths of the Curves



topics:

1 Polar Coordinates

2 Curve Tracing

3 Faster Graphing

4 Areas and Lengths of the Curves

5 Conic Sections


Section 11.3

Polar Coordinates


What are cartesian coordinates of a point in the

plane?


What are cartesian coordinates of a point in the

plane?

What are polar coordinates of a point in the plane?


Cartesian Coordinates

To define cartesian coordinates in a plane, we start

with two perpendicular lines (called x-axis and

y-axis respectively).


Polar Coordinates

To define polar coordinates in a plane, we start with

an origin (called the pole) and an initial ray.


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.


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 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.


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.


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.


The Point (r,θ) Has Infinitely Many Polar

Coordinate Pairs

The different polar coordinates of a point (r,θ) are:



Coordinate Pairs


(r,θ+2nπ), n= 0,±1,±2, . . . . . .



Coordinate Pairs


(r,θ+2nπ), n= 0,±1,±2, . . . . . .

(−r,θ+ (2n+1)π), n= 0,±1,±2, . . . . . ..



Coordinate Pairs


(r,θ+2nπ), n= 0,±1,±2, . . . . . .

(−r,θ+ (2n+1)π), n= 0,±1,±2, . . . . . ..

(0,0)= (0,θ), for all θ.


Figure: The point(

2, π6

)

has infinitely many polar coordinate

pairs


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).


The Polar Equation of a Circle

Figure: r = a is the equation of a circle of radius |a| centered at

O.


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.


The Graph of Some Inequalities


Homework� Graph the set of points whose polar

coordinates satisfy given conditions:

r = 1

0É r É 1

r Ê 1


Q:14� Graph the set of points whose polar

coordinates satisfy the inequality 1É r É 2.


Q:14� Graph the set of points whose polar

coordinates satisfy the inequality 1É r É 2.

Sol.

Figure: A finite region: An annulus


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.


Q:15� 0É θ É π

6, r Ê 0.


Q:15� 0É θ É π

6, r Ê 0.

Sol.

Figure: An infinite region


Q:17� θ = π

3, −1É r É 3.


Q:17� θ = π

3, −1É r É 3.

Sol.

Figure: A line segment


Equation Relating Polar and Cartesian Coordinates

Figure: The usual way to relate polar and cartesian

coordinates


From figure, we have

x= r cosθ, y= rsinθ,

on squaring and adding:

x2+ y2 = r2,

on dividing:

θ = tan−1 y

x.


Q:37� Replace r = 5sinθ−2cosθ

by an equivalent

cartesian equation.


Q:37� Replace r = 5sinθ−2cosθ

by an equivalent

cartesian equation.

Sol.

r =5

sinθ−2cosθ⇒ rsinθ−2r cosθ = 5

⇒ y−2x= 5.


Example

Replace r2(1+sin2θ)= 1 by an equivalent cartesian

equation.


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.


Q:64� Replace (x−5)2+ y2 = 25 by an equivalent

polar equation.


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.


Section 11.4

Graphing in Polar

Coordinates


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.


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.


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 Test

Symmetry about the x-axis: If the point (r,θ)

lies on the graph, the point (r,−θ) or (−r,π−θ)


Symmetry about the y-axis: If the point (r,θ)

lies on the graph, the point (−r,−θ) or (r,π−θ)



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 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

)



Figure: Symmetry about x-axis and y-axis


Figure: Symmetry about the pole and the line y= x


Slope of a Polar Curve

The parametric equations of r = f (θ) are

x= r cosθ, y= rsinθ.


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).


Example

Find the slope of the curve r = 1+ cosθ at θ = π

2.


Example


2.

Sol.

dy

dx

∣

∣

∣

∣

(r,θ)



=−sinθsinθ+ (1+ cosθ)cosθ

−sinθcosθ− (1+ cosθ)sinθ.

Example


2.

Sol.

dy

dx

∣

∣

∣

∣

(r,θ)



=−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.


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).



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.


Step 3. drdθ

=−sinθ.


Step 3. drdθ

=−sinθ.

�drdθ

> 0⇒ sinθ < 0, thus no value of θ.


Step 3. drdθ

=−sinθ.

�drdθ


�drdθ

< 0⇒ sinθ > 0⇒ 0< θ <π, thus r

decreases in the interval [0,π].


Step 3. drdθ

=−sinθ.

�drdθ


�drdθ

< 0⇒ sinθ > 0⇒ 0< θ <π, thus r


Clearly max r = 2 at θ = 0 and min r = 0 at

θ =π.


Step 3. drdθ

=−sinθ.

�drdθ


�drdθ

< 0⇒ sinθ > 0⇒ 0< θ <π, thus r


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.


Step 5. Table θ vs r:

θ 0 π

4π

3π

22π3

3π4

π

r 2 1+ 1p2

32

1 12

1− 1p2

0


Figure: r = 1+ cosθ: The cardioid


Q:22� Trace the curve r = 1− cosθ.


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.


Figure: r = 1− cosθ: The cardioid


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.



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.


Q:20� Trace the curve r = cos2θ.



Sol.

Step 1. 1 Since (r,−θ) lies on the curve, so

symmetric about x-axis.



Sol.


symmetric about x-axis.2 Since (r,π−θ) lies on the curve, so

symmetric about y-axis.



Sol.



symmetric about y-axis.3 Since (−r,−π

2−θ) lies on the curve, so

symmetric about the line y= x.



Sol.



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.


Step 2. r = 0 gives cos2θ = 0 so that θ = π

4. Thus

θ = π

4is tangent to the curve at pole.



4. Thus

θ = π


Step 3. drdθ

=−2sin2θ.



4. Thus

θ = π


Step 3. drdθ

=−2sin2θ.

�drdθ

> 0⇒ sin2θ < 0, thus no value of θ.



4. Thus

θ = π


Step 3. drdθ

=−2sin2θ.

�drdθ


�drdθ

< 0⇒ sin2θ > 0⇒ 0< θ < π

2, thus r

decreases in the interval[

0, π4

]

.



4. Thus

θ = π


Step 3. drdθ

=−2sin2θ.

�drdθ


�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.


Step 4.dy

dx

∣

∣

∣

∣

θ=0

=∞.

Thus the tangent to the curve at θ = 0 is

perpendicular to x-axis.


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


Figure: r = cos2θ: Four leaved rose


Documents

Maths Lecture 6-7