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Polar and cartesian form

  1. Jun 18, 2007 #1
    1. The problem statement, all variables and given/known data
    i) Express the equation r = 1 + cos(theta) in Cartesian form.

    ii) Sketch the curve whose equation in polar form is r = cos(theta)

    2. Relevant equations

    x = r cos (theta)
    y = r sin (theta)

    3. The attempt at a solution

    i) x = r cos (theta) = (1 + cos(theta)) cos (theta)
    y = r cos (theta) = (1 + cos(theta)) cos (theta)

    i'm not sure thats right???

    ii) i have no idea how to sketch it, because there are no figures to work with!

    thanks
     
    Last edited: Jun 18, 2007
  2. jcsd
  3. Jun 18, 2007 #2

    berkeman

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    Staff: Mentor

    I'm not sure that we are getting the whole problem statement here, but hopefully I can at least help you with sketching it.

    Let's sketch the polar form of the equation. Draw the traditional x,y axis cross, with the y axis vertical (+ up) and the x axis horizontal (+ to the right). Now remember that in polar coordinates, r is the distance away from the origin, and theta is the angle formed to that radius r from the + x axis, measured in the counterclockwise direction from the x axis over to the radius r line.

    Now, the equation has a constant offset r = 1, which then gets the cos(theta) term added to it, depending on the angle theta as it sweeps around from zero to 2PI.

    So what is the value of r for these main points in the sketch?

    theta = 0
    theta = PI/2 (90 degrees, which is on the y-axis, right?)
    theta = PI (180 degrees, along the - x axis)
    theta = 3PI/2 (270 degrees, along the - y axis)

    Now that you have those 4 points on the curve, you can start to see what it will look like. Now fill in these next 4 points to get a better idea:

    theta = PI/4 (45 degrees)
    theta = 3PI/4
    theta = 5PI/4
    theta = 7PI/4

    Do you see what the curve looks like? Now, can you start to see how to use the polar --> retangular conversion equations to express this curve in rectangular form? You will end up with two values of y for each x, it would seem.
     
    Last edited: Jun 18, 2007
  4. Jun 18, 2007 #3
    ok thanks, when doing cos(theta) woud theta be in radians or degrees?? thanks
     
  5. Jun 18, 2007 #4

    berkeman

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    Staff: Mentor

    You can do either; that's why I listed both. It depends on the situation, but it's usually safest to just work in radians, using the fractions of PI in the arguments. Just be consistent, and check to be sure that your calculator is using the same system of angular units as you are in your head.
     
  6. Jun 18, 2007 #5
    I just tried to sketch the curve whos form is r = cos(theta)

    but there is not curve!!??
    when theta = 0, r = 1
    when theta = (pi/2), r = 0
    when theta = (pi), r = -1
    ect

    so there is no curve!!! the point im finding are just on the x or y axis, or on the origin!

    am i doing something wrong.
     
  7. Jun 18, 2007 #6

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Well, choosing theta to be a multiple of pi/2, of course the points you get are on the x and y axes- they correspond to theta= 0, pi/2, pi, 3pi/2.
    Try thet= pi/4, pi/3, pi/2, etc.

    By the way, if you want to convert the equations in r and theta to x and y you might try multiplying your equation by r: r2= r + rcos(theta).
    Of course, [itex]r= \sqrt{x^2+ y^2}[/itex].
     
  8. Jun 18, 2007 #7

    berkeman

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    Staff: Mentor

    As HallsofIvy says, you need to fill in more points to see the curve.

    Try this -- draw a different set of axes first: draw crossed axes, with r as the vertical axis and theta as the horizontal axis. Now draw r = cos(theta) on that plot. You know, where r = 1 at theta = 0, and the curve goes to the right and comes down through 0 at theta = PI/2, then goes to -1 at theta = PI, etc., and finishes back at r = 1 when theta = 2PI. That's the traditional plot of cos(theta), right?

    Now let's go back to the polar plot that you were trying to sketch -- the simplified one where r = cos(theta) without the offset of 1 yet. Picture a line in your mind on the polar plot that starts out pointing along the x axis, and pivots at the origin. You swing that line around the full circle of the polar plot, and the "r" distance of the point that you are plotting at any theta is just given by the cos(theta) number from your linear plot that you just made.

    So assuming that you are right-handed, draw with your right hand on the polar plot as you move your left finger along the linear cos(theta) plot to see what the value of cos(theta) is for each theta. Mentally picture the line that pivots at the origin as it swings up and around through each angle theta, and look at the linear cos(theta) plot to see what value cos(theta) has at that angle, and plot that point on the polar plot.

    What kind of figure do you get?

    Now draw another linear plot, this time of 1 + cos(theta). What does that do to the linear plot? Now draw another polar sketch, this time of r = 1 + cos(theta). What did the offset of 1 do to the polar plot?
     
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