Sketching/solving polar: r^2=a^2*cos2t

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The discussion focuses on sketching the polar equation r^2 = a^2 * cos(2t) in the x-y plane. Participants emphasize the effectiveness of plotting points directly in polar coordinates rather than converting to Cartesian coordinates, suggesting specific angles like θ = 0, π/4, and π/2 for evaluation. They recommend calculating the corresponding r values for these angles and plotting them to visualize the relationship. The importance of the parameter 'a' is highlighted, as it affects the graph's shape. Overall, the consensus is that sketching in polar coordinates provides clearer insights than Cartesian transformations.
ponjavic
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Sketch (in the x-y plane): r^2=a^2*cos2t where r and t are polar coordinates.

I simply am not able to convert this formula to x and y.
I have gotten as far as:
(x^2+y^2)^2=a^2(x^2-y^2)
or r^4=a^2(r^2*cost^2-r^2*sint^2)
using r = x^2+y^2, cos2t=1-sint^2, x = rcost and y = rsint but I simply can not get any further.

Please give any hints you think might help me solve this.
 
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I don't see why you need to convert to Cartesian coordinates. You can make a sketch of the relation directly. \theta is the angle relative to the x-axis and r is the distance from the origin.
 
Try this: take \theta= 0, \pi/4, \pi/2, \pi/3, etc. and see what you get for r: graph those points in polar coordinates.
 
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well, there is now way to isolate y from x using cartesian coordianres, its really easier to see what's going on in polar coordinates.
the best way is to do as suggested, and plot what you get on x-y plane.
 
Tide said:
I don't see why you need to convert to Cartesian coordinates. You can make a sketch of the relation directly. \theta is the angle relative to the x-axis and r is the distance from the origin.
I don't know how to use polar coordinates to sketch, If I have it in cartesians I could do y=0, y'=0, x=0, find asymptotes and such.
How can I find this to help me sketch in polar coordinates?

\theta= 0, \pi/4, \pi/2, \pi/3,
I'll try to work around with this and see what it gets me, what's a^2 though?
 
Last edited:
pon,

Do as Halls suggested. For example, when \theta = 0 you know that the point lies on the x axis. Evaluate the expresstion when \theta = 0 to find out how far from the origin the corresponding point is. Then place a point r units from the origin and on the x-axis corresponding to that point.

Next try \theta = \pi / 4 which you know lies along a line at 45 degrees above the x axis. Find the distance to that point using your formula and place a point that far from the origin and along the line y = x on your graph. Do this for several values of \theta.

Also, your graph will depend on the parameter a. Do all of the above for different values of a like a = 1, a = 1/2, a = 2 etc.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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