
#1
Sep2308, 12:38 AM

P: 161

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 3D 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. 



#2
Sep2308, 07:12 AM

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P: 26,167

(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 = cos²θsin²θ, it's cos2θ) 



#3
Sep2308, 07:34 AM

Math
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PF Gold
P: 38,879

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 zaxis, 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 xyplane 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 xyplane. 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? 


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