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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