# Graph of polar coordinate

shayaan_musta
How will you graph θ=π/4?
In polar coordinates you can graph an equation like, r=a(1-sinθ) which gives a cardioid graph.
But when my teacher giving me a class of sketching in polar coordinates then he ask to sketch θ=π/4.
But as we know that in polar coordinates (r,θ).
so in θ=π/4, there is no "r", so how we can graph it?
I think there must be some "r" which could make angle with +x-axis, if we take "r=0" then how could you make inclination with x-axis by π/4?

How will do this?

## Answers and Replies

Science Advisor
Homework Helper
hi shayaan_musta!
How will you graph θ=π/4?

how would you graph y = 4 ?

all points (x,y) with y = 4 (and x = anything)!​

how would you graph θ = π/4 ?

all points (r,θ) with θ = π/4 (and r = anything)!

shayaan_musta
hi shayaan_musta!

how would you graph y = 4 ?

all points (x,y) with y = 4 (and x = anything)!​

how would you graph θ = π/4 ?

all points (r,θ) with θ = π/4 (and r = anything)!

you mean we have not given r in θ = π/4 but here is r as r=any value?
Am I right?
If it is so then how will plot this as a polar coordinate?

Science Advisor
Homework Helper
hi shayaan_musta!
you mean we have not given r in θ = π/4 but here is r as r=any value?
Am I right?

yes
If it is so then how will plot this as a polar coordinate?

(you mean "in polar coordinates")

a straight line from the origin at an angle of π/4

shayaan_musta
(you mean "in polar coordinates")

yes I mean.

a straight line from the origin at an angle of π/4

but a straight line(as you said) shows that here is r=any value but not "0(zero)". Is it?

Science Advisor
Homework Helper
hmm … it would look the same, whether you included r = 0 or not

i think technically θ is undefined at r = 0, so if the equation is θ = π/4, then you're correct: r = 0 should be excluded

shayaan_musta
hmm … it would look the same, whether you included r = 0 or not

i think technically θ is undefined at r = 0, so if the equation is θ = π/4, then you're correct: r = 0 should be excluded

ok. thanks. great help dude.