How do I draw this r=2cos(Ө)

is there something I could use instead of graphing calculator?

I don't own one and its a big problem for me and I don't even know how to draw these

This post was edited by twiiztedacer on Mar 16 2015 12:01pm

type 2cosx into gooogle

Since r = 2 cos θ, it follows that r^2= 2r cos θ. But r^2 = x^2 + y^2 and r cos θ = x, therefore
x^2 + y^2 = 2x. This can be rewritten as (x − 1)^2 + y^2 = 1 which is a circle of radius 1 with the center at (1, 0).

If you need further help with polar graphs, you can refer here: http://cims.nyu.edu/~kiryl/Precalculus/Section_8.2-Graphs%20of%20Polar%20Equations/Graphs%20of%20Polar%20Equations.pdf

does that rly work?

i know a pilot with a polar rout, but i get the feeling that's not what you mean.

Google works for everything.

In this case, it's easier to use wolframalpha.com for your math questions.

its suppose to be a circle

Yah, you need to do it how I explained it to you. If you want extra help, feel free to send me a personal message. If you would like a program that helps you graph polar plots online, use this resource:

http://fooplot.com/#W3sidHlwZSI6MSwiZXEiOiIyY29zKHRoZXRhKSIsImNvbG9yIjoiIzAwODBjYyIsInRoZXRhbWluIjoiMCIsInRoZXRhbWF4IjoiMnBpIiwidGhldGFzdGVwIjoiLjAxIn0seyJ0eXBlIjoxMDAwfV0-

Jeffry has your correct response.



This is in Euclidean coordinates, not Polar coordinates.

This is incorrect.

Remember, polar coordinates give give r (DISTANCE from origin) as a function of theta (the angle). It should form a circle.

There are multiple ways to draw this graph. You can plug in your own values for theta (I would advise that you use values from the unit circle) or you can convert into (x,y) cartesian coordinates.

You will indeed see that as theta goes up, the radius stays constant and you will get a circle.

/e looks like Jeffry has you covered. You can always pm me if you want more polar coordinate help (he can probably help you as well, but don't pm him unless he gives you permission of course).

This post was edited by Casey on Mar 17 2015 04:04pm