7/31/2019 Lecture 10 (Calculus of Polar Curves)

1/184

Tangent Lines to Polar Curves

Calculus of Polar Curves

Institute of Mathematics, University of the Philippines Diliman

Mathematics 54 (Elementary Analysis 2)

Polar Curves 1/ 18

7/31/2019 Lecture 10 (Calculus of Polar Curves)

2/184

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

Polar Curves 2/ 18

7/31/2019 Lecture 10 (Calculus of Polar Curves)

3/184

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

Polar Curves 2/ 18

7/31/2019 Lecture 10 (Calculus of Polar Curves)

4/184

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

Polar Curves 2/ 18

l

7/31/2019 Lecture 10 (Calculus of Polar Curves)

5/184

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

Polar Curves 2/ 18

T t Li t P l C

7/31/2019 Lecture 10 (Calculus of Polar Curves)

6/184

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

Polar Curves 2/ 18

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

7/184

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

Goal: obtain slopes of tangent lines to polar curves of form r=f()

Polar Curves 2/ 18

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

8/184

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

Parametrization of a Polar Curve

A polar curve r=f() can be parametrized as

Polar Curves 3/ 18

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

9/184

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

Parametrization of a Polar Curve

A polar curve r=f() can be parametrized asx = rcos

Polar Curves 3/ 18

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

10/184

a ge t es to o a Cu es

Tangent Lines to Polar Curves

Parametrization of a Polar Curve

A polar curve r=f() can be parametrized asx = rcos = f()cos

Polar Curves 3/ 18

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

11/184

g

Tangent Lines to Polar Curves

Parametrization of a Polar Curve

A polar curve r=f() can be parametrized asx = rcos = f()cosy = rsin

Polar Curves 3/ 18

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

12/184

Tangent Lines to Polar Curves

Parametrization of a Polar Curve

A polar curve r=f() can be parametrized asx = rcos = f()cosy = rsin = f()sin

Polar Curves 3/ 18

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

13/184

Tangent Lines to Polar Curves

Parametrization of a Polar Curve

A polar curve r=f() can be parametrized asx = rcos = f()cosy = rsin = f()sin

Recall. slope of a parametric curvedydx

=

dyddx

d

Polar Curves 3/ 18

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

14/184

Tangent Lines to Polar Curves

Parametrization of a Polar Curve

A polar curve r=f() can be parametrized asx = rcos = f()cosy = rsin = f()sin

Recall. slope of a parametric curvedydx

=

dyddx

d

dx

d=f() cosf()sin,

Polar Curves 3/ 18

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

15/184

Tangent Lines to Polar Curves

Parametrization of a Polar Curve

A polar curve r=f() can be parametrized asx = rcos = f()cosy = rsin = f()sin

Recall. slope of a parametric curvedydx

=

dyddx

d

dx

d=f() cosf()sin, dy

d=f() sin+f()cos

Polar Curves 3/ 18

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

16/184

Tangent Lines to Polar Curves

Parametrization of a Polar Curve

A polar curve r=f() can be parametrized asx = rcos = f()cosy = rsin = f()sin

Recall. slope of a parametric curvedydx

=

dyddx

d

dx

d=f() cosf()sin, dy

d=f() sin+f()cos

Slope of a Tangent Line to a Polar Curve

Given that dy/d and dx/d are continuous and dx/d = 0, then the slope of thepolar curve r=f()is

dy

dx=

drd

sin+ rcosdrd

cos rsin

Polar Curves 3/ 18

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

17/184

Tangent Lines to Polar Curves

Example.

Find the (Cartesian) equation of the tangent line to the cardioid r

=1

+sin at the

point where = 3 .

Polar Curves 4/ 18

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

18/184

Tangent Lines to Polar Curves

Example.

Find the (Cartesian) equation of the tangent line to the cardioid r

=1

+sin at the

point where = 3 .

Solution. Recall thatdydx

=drd

sin+rcosdrd

cosrsin

Polar Curves 4/ 18

Tangent Lines to Polar Curves

l

7/31/2019 Lecture 10 (Calculus of Polar Curves)

19/184

Tangent Lines to Polar Curves

Example.

Find the (Cartesian) equation of the tangent line to the cardioid r

=1

+sin at the

point where = 3 .

Solution. Recall thatdydx

=drd

sin+rcosdrd

cosrsin

Meanwhile, we have the following:

r= 1 +sin 3

Polar Curves 4/ 18

Tangent Lines to Polar Curves

T Li P l C

7/31/2019 Lecture 10 (Calculus of Polar Curves)

20/184

Tangent Lines to Polar Curves

Example.

Find the (Cartesian) equation of the tangent line to the cardioid r

=1

+sin at the

point where = 3 .

Solution. Recall thatdydx

=drd

sin+rcosdrd

cosrsin

Meanwhile, we have the following:

r= 1 +sin 3

= 1+ 32

Polar Curves 4/ 18

Tangent Lines to Polar Curves

T t Li t P l C

7/31/2019 Lecture 10 (Calculus of Polar Curves)

21/184

Tangent Lines to Polar Curves

Example.

Find the (Cartesian) equation of the tangent line to the cardioid r

=1

+sin at the

point where = 3 .

Solution. Recall thatdydx

=drd

sin+rcosdrd

cosrsin

Meanwhile, we have the following:

r= 1 +sin 3

= 1+ 32 = 2+32

Polar Curves 4/ 18

Tangent Lines to Polar Curves

T t Li t P l C

7/31/2019 Lecture 10 (Calculus of Polar Curves)

22/184

Tangent Lines to Polar Curves

Example.

Find the (Cartesian) equation of the tangent line to the cardioid r= 1 +sin at thepoint where = 3 .

Solution. Recall thatdydx

=drd

sin+rcosdrd

cosrsin

Meanwhile, we have the following:

r= 1 +sin 3

= 1+ 32 = 2+32drd

Polar Curves 4/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

23/184

Tangent Lines to Polar Curves

Example.

Find the (Cartesian) equation of the tangent line to the cardioid r= 1 +sin at thepoint where = 3 .

Solution. Recall thatdydx

=drd

sin+rcosdrd

cosrsin

Meanwhile, we have the following:

r= 1 +sin 3

= 1+ 32 = 2+32drd

= cos

3

Polar Curves 4/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

24/184

Tangent Lines to Polar Curves

Example.

Find the (Cartesian) equation of the tangent line to the cardioid r= 1 +sin at thepoint where = 3 .

Solution. Recall thatdydx

=drd

sin+rcosdrd

cosrsin

Meanwhile, we have the following:

r= 1 +sin 3

= 1+ 32 = 2+32drd

= cos

3

= 12

Polar Curves 4/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

25/184

Tangent Lines to Polar Curves

Example.

Find the (Cartesian) equation of the tangent line to the cardioid r= 1 +sin at thepoint where = 3 .

Solution. Recall thatdydx

=drd

sin+rcosdrd

cosrsin

Meanwhile, we have the following:

r= 1 +sin 3

= 1+ 32 = 2+32drd

= cos

3

= 12

cos3

Polar Curves 4/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

26/184

Tangent Lines to Polar Curves

Example.

Find the (Cartesian) equation of the tangent line to the cardioid r= 1 +sin at thepoint where = 3 .

Solution. Recall thatdydx

=drd

sin+rcosdrd

cosrsin

Meanwhile, we have the following:

r= 1 +sin 3

= 1+ 32 = 2+32drd

= cos

3

= 12

cos3

= 12

Polar Curves 4/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

27/184

Tangent Lines to Polar Curves

Example.

Find the (Cartesian) equation of the tangent line to the cardioid r= 1 +sin at thepoint where = 3 .

Solution. Recall thatdydx

=drd

sin+rcosdrd

cosrsin

Meanwhile, we have the following:

r= 1 +sin 3

= 1+ 32 = 2+32drd

= cos

3

= 12

cos3

= 12

sin

3=

Polar Curves 4/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

28/184

Tangent Lines to Polar Curves

Example.

Find the (Cartesian) equation of the tangent line to the cardioid r= 1 +sin at thepoint where = 3 .

Solution. Recall thatdydx

=drd

sin+rcosdrd

cosrsin

Meanwhile, we have the following:

r= 1 +sin 3

= 1+ 32 = 2+32drd

= cos

3

= 12

cos3

= 12

sin

3=

32

Polar Curves 4/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

29/184

Tangent Lines to Polar Curves

Example.

Find the (Cartesian) equation of the tangent line to the cardioid r= 1 +sin at thepoint where =

3 .

Solution. Recall thatdydx

=drd

sin+rcosdrd

cosrsin

Meanwhile, we have the following:

r= 1 +sin 3

= 1+ 32 = 2+32drd

= cos

3

= 12

cos3

= 12

sin

3=

32

Hence,

dy

dx=

Polar Curves 4/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

30/184

g C

Example.

Find the (Cartesian) equation of the tangent line to the cardioid r= 1 +sin at thepoint where =

3 .

Solution. Recall thatdydx

=drd

sin+rcosdrd

cosrsin

Meanwhile, we have the following:

r= 1 +sin 3

= 1+ 32 = 2+32drd

= cos

3

= 12

cos3

= 12

sin

3=

32

Hence,

dy

dx=

12

Polar Curves 4/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

31/184

g

Example.

Find the (Cartesian) equation of the tangent line to the cardioid r= 1 +sin at thepoint where =

3 .

Solution. Recall thatdydx

=drd

sin+rcosdrd

cosrsin

Meanwhile, we have the following:

r= 1 +sin 3

= 1+ 32 = 2+32drd

= cos

3

= 12

cos3

= 12

sin

3=

32

Hence,

dy

dx=

12

3

2

+

Polar Curves 4/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

32/184

g

Example.

Find the (Cartesian) equation of the tangent line to the cardioid r= 1 +sin at thepoint where =

3 .

Solution. Recall thatdydx

=drd

sin+rcosdrd

cosrsin

Meanwhile, we have the following:

r= 1 +sin 3

= 1+ 32 = 2+32drd

= cos

3

= 12

cos3

= 12

sin

3=

3

2

Hence,

dy

dx=

12

3

2

+ 2+

3

2

Polar Curves 4/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

33/184

g

Example.

Find the (Cartesian) equation of the tangent line to the cardioid r= 1 +sin at thepoint where =

3 .

Solution. Recall thatdydx

=drd

sin+rcosdrd

cosrsin

Meanwhile, we have the following:

r= 1 +sin 3

= 1+ 32 = 2+32drd

= cos

3

= 12

cos3

= 12

sin

3=

3

2

Hence,

dy

dx=

12

3

2

+ 2+

3

2

12

Polar Curves 4/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

34/184

Example.

Find the (Cartesian) equation of the tangent line to the cardioid r= 1 +sin at thepoint where =

3 .

Solution. Recall thatdydx

=drd

sin+rcosdrd

cosrsin

Meanwhile, we have the following:

r= 1 +sin 3

= 1+ 32 = 2+32drd

= cos

3

= 12

cos3

= 12

sin

3=

3

2

Hence,

dy

dx=

12

3

2

+ 2+

3

2

12

12

Polar Curves 4/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

35/184

Example.

Find the (Cartesian) equation of the tangent line to the cardioid r= 1 +sin at thepoint where =

3 .

Solution. Recall thatdydx

=drd

sin+rcosdrd

cosrsin

Meanwhile, we have the following:

r= 1 +sin 3

= 1+ 32 = 2+32drd

= cos

3

= 12

cos3

= 12

sin3

=

32

Hence,

dy

dx=

12

3

2

+ 2+

3

2

12

12

12

Polar Curves 4/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

36/184

Example.

Find the (Cartesian) equation of the tangent line to the cardioid r= 1 +sin at thepoint where =

3 .

Solution. Recall thatdydx

=drd

sin+rcosdrd

cosrsin

Meanwhile, we have the following:

r= 1 +sin 3

= 1+ 32 = 2+32drd

= cos

3

= 12

cos3

= 12

sin3

=

32

Hence,

dy

dx=

12

3

2

+ 2+

3

2

12

12

12

2+

32

Polar Curves 4/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

37/184

Example.

Find the (Cartesian) equation of the tangent line to the cardioid r= 1 +sin at thepoint where =

3 .

Solution. Recall thatdydx

=drd

sin+rcosdrd

cosrsin

Meanwhile, we have the following:

r= 1 +sin 3

= 1+ 32 = 2+32drd

= cos

3

= 12

cos3

= 12

sin3

=

32

Hence,

dy

dx=

12

3

2

+ 2+

3

2

12

12

12

2+

32

3

2 Polar Curves 4/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

38/184

Example.

Find the (Cartesian) equation of the tangent line to the cardioid r= 1 +sin at thepoint where =

3 .

Solution. Recall thatdydx

=drd

sin+rcosdrd

cosrsin

Meanwhile, we have the following:

r= 1 +sin 3

= 1+ 32 = 2+32drd

= cos

3

= 12

cos3

= 12

sin3

=

32

Hence,

dy

dx=

12

3

2

+ 2+

3

2

12

12

12

2+

3

2

3

2

= 1+

32

1

32

Polar Curves 4/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

39/184

Example.

Find the (Cartesian) equation of the tangent line to the cardioid r= 1 +sin at thepoint where =

3 .

Solution. Recall thatdydx

=drd

sin+rcosdrd

cosrsin

Meanwhile, we have the following:

r= 1 +sin 3

= 1+

32 = 2+

32

drd

= cos

3

= 12

cos3

= 12

sin3

=

32

Hence,

dy

dx=

12

3

2

+ 2+

3

2

12

12

12

2+

3

2

3

2

= 1+

32

1

32

= 1

Polar Curves 4/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

40/184

Example.

Find the (Cartesian) equation of the tangent line to the cardioid r= 1 +sin at thepoint where =

3 .

Solution. Recall thatdydx

=drd

sin+rcosdrd

cosrsin

Meanwhile, we have the following:

r= 1 +sin 3

= 1+

32 = 2+

32

drd

= cos

3

= 12

cos3

= 12

sin3

=

32

Hence,

dy

dx=

12

3

2

+ 2+

3

2

12

12

12

2+

3

2

3

2

= 1+

32

1

32

= 1

The point of tangency

2+

32 ,

3

in

Cartesian is

Polar Curves 4/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

41/184

Example.

Find the (Cartesian) equation of the tangent line to the cardioid r= 1 +sin at the

point where =

3 .

Solution. Recall thatdydx

=drd

sin+rcosdrd

cosrsin

Meanwhile, we have the following:

r= 1 +sin 3

= 1+

32 = 2+

32

drd

= cos

3

= 12

cos3

= 12

sin3 =

3

2

Hence,

dy

dx=

12

3

2

+ 2+

3

2

12

12

12

2+

3

2

3

2

= 1+

32

1

32

= 1

The point of tangency

2+

32 ,

3

in

Cartesian is

2+

34 ,

3+2

34

.

Polar Curves 4/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

42/184

Example.

Find the (Cartesian) equation of the tangent line to the cardioid r= 1 +sin at the

point where =

3 .

Solution. Recall thatdydx

=drd

sin+rcosdrd

cosrsin

Meanwhile, we have the following:

r= 1 +sin 3

= 1+

32 = 2+

32

drd

= cos

3

= 12

cos3

= 12

sin3 =

3

2

Hence,

dy

dx=

12

3

2

+ 2+

3

2

12

12

12

2+

3

2

3

2

= 1+

32

1

32

= 1

The point of tangency

2+

32 ,

3

in

Cartesian is

2+

34 ,

3+2

34

.

Thus, the equation is

y

3+2

3

4= x

2+

34

Polar Curves 4/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

43/184

Example.

Find the (Cartesian) equation of the tangent line to the cardioid r= 1 +sin at the

point where =

3 .

Solution. Recall thatdydx

=drd

sin+rcosdrd

cosrsin

Meanwhile, we have the following:

r= 1 +sin 3

= 1+

32 = 2+

32

drd

= cos

3

= 12

cos3

= 12

sin3 =

3

2

Hence,

dy

dx=

12

3

2

+ 2+

3

2

12

12

12

2+

3

2

3

2

= 1+

32

1

32

= 1

The point of tangency

2+

32 ,

3

in

Cartesian is

2+

34 ,

3+2

34

.

Thus, the equation is

y

3+2

3

4= x

2+

34

Polar Curves 4/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

44/184

Example.

Determine the points on the cardioid r= 1 coswhere the tangent line ishorizontal, or vertical.

Polar Curves 5/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

45/184

Example.

Determine the points on the cardioid r= 1 coswhere the tangent line ishorizontal, or vertical.

Recall. For a parametric curve

C : x

=x(), y

=y(),

dyd

= 0

Polar Curves 5/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

46/184

Example.

Determine the points on the cardioid r= 1 coswhere the tangent line ishorizontal, or vertical.

Recall. For a parametric curve

C : x

=x(), y

=y(),

dyd

= 0 = horizontal tangent line

Polar Curves 5/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

47/184

Example.

Determine the points on the cardioid r= 1 coswhere the tangent line ishorizontal, or vertical.

Recall. For a parametric curve

C : x

=x(), y

=y(),

dyd

= 0 = horizontal tangent linedxd

= 0

Polar Curves 5/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

48/184

Example.

Determine the points on the cardioid r= 1 coswhere the tangent line ishorizontal, or vertical.

Recall. For a parametric curve

C : x

=x(), y

=y(),

dyd

= 0 = horizontal tangent linedxd

= 0 = vertical tangent line

Polar Curves 5/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

49/184

Example.

Determine the points on the cardioid r= 1 coswhere the tangent line ishorizontal, or vertical.

Recall. For a parametric curve

C : x

=x(), y

=y(),

dyd

= 0 = horizontal tangent linedxd

= 0 = vertical tangent lineprovided that they are not simultaneously zero.

Polar Curves 5/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

50/184

Example.

Determine the points on the cardioid r= 1 coswhere the tangent line ishorizontal, or vertical.

Recall. For a parametric curve

C : x

=x(), y

=y(),

dyd

= 0 = horizontal tangent linedxd

= 0 = vertical tangent lineprovided that they are not simultaneously zero.

Note that for the above curve, we have the following parametrization:

Polar Curves 5/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

51/184

Example.

Determine the points on the cardioid r= 1 coswhere the tangent line ishorizontal, or vertical.

Recall. For a parametric curve

C : x

=x(), y

=y(),

dyd

= 0 = horizontal tangent linedxd

= 0 = vertical tangent lineprovided that they are not simultaneously zero.

Note that for the above curve, we have the following parametrization:

x = rcos

Polar Curves 5/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

52/184

Example.

Determine the points on the cardioid r= 1 coswhere the tangent line ishorizontal, or vertical.

Recall. For a parametric curve

C : x

=x(), y

=y(),

dyd

= 0 = horizontal tangent linedxd

= 0 = vertical tangent lineprovided that they are not simultaneously zero.

Note that for the above curve, we have the following parametrization:

x = rcos = (1 cos)cos

Polar Curves 5/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

53/184

Example.

Determine the points on the cardioid r= 1 coswhere the tangent line ishorizontal, or vertical.

Recall. For a parametric curve

C : x

=x(), y

=y(),

dyd

= 0 = horizontal tangent linedxd

= 0 = vertical tangent lineprovided that they are not simultaneously zero.

Note that for the above curve, we have the following parametrization:

x = rcos = (1 cos)cos= coscos2

Polar Curves 5/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

54/184

Example.

Determine the points on the cardioid r= 1 coswhere the tangent line ishorizontal, or vertical.

Recall. For a parametric curve

C : x

=x(), y

=y(),

dyd

= 0 = horizontal tangent linedxd

= 0 = vertical tangent lineprovided that they are not simultaneously zero.

Note that for the above curve, we have the following parametrization:

x = rcos = (1 cos)cos= coscos2y = rsin

Polar Curves 5/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

55/184

Example.

Determine the points on the cardioid r= 1 coswhere the tangent line ishorizontal, or vertical.

Recall. For a parametric curve

C : x= x(), y=y(),dyd

= 0 = horizontal tangent linedxd

= 0 = vertical tangent lineprovided that they are not simultaneously zero.

Note that for the above curve, we have the following parametrization:

x = rcos = (1 cos)cos= coscos2y = rsin= (1 cos)sin

Polar Curves 5/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

56/184

Example.

Determine the points on the cardioid r= 1 coswhere the tangent line ishorizontal, or vertical.

Recall. For a parametric curve

C : x= x(), y=y(),dyd

= 0 = horizontal tangent linedxd

= 0 = vertical tangent lineprovided that they are not simultaneously zero.

Note that for the above curve, we have the following parametrization:

x = rcos = (1 cos)cos= coscos2y = rsin= (1 cos)sin= sinsincos

Polar Curves 5/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

E l ti d

7/31/2019 Lecture 10 (Calculus of Polar Curves)

57/184

Example. continued...

Determine the points on the cardioid C : x= coscos2, y= sin sincoswhere the tangent line is horizontal, or vertical.

dx

d=

Polar Curves 6/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

Example continued

7/31/2019 Lecture 10 (Calculus of Polar Curves)

58/184

Example. continued...

Determine the points on the cardioid C : x= coscos2, y= sin sincoswhere the tangent line is horizontal, or vertical.

dx

d= sin+2cos sin

Polar Curves 6/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

Example continued

7/31/2019 Lecture 10 (Calculus of Polar Curves)

59/184

Example. continued...

Determine the points on the cardioid C : x= coscos2, y= sin sincoswhere the tangent line is horizontal, or vertical.

dx

d= sin+2cos sin= sin(2cos1)

Polar Curves 6/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

Example continued

7/31/2019 Lecture 10 (Calculus of Polar Curves)

60/184

Example. continued...

Determine the points on the cardioid C : x= coscos2, y= sin sincoswhere the tangent line is horizontal, or vertical.

dx

d= sin+2cos sin= sin(2cos1)

=0

Polar Curves 6/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

Example continued

7/31/2019 Lecture 10 (Calculus of Polar Curves)

61/184

Example. continued...

Determine the points on the cardioid C : x= coscos2, y= sin sincoswhere the tangent line is horizontal, or vertical.

dx

d= sin+2cos sin= sin(2cos1)

=0

=sin

=0 or cos

=1

2

Polar Curves 6/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

Example continued

7/31/2019 Lecture 10 (Calculus of Polar Curves)

62/184

Example. continued...

Determine the points on the cardioid C : x= coscos2, y= sin sincoswhere the tangent line is horizontal, or vertical.

dx

d= sin+2cos sin= sin(2cos1)

=0

=sin

=0 or cos

=1

2 =

=0,

Polar Curves 6/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

Example. continued...

7/31/2019 Lecture 10 (Calculus of Polar Curves)

63/184

Example. continued...

Determine the points on the cardioid C : x= coscos2, y= sin sincoswhere the tangent line is horizontal, or vertical.

dx

d= sin+2cos sin= sin(2cos1)

=0

=sin

=0 or cos

=1

2 =

=0,,

Polar Curves 6/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

Example. continued...

7/31/2019 Lecture 10 (Calculus of Polar Curves)

64/184

p

Determine the points on the cardioid C : x= coscos2, y= sin sincoswhere the tangent line is horizontal, or vertical.

dx

d= sin+2cos sin= sin(2cos1)

=0

=sin

=0 or cos

=1

2 =

=0,,

3

,

Polar Curves 6/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

Example. continued...

7/31/2019 Lecture 10 (Calculus of Polar Curves)

65/184

p

Determine the points on the cardioid C : x= coscos2, y= sin sincoswhere the tangent line is horizontal, or vertical.

dx

d= sin+2cos sin= sin(2cos1)

=0

=sin

=0 or cos

=1

2 =

=0,,

3

,5

3

Polar Curves 6/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

Example. continued...

7/31/2019 Lecture 10 (Calculus of Polar Curves)

66/184

p

Determine the points on the cardioid C : x= coscos2, y= sin sincoswhere the tangent line is horizontal, or vertical.

dx

d= sin+2cos sin= sin(2cos1)

=0

=sin

=0 or cos

=1

2 =

=0,,

3

,5

3

dy

d=

Polar Curves 6/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

Example. continued...

7/31/2019 Lecture 10 (Calculus of Polar Curves)

67/184

Determine the points on the cardioid C : x= coscos2, y= sin sincoswhere the tangent line is horizontal, or vertical.

dx

d= sin+2cos sin= sin(2cos1)

=0

=sin

=0 or cos

=1

2 =

=0,,

3

,5

3

dy

d= cos (cos2 sin2)

Polar Curves 6/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

Example. continued...

7/31/2019 Lecture 10 (Calculus of Polar Curves)

68/184

Determine the points on the cardioid C : x= coscos2, y= sin sincoswhere the tangent line is horizontal, or vertical.

dx

d= sin+2cos sin= sin(2cos1)

=0

=sin

=0 or cos

=

1

2 =

=0,,

3

,5

3

dy

d= cos (cos2 sin2)

= cos 2cos2+1

Polar Curves 6/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

Example. continued...

7/31/2019 Lecture 10 (Calculus of Polar Curves)

69/184

Determine the points on the cardioid C : x= coscos2, y= sin sincoswhere the tangent line is horizontal, or vertical.

dx

d= sin+2cos sin= sin(2cos1)

=0

=sin

=0 or cos

=

1

2 =

=0,,

3

,5

3

dy

d= cos (cos2 sin2)

= cos 2cos2+1

=(1

cos)(1

+2cos)

Polar Curves 6/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

Example. continued...

7/31/2019 Lecture 10 (Calculus of Polar Curves)

70/184

Determine the points on the cardioid C : x= coscos2, y= sin sincoswhere the tangent line is horizontal, or vertical.

dx

d= sin+2cos sin= sin(2cos1)

=0

=sin

=0 or cos

=

1

2 =

=0,,

3

,5

3

dy

d= cos (cos2 sin2)

= cos 2cos2+1

=(1

cos)(1

+2cos)

= 0

Polar Curves 6/ 18

7/31/2019 Lecture 10 (Calculus of Polar Curves)

71/184

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

Example. continued...

7/31/2019 Lecture 10 (Calculus of Polar Curves)

72/184

Determine the points on the cardioid C : x= coscos2, y= sin sincoswhere the tangent line is horizontal, or vertical.

dx

d= sin+2cos sin= sin(2cos1)

=0

=sin

=0 or cos

=

1

2 =

=0,,

3

,5

3

dy

d= cos (cos2 sin2)

= cos 2cos2+1

=(1

cos)(1

+2cos)

= 0 = cos= 1 or cos = 12

= = 0,

Polar Curves 6/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

Example. continued...

7/31/2019 Lecture 10 (Calculus of Polar Curves)

73/184

Determine the points on the cardioid C : x= coscos2, y= sin sincoswhere the tangent line is horizontal, or vertical.

dx

d= sin+2cos sin= sin(2cos1)

=0

=sin

=0 or cos

=

1

2 =

=0,,

3

,5

3

dy

d= cos (cos2 sin2)

= cos 2cos2+1

=(1

cos)(1

+2cos)

= 0 = cos= 1 or cos = 12

= = 0, 23

,

Polar Curves 6/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

Example. continued...

7/31/2019 Lecture 10 (Calculus of Polar Curves)

74/184

Determine the points on the cardioid C : x= coscos2, y= sin sincoswhere the tangent line is horizontal, or vertical.

dx

d= sin+2cos sin= sin(2cos1)

=0

=sin

=0 or cos

=

1

2 =

=0,,

3

,5

3

dy

d= cos (cos2 sin2)

= cos 2cos2+1

=(1

cos)(1

+2cos)

= 0 = cos= 1 or cos = 12

= = 0, 23

, 43

Polar Curves 6/ 18

Tangent Lines to Polar Curves

Tangent Lines to Polar Curves

Example. continued...

7/31/2019 Lecture 10 (Calculus of Polar Curves)

75/184

Determine the points on the cardioid C : x= coscos2, y= sin sincoswhere the tangent line is horizontal, or vertical.

dx

d= sin+2cos sin= sin(2cos1)

=0

=sin

=0 or cos

=

1

2 =

=0,,

3

,5

3

dy

d= cos (cos2 sin2)

= cos 2cos2+1

=(1

cos)(1

+2cos)

= 0 = cos= 1 or cos = 12

= = 0, 23

, 43

Thus, tangent lines are horizontal at

32 ,

23

and

32 ,

43

, and are vertical at

12 ,

3

,

12 ,

53 and (2,).

Polar Curves 6/ 18

Tangent Lines to Polar Curves

Arc Length for Polar Curves

Recall. arc length for a parametric curve

7/31/2019 Lecture 10 (Calculus of Polar Curves)

76/184

Polar Curves 7/ 18

Tangent Lines to Polar Curves

Arc Length for Polar Curves

Recall. arc length for a parametric curve

7/31/2019 Lecture 10 (Calculus of Polar Curves)

77/184

L=b

adx

dt2 +

dy

dt2

dt.

Polar Curves 7/ 18

Tangent Lines to Polar Curves

Arc Length for Polar Curves

Recall. arc length for a parametric curve

7/31/2019 Lecture 10 (Calculus of Polar Curves)

78/184

L=b

adx

dt2 +

dy

dt2

dt.

For a polar curve r=f(), parametrized as x=f()cos, y=f()sin,

dx

d 2

Polar Curves 7/ 18

Tangent Lines to Polar Curves

Arc Length for Polar Curves

Recall. arc length for a parametric curve

7/31/2019 Lecture 10 (Calculus of Polar Curves)

79/184

L=

b

adx

dt2 +

dy

dt2 dt.

For a polar curve r=f(), parametrized as x=f()cos, y=f()sin,

dx

d 2

= dr

dcos rsin

2

Polar Curves 7/ 18

Tangent Lines to Polar Curves

Arc Length for Polar Curves

Recall. arc length for a parametric curve

7/31/2019 Lecture 10 (Calculus of Polar Curves)

80/184

L=

b

adx

dt2 +

dy

dt2 dt.

For a polar curve r=f(), parametrized as x=f()cos, y=f()sin,

dx

d 2

= dr

dcos rsin

2

=

dr

d

2cos22rdr

dcos sin+ r2 sin2

Polar Curves 7/ 18

7/31/2019 Lecture 10 (Calculus of Polar Curves)

81/184

7/31/2019 Lecture 10 (Calculus of Polar Curves)

82/184

Tangent Lines to Polar Curves

Arc Length for Polar Curves

Recall. arc length for a parametric curve

7/31/2019 Lecture 10 (Calculus of Polar Curves)

83/184

L=

b

adx

dt2 +

dy

dt2 dt.

For a polar curve r=f(), parametrized as x=f()cos, y=f()sin,

dx

d 2

= dr

dcos rsin

2

=

dr

d

2cos22rdr

dcos sin+ r2 sin2

dy

d

2=

dr

dsin+ rcos

2

= drd

2sin2+2r

dr

d cos sin+ r2 cos2

Polar Curves 7/ 18

Tangent Lines to Polar Curves

Arc Length for Polar Curves

Recall. arc length for a parametric curve

7/31/2019 Lecture 10 (Calculus of Polar Curves)

84/184

L=

b

adx

dt

2

+dy

dt

2

dt.

For a polar curve r=f(), parametrized as x=f()cos, y=f()sin,

dx

d 2

= dr

dcos rsin

2

=

dr

d

2cos22rdr

dcos sin+ r2 sin2

dy

d

2=

dr

dsin+ rcos

2

= drd

2sin2+2r

dr

d cos sin+ r2 cos2

Thus,

dxd

2 + dyd

2=

Polar Curves 7/ 18

Tangent Lines to Polar Curves

Arc Length for Polar Curves

Recall. arc length for a parametric curve

7/31/2019 Lecture 10 (Calculus of Polar Curves)

85/184

L=

b

adx

dt

2

+dy

dt

2

dt.

For a polar curve r=f(), parametrized as x=f()cos, y=f()sin,

dx

d 2

= dr

dcos rsin

2

=

dr

d

2cos22rdr

dcos sin+ r2 sin2

dy

d

2=

dr

dsin+ rcos

2

= drd 2

sin2+2rdr

d cos sin+ r2 cos2

Thus,

dxd

2 + dyd

2= r2 +

drd

2

Polar Curves 7/ 18

7/31/2019 Lecture 10 (Calculus of Polar Curves)

86/184

7/31/2019 Lecture 10 (Calculus of Polar Curves)

87/184

7/31/2019 Lecture 10 (Calculus of Polar Curves)

88/184

Tangent Lines to Polar Curves

Arc Length for Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

89/184

Example.

Find the length of the circle r= 4cos.

As varies from 0 to 2

Polar Curves 9/ 18

7/31/2019 Lecture 10 (Calculus of Polar Curves)

90/184

Tangent Lines to Polar Curves

Arc Length for Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

91/184

Example.

Find the length of the circle r= 4cos.

As varies from 2 to

Polar Curves 9/ 18 Tangent Lines to Polar Curves

Arc Length for Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

92/184

Example.

Find the length of the circle r= 4cos.

must vary from 0 to L=

0 (4cos)2 + (4sin)2 d

Polar Curves 9/ 18 Tangent Lines to Polar Curves

Arc Length for Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

93/184

Example.

Find the length of the circle r= 4cos.

must vary from 0 to L=

0 (4cos)2 + (4sin)2 d

=

0

16cos2+ 16sin2d

Polar Curves 9/ 18

7/31/2019 Lecture 10 (Calculus of Polar Curves)

94/184

Tangent Lines to Polar Curves

Arc Length for Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

95/184

Example.

Find the length of the circle r= 4cos.

must vary from 0 to L=

0 (4cos)2 + (4sin)2 d

=

0

16cos2+ 16sin2d

=

04 d

= 4

Polar Curves 9/ 18 Tangent Lines to Polar Curves

Arc Length for Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

96/184

Example.Setup the integral that yields the length of one petal of the rose r= sin 2.

Polar Curves 10/ 18 Tangent Lines to Polar Curves

Arc Length for Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

97/184

Example.Setup the integral that yields the length of one petal of the rose r= sin 2.

As varies from 0 to 4

Polar Curves 10/ 18

7/31/2019 Lecture 10 (Calculus of Polar Curves)

98/184

Tangent Lines to Polar Curves

Arc Length for Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

99/184

Example.

Setup the integral that yields the length of one petal of the rose r= sin 2.

As varies from 4 to2

Polar Curves 10/ 18 Tangent Lines to Polar Curves

Arc Length for Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

100/184

Example.

Setup the integral that yields the length of one petal of the rose r= sin 2.

As varies from 4 to2

Polar Curves 10/ 18

7/31/2019 Lecture 10 (Calculus of Polar Curves)

101/184

Tangent Lines to Polar Curves

Arc Length for Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

102/184

Example.

Setup the integral that yields the length of one petal of the rose r= sin 2.

must vary from 0 to 2L=

/20

(sin 2)2 + (2cos 2)2 d

=/20

sin2 2+4cos2 2d

Polar Curves 10/ 18

Tangent Lines to Polar Curves

Arc Length for Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

103/184

Example.

Setup the integral that yields the length of one petal of the rose r= sin 2.

must vary from 0 to 2L=

/20

(sin 2)2 + (2cos 2)2 d

=/20

sin2 2+4cos2 2d

=/2

0

1 +3cos2 2d

Polar Curves 10/ 18

Tangent Lines to Polar Curves

Arc Length for Polar Curves

7/31/2019 Lecture 10 (Calculus of Polar Curves)

104/184

Example.

Setup the integral that yields the length of one petal of the rose r= sin 2.

must vary from 0 to 2L=

/20

(sin 2)2 + (2cos 2)2 d

=/20

sin2 2+4cos2 2d

=/2

0

1 +3cos2 2d

or, by symmetry,

=2

/4

01 + 3cos

2 2d

Polar Curves 10/ 18

Tangent Lines to Polar Curves

Area in Polar Coordinates

Consider the following polar region.

7/31/2019 Lecture 10 (Calculus of Polar Curves)

105/184

Polar Curves 11/ 18

7/31/2019 Lecture 10 (Calculus of Polar Curves)

106/184

7/31/2019 Lecture 10 (Calculus of Polar Curves)

107/184

Tangent Lines to Polar Curves

Area in Polar Coordinates

Consider the following polar region.

7/31/2019 Lecture 10 (Calculus of Polar Curves)

108/184

Assume that + 2.

Polar Curves 11/ 18

7/31/2019 Lecture 10 (Calculus of Polar Curves)

109/184

7/31/2019 Lecture 10 (Calculus of Polar Curves)

110/184

7/31/2019 Lecture 10 (Calculus of Polar Curves)

111/184

Tangent Lines to Polar Curves

Area in Polar Coordinates

Consider the following polar region.

7/31/2019 Lecture 10 (Calculus of Polar Curves)

112/184

Assume that + 2.Divide R into n wedgesusing the rays

=1,

=2, .. .,

=n

1.

Let A1, A2, .. ., An be the areas of the wedges.

Polar Curves 11/ 18

Tangent Lines to Polar Curves

Area in Polar CoordinatesConsider the following polar region.

7/31/2019 Lecture 10 (Calculus of Polar Curves)

113/184

Assume that + 2.Divide R into n wedgesusing the rays

=1,

=2, .. .,

=n

1.

Let A1, A2, .. ., An be the areas of the wedges.

Let 1, 2, .. ., n be the central angles of the wedges.

Polar Curves 11/ 18

7/31/2019 Lecture 10 (Calculus of Polar Curves)

114/184

Tangent Lines to Polar Curves

Area in Polar CoordinatesConsider the following polar region.

7/31/2019 Lecture 10 (Calculus of Polar Curves)

115/184

Assume that + 2.Divide R into n wedgesusing the rays

=1,

=2, .. .,

=n

1.

Let A1, A2, .. ., An be the areas of the wedges.

Let 1, 2, .. ., n be the central angles of the wedges.

The area Aof the region is given byA=A1 +A2 + +An=n

i=1Ai.

Polar Curves 11/ 18

Tangent Lines to Polar Curves

Area in Polar Coordinates

Approximate the area of the ith wedge by that of a sector with central angle i,

and radius f(i

), where i

[i1,i].

7/31/2019 Lecture 10 (Calculus of Polar Curves)

116/184

Polar Curves 12/ 18

Tangent Lines to Polar Curves

Area in Polar Coordinates

Approximate the area of the ith wedge by that of a sector with central angle i,

and radius f(i

), where i

[i1,i].

7/31/2019 Lecture 10 (Calculus of Polar Curves)

117/184

Polar Curves 12/ 18

7/31/2019 Lecture 10 (Calculus of Polar Curves)

118/184

Tangent Lines to Polar Curves

Area in Polar Coordinates

Approximate the area of the ith wedge by that of a sector with central angle i,

and radius f(i

), where i

[i1,i].

7/31/2019 Lecture 10 (Calculus of Polar Curves)

119/184

Thus, Ai

Polar Curves 12/ 18

Tangent Lines to Polar Curves

Area in Polar Coordinates

Approximate the area of the ith wedge by that of a sector with central angle i,

and radius f(i

), where i

[i1,i].

7/31/2019 Lecture 10 (Calculus of Polar Curves)

120/184

Thus, Ai1

2f(i )

2i.

Polar Curves 12/ 18

Tangent Lines to Polar Curves

Area in Polar Coordinates

Approximate the area of the ith wedge by that of a sector with central angle i,

and radius f(i

), where i

[i1,i].

7/31/2019 Lecture 10 (Calculus of Polar Curves)

121/184

Thus, Ai1

2f(i )

2i. And A=

ni=1Ai

ni=1

1

2f(i )2i.

Polar Curves 12/ 18

Tangent Lines to Polar Curves

Area in Polar Coordinates

Approximate the area of the ith wedge by that of a sector with central angle i,

and radius f(i

), where i

[i1,i].

7/31/2019 Lecture 10 (Calculus of Polar Curves)

122/184

Thus, Ai1

2f(i )

2i. And A=

ni=1Ai

ni=1

1

2f(i )2i.

If we increase nin such a way that i 0, for everyi= 1,2,..., n, then

Polar Curves 12/ 18

7/31/2019 Lecture 10 (Calculus of Polar Curves)

123/184

Tangent Lines to Polar Curves

Area in Polar Coordinates

Approximate the area of the ith wedge by that of a sector with central angle i,

and radius f(i

), where i

[i1,i].

7/31/2019 Lecture 10 (Calculus of Polar Curves)

124/184

Thus, Ai1

2f(i )

2

i. And A=n

i=1Ain

i=1

1

2f(i )2i.

If we increase nin such a way that i 0, for everyi= 1,2,..., n, thenA= lim

nn

i=1

1

2

f(i )

2i=

1

2

f()

2d.

Polar Curves 12/ 18

Tangent Lines to Polar Curves

Area in Polar Coordinates

Area of a Polar Region.

Let and be angles such that +2. Let r=f() be continuous on [,].The area of the region

7/31/2019 Lecture 10 (Calculus of Polar Curves)

125/184

is given by

A=

1

2[f()]2d =

1

2r2d.

Polar Curves 13/ 18

Tangent Lines to Polar Curves

Area in Polar Coordinates

Example.

Determine the area of the region in the first quadrant inside the cardioid

r= 1 cos.

7/31/2019 Lecture 10 (Calculus of Polar Curves)

126/184

Polar Curves 14/ 18

7/31/2019 Lecture 10 (Calculus of Polar Curves)

127/184

7/31/2019 Lecture 10 (Calculus of Polar Curves)

128/184

7/31/2019 Lecture 10 (Calculus of Polar Curves)

129/184

Tangent Lines to Polar Curves

Area in Polar Coordinates

Example.

Determine the area of the region in the first quadrant inside the cardioid

r= 1 cos.

Solution The region is

7/31/2019 Lecture 10 (Calculus of Polar Curves)

130/184

Solution. The region is

The limits of integration are determined by the radial lines, or rays, that bound the

region.

Polar Curves 14/ 18

Tangent Lines to Polar Curves

Area in Polar Coordinates

Example.

Determine the area of the region in the first quadrant inside the cardioid

r= 1 cos.

Solution The region is

7/31/2019 Lecture 10 (Calculus of Polar Curves)

131/184

Solution. The region is

The limits of integration are determined by the radial lines, or rays, that bound the

region. Here, must vary from = 0

Polar Curves 14/ 18

Tangent Lines to Polar Curves

Area in Polar Coordinates

Example.

Determine the area of the region in the first quadrant inside the cardioid

r= 1 cos.

Solution The region is

7/31/2019 Lecture 10 (Calculus of Polar Curves)

132/184

Solution. The region is

The limits of integration are determined by the radial lines, or rays, that bound the

region. Here, must vary from = 0 to =/2 to sweep out the region.

Polar Curves 14/ 18

Tangent Lines to Polar Curves

Area in Polar Coordinates

Example.

Determine the area of the region in the first quadrant inside the cardioid

r= 1 cos.

Solution. The region is

7/31/2019 Lecture 10 (Calculus of Polar Curves)

133/184

Solution. The region is

The limits of integration are determined by the radial lines, or rays, that bound the

region. Here, must vary from = 0 to =/2 to sweep out the region. Thus

A =/2

012

(1 cos)2 d

Polar Curves 14/ 18

Tangent Lines to Polar Curves

Area in Polar Coordinates

Example.

Determine the area of the region in the first quadrant inside the cardioid

r= 1 cos.

Solution. The region is

7/31/2019 Lecture 10 (Calculus of Polar Curves)

134/184

Solution. The region is

The limits of integration are determined by the radial lines, or rays, that bound the

region. Here, must vary from = 0 to =/2 to sweep out the region. Thus

A = /20

12

(1 cos)2 d=/20

12

(1 2cos+cos2) d

Polar Curves 14/ 18

Tangent Lines to Polar Curves

Area in Polar Coordinates

Example.

Determine the area of the region in the first quadrant inside the cardioid

r= 1 cos.

Solution. The region is

7/31/2019 Lecture 10 (Calculus of Polar Curves)

135/184

g

The limits of integration are determined by the radial lines, or rays, that bound the

region. Here, must vary from = 0 to =/2 to sweep out the region. Thus

A = /20

12

(1 cos)2 d=/20

12

(1 2cos+cos2) d

= 12

/20

3

2 2cos+ 1

2cos2

d

Polar Curves 14/ 18

Tangent Lines to Polar Curves

Area in Polar Coordinates

Example.

Determine the area of the region in the first quadrant inside the cardioid

r= 1 cos.

Solution. The region is

7/31/2019 Lecture 10 (Calculus of Polar Curves)

136/184

g

The limits of integration are determined by the radial lines, or rays, that bound the

region. Here, must vary from = 0 to =/2 to sweep out the region. Thus

A = /20

12

(1 cos)2 d=/20

12

(1 2cos+cos2) d

= 12

/20

3

2 2cos+ 1

2cos2

d= 1

2

3

22sin+ 1

4sin2

/20

Polar Curves 14/ 18

Tangent Lines to Polar Curves

Area in Polar Coordinates

Example.

Determine the area of the region in the first quadrant inside the cardioid

r= 1 cos.

Solution. The region is

7/31/2019 Lecture 10 (Calculus of Polar Curves)

137/184

g

The limits of integration are determined by the radial lines, or rays, that bound the

region. Here, must vary from = 0 to =/2 to sweep out the region. Thus

A = /20

12

(1 cos)2 d=/20

12

(1 2cos+cos2) d

= 12

/20

3

2 2cos+ 1

2cos2

d= 1

2

3

22sin+ 1

4sin2

/20

= 38

1

Polar Curves 14/ 18

Tangent Lines to Polar Curves

Area in Polar Coordinates

Example.

Find the area of the region enclosed by the rose curve r= 3cos2.

7/31/2019 Lecture 10 (Calculus of Polar Curves)

138/184

Polar Curves 15/ 18

7/31/2019 Lecture 10 (Calculus of Polar Curves)

139/184

7/31/2019 Lecture 10 (Calculus of Polar Curves)

140/184

Tangent Lines to Polar Curves

Area in Polar Coordinates

Example.

Find the area of the region enclosed by the rose curve r= 3cos2.

Solution. The region is

7/31/2019 Lecture 10 (Calculus of Polar Curves)

141/184

Polar Curves 15/ 18

Tangent Lines to Polar Curves

Area in Polar Coordinates

Example.

Find the area of the region enclosed by the rose curve r= 3cos2.

Solution. The region is

We use s mmetr

7/31/2019 Lecture 10 (Calculus of Polar Curves)

142/184

We use symmetry.

Polar Curves 15/ 18

Tangent Lines to Polar Curves

Area in Polar Coordinates

Example.

Find the area of the region enclosed by the rose curve r= 3cos2.

Solution. The region is

We use symmetry

7/31/2019 Lecture 10 (Calculus of Polar Curves)

143/184

We use symmetry.

Here, must vary from = 0

Polar Curves 15/ 18

Tangent Lines to Polar Curves

Area in Polar Coordinates

Example.

Find the area of the region enclosed by the rose curve r= 3cos2.

Solution. The region is

We use symmetry

7/31/2019 Lecture 10 (Calculus of Polar Curves)

144/184

We use symmetry.

Here, must vary from = 0 to =/4to sweep out the region.

Polar Curves 15/ 18

Tangent Lines to Polar Curves

Area in Polar Coordinates

Example.

Find the area of the region enclosed by the rose curve r= 3cos2.

Solution. The region is

We use symmetry

7/31/2019 Lecture 10 (Calculus of Polar Curves)

145/184

We use symmetry.

Here, must vary from = 0 to =/4to sweep out the region. Thus

A = 8/4

0

1

2(3cos2)2 d

Polar Curves 15/ 18

Tangent Lines to Polar Curves

Area in Polar Coordinates

Example.

Find the area of the region enclosed by the rose curve r= 3cos2.

Solution. The region is

We use symmetry

7/31/2019 Lecture 10 (Calculus of Polar Curves)

146/184

We use symmetry.

Here, must vary from = 0 to =/4to sweep out the region. Thus

A = 8/4

0

1

2(3cos2)2 d= 36

/40

1

2(1 +cos4) d

Polar Curves 15/ 18

7/31/2019 Lecture 10 (Calculus of Polar Curves)

147/184

Tangent Lines to Polar Curves

Area in Polar Coordinates

Example.

Find the area of the region enclosed by the rose curve r= 3cos2.

Solution. The region is

We use symmetry.

7/31/2019 Lecture 10 (Calculus of Polar Curves)

148/184

e use sy et y.

Here, must vary from = 0 to =/4to sweep out the region. Thus

A = 8/4

0

1

2(3cos2)2 d= 36

/40

1

2(1 +cos4) d

= 18/40

(1 +cos4)d = 18+ 14

sin4/4

0

Polar Curves 15/ 18

Tangent Lines to Polar Curves

Area in Polar Coordinates

Example.

Find the area of the region enclosed by the rose curve r= 3cos2.

Solution. The region is

We use symmetry.

7/31/2019 Lecture 10 (Calculus of Polar Curves)

149/184

y y

Here, must vary from = 0 to =/4to sweep out the region. Thus

A = 8/4

0

1

2(3cos2)2 d= 36

/40

1

2(1 +cos4) d

= 18/40

(1 +cos4)d = 18+ 14

sin4/4

0= 9

2

Polar Curves 15/ 18

Tangent Lines to Polar Curves

Area in Polar Coordinates

Example.

Find the area of the region enclosed by the rose curve r= 3cos2.

Solution. The region is

We use symmetry.

7/31/2019 Lecture 10 (Calculus of Polar Curves)

150/184

y y

Here, must vary from = 0 to =/4to sweep out the region. Thus

A = 8/4

0

1

2(3cos2)2 d= 36

/40

1

2(1 +cos4) d

= 18/40

(1 +cos4)d = 18+ 14

sin4/4

0= 9

2

or = 4/4/4

1

2(3cos2)2 d

Polar Curves 15/ 18

Tangent Lines to Polar Curves

Area in Polar Coordinates

Example.

Find the area of the region inside the cardioid r= 4 + 4cos and outside the circler= 6.

7/31/2019 Lecture 10 (Calculus of Polar Curves)

151/184

Polar Curves 16/ 18

7/31/2019 Lecture 10 (Calculus of Polar Curves)

152/184

Tangent Lines to Polar Curves

Area in Polar Coordinates

Example.

Find the area of the region inside the cardioid r= 4 + 4cos and outside the circler= 6.

Solution. The region is

7/31/2019 Lecture 10 (Calculus of Polar Curves)

153/184

Polar Curves 16/ 18

7/31/2019 Lecture 10 (Calculus of Polar Curves)

154/184

Tangent Lines to Polar Curves

Area in Polar Coordinates

Example.

Find the area of the region inside the cardioid r= 4 + 4cos and outside the circler= 6.

Solution. The region is

7/31/2019 Lecture 10 (Calculus of Polar Curves)

155/184

We use symmetry.

Polar Curves 16/ 18

7/31/2019 Lecture 10 (Calculus of Polar Curves)

156/184

Tangent Lines to Polar Curves

Area in Polar Coordinates

Example.

Find the area of the region inside the cardioid r= 4 + 4cos and outside the circler= 6.

Solution. The region is

7/31/2019 Lecture 10 (Calculus of Polar Curves)

157/184

We use symmetry. Here, must vary from = 0 to =.

Polar Curves 16/ 18

Tangent Lines to Polar Curves

Area in Polar Coordinates

Example.

Find the area of the region inside the cardioid r= 4 + 4cos and outside the circler= 6.

Solution. The region is

7/31/2019 Lecture 10 (Calculus of Polar Curves)

158/184

We use symmetry. Here, must vary from = 0 to =.Solving for from the intersection of the two curves,

Polar Curves 16/ 18

Tangent Lines to Polar Curves

Area in Polar Coordinates

Example.

Find the area of the region inside the cardioid r= 4 + 4cos and outside the circler= 6.

Solution. The region is

7/31/2019 Lecture 10 (Calculus of Polar Curves)

159/184

We use symmetry. Here, must vary from = 0 to =.Solving for from the intersection of the two curves,

4 +4cos= 6

Polar Curves 16/ 18

Tangent Lines to Polar Curves

Area in Polar Coordinates

Example.

Find the area of the region inside the cardioid r= 4 + 4cos and outside the circler= 6.

Solution. The region is

7/31/2019 Lecture 10 (Calculus of Polar Curves)

160/184

We use symmetry. Here, must vary from = 0 to =.Solving for from the intersection of the two curves,

4 +4cos= 6 = cos = 12Polar Curves 16/ 18

Tangent Lines to Polar Curves

Area in Polar Coordinates

Example.

Find the area of the region inside the cardioid r= 4 + 4cos and outside the circler= 6.

Solution. The region is

7/31/2019 Lecture 10 (Calculus of Polar Curves)

161/184

We use symmetry. Here, must vary from = 0 to =.Solving for from the intersection of the two curves,

4 +4cos= 6 = cos = 12 = = 3 , 53Polar Curves 16/ 18

Tangent Lines to Polar Curves

Area in Polar Coordinates

Example.

Find the area of the region inside the cardioid r= 4 + 4cos and outside the circler= 6.

Solution. The region is

7/31/2019 Lecture 10 (Calculus of Polar Curves)

162/184

We use symmetry. Here, must vary from = 0 to =.Solving for from the intersection of the two curves,

4 +4cos= 6 = cos = 12 = = 3 , 53 = = 3Polar Curves 16/ 18

Tangent Lines to Polar Curves

Area in Polar Coordinates

Example.

Find the area of the region inside the cardioid r= 4 + 4cos and outside the circler= 6.

Also, notice that the region can be obtained as follows.

7/31/2019 Lecture 10 (Calculus of Polar Curves)

163/184

Polar Curves 17/ 18

Tangent Lines to Polar Curves

Area in Polar Coordinates

Example.

Find the area of the region inside the cardioid r= 4 + 4cos and outside the circler= 6.

Also, notice that the region can be obtained as follows.

7/31/2019 Lecture 10 (Calculus of Polar Curves)

164/184

=

Polar Curves 17/ 18

Tangent Lines to Polar Curves

Area in Polar Coordinates

Example.

Find the area of the region inside the cardioid r= 4 + 4cos and outside the circler= 6.

Also, notice that the region can be obtained as follows.

7/31/2019 Lecture 10 (Calculus of Polar Curves)

165/184

=

Polar Curves 17/ 18

Tangent Lines to Polar Curves

Area in Polar Coordinates

Example.

Find the area of the region inside the cardioid r= 4 + 4cos and outside the circler= 6.

Also, notice that the region can be obtained as follows.

7/31/2019 Lecture 10 (Calculus of Polar Curves)

166/184

= Hence, the area is

Polar Curves 17/ 18

Tangent Lines to Polar Curves

Area in Polar Coordinates

Example.

Find the area of the region inside the cardioid r= 4 + 4cos and outside the circler= 6.

Also, notice that the region can be obtained as follows.

7/31/2019 Lecture 10 (Calculus of Polar Curves)

167/184

= Hence, the area is

A = 2/3

0

1

2(4+ 4cos)2 d

Polar Curves 17/ 18

Tangent Lines to Polar Curves

Area in Polar Coordinates

Example.

Find the area of the region inside the cardioid r= 4 + 4cos and outside the circler= 6.

Also, notice that the region can be obtained as follows.

7/31/2019 Lecture 10 (Calculus of Polar Curves)

168/184

= Hence, the area is

A = 2/3

0

1

2(4+ 4cos)2 d 2

/30

1

2(6)2 d

Polar Curves 17/ 18

Tangent Lines to Polar Curves

Area in Polar Coordinates

Example.

Find the area of the region inside the cardioid r= 4 + 4cos and outside the circler= 6.

Also, notice that the region can be obtained as follows.

7/31/2019 Lecture 10 (Calculus of Polar Curves)

169/184

= Hence, the area is

A = 2/3

0

1

2(4+ 4cos)2 d 2

/30

1

2(6)2 d =

/30

(16+ 32cos+ 16cos2 36)d

Polar Curves 17/ 18

Tangent Lines to Polar Curves

Area in Polar Coordinates

Example.

Find the area of the region inside the cardioid r= 4 + 4cos and outside the circler= 6.

Also, notice that the region can be obtained as follows.

7/31/2019 Lecture 10 (Calculus of Polar Curves)

170/184

= Hence, the area is

A = 2/3

0

1

2(4+ 4cos)2 d 2

/30

1

2(6)2 d =

/30

(16+ 32cos+ 16cos2 36)d

= /3

0

(16cos2

+32cos

20)d

Polar Curves 17/ 18

Tangent Lines to Polar Curves

Area in Polar Coordinates

Example.

Find the area of the region inside the cardioid r= 4 + 4cos and outside the circler= 6.

Also, notice that the region can be obtained as follows.

7/31/2019 Lecture 10 (Calculus of Polar Curves)

171/184

= Hence, the area is

A = 2/3

0

1

2(4+ 4cos)2 d 2

/30

1

2(6)2 d =

/30

(16+ 32cos+ 16cos2 36)d

= /3

0

(16cos2

+32cos

20)d

=18

3

4

Polar Curves 17/ 18

Tangent Lines to Polar Curves

Area in Polar Coordinates

Example.

Find the area of the region inside the cardioid r= 4 + 4cos and outside the circler= 6.

Also, notice that the region can be obtained as follows.

7/31/2019 Lecture 10 (Calculus of Polar Curves)

172/184

= Hence, the area is

A = 2/3

0

1

2(4+ 4cos)2 d 2

/30

1

2(6)2 d =

/30

(16+ 32cos+ 16cos2 36)d

= /3

0

(16cos2

+32cos

20)d

=18

3

4

or =/3/3

1

2(16cos2 + 32cos 20) d (without using symmetry)

Polar Curves 17/ 18

Tangent Lines to Polar Curves

Intersection of Polar Curves

Remark.

Intersection points of polar curves are naturally obtained by equating the

expressions for r.

7/31/2019 Lecture 10 (Calculus of Polar Curves)

173/184

Polar Curves 18/ 18

Tangent Lines to Polar Curves

Intersection of Polar Curves

Remark.

Intersection points of polar curves are naturally obtained by equating the

expressions for r. However, this method will not always produce all the

intersection points because of the many ways of representing a point in polarcoordinates.

7/31/2019 Lecture 10 (Calculus of Polar Curves)

174/184

Polar Curves 18/ 18

7/31/2019 Lecture 10 (Calculus of Polar Curves)

175/184

Tangent Lines to Polar Curves

Intersection of Polar Curves

Remark.

Intersection points of polar curves are naturally obtained by equating the

expressions for r. However, this method will not always produce all the

intersection points because of the many ways of representing a point in polarcoordinates.

Consider the cardioids r= 1 cos and r= 1 + cos. Equating the expressions,

7/31/2019 Lecture 10 (Calculus of Polar Curves)

176/184

q g p ,

1 cos= 1 +cos

Polar Curves 18/ 18

Tangent Lines to Polar Curves

Intersection of Polar Curves

Remark.

Intersection points of polar curves are naturally obtained by equating the

expressions for r. However, this method will not always produce all the

intersection points because of the many ways of representing a point in polar

coordinates.

Consider the cardioids r= 1 cos and r= 1 + cos. Equating the expressions,

7/31/2019 Lecture 10 (Calculus of Polar Curves)

177/184

q g p

1 cos= 1 +cos = cos= 0

Polar Curves 18/ 18

Tangent Lines to Polar Curves

Intersection of Polar Curves

Remark.

Intersection points of polar curves are naturally obtained by equating the

expressions for r. However, this method will not always produce all the

intersection points because of the many ways of representing a point in polar

coordinates.

Consider the cardioids r= 1 cos and r= 1 + cos. Equating the expressions,

7/31/2019 Lecture 10 (Calculus of Polar Curves)

178/184

g

1 cos= 1 +cos = cos= 0 = = 2

Polar Curves 18/ 18

Tangent Lines to Polar Curves

Intersection of Polar Curves

Remark.

Intersection points of polar curves are naturally obtained by equating the

expressions for r. However, this method will not always produce all the

intersection points because of the many ways of representing a point in polar

coordinates.

Consider the cardioids r= 1 cos and r= 1 + cos. Equating the expressions,

7/31/2019 Lecture 10 (Calculus of Polar Curves)

179/184

1 cos= 1 +cos = cos= 0 = = 2

+k2

, kZ.

Polar Curves 18/ 18

Tangent Lines to Polar Curves

Intersection of Polar Curves

Remark.

Intersection points of polar curves are naturally obtained by equating the

expressions for r. However, this method will not always produce all the

intersection points because of the many ways of representing a point in polar

coordinates.

Consider the cardioids r= 1 cos and r= 1 + cos. Equating the expressions,

7/31/2019 Lecture 10 (Calculus of Polar Curves)

180/184

1 cos= 1 +cos = cos= 0 = = 2

+k2

, kZ.

These values of give us only two points1, 2

and

1, 32

.

Polar Curves 18/ 18

Tangent Lines to Polar Curves

Intersection of Polar Curves

Remark.

Intersection points of polar curves are naturally obtained by equating the

expressions for r. However, this method will not always produce all the

intersection points because of the many ways of representing a point in polar

coordinates.

Consider the cardioids r= 1 cos and r= 1 + cos. Equating the expressions,

7/31/2019 Lecture 10 (Calculus of Polar Curves)

181/184

1 cos= 1 +cos = cos= 0 = = 2

+k2

, kZ.

These values of give us only two points1, 2

and

1, 32

.

Polar Curves 18/ 18

Tangent Lines to Polar Curves

Intersection of Polar Curves

Remark.

Intersection points of polar curves are naturally obtained by equating the

expressions for r. However, this method will not always produce all the

intersection points because of the many ways of representing a point in polar

coordinates.

Consider the cardioids r= 1 cos and r= 1 + cos. Equating the expressions,

7/31/2019 Lecture 10 (Calculus of Polar Curves)

182/184

1 cos= 1 +cos = cos= 0 = = 2

+k2

, kZ.

These values of give us only two points1, 2

and

1, 32

.

Clearly, the pole has been missed out.

Polar Curves 18/ 18

Tangent Lines to Polar Curves

Intersection of Polar Curves

Remark.

Intersection points of polar curves are naturally obtained by equating the

expressions for r. However, this method will not always produce all the

intersection points because of the many ways of representing a point in polar

coordinates.

Consider the cardioids r= 1 cos and r= 1 + cos. Equating the expressions,

7/31/2019 Lecture 10 (Calculus of Polar Curves)

183/184

1 cos= 1 +cos = cos= 0 = = 2

+k2

, kZ.

These values of give us only two points1, 2

and

1, 32

.

Clearly, the pole has been missed out. This is

because the cardioids pass through the pole

at different values of.

Polar Curves 18/ 18

Tangent Lines to Polar Curves

Intersection of Polar Curves

Remark.

Intersection points of polar curves are naturally obtained by equating the

expressions for r. However, this method will not always produce all the

intersection points because of the many ways of representing a point in polar

coordinates.

Consider the cardioids r= 1 cos and r= 1 + cos. Equating the expressions,

7/31/2019 Lecture 10 (Calculus of Polar Curves)

184/184

1 cos= 1 +cos = cos= 0 = = 2

+k2

, kZ.

These values of give us only two points1, 2

and

1, 32

.

Clearly, the pole has been missed out. This is

because the cardioids pass through the pole

at different values of.

In general, it is a good idea to graph thecurves to determine exactly how many

intersections points should there be.

Polar Curves 18/ 18