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

  • Thread starter Thread starter ponjavic
  • Start date Start date
  • Tags Tags
    Polar
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
5 replies · 3K views
ponjavic
Messages
221
Reaction score
0
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.
 
Physics news on Phys.org
I don't see why you need to convert to Cartesian coordinates. You can make a sketch of the relation directly. [itex]\theta[/itex] is the angle relative to the x-axis and r is the distance from the origin.
 
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. [itex]\theta[/itex] 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?

[itex]\theta= 0, \pi/4, \pi/2, \pi/3,[/itex]
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 [itex]\theta = 0[/itex] you know that the point lies on the x axis. Evaluate the expresstion when [itex]\theta = 0[/itex] 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 [itex]\theta = \pi / 4[/itex] 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 [itex]\theta[/itex].

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.