What is the Cartesian form of 4-r = sin(theta) - 2 cos^2(theta) ?

1 Answer
Jun 26, 2018

6x^2+4y^2=(x^2+y+y^2)sqrt(x^2+y^2)

Explanation:

Polar to Cartesian relations:
r^2=x^2+y^2
theta=arctan(y/x)

Substitute these in:
4-r=sintheta-2cos^2theta
4-sqrt(x^2+y^2)=sinarctan(y/x)-2cos^2arctan(y/x)

We can simplify the RHS; we need to express sine and cosine in terms of tangent.

Sine

tanx=sinx/cosx=sinx/sqrt(1-sin^2x)
(1-sin^2x)tan^2x=sin^2x
tan^2x=sin^2x(1+tan^2x)
sinx=tanx/sqrt(1+tan^2x)

So sinarctanx=x/sqrt(1+x^2)

Cosine

tanx=sinx/cosx=sqrt(1-cos^2x)/cosx
cos^2xtan^2x=1-cos^2x
cos^2x(1+tan^2x)=1
cosx=1/sqrt(1+tan^2x)

So cosarctanx=1/sqrt(1+x^2)

Substitute

So
sinarctan(y/x)=(y/x)/sqrt(1+y^2/x^2)=y/sqrt(x^2+y^2)
and
cos^2arctan(y/x)=[1/sqrt(1+y^2/x^2)]^2=[x/sqrt(x^2+y^2)]^2=x^2/(x^2+y^2)

We have after Polar to Cartesian substitution
4-sqrt(x^2+y^2)=sinarctan(y/x)-2cos^2arctan(y/x)

which then becomes
4-sqrt(x^2+y^2)=y/sqrt(x^2+y^2)-(2x^2)/(x^2+y^2)

Put over a common denominator:
4(x^2+y^2)-(x^2+y^2)^(3/2)=ysqrt(x^2+y^2)-2x^2
6x^2+4y^2=(x^2+y+y^2)sqrt(x^2+y^2)

We can square out the root to create a polynomial in x and y, which is a more conventional way to express this, but less tidy. It also risks creating surplus roots to the equation that weren't roots of the original.

(6x^2+4y^2)^2=(x^2+y+y^2)^2(x^2+y^2)

36x^4+48x^2y^2+16y^2=(x^4+2x^2y+2x^2y^2+y^2+2y^3+y^4)(x^2+y^2)

36x^4+48x^2y^2+16y^2= x^6+2x^4y+2x^4y^2+x^2y^2+2x^2y^3+x^2y^4+ x^4y^2+2x^2y^3+2x^2y^4+y^4+2y^5+y^6

Collect terms and group them according to the degree of each monomial:

(x^6+3x^4y^2+3x^2y^4+y^6)+(2x^4y+4x^2y^3+2y^5)+(-36x^4-47x^2y^2+y^4)+(-16y^2)=0

Notice that we can factorise a couple of the groups, a process that does some returning along the path taken:

(x^2+y^2)^3+2y(x^2+y^2)^2-36x^4-47x^2y^2+y^4-16y^2=0