What is the Cartesian form of r = -sintheta+4cos^2theta r=sinθ+4cos2θ?

1 Answer

(x^2+y^2+y)sqrt(x^2+y^2)=4x^2(x2+y2+y)x2+y2=4x2

Explanation:

graph{(x^2+y^2+y)sqrt(x^2+y^2)=4x^2 [-8.046, 8.046, -4.02, 4.024]}

From the given

r=-sin theta+4*cos^2 thetar=sinθ+4cos2θ

the conversion equations:

r=sqrt(x^2+y^2)r=x2+y2 and theta=tan^-1(y/x)θ=tan1(yx)

tan theta=y/xtanθ=yx and sin theta=y/sqrt(x^2+y^2)sinθ=yx2+y2 and cos theta=x/sqrt(x^2+y^2)cosθ=xx2+y2

Convert now, the given equation becomes

sqrt(x^2+y^2)=-y/sqrt(x^2+y^2)+(4*x^2)/(x^2+y^2)x2+y2=yx2+y2+4x2x2+y2

multiply both sides of the equation by (x^2+y^2)(x2+y2)

(x^2+y^2)sqrt(x^2+y^2)=-(y*(x^2+y^2))/sqrt(x^2+y^2)+((x^2+y^2)(4*x^2))/(x^2+y^2)(x2+y2)x2+y2=y(x2+y2)x2+y2+(x2+y2)(4x2)x2+y2

(x^2+y^2)sqrt(x^2+y^2)=-(ycancel((x^2+y^2)))/cancelsqrt(x^2+y^2)+(cancel((x^2+y^2))(4*x^2))/cancel(x^2+y^2)

(x^2+y^2)sqrt(x^2+y^2)=-y*sqrt(x^2+y^2)+4*x^2

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

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