[SOLVED] Conversion of Polar and Rectangular Equations

**Nerdfighter** · June 6th 2010, 02:19 PM

I need to convert this rectangular equation into polar form:
$x^4 + 2x^2y^2 + y^4 = 2xy$

I know that $x = rcos\theta$ and $y = rsin\theta$ . I've tried a little substitution but have not had a lot of success in simplifying it.

I also need to convert this polar equation into rectangular form:
$r^2 = \frac{36}{9cos^2\theta - 4sin^2\theta}$

I have a suspicion that $r^2 = x^2 + y^2$ is a useful tool, but I am having trouble with the conversion still. I used it to create this simplified equation in rectangular form:

$y = \frac{6}{3cos\theta - 2sin\theta} -x$ Is this right? I simply substituted $r^2$ , subtracted $x^2$ and took the square root of everything. Is this correct?

**TheEmptySet** · June 6th 2010, 02:29 PM

Originally Posted by Nerdfighter

I need to convert this rectangular equation into polar form:
$x^4 + 2x^2y^2 + y^4 = 2xy$

I know that $x = rcos\theta$ and $y = rsin\theta$ . I've tried a little substitution but have not had a lot of success in simplifying it.

I also need to convert this polar equation into rectangular form:
$r^2 = \frac{36}{9cos^2\theta - 4sin^2\theta}$

I have a suspicion that $r^2 = x^2 + y^2$ is a useful tool, but I am having trouble with the conversion still. I used it to create this simplified equation in rectangular form:

$y = \frac{6}{3cos\theta - 2sin\theta} -x$ Is this right? I simply substituted $r^2$ , subtracted $x^2$ and took the square root of everything. Is this correct?

For the first one note that
$x^4+2x^2y^2+y^4=(x^2+y^2)^2$

Then use $x^2+y^2=r^2$

For the next one clear the fractions to get

$r^2(9cos^2\theta - 4sin^2\theta) = 36 \iff 9(r\cos(\theta))^2-4(r\sin(\theta))^2=36$

See if this can get you started

**SpringFan25** · June 6th 2010, 02:32 PM

Question 1:

$x^4 + 2x^2y^2 + y^4 = 2xy$

Make the substitutions you suggested

$r^4cos^4(\theta) +2r^4cos^2(\theta)\sin^2(\theta) + r^4sin^4(\theta) = 2r^2\sin(\theta)\cos(\theta)$

Divide by r^4

$cos^4(\theta) + sin^4(\theta) + 2cos^2(\theta)\sin^2(\theta)= 2r^{-2}\sin(\theta)\cos(\theta)$

You can factorise the left hand side

$\left(sin^2(\theta) + cos^2(\theta) \right)^2 = 2r^{-2}\sin(\theta)\cos(\theta)$

But you know sin^2 + cos^2 = 1

$1^2 = 2r^{-2}\sin(\theta)\cos(\theta)$

Tidy up to finish.

editToo slow...again

**Nerdfighter** · June 6th 2010, 02:39 PM

Thank you for the tips and help everyone, this is really useful. I can't believe I didn't see that factoring from the first problem, it's so obvious now!

**studentathlete24** · June 6th 2010, 08:37 PM

I have the same questions.
For the first conversion I ended up with:

$r^3 = 2cosOsinO$
Is this a valid polar equation, or do I need to simplify it further?

For the second conversion my answer is:
$3x - 2y = 6$ OR $y=2/3x-3$

Do these look accurate?

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