Determine Polar Coordinates From Graph...1

[nycmathguy]

 

[MathLover1]

26

r=2/cos(θ-π/3)

by definition
x=rcos(θ)
y=r sin(θ)


(r,θ)

point A is at (r,θ) where θ=0
A (r,0)
find r
r=2/cos(0-π/3)
r=2/cos(-π/3)
r=2/(1/2)
r=4

=>A (4,0)

point B is at (r,θ) where θ=π/3
B (r,pi/3)
find r
r=2/cos(π/3-π/3)
r=2/cos(0)
r=2

=>B (2,π/3)


point C is at (r,θ) where θ=π/2
C (r,pi/2)
find r
r=2/cos(π/2-π/3)
r=2/(sqrt(3)/2)
r=4/sqrt(3)

=>C (4/sqrt(3) ,π/2)

point D is at (r,θ) where θ=2π/3
D (r,2π/3)

find r
r=2/cos(2π/3-π/3)
r=2/cos(π/3)
r=4
D (4,2π/3)

27.
r=-1- sin(θ)

point A is at (r,θ) where θ=3π/4
A (r,3π/4)
find r
r=-1- sin(3π/4)
r=-1 - 1/sqrt(2) exact value
r=-1.71 rounded
A (-1.71,3π/4)

similarly, you can do the rest

 

[nycmathguy]

26

r=2/cos(θ-π/3)

by definition
x=rcos(θ)
y=r sin(θ)


(r,θ)

point A is at (r,θ) where θ=0
A (r,0)
find r
r=2/cos(0-π/3)
r=2/cos(-π/3)
r=2/(1/2)
r=4

=>A (4,0)

point B is at (r,θ) where θ=π/3
B (r,pi/3)
find r
r=2/cos(π/3-π/3)
r=2/cos(0)
r=2

=>B (2,π/3)


point C is at (r,θ) where θ=π/2
C (r,pi/2)
find r
r=2/cos(π/2-π/3)
r=2/(sqrt(3)/2)
r=4/sqrt(3)

=>C (4/sqrt(3) ,π/2)

point D is at (r,θ) where θ=2π/3
D (r,2π/3)

find r
r=2/cos(2π/3-π/3)
r=2/cos(π/3)
r=4
D (4,2π/3)

27.
r=-1- sin(θ)

point A is at (r,θ) where θ=3π/4
A (r,3π/4)
find r
r=-1- sin(3π/4)
r=-1 - 1/sqrt(2) exact value
r=-1.71 rounded
A (-1.71,3π/4)

similarly, you can do the rest

Yes, I can and will do the rest and a few other questions when time allows.

 

[nycmathguy]

26

r=2/cos(θ-π/3)

by definition
x=rcos(θ)
y=r sin(θ)


(r,θ)

point A is at (r,θ) where θ=0
A (r,0)
find r
r=2/cos(0-π/3)
r=2/cos(-π/3)
r=2/(1/2)
r=4

=>A (4,0)

point B is at (r,θ) where θ=π/3
B (r,pi/3)
find r
r=2/cos(π/3-π/3)
r=2/cos(0)
r=2

=>B (2,π/3)


point C is at (r,θ) where θ=π/2
C (r,pi/2)
find r
r=2/cos(π/2-π/3)
r=2/(sqrt(3)/2)
r=4/sqrt(3)

=>C (4/sqrt(3) ,π/2)

point D is at (r,θ) where θ=2π/3
D (r,2π/3)

find r
r=2/cos(2π/3-π/3)
r=2/cos(π/3)
r=4
D (4,2π/3)

27.
r=-1- sin(θ)

point A is at (r,θ) where θ=3π/4
A (r,3π/4)
find r
r=-1- sin(3π/4)
r=-1 - 1/sqrt(2) exact value
r=-1.71 rounded
A (-1.71,3π/4)

similarly, you can do the rest

 

[MathLover1]

good, now graph it

 

[nycmathguy]

good, now graph it

The graph has already been given. Graph what?

 

[MathLover1]

The graph has already been given. Graph what?

I saw it now

 

[nycmathguy]

I saw it now

I will graph 3 polar equations today to end our Precalculus journey. We then begin College Algebra one week and Calculus 1 the next and keep going that way.