I've got r = yx/r^2 and r^2 = x^2 + y^2, but how do you clear out the r on the left?
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You're almost there.
using r = yx/r^2 that you had:
r^2 = (y^2 * x^2) / r^4 = (y^2 * x^2) / (r^2)^2
x^2 + y^2 = (y^2 * x^2) / (y^2 + x^2)^2
another way is to use r = sqrt( x^2 + y^2) and substitute that into r of your first equation which will get you:
sqrt(x^2 + y^2) = (x * y) / (x^2 + y^2)
both answers are the same
using r = yx/r^2 that you had:
r^2 = (y^2 * x^2) / r^4 = (y^2 * x^2) / (r^2)^2
x^2 + y^2 = (y^2 * x^2) / (y^2 + x^2)^2
another way is to use r = sqrt( x^2 + y^2) and substitute that into r of your first equation which will get you:
sqrt(x^2 + y^2) = (x * y) / (x^2 + y^2)
both answers are the same
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Well, there's just a few things you need to know for all conversions from polar to rectangular or the other way around.
1. r²=x²+y²
2. x=rcosθ
3. y=rsinθ
From those 3 things, you should be able to derive anything else you may need.
Ok, here's the problem.
r=sinθcosθ
Instead of using the fact that sinθ=y/r and cosθ=x/r, I'm going to use the fact that x=rcosθ and y=rsinθ. I'll need to get some 'r's in there to do that, so I'll multiply each side by r².
r³=rsinθrcosθ
r³=xy
r(r²)=xy
r(x²+y²)=xy
I still have an r that is alone, so I'll have to take one of the equations from above and alter it.
Take: r²=x²+y². If you take the root of both sides, r will be isolated. So, r=√(x²+y²). Now substitute that in. (Sorry for the awkward spaces after superscripts bigger than 3)
√(x²+y²)(x²+y²)=xy
Since rectangular equations are usually solved for y, I'd normally solve for y, but that's too gross for me. You can look at where it led if you want to. I'd just leave the answer as it is above.
(x²+y²)(x²+y²)²=x²y²
(x²+y²)(x⁴+2x²y²+y⁴)=x²y²
x⁶+3x⁴y²+3x²y⁴+y⁶=x²y²
y⁶+3x²y⁴+3x⁴y²-x²y²-x⁶=0
y⁶+3x²y⁴+y²(3x⁴-x²)-x⁶=0
1. r²=x²+y²
2. x=rcosθ
3. y=rsinθ
From those 3 things, you should be able to derive anything else you may need.
Ok, here's the problem.
r=sinθcosθ
Instead of using the fact that sinθ=y/r and cosθ=x/r, I'm going to use the fact that x=rcosθ and y=rsinθ. I'll need to get some 'r's in there to do that, so I'll multiply each side by r².
r³=rsinθrcosθ
r³=xy
r(r²)=xy
r(x²+y²)=xy
I still have an r that is alone, so I'll have to take one of the equations from above and alter it.
Take: r²=x²+y². If you take the root of both sides, r will be isolated. So, r=√(x²+y²). Now substitute that in. (Sorry for the awkward spaces after superscripts bigger than 3)
√(x²+y²)(x²+y²)=xy
Since rectangular equations are usually solved for y, I'd normally solve for y, but that's too gross for me. You can look at where it led if you want to. I'd just leave the answer as it is above.
(x²+y²)(x²+y²)²=x²y²
(x²+y²)(x⁴+2x²y²+y⁴)=x²y²
x⁶+3x⁴y²+3x²y⁴+y⁶=x²y²
y⁶+3x²y⁴+3x⁴y²-x²y²-x⁶=0
y⁶+3x²y⁴+y²(3x⁴-x²)-x⁶=0
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I think r^2=x^2+y^2 is a rectangular equation for a circle correct me if I am wrong but that looks pretty much the same as the equation of a circle