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Cylindrical coordinates

  1. Mar 12, 2009 #1
    Convert r=2cos(theta) from cylindrical coordinates to rectangular coordinates

    I have tried squaring both sides so that it will be equal to x squared plus y squared, and then solving for a variable. No matter what I do though I am left with two variables so I feel like I am taking the wrong approach. Where do I even start?
  2. jcsd
  3. Mar 12, 2009 #2
    I believe this is polar coordinates. I think you would benefit from drawing a picture of this on top of a rectangular plane. You can draw a right triangle with an acute angle theta at the origin, one leg on the x-axis, and a hypotenuse of r. You should be able to find x and y in terms of theta. From there, you can find y in terms of x.
  4. Mar 12, 2009 #3


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    Staff: Mentor

    Are you given a range of theta? If it is 0 to 2PI, the y=f(x) plot is not single-valued for x.

    Interesting plot, BTW. Have you sketched the function?
  5. Mar 12, 2009 #4
    I was not given a range, and no I have not tried to sketch it yet.
    Last edited: Mar 12, 2009
  6. Mar 12, 2009 #5
    How do I convert that to rectangular coordinates though? I mean I have tried plugging r into x=rcos(theta), and all the other various equations I could think of, and always ended up with multiple variables. The cos in there really throws me off.
    Last edited: Mar 12, 2009
  7. Mar 12, 2009 #6
    x=rcos(@), y=rsin(@) , z=z, (@=theta)

    r=2cos(@)=> r/2=cos(@)=> r/2=x/r, and we know in general that : r^2=x^2+y^2

    Now try to combine these expressions, and see what you get...
    Last edited: Mar 12, 2009
  8. Mar 12, 2009 #7
    You can try the method that I tried to explain earlier. Here is the diagram I was talking about to further illustrate.


    You make a right triangle. I am drawing it at an arbitrary theta, the math should hold up for all theta though. As you can see, one side of the hypotenuse is at the origin and the other end is at the arbitrary point (r, theta) which is part of the polar function r = 2cos(theta). You should also notice that the two legs of the triangle are the horizontal and vertical distance from the origin or (x, y). You can use trig to solve for x and y in terms of theta. Start with that.
  9. Mar 12, 2009 #8
    I guess i just ruined the party!
  10. Mar 12, 2009 #9
    I'm sorry sutupidmath, I'm working with algebra II math, so I'm not completely sure what you did, but you got the same answer as me. I started writing my previous thread before you posted your response, I didn't mean to contradict you or anything.
  11. Mar 12, 2009 #10
    No, no, don't worry. It's just that i shouldn't have given the whole answer to the OP. I'm debating whether i should delete it or not!...lol...
  12. Mar 12, 2009 #11
    The difference between polar and cylindrical coordinates, is that in cylindirical coordinates we have the third dimension,(z), but this doesn't cause us any trouble since z=z(i.e it has the same value in cylindrical and rectangular coordinates) SO as a result it will change how we interpret our result.

    From here, i suspect, you would be inclined to interpret the final result as a circle with radius 1 and center (1,0), wouldn't you?
  13. Mar 12, 2009 #12


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    Staff: Mentor

    Yeah, too much help, IMO. I'm too tired to delete it though -- can you still edit it into more of a hint?
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