SUTCLIFFE’S NOTES: CALCULUS 2 SWOKOWSKI’S CHAPTER 13 PLANE CURVES AND POLAR COORDINATES

13.3 Polar Coordinates

Relationships between Polar Coordinates ,r and Rectangular Coordinates

,x y

2 2 2

cos ,

sin ,

;

tan , if 0

x r

y r

x y r

yx

x

Polar Graphs 0, 0, 0a b p

Line Spiral Lemniscates

k r 2 cos2r a

2 sin2r a

Circles

r a 2 cosr a 2 sinr a

Cardioids

cosr a a cosr a a sinr a a sinr a a

Limacons with Loop a b

cosr a b cosr a b sinr a b sinr a b

Limacons with no Loop a b

cosr a b cosr a b sinr a b sinr a b

Roses

sin3r a cos3r a sin2r a cos2r a

( )r f

Tests for symmetry

(i) The graph of r f is symmetric with respect to the polar axis if substitution of

for leads to an equivalent equation.

(ii) The graph of r f is symmetric with respect to the vertical line 2 if substitution

of either (a) for or (b) for and for r r leads to an equivalent equation.

(iii) The graph of r f is symmetric with respect to the pole if substitution of either (a)

for r r or (b) for leads to an equivalent equation.

Note:

1. Unlike the graph of an equation in x and y, the graph of a polar equation r f can be

symmetric with respect to the polar axis, the vertical line 2 , or the pole without

satisfying one of the tests for symmetry. The reason for this is there are many different ways of specifying a point in polar coordinates.

2. The points of intersection of two polar graphs cannot always be found by solving the

polar equations simultaneously. For example, 4cosr and 4sinr intersect at the

pole or origin but this point of intersection can only be found by graphing the two polar curves.

Slope of a Tangent Line to a Polar Curve

Theorem The slope m of the tangent line to the polar curve r f at the point ,P r is

sin cos

cos sin

drr

dmdr

rd

.

Proof:

If ,x y are the rectangular coordinates of ,P r , then

cos cos

sin sin .

x r f

y r f

These may be considered as parametric equations for the graph with parameter . Thus,

cos sin sin cos.

sin cos cos sin

f f f fdy dy d

dx dx d f f f f

Note:

1. Horizontal tangent lines occur if the numerator is 0 and the denominator is not 0. 2. Vertical tangent lines occur if the denominator is 0 and the numerator is not 0. 3. The case 0/0 requires further investigation. 4. To find the slopes of the tangent lines at the pole, we must determine the values of for

which 0r f . For such values, the formula reduces to tanm .

Exercises Sketch the graph of the polar equation.

#2 2r Solution:

2r no matter what is so this will graph as a circle with radius 2 and center at the pole or

origin. One can also see this by writing the Cartesian equation 2 2 4x y .

#4 4

Solution:

4

no matter what r is. This will graph as a line through the origin with slope 1. One can also

see this by writing the Cartesian equation:

tan tan 14 4

yy x

x

#6 2sinr

Solution:

Note that there is symmetry with respect to the y-axis or pi/2 axis (replace by and the

equation is unchanged. So we can assign values for ranging from to 2 2

and reflect the

graph we get along the pi/2 axis to get the full graph.

Range of to 0

2

0 to

2

Range of r 2 to 0 0 to -2

Note that we actually get the entire graph (circle with radius 1 and center at (0,-1)) already when we do this.

#8 6 1 cosr

Solution:

Note that there is symmetry with respect to the polar axis (replace by and the equation is

unchanged) so it is enough to assign values for in the range 0 to and reflect this graph along

the polar axis.

Range of 0 to

2

2

to

Range of r -12 to -6 -6 to 0

We are able to draw the bottom half and reflecting this on the polar axis will give us the entire

graph of the cardioid.

#10 1 2cosr

Solution: There is again symmetry with respect to the polar axis.

Range of 0 to

2

2

to

Range of r 3 to 1 1 to -1

Because the signs changed, we probably should check when the graph crosses the pole so let’s set r to 0:

1 21 2cos 0 cos

2 3r

This means that from

20 to

3

we generate the top half

of the big loop and from2

to 3

we generate the bottom half of the small loop. Reflecting these

along the polar axis will give us the entire graph of the limacon with a loop.

#12 5 3sinr

Solution: The graph is symmetric with respect to the pi/2 axis.

Range of to 0

2

0 to

2

Range of r 2 to 5 5 to 8

We generate the right half of the limacon without a loop and reflecting this along the pi/2 axis will give us the entire graph.

#14 3secr

Solution:

33sec cos 3 3

cosr r r x

#16 2sin4r

Solution: The graph will be a rose with 2(4)=8 petals. Check for symmetry:

(1) Polar axis

by and by :

2sin4 2sin 4 4 2 sin4 cos4 cos4 sin4 2 0 sin4 2sin4

2sin4 Yes

r r

r r

r

(2) Pi/2 axis

by : 2sin4 2sin 4 4 2 sin4 cos4 cos4 sin4 2 0 sin4 2sin4r r

No conclusion.

by - and by : 2sin4 2sin4 2sin4r r r r r YES

(3) Pole

by : 2sin4 2sin 4 4 2 sin4 cos4 cos4 sin4 2 0 sin4 2sin4r r

Yes

Range of 4 0 to

2

2

to to

3

2

3

2

to

2

Range of 0 to

8

8

to

4

4

to

3

8

3

8

to

2

Range of r 0 to 2 2 to 0 0 to -2 -2 to 0

The first 2 columns will generate the lower petal in the first quadrant. By symmetry with respect to the polar axis, we can draw the petal right below it; by symmetry with respect to the pi/2 axis we can draw the lower petal in quadrant 2; by symmetry with respect to the pole we can draw the petal right under the petal that we drew in quadrant 2. Thus, we have 4 petals. The next two will generate the petal close to the 3pi/2 axis in the 3rd quadrant. We can draw the rest by using the symmetries.

#18 8cos5r

Solution: The graph will be a 5-leaf rose. The only symmetry is with respect to the polar axis.

Range of 5 0 to

2

to

2

3 to

2

3 to 2

2

52 to

2

Range of 0 to

10

10

to

5

3 to

5 10

3 2 to

10 5

2 to

5 2

Range of r 8 to 0 0 to -8 -8 to 0 0 to 8 8 to 0

The first column will generate the top half of the petal on the polar axis; reflect this on the polar axis to get the entire petal. The next two columns will generate the petal in quadrant 3 and using symmetry with respect to the polar axis, we get the petal in quadrant 2. The last 2 columns will give us the petal in quadrant 1 and using symmetry with respect to the polar axis, we will get the petal in quadrant 4.

#20 2 16sin2r

Solution: This will graph as a lemniscate that is symmetric with respect to the pole.

Range of 2 0 to

2

to

2

3 to

2

3 to 2

2

Range of 0 to

4

to

4 2

3 to

2 4

3 to

4 2

Range of r undefined und to 0 0 to 4 4 to 0

The third column will give us the “top” half of the graph in quadrant 2 and the “bottom” half of the graph in quadrant 4. The 4th column will complete the graph.

#24 1

1r r

Solution: Plug in values for starting from 0 and use a calculator!

Find a polar equation that has the same graph as the equation in x and y.

#28 2y

Solution:

2

sin 2 2csc

y

r r

#30 2 8x y

Solution:

2

2 2 2

2

8

8sincos 8 sin cos 8sin 8 tan sec

cos

x y

r r r r r

#32 6y x

Solution:

1

6

sin 6 cos tan 6 tan 6

y x

r r

#34 8xy

Solution:

2 2

8

8cos sin 8 16csc 2

cos sin

xy

r r r r

#36 3 3 3 0x y axy (Folium of Descartes)

Solution:

3 3

3 3 3 3 2

3 3

3 3

3 0

cos sin 3 cos sin 0

cos sin 3 cos sin 0 OR 0

3 cos sin

cos sin

x y axy

r r ar

r r a r

ar

Find an equation in x and y that has the same graph as the polar

#38 sin 2r

Solution:

sin 2

2

r

y

#40 4secr

Solution:

4sec cos 4 4

undefined at 2

r r x

n

#42 2 sin2 4r

Solution:

2 2sin2 4 2sin cos 4 sin cos 2 2r r r r xy

#44 3cos 4sin 12r

Solution:

3cos 4sin 12 3 cos 4 sin 12 3 4 12r r r x y

#46 2sin cos 3r r

Solution:

2 2 2 2 2sin cos 3 sin cos 3 3 3r r r r y x x y

#48 2cos 4sinr

Solution:

2

2 2

2 2

2 2

2cos 4sin

2 cos 4 sin

2 4

2 1 4 4 1 4

1 2 5

r

r r r

x y x y

x x y y

x y

#50 6cotr

Solution:

22 2 2 2 2 2 4 2

2

42 2 2 4 2

2

6cot 36cot 36 36

3636

xr r x y x y y x

y

yx x y y x

y

Find the slope of the tangent line to the graph of the polar equation at the point corresponding to the given value of .

#52 2sin ; 6

r

Solution:

2 2

2sin 2cos

sin cos 2cos sin 2sin cos 4cos sin sin2tan2

cos sin 2cos cos 2sin sin cos22 sin cos

At , tan 36 3

r r

r rm

r r

m

#54 1 2cos ; 2

r

Solution:

2 2

1 2cos 2sin

2sin sin 1 2cos cossin cos 2cos 2sin cos

cos sin 2sin cos 1 2cos sin 4sin cos sin

2 0 2 1 0At , 2

2 4 1 0 1

r r

r rm

r r

m

#56 2sin4 ; 4

r

Solution:

2sin4 8cos4

sin cos 8cos4 sin 2sin4 cos

cos sin 8cos4 cos 2sin4 sin

2 28 1 2 0

2 2At , 1

4 2 28 1 2 0

2 2

r r

r rm

r r

m

Note: For all the above, you can also evaluate r and r at the given value of first.

#62 If a and b are nonzero real numbers, prove that the graph of sin cosr a b is a circle,

and find its center and radius. Proof:

2 2

2 2 2 2

2 2

2 2

2 2 2 22 2

2 2 2 2

2 2

sin cos

0

4 4 4

2 2 4

circle with center at , and radius 2 2 2

r a b

y xx y a b

x y x y

x y ay bx

x bx y ay

b a a bx bx y ay

b a a bx y

b a a b