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10.3 Polar Coordinates

2 3,2

3π

4π

x2 + y2 =25

−5 3 cot t−5 3 cot 2

0,5( ), 0,−5( )3,0( ), −3,0( )

≈12.763

Converting Polar to Rectangular

Use the polar-rectangular conversion formulas to show that the polar graph of r = 4 sin is a circle.

θ

Converting Polar to Rectangular

Use the polar-rectangular conversion formulas to show that the polar graph of r = 4 sin is a circle.

r = 4 sin θ

r2 = 4 r sin θ Multiply by r.

x2 + y2 = 4y Polar-rectangular conversion.

x2 + y2- 4y = 0

x2 + y2- 4y +4 = 4 Completing the square.

x2 + y - 2( )2 = 22 Circle in standard form.

θ

One way to give someone directions is to tell them to go three blocks East and five blocks South.

Another way to give directions is to point and say “Go a half mile in that direction.”

Polar graphing is like the second method of giving directions. Each point is determined by a distance and an angle.

θInitial ray

r A polar coordinate pair

determines the location of a point.

( ),r θ

→

1 2 02

rπθ≤ ≤ ≤ ≤

r a=

oθ θ=

(Circle centered at the origin)

(Line through the origin)

Some curves are easier to describe with polar coordinates:

→

30o

2

More than one coordinate pair can refer to the same point.

( )2,30o

( )2, 210o= −

( )2, 150o= − −

210o

150o−

All of the polar coordinates of this point are:

( )( )2,30 360

2, 150 360 0, 1, 2 ...

o o

o o

n

n n

+ ⋅

− − + ⋅ = ± ±→

Tests for Symmetry:

x-axis: If (r, θ) is on the graph,

-1

0

1

1 2

θ

r

2cosr θ=θ−

r

so is (r, -θ).

→

Tests for Symmetry:

y-axis: If (r, θ) is on the graph,

θ

r

2sinr θ= π θ−

r

so is (r, π-θ)

0

1

2

-1 1

θ−

or (-r, -θ).

→

Tests for Symmetry:

origin: If (r, θ) is on the graph,

θ

r

θ π+r

so is (-r, θ) or (r, θ+π) .

-2

-1

0

1

2

-2 -1 1 2

tan

cosr

θθ

=±

→

Tests for Symmetry:

If a graph has two symmetries, then it has all three:

-2

-1

0

1

2

-2 -1 1 2

( )2cos 2r θ=

π

Try graphing this on the TI-89.

( )2sin 2.15

0 16

r θ

θ π

=

≤ ≤

To find the slope of a polar curve:

dy

dy ddxdxd

θ

θ

= sin

cos

dr

ddr

d

θθ

θθ

=sin cos

cos sin

r r

r r

θ θθ θ

′ +=

′ −

We use the product rule here.

To find the slope of a polar curve:

dy

dy ddxdxd

θ

θ

= sin

cos

dr

ddr

d

θθ

θθ

=sin cos

cos sin

r r

r r

θ θθ θ

′ +=

′ −

sin cos

cos sin

dy r r

dx r r

θ θθ θ

′ +=

′ −

→

Example: 1 cosr θ= − sinr θ′=

( )( )

sin sin 1 cos cosSlope

sin cos 1 cos sin

θ θ θ θ

θ θ θ θ

+ −=

− −

2 2sin cos cos

sin cos sin sin cos

θ θ θθ θ θ θ θ

+ −=

− +2 2sin cos cos

2sin cos sin

θ θ θθ θ θ− +

=−

cos 2 cos

sin 2 sin

θ θθ θ

− +=

−

→

Find the slope of the rose curve r = 2 sin 3 at the point where = π/6 and use it to find the equation of the tangent line.

θθ

Finding slope of a polar curve

Finding slope of a polar curve

dy

dxθ=

π6

= dy

dθdx

dθ θ=π6

=

ddθ

2 sin 3θ sin θ( )

ddθ

2 sin 3θ cos θ( )θ=

π6

= - 3

At θ =π6, x = 2 sin π

2( ) cos π6( ) = 3 and

y = 2 sin π2( ) sin π

6( ) = 1

The equation of the tangent line is:

y - 1 = - 3 x - 3( )

Find the slope of the rose curve r = 2 sin 3 at the point where = π/6 and use it to find the equation of the tangent line.

θθ

The length of an arc (in a circle) is given by r.θ when θ is given in radians.

Area Inside a Polar Graph:

For a very small θ, the curve could be approximated by a straight line and the area could be found using the triangle formula: 1

2A bh=

r dθ⋅r

( ) 21 1

2 2dA rd r r dθ θ= ⋅ =

→

We can use this to find the area inside a polar graph.

21

2dA r dθ=

21

2dA r dθ=

21

2A r d

β

αθ=∫

→

Example: Find the area enclosed by: ( )2 1 cosr θ= +

-2

-1

0

1

2

1 2 3 4

2 2

0

1

2r d

θ π

θθ

=

=∫( )

2 2

0

14 1 cos

2d

πθ θ= ⋅ +∫

( )2 2

02 1 2cos cos d

πθ θ θ= + +∫

2

0

1 cos 22 4cos 2

2d

π θθ θ+= + + ⋅∫

→

2

0

1 cos 22 4cos 2

2d

π θθ θ+= + + ⋅∫

2

03 4cos cos 2 d

πθ θ θ= + +∫

2

0

13 4sin sin 2

2

π

θ θ θ= + +

6 0π= −

6π=

→

Notes:

To find the area between curves, subtract:

2 21

2A R r d

β

αθ= −∫

Just like finding the areas between Cartesian curves, establish limits of integration where the curves cross.

→

Finding Area Between Curves

Find the area of the region that lies inside the circle r = 1 and outside the cardioid r = 1 – cos Ø.

Finding Area Between Curves

Find the area of the region that lies inside the circle r = 1 and outside the cardioid r = 1 – cos .

A = 1

2-π

2

π2

∫ r22 - r1

2( ) dθ

= 2 120

π2

∫ r22 - r1

2( ) dθ

= 12 - 1 - cos θ( )2

( )0

π2

∫ dθ

= 1 - 1 - 2 cos θ + cos2 θ( )( )0

π2

∫ dθ

= 2 cos θ - cos2 θ( )0

π2

∫ dθ ≈ 1.215 or 2 - π4

θ

When finding area, negative values of r cancel out:

-1

0

1

-1 1( )2sin 2r θ=

( ) 22

0

14 2sin 2

2A d

π

θ θ= ⋅ ⎡ ⎤⎣ ⎦∫

Area of one leaf times 4:

2A π=

Area of four leaves:

( )2 2

0

12sin 2

2A d

πθ θ= ⎡ ⎤⎣ ⎦∫

2A π=

→

To find the length of a curve:

Remember: 2 2ds dx dy= +

For polar graphs: cos sinx r y rθ θ= =

If we find derivatives and plug them into the formula, we (eventually) get:

22 dr

ds r dd

θθ

⎛ ⎞= +⎜ ⎟⎝ ⎠

So: 22Length

drr d

d

β

αθ

θ⎛ ⎞= +⎜ ⎟⎝ ⎠∫

→

22Length

drr d

d

β

αθ

θ⎛ ⎞= +⎜ ⎟⎝ ⎠∫

There is also a surface area equation similar to the others we are already familiar with:

22S 2

dry r d

d

β

απ θ

θ⎛ ⎞= +⎜ ⎟⎝ ⎠∫

When rotated about the x-axis:

22S 2 sin

drr r d

d

β

απ θ θ

θ⎛ ⎞= +⎜ ⎟⎝ ⎠∫

π