Polar equation to cartesian

In summary, the equation r=\sqrt{1+sin2\theta} can be converted to a cartesian equation, consisting of two circles. One can manipulate the equation to get x^6 + 3x^4 y^2 + 3 x^2 y^4 + y^6 - x^4 -6x^2 y^2 - 4x^3 y -y^4 - 4y^3 x=0, which can then be simplified into the required form of a fourth order equation in x and y. Using the hint \sin \alpha = \cos \left (\frac {\pi}{2} - \alpha\right), the equation can be rewritten as (x^{2}
  • #1
thenewbosco
187
0
i have this equation: [tex]r=\sqrt{1+sin2\theta}[/tex]
and am to convert to cartesian equation and from the equation see that it consists of two circles and directly note the radii of the cirlcles from the equation.

so far i have manipulated it and gotten:

[tex]x^6 + 3x^4 y^2 + 3 x^2 y^4 + y^6 - x^4 -6x^2 y^2 - 4x^3 y -y^4 - 4y^3 x=0[/tex]

any suggestions on how to get it into the required form?
thanks
 
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  • #2
HINT: [tex]\sin \alpha = \cos \left (\frac {\pi}{2} - \alpha\right)[/tex]
 
  • #3
It should be a fourth order equation in x and y. I get:

[tex]\left[ \left(x+\frac{1}{2}\right) ^2 + \left(y+\frac{1}{2}\right) ^2 -\frac{1}{2}\right] \left[ \left(x-\frac{1}{2}\right) ^2 + \left(y-\frac{1}{2}\right) ^2 -\frac{1}{2}\right] =0[/tex]
 
  • #4
Well, i get something like

[tex] x^2 +y^2 =1+\frac{2xy}{x^2 +y^2} [/tex],

from which the conclusion follows easily.

Daniel.
 
  • #5
dextercioby said:
Well, i get something like
[tex] x^2 +y^2 =1+\frac{2xy}{x^2 +y^2} [/tex],
from which the conclusion follows easily.
Daniel.

yes, i too found this but can you explain how the conclusion follows easily from this?
 
  • #6
For starters, rewrite it in the form:
[tex](x^{2}+y^{2})^{2}=(x+y)^{2}[/tex]
 
  • #7
benorin said:
then factor it using difference of squares
That was to be my next hint..:frown:

Then, the grand finale would have been to refer to your earlier solution to this problem. I won't do that now.
 
  • #8
There :rolleyes: , or rather not there: all better?
 

What is the difference between polar and Cartesian coordinates?

Polar coordinates use a distance from a fixed point and an angle from a fixed line to locate a point, while Cartesian coordinates use an x and y coordinate on a grid to locate a point.

How do you convert a polar equation to a Cartesian equation?

To convert a polar equation to a Cartesian equation, you can use the equations x = rcosθ and y = rsinθ, where r is the distance and θ is the angle.

What is the purpose of using polar equations?

Polar equations are useful for describing circular and symmetric patterns, which can be difficult to express using Cartesian coordinates. They are also commonly used in physics and engineering.

Can polar equations be graphed on a Cartesian plane?

Yes, polar equations can be graphed on a Cartesian plane by converting them into Cartesian equations using the conversion equations mentioned earlier.

What is the significance of the polar axis in polar equations?

The polar axis is the fixed line in polar coordinates, which is equivalent to the x-axis in Cartesian coordinates. It is used to measure the angle from a reference point to locate a point in the polar coordinate system.

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