3-D cylindrical equation graphing

NBAJam100

1. The problem statement, all variables and given/known data

Represent the surface in space, identify the surface-

r=2cos(theta)



2. Relevant equations

Uhh....



3. The attempt at a solution

My main thing here is this... how does r tell me anything about the graph of the function in a 3-D plane? i see that r=2cos(theta) but what does it tell me?? i know that 2cos(theta) = cos^2(t)-sin^2(t), so r= that... now what>??? im honestly stumped here, thats really as far as i can go with what i know.

tiny-tim

Hi NBAJam100!

(have a theta: θ and a squared: ² )

Can you check the equation?

r can't be negative, and cosθ can be negative, so how can it be r = 2cosθ?

(also, it isn't 2cosθ = cos²θ-sin²θ, it's cos2θ)

HallsofIvy

First, forget "2cos(theta)= cos^2(t)- sin^2(t)". It isn't true. You may be thinking about "cos(2 theta)= cos^2(theta)- sin^2(theta)" but that doesn't help here.

What does r= 2cos(theta) tell you? It tells you the r coordinate: the distance from the point to the z-axis, for each angle theta, of course.

Because this is in cylindrical coordinates, the first thing you should do is think about exactly what cylindrical coordinates are: there are 3 coordinates, of course, r, theta, and z. z measures the height above the xy-plane just as in Cartesian coordinates and r and theta are just the usual polar coordinates except that they apply to each plane parallel to the xy-plane. In particular, do you notice that there is no "z" in the formula? That tells you that the surface looks exactly the same for all different z!

You should also think about the equations connecting cylindrical and Cartesian coordinates: x= r cos(theta), y= r sin(theta), z= z or (inverting) r= [itex]\sqrt{x^2+ y^2}, theta= arctan(y/x), z= z. Your equation is r= 2 cos(theta). What do you get if you multiply both sides of the equation by r?