Polar Coordinates

Differences: Polar vs. RectangularPOLAR RECTANGULAR

• (0,0) is called the pole• Coordinates are in form (r, θ)

• (0,0) is called the origin

• Coordinates are in form (x,y)

How to Graph Polar Coordinates

•Given: (3, л/3)

Answer- STEP ONE

•Look at r and move that number of circles out

•Move 3 units out (highlighted in red) 1

2 3

Answer- STEP TWO

•Look at θ- this tells you the direction/angle of the line

•Place a point where the r is on that angle.

•In this case, the angle is л/3

1 2 3

Answer: STEP THREE

•Draw a line from the origin through the point

Converting Coordinates•Remember: The hypotenuse has a length of r. The sides are x and y.

•By using these properties, we get that:

x = rcosθy=rsinθtanθ=y/xr2=x2+y2

3, л/3

ry

x

CONVERT: Polar to Rectangle: (3, л/3)•x=3cos (л/3) x=3cos(60) 1.5

•y=3sin(л/3) x=3sin(60) 2.6

•New coordinates are (1.5, 2.6)***x = rcosθ***y=rsinθtanθ=y/xr2=x2+y2

CONVERT: Rectangular to Polar: (1, 1)•Find Angle: tanθ= y/x tanθ= 1

tan-1(1)= л/4

•Find r by using the equation r2=x2+y2

• r2=12+12

•r= √2•New Coordinates are (√2, л/4)

(You could also find r by recognizing this is a 45-45-90 right triangle)

Finding points of intersection

1 2cos 1

1 1 2cos

3,

2 2

r and r

Third point does not show up.

On r = 1-2 cos θ, point is (-1, 0)

On r = 1, point is (1, π)

43210

TANGENTS TO POLAR CURVES

•To find a tangent line to a polar curve r = f(θ), we regard θ as a parameter and write its parametric equations as:

x = r cos θ = f (θ) cos θy = r sin θ = f (θ) sin θ

Slope of a polar curve

•Where x = r cos θ = f(θ) cos θ•And y = r sin θ = f(θ) sin θ

( )sin ( )cos

( )cos ( )sin

dydy d

dd fx

f fxd

f

Horizontal tangent where dy/dθ = 0 and dx/dθ≠0

Vertical tangent where dx/dθ = 0 and dy/dθ≠0

TANGENTS TO POLAR CURVES•We locate horizontal tangents by finding the points where dy/dθ = 0 (provided that dx/dθ ≠ 0).

•Likewise, we locate vertical tangents at the points where dx/dθ = 0 (provided that dy/dθ ≠ 0).

TANGENTS TO POLAR CURVES

For the cardioid r = 1 + sin θ Find the points on the cardioid where the tangent line is horizontal or vertical.

TANGENTS TO POLAR CURVES

•Observe that:

dy

dcos(1 2sin) 0 when

2

,32

,76

,11

6

dx

d(1 sin)(1 2sin) 0 when

32

,6

,56

TANGENTS TO POLAR CURVES

•Using Equation 3 with r = 1 + sin θ, we have:

2

sin cos

cos sin

cos sin (1 sin )cos

cos cos (1 sin )sin

cos (1 2sin ) cos (1 2sin )

1 2sin sin (1 sin )(1 2sin )

drrdy d

drdx rd

TANGENTS TO POLAR CURVES

•Hence, there are horizontal tangents at the points

(2, π/2), (½, 7π/6), (½, 11π/6) and vertical tangents at

(3/2, π/6), (3/2, 5π/6)

▫ When θ = 3π/2, both dy/dθ and dx/dθ are 0.▫ So, we must be careful.

Figure 11.32

Length of a Curve in Polar Coordinates

2 2( ) ( )L f f d

2

2 2

0

2

0

2 2cos 2sin

2 2 1 cos

L d

d

Find the length of the arc for r = 2 – 2cosθ

2 22

0 0

2 2 2sin 4 sin2 2d d

2

0

8 cos 8(1 1) 162

sin2A =(1-cos2A)/22 sin2A =1-cos2A2 sin2 (1/2θ) =1-cosθ

Area of region 221 1( )

2 2A r d f d

Find Area of region inside smaller loop

2cos 1 0

2 4,

3 3

4

322

2 2

3 3

12cos 1

2A r d d

2

2

3

4cos 4cos 1A d

2 2

3 3

4cos 1 3 2cos2 4(2c 2 2 os co s)A d d

2

3

3 4 3 3 33 sin 2 4sin (3 ) (2 )

2 2 2

Area of a Surface

1.aculty.essex.edu/~wang/221/Chap10_Sec32. Nate Long3. Pearson

SOURCES (LINKS AND NAMES)