Lecture 8 10

Mathematics I

Chapter 11

Dr. Devendra Kumar

Department of Mathematics

Birla Institute of Technology & Science, Pilani

2015–2016

Devendra Kumar BITS, Pilani Mathematics I

Deceptive Point

A point (r,θ) is said to be deceptive if it lies on the

curve but does not satisfy the equation of the curve.


Deceptive Point



Example

The point (0,0) lies on the curve r = 2cosθ but does

not satisfy the equation.


Deceptive Point



Example

The point (0,0) lies on the curve r = 2cosθ but does


Sol. Another representation(

0, π2

)

of (0,0) satisfies

the equation so does lie on the curve and hence is a

deceptive point.


Example

The point(

2, 3π4

)

lies on the curve r = 2sin2θ but does



Example

The point(

2, 3π4

)

lies on the curve r = 2sin2θ but does


Sol. Substituting the given point in r = 2sin2θ, we

obtain 2= 2sin 3π2

which gives 2=−2. Thus the point

does not satisfy the given equation.

Now consider the another representation of(

2, 3π4

)

as(

2, 3π4

)

=(

−2, 3π4−π

)

=(

−2,−π4

)

.


Now substituting the point(

−2,−π4

)

in r = 2sin2θ,

we obtain

−2= 2sin2(

−π

4

)

= 2sin(

−π

2

)

=−2.


Now substituting the point(

−2,−π4

)

in r = 2sin2θ,

we obtain

−2= 2sin2(

−π

4

)

= 2sin(

−π

2

)

=−2.

Thus(

2, 3π4

)

lies on the curve and hence is a

deceptive point.


Methods to find the points of intersection of two

given curves

To find the point of intersection of two given curves



given curves


Step 1 : Solve the simultaneous equations by

eliminating one variable (preferably r).



given curves




Step 2 : Step 1 may not give all points of

intersection (because of deceptive points).



given curves




Step 2 : Step 1 may not give all points of

intersection (because of deceptive points).

Plot the graphs simultaneously and see the

missing points of intersection in Step 1.


Example

Find the points of intersection of r = 1+ cosθ and

r = 1− cosθ.


Example

Find the points of intersection of r = 1+ cosθ and

r = 1− cosθ.

Sol. Step 1 : Solve the simultaneous equations:

Putting both values of r equal and solve for θ:

1+ cosθ = 1− cosθ,

which gives

cosθ = 0 and so θ =π

2,

3π

2.

Now substitute θ = π2

in first equation to get r = 1

(you can substitute in second equation also).Devendra Kumar BITS, Pilani Mathematics I

Thus one point of intersection is(

1, π2

)

.



1, π2

)

.

Now substitute θ = 3π2

in first equation to get r = 1.

Thus the second point of intersection is(

1, 3π2

)

.



1, π2

)

.




1, 3π2

)

.

Step 2: Plot the graphs simultaneously to find

the missing points of intersection in Step 1�



1, π2

)

.




1, 3π2

)

.


the missing points of intersection in Step 1� The

point of intersection (0,0) is found by graphing.



1, π2

)

.




1, 3π2

)

.



point of intersection (0,0) is found by graphing.

Thus all points of intersection are (0,0),(

1, π2

)

,(

1, 3π2

)

.


Example

Find the points of intersection of r = 1 and

r = 2sin2θ.


Example

Find the points of intersection of r = 1 and

r = 2sin2θ.

Sol. Step 1 : Solve the simultaneous equations:

Putting both values of r equal and solve for θ:

2sin2θ = 1

⇒ sin2θ =1

2

⇒ 2θ =π

6,

5π

6,

13π

6,

17π

6

⇒ θ =π

12,

5π

12,

13π

12,

17π

12.


The points of intersection are (as r = 1 for all these θ)

(

1,π

12

)

,

(

1,5π

12

)

,

(

1,13π

12

)

,

(

1,17π

12

)

.



(

1,π

12

)

,

(

1,5π

12

)

,

(

1,13π

12

)

,

(

1,17π

12

)

.


the missing points of intersection in Step 1�



(

1,π

12

)

,

(

1,5π

12

)

,

(

1,13π

12

)

,

(

1,17π

12

)

.



points of intersection(

1,7π

12

)

,

(

1,11π

12

)

,

(

1,19π

12

)

,

(

1,23π

12

)

,

are found by graphing and symmetry.


Faster Graphing


Faster Graphing Method

The polar equation can quickly be captured by

plotting r = f (θ) in cartesian θr-plane.





First graph r = f (θ) in the cartesian θr-plane

(that is, x-axis for θ and y-axis for r) in the

interval [0,2π].







interval [0,2π].

Use the cartesian graph as a table and guide to

sketch the polar coordinate graph.







interval [0,2π].

Use the cartesian graph as a table and guide to

sketch the polar coordinate graph.

The value of θ where the curve crosses (or

touches) θ-axis is a tangent to the curve at pole.


For the equation of the form r2 = f (θ) first trace

r2 = f (θ) in the cartesian θr2-plane (that is, x-axis

for θ and y-axis for r2) in the interval [0,2π] and

then graph r = f (θ) in the cartesian θr-plane (that

is, x-axis for θ and y-axis for r) in the interval

[0,2π].


Example

Trace the curve r = cos2θ using faster graphing.

Sol.

Figure: r = cos2θ: Four leaved roseDevendra Kumar BITS, Pilani Mathematics I

Example

Trace the curve r = cos3θ using faster graphing.

Sol.


Example

Trace the curve r = sin3θ using faster graphing.

Sol.


Example

Trace the curve r2 = sin2θ using faster graphing.

Sol.

Figure: r = cos3θ: Four leaved rose


Section 11.5

Areas and Lengths in

Polar Coordinates


To derive a formula for the area A of region OTS, we

approximate the region with n nonoverlapping fan

shaped circular sectors based on a partition P of

angle TOS. The typical sector has radius rk and

central angle ∆θk.


To derive a formula for the area A of region OTS, we

approximate the region with n nonoverlapping fan

shaped circular sectors based on a partition P of

angle TOS. The typical sector has radius rk and

central angle ∆θk.

Therefore the circular sector (with an angle ∆θk) is a

part of circle of radius rk. Thus the area of this

sector is

Ak =∆θk

2π(πr2

k)=1

2r2

k∆θk.


Then the area of region OTS is approximately

n∑

k=1

Ak =n∑

k=1

1

2r2

k∆θk.



n∑

k=1

Ak =n∑

k=1

1

2r2

k∆θk.

Thus the area A of OTS can be given by taking

n→∞. Thus



n∑

k=1

Ak =n∑

k=1

1

2r2

k∆θk.

Thus the area A of OTS can be given by taking

n→∞. Thus

A = limn→∞

n∑

k=1

Ak = limn→∞

n∑

k=1

1

2r2

k∆θk =1

2

∫β

αr2 dθ.


Then the area A of fan shaped region OTS between

the origin and the curve r = f (θ) and also between

the rays θ =α and θ =β is given by (assuming αÉβ)

A =1

2

∫β

αr2 dθ.


Steps to Evaluate Area

Plot the polar graphs correctly and clearly.




Label the relevant curves by their equations.





Shade the area required by marking its angular

spread and radial spread. You may require to find

intersections or tangents at pole.





Shade the area required by marking its angular

spread and radial spread. You may require to find

intersections or tangents at pole.

See that area is covered exactly once by radial

segment. Justify symmetries if used.


Example

Find the area of the region inside the lemniscate

r2 = 2a2 cos2θ, a> 0.


Example


r2 = 2a2 cos2θ, a> 0.

Sol.


Using symmetries, the required area is given by

A = 4×Area in first quadrant

= 4×1

2

∫π4

0r2 dθ

= 2

∫π4

02a2 cos2θ dθ

= 2a2.


Example


r2 = sin2θ.


Example


r2 = sin2θ.

Sol.


Using symmetries, the required area is given by

A = 4×Area lying between 0 and π/4

= 4×1

2

∫π4

0r2 dθ

= 2

∫π4

0sin2θ dθ

= 1.


Area Shared by Two Curves


Area Shared by Two Curves

Area of the region 0É r1(θ)É r É r2(θ), αÉ θ Éβ is

given by

A =1

2

∫β

α(r2

2− r21) dθ.


Q:10� Find the area of the region shared by the

circles r = 1 and r = 2sinθ.




Sol.

(i) The points of intersections are θ = π6, 5π

6.




Sol.


6.

(ii) Both the curves are symmetric about y-axis.




Sol.


6.

(ii) Both the curves are symmetric about y-axis.

(iii) Required area is the shaded area.


A1 =1

2

∫π6

0(2sinθ)2 dθ =

π

6−

p3

4.

A2 =1

2

∫π2

π6

(1)2 dθ =π

6.


A1 =1

2

∫π6

0(2sinθ)2 dθ =

π

6−

p3

4.

A2 =1

2

∫π2

π6

(1)2 dθ =π

6.

The area in first quadrant = A1+ A2 = π3−

p3

4.


A1 =1

2

∫π6

0(2sinθ)2 dθ =

π

6−

p3

4.

A2 =1

2

∫π2

π6

(1)2 dθ =π

6.

The area in first quadrant = A1+ A2 = π3−

p3

4.

Thus the required area A = 2(A1+ A2)= 2(

π3−

p3

4

)

.


Q:14� Find the area of the region inside the circle

r = 3acosθ and outside the cardioid

r = a(1+ cosθ), a> 0.




r = a(1+ cosθ), a> 0.

Sol.

(i) The points of intersections are θ =−π3, π

3.




r = a(1+ cosθ), a> 0.

Sol.


3.

(ii) Both the curves are symmetric about x-axis.




r = a(1+ cosθ), a> 0.

Sol.


3.


(iii) The graph also gives the point of

intersection (0,0).




r = a(1+ cosθ), a> 0.

Sol.


3.


(iii) The graph also gives the point of

intersection (0,0).

(iv) Required area is the shaded area.


A = 2×Area in first quadrant

= 2×1

2

∫π3

0[(3acosθ)2−a2(1+ cosθ)2] dθ

=πa2.


Q:20� Find the area of the region inside the cardioid

r = 1+ cosθ and outside the circle r = cosθ.


Q:20� Find the area of the region inside the cardioid

r = 1+ cosθ and outside the circle r = cosθ.

Sol.


Let A1 and A2 are the areas in first and second

quadrant respectively. Then


Let A1 and A2 are the areas in first and second

quadrant respectively. Then

A1 =1

2

∫π/2

0[(1+ cosθ)2− (cosθ)2] dθ

=1

2

(π

2+2

)

.

A2 =1

2

∫π

π/2(1+ cosθ)2 dθ

=1

2

(

3π

4−2

)

.


Thus the required area

A = 2(A1+ A2)

=(π

2+2

)

+(

3π

4−2

)

=5π

4.


Length of a Polar Curve

Let r = f (θ) has a continuous first derivative for

αÉ θ Éβ and the point P(r,θ) traces the curve

exactly once as θ varies from α to β. Then the length

of the curve from θ =α to θ =β is given by

L =∫β

α

√

r2+(

dr

dθ

)2

dθ.


Q:21� Find the length of the spiral r = θ2, 0É θ Ép

5.


Q:21� Find the length of the spiral r = θ2, 0É θ Ép

5.

Sol. The required length is

L =∫

p5

0

√

(θ2)2+ (2θ)2 dθ

=∫

p5

0

√

θ4+4θ2 dθ

=∫

p5

0|θ|

√

θ2+4 dθ

=∫

p5

0θ

√

θ2+4 dθ


Substitute u = θ2+4

L =1

2

∫9

4

pu du

=19

3.


Q:24� Find the length of the curve

r = asin2 θ

2, 0É θ Éπ, a> 0.


Q:24� Find the length of the curve

r = asin2 θ

2, 0É θ Éπ, a> 0.

Sol. Note that the curve is a cardioid. The required

length is

L =∫π

0

√

(

asin2 θ

2

)2

+(

asinθ

2cos

θ

2

)2

dθ

=∫π

0

∣

∣

∣

∣

asinθ

2

∣

∣

∣

∣

dθ


L =∫π

0asin

θ

2dθ

=−2a

(

cosθ

2

)π

0

= 2a.


Example

Find the length of the part of cardioid r = 1+ cosθ

that lies outside the circle r = sinθ.


Example

Find the length of the part of cardioid r = 1+ cosθ

that lies outside the circle r = sinθ.

Sol. The points of intersection are (0,0) and (1, π2).


The required length is the length of the cardioid from

−π to π2. Thus


The required length is the length of the cardioid from

−π to π2. Thus

L =∫π

2

−π

√

(1+ cosθ)2+ (−sinθ)2 dθ

=p

2

∫π2

−π

p1+ cosθ dθ

= 2

∫π2

−π

∣

∣

∣

∣

cosθ

2

∣

∣

∣

∣

dθ

= 2

∫π2

−πcos

θ

2dθ

= 2p

2(p

2+1).Devendra Kumar BITS, Pilani Mathematics I

Section 11.7

Conics in Polar

Coordinates


Polar Equation of a Straight Line

If the point P(r0,θ0) is the foot of the perpendicular

from the pole to the line L and r0 Ê 0, then the

equation of L is

r cos(θ−θ0)= r0.


Q:47� Sketch the line and find the cartesian equation

for r cos(

θ− 2π3

)

= 3.


Q:47� Sketch the line and find the cartesian equation

for r cos(

θ− 2π3

)

= 3.

Sol.

r cos

(

θ−2π

3

)

= 3

r

(

cosθcos2π

3+sinθsin

2π

3

)

= 3

−1

2x+

p3

2y= 3

−x+p

3y= 6.


Q:50� Find a polar equation forp

3x− y= 1 in the

form r cos(θ−θ0)= r0.


Q:50� Find a polar equation forp

3x− y= 1 in the

form r cos(θ−θ0)= r0.

Sol.p

3x− y= 1p

3r cosθ− rsinθ = 1

r

( p3

2cosθ−

1

2sinθ

)

=1

2

r(

cosπ

6cosθ−sin

π

6sinθ

)

=1

2

r cos(

θ+π

6

)

=1

2.


Polar Equation of a Circle

The polar equation of a circle of radius a and

centered at (r0,θ0) is (using cosines law)

r2+ r20−2rr0 cos(θ−θ0)= a2.


Circle passes through the pole

If the circle passes through the origin, then r0 = a

and the equation simplifies to

r = 2acos(θ−θ0).


Special Cases

1 Equation of a circle centered at (a,0) and

radius a. If the center lies on positive x-axis then

the equation becomes

r = 2acosθ.


Special Cases

1 Equation of a circle centered at (a,0) and

radius a. If the center lies on positive x-axis then


r = 2acosθ.

2 Equation of a circle centered at (a, π2) and

radius a. If the center lies on positive y-axis then


r = 2asinθ.


3 Equation of a circle centered at (−a,0) and

radius a. If the center lies on negative x-axis

then the equation becomes

r =−2acosθ.


3 Equation of a circle centered at (−a,0) and

radius a. If the center lies on negative x-axis


r =−2acosθ.

4 Equation of a circle centered at (−a, π2) and

radius a. If the center lies on negative y-axis


r =−2asinθ.


Q:54� Sketch the circle r = 6sinθ. Find polar

coordinate of the center and identify the radius.


Q:54� Sketch the circle r = 6sinθ. Find polar

coordinate of the center and identify the radius.

Sol. Compare with r = 2asinθ, we get a= 3.

Therefore the polar coordinate of the center is(

3, π2

)

and radius is 3.


Q:60� Find polar equation for the circle

x2+ (y+7)2 = 49. Sketch the circle and label it with

both its cartesian and polar equations.



x2+ (y+7)2 = 49. Sketch the circle and label it with


Sol. Compare with (x− x0)2+ (y− y0)2 = a2. The

center is (0,−7) and radius is a= 7. Therefore the

polar equation is r =−14sinθ.



x2+ y2− 43

y= 0. Sketch the circle and label it with




x2+ y2− 43

y= 0. Sketch the circle and label it with


Sol. Rewrite the equation as x2+(

y− 23

)2 =(

23

)2.

Compare with (x− x0)2+ (y− y0)2 = a2. The center is(

0, 23

)

and radius is a= 23. Therefore the polar

equation is r = 43

sinθ or one can find by changing

into polar form as follows: x2+ y2− 43

y= 0 gives

r2− 43rsinθ = 0 or r = 4

3sinθ.


Conic Sections

Conic Section

The curve obtained from the intersection of a double

cone with a plane is called conic section.


Conic Sections

Conic Section

The curve obtained from the intersection of a double

cone with a plane is called conic section.

There are three proper conic sections:

1 Parabola

2 Ellipse

3 Hyperbola

Each conic section can be defined in several ways.


Directrix

A directrix of a conic section is a line which together

with the focus (a fixed point) define a conic section as

the locus of the points whose distance from the focus

is proportional to the distance from the directrix.


Directrix

A directrix of a conic section is a line which together

with the focus (a fixed point) define a conic section as

the locus of the points whose distance from the focus

is proportional to the distance from the directrix.

The constant of proportionality is called the

eccentricity of the conic section.


Parabola

A parabola is a locus of points equidistant from a

fixed point, called the focus of the parabola, and a

line, called the directrix of the parabola.


For ellipse and hyperbola, there are two special

points - their foci - in terms of which the definitions

are set. Denote the foci F1 and F2.


For ellipse and hyperbola, there are two special

points - their foci - in terms of which the definitions

are set. Denote the foci F1 and F2.

Ellipse

An ellipse is a locus of points the sum of whose

distances to F1 and F2 is a constant.


Hyperbola

Hyperbola is a locus of points for which the absolute

value of the difference of the distances to F1 and F2

is a constant.


Remark.

For parabola there is one focus and one directrix

while for ellipse and hyperbola there are two foci and

two directrices.


Let e > 0 be the eccentricity. Then we have



e < 1 : ellipse



e < 1 : ellipse

e = 1 : parabola



e < 1 : ellipse

e = 1 : parabola

e > 1 : hyperbola



e < 1 : ellipse

e = 1 : parabola

e > 1 : hyperbola

Remark.

In all the problems we consider one focus of the conic

section at the origin.


Polar Equation of Conic Section: Case I

If the directrix be x= k, (k > 0) (vertical, to the right

of the origin). Then the polar equation of the conic

section is

r =ke

1+ ecosθ.


Case II

If the directrix be x=−k, (k > 0) (vertical, to the left

of the origin). Then the polar equation of the conic

section is

r =ke

1− ecosθ.


Case III

If the directrix be y= k, (k > 0) (horizontal, above to

the origin). Then the polar equation of the conic

section is

r =ke

1+ esinθ.


Case IV

If the directrix be y=−k, (k > 0) (horizontal, below to

the origin). Then the polar equation of the conic

section is

r =ke

1− esinθ.


Polar equation of an ellipse

For an ellipse with semi-major axis a and

eccentricity e (with focus at the origin), we have

k =a

e− ea

ke = a(1− e2).

Hence from case I,

r =a(1− e2)

1+ ecosθ.


Q:31� If e = 5, and y=−6, then find equation of the

conic section (Assume one focus at the origin).


Q:31� If e = 5, and y=−6, then find equation of the


Sol.

r =ke

1− esinθ

r =6(5)

1−5sinθ

r =30

1−5sinθ.


Q:34� If e = 14, and x=−2, then find equation of the



Q:34� If e = 14, and x=−2, then find equation of the


Sol.

r =ke

1− ecosθ

r =(2)

(

14

)

1− 14

cosθ

r =2

4− cosθ.


Q:36� If e = 13, and y= 6, then find equation of the



Q:36� If e = 13, and y= 6, then find equation of the


Sol.

r =ke

1+ esinθ

r =(6)

(

13

)

1+ 13

sinθ

r =6

3+sinθ.


Q:39� Sketch r = 2510−5cosθ

. Include the directrix that

corresponds to the focus at the origin. Level the

vertices with appropriate polar coordinates. Label

the center in the case of ellipse.


Q:39� Sketch r = 2510−5cosθ





Sol.

r =25

10−5cosθ

r =52

1− 12

cosθ.

So e = 12

and ke = 52⇒ k = 5 and so x=−5 is the

directrix. Since e < 1 so the curve is an ellipse.


Now for the ellipse

ke = a(1− e2)

⇒5

2= a

(

1−1

4

)

⇒ a=10

3

⇒ ae =5

3.

Thus center =(

53,0

)

.


Q:42� Sketch r = 123+3sinθ






Q:42� Sketch r = 123+3sinθ





Sol.

r =12

3+3sinθ

r =4

1+sinθ.

So e = 1 and ke = 4⇒ k = 4 and so y= 4 is the

directrix. Since e = 1 so the curve is a parabola.


Q:44� Sketch r = 42−sinθ






Q:44� Sketch r = 42−sinθ





Sol.

r =4

2−sinθ

r =2

1− 12

sinθ.

So e = 12

and ke = 2⇒ k = 4 and so y=−4 is the

directrix. Since e < 1 so the curve is an ellipse.


Now for the ellipse

ke = a(1− e2)

⇒ 2= a

(

1−1

4

)

⇒ a=8

3

⇒ ae =4

3.

Thus center =(

43, π

2

)

.


Lecture 8 10

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