42 tangent and arc length in polar coordinates

Tangent and Arc Length in Polar Equations


Given the polar function r = f(), it may be realized as

the parametric equations:

x() = f()cos()

y() = f()sin().


Given the parametric equations

x = x()

y = y()

the dervative . dy

dx =dy/d

dx/d



x() = f()cos()

y() = f()sin().



x = x()

y = y()

the dervative .

Hence for r = f(), the slope at the point (r=f(), ) isdy

dx =dy/d

dx/d=

d(f()cos())/d

d(f()sin())/d

dy

dx =dy/d

dx/d



x() = f()cos()

y() = f()sin().



x = x()

y = y()

the dervative .



x() = f()cos()

y() = f()sin().

Hence for r = f(), the slope at the point (r=f(), ) isdy

dx =dy/d

dx/d=

d(f()cos())/d

d(f()sin())/d=

f '()sin() + f()cos()

f '()cos() – f()sin()

dy

dx =dy/d

dx/d

Slopes in Polar Equations

Example: a. Find the slope at r = cos(2) at = /6.


(1/2,/6)



We have x() = cos(2)cos() and y() = cos(2)sin(),

(1/2,/6)



We have x() = cos(2)cos() and y() = cos(2)sin(), hence

dy

dx =dy/d

dx/d=

(1/2,/6)


-2sin(2)sin() + cos(2)cos()

(1/2,/6)




dy

dx =dy/d

dx/d=

=-2sin(2)cos() – cos(2)sin()

(1/2,/6)





dy

dx =dy/d

dx/d=

at = /6,dy

dx=

-2sin(/3)sin(/6) + cos(/3)cos(/6)

-2sin(/3)cos(/6) – cos(/3)sin(/6)

(1/2,/6)






dy

dx =dy/d

dx/d=

=-3 * ½ + ½ * 3/2

-3* 3/2 – ½ * ½ (1/2,/6)


at = /6,dy

dx=

-2sin(/3)sin(/6) + cos(/3)cos(/6)

-2sin(/3)cos(/6) – cos(/3)sin(/6)





dy

dx =dy/d

dx/d=

=-3 * ½ + ½ * 3/2

-3* 3/2 – ½ * ½

=-23 + 3

-6 – 1

=3 7

(1/2,/6)


at = /6,dy

dx=

-2sin(/3)sin(/6) + cos(/3)cos(/6)

-2sin(/3)cos(/6) – cos(/3)sin(/6)





dy

dx =dy/d

dx/d=

Example: b. Find where the slope is 0 for r = 1+ cos().





We have x() = (1+cos())cos() and y() = (1+cos())sin().


dy

dx=

dy/d

dx/dWe want = 0, or

dy

d = 0.




-sin()sin() +(1+ cos())cos() = 0dy

d=Set


dy

dx=

dy/d

dx/dWe want = 0, or

dy

d = 0.



-S2 + C + C2 = 0



d=Set

dy

dx=

dy/d

dx/dWe want = 0, or

dy

d = 0.



-S2 + C + C2 = 0

-(1 – C2) + C + C2 = 0



d=Set

dy

dx=

dy/d

dx/dWe want = 0, or

dy

d = 0.



d-S2 + C + C2 = 0

-(1 – C2) + C + C2 = 0

2C2 + C – 1 = 0

(2C – 1)(C + 1) = 0



d=Set

dy

dx=

dy/d

dx/dWe want = 0, or

dy

d = 0.



-S2 + C + C2 = 0

-(1 – C2) + C + C2 = 0

2C2 + C – 1 = 0

(2C – 1)(C + 1) = 0

cos() = ½ = ±/3



d=Set

dy

dx=

dy/d

dx/dWe want = 0, or

dy

d = 0.



-S2 + C + C2 = 0

-(1 – C2) + C + C2 = 0

2C2 + C – 1 = 0

(2C – 1)(C + 1) = 0

cos() = ½ = ±/3

cos() = -1 = -π



d=Set

dy

dx=

dy/d

dx/dWe want = 0, or

dy

d = 0.



d-S2 + C + C2 = 0

-(1 – C2) + C + C2 = 0

2C2 + C – 1 = 0

(2C – 1)(C + 1) = 0

cos() = ½ = ±/3

cos() = -1 = 3/2

However, 1+cos() is not

differentiable at -π since

dx/d = 0. Hence = ±/3.



d=Set

dy

dx=

dy/d

dx/dWe want = 0, or

dy

d = 0.



d-S2 + C + C2 = 0

-(1 – C2) + C + C2 = 0

2C2 + C – 1 = 0

(2C – 1)(C + 1) = 0

cos() = ½ = ±/3

cos() = -1 = 3/2

However, 1+cos() is not

differentiable at -π since

dx/d = 0. Hence = ±/3.



d=Set

dy

dx=

dy/d

dx/dWe want = 0, or

dy

d = 0.



(1+3/2, /3)

Arc Length for Polar CurvesGiven the polar equation r = f() in parametric form

x = r*cos(), y = r*sin(). We may easily verify that

(dx/d)2+(dy/d)2 = r2 + (dr/d)2.

Hence the arc-length from =a to =b is:



(dx/d)2+(dy/d)2 = r2 + (dr/d)2.


drrdddyddx

b

a

b

a

2222 ))('())/()/(



(dx/d)2+(dy/d)2 = r2 + (dr/d)2.


drrdddyddx

b

a

b

a

2222 ))('())/()/(

Example: Let r = 2sin(). Find the arc-length for = 0 to =2π.



(dx/d)2+(dy/d)2 = r2 + (dr/d)2.


drrdddyddx

b

a

b

a

2222 ))('())/()/(


r =1



(dx/d)2+(dy/d)2 = r2 + (dr/d)2.


drrdddyddx

b

a

b

a

2222 ))('())/()/(


r2 + (r')2 = (2sin())2 + (2cos())2

= 4sin2()+4cos()2 = 4. Hence the

arc-length is: r =1



(dx/d)2+(dy/d)2 = r2 + (dr/d)2.


drrdddyddx

b

a

b

a

2222 ))('())/()/(


424))('(

2

0

2

0

2

0

22

dddrr

r =1

r2 + (r')2 = (2sin())2 + (2cos())2


arc-length is:



(dx/d)2+(dy/d)2 = r2 + (dr/d)2.


drrdddyddx

b

a

b

a

2222 ))('())/()/(


424))('(

2

0

2

0

2

0

22

dddrr

r =1

Note that we get the circumference of two circles since the

Graphs traced out two cicles as goes from 0 to 2π.

r2 + (r')2 = (2sin())2 + (2cos())2


arc-length is:

Arc Length for Polar CurvesExample: Find the arc-length of one round of r = 1 – cos().


The graph is a cardiod.

r = 1 – cos()



r2 + (r')2 = (1 – cos())2 + (sin())2

r = 1 – cos()



r2 + (r')2 = (1 – cos())2 + (sin())2

= 1 – 2c + c2 + s2

r = 1 – cos()



r2 + (r')2 = (1 – cos())2 + (sin())2

= 1 – 2c + c2 + s2

= 2 – 2c.

r = 1 – cos()



r2 + (r')2 = (1 – cos())2 + (sin())2

= 1 – 2c + c2 + s2

= 2 – 2c. We can't integrate 2 – 2c as is.

r = 1 – cos()



r2 + (r')2 = (1 – cos())2 + (sin())2

= 1 – 2c + c2 + s2


Use the double formula to remove the root:

r = 1 – cos()1 – cos(2A)

2sin2( ) =A or 1 – cos()2sin2 ( ) =

2



r2 + (r')2 = (1 – cos())2 + (sin())2

= 1 – 2c + c2 + s2



r = 1 – cos()1 – cos(2A)

2sin2( ) =A or 1 – cos()2sin2 ( ) =

2

Use symmetry, we double the integral from 0 to π:



r2 + (r')2 = (1 – cos())2 + (sin())2

= 1 – 2c + c2 + s2



ddrr

b

a

0

22 ))cos(1(22))('(

r = 1 – cos()1 – cos(2A)

2sin2( ) =A or 1 – cos()2sin2 ( ) =

2




r2 + (r')2 = (1 – cos())2 + (sin())2

= 1 – 2c + c2 + s2



dddrr

b

a

0

2

2

0

22 )(sin2*22))cos(1(22))('(

r = 1 – cos()1 – cos(2A)

2sin2( ) =A or 1 – cos()2sin2 ( ) =

2




r2 + (r')2 = (1 – cos())2 + (sin())2

= 1 – 2c + c2 + s2



d

dddrr

b

a

0

2

0

2

2

0

22

)sin(4

)(sin2*22))cos(1(22))('(

r = 1 – cos()1 – cos(2A)

2sin2( ) =A or 1 – cos()2sin2 ( ) =

2




r2 + (r')2 = (1 – cos())2 + (sin())2

= 1 – 2c + c2 + s2



8)8(0|)cos(8)sin(4

)(sin2*22))cos(1(22))('(

02

0

2

0

2

2

0

22

d

dddrr

b

a

r = 1 – cos()1 – cos(2A)

2sin2( ) =A or 1 – cos()2sin2 ( ) =

2


Education

42 tangent and arc length in polar coordinates