Question

Find an equation equivalent to r = 1 + 2 sin in rectangular coordinates.

Answer 1

Answer:

$x^2+y^2=(x^2+y^2-2y)^2$

Step-by-step explanation:

Here we are given that

r=1+2sin$\theta$

we know that the polar cordinates are represented as

x=rsin$\theta$

y=rcos$\theta$

Also squaring both and adding them gives you ,

$x^2+y^2=r^2sin^{2}\theta + r^2cos^{2}\theta\\x^2+y^2=r^2(sin^{2}\theta + cos^{2}\theta)\\x^2+y^2= r^2 \\As (sin^{2}\theta + cos^{2}\theta) = 1\\\\Hence \\\\x^2+y^2=r^2$

Hence

$r=\sqrt{x^2+y^2}$

also r=1+2sin$\theta$

1+2sin$\theta$ = $r=\sqrt{x^2+y^2}$

Also ,

r = r=1+2sin$\theta$

also sin$\theta$ = $\frac{y}{\sqrt{x^2+y^2}}$

Hence replacing sin$\theta$ we get

$\sqrt{x^2+y^2} = 1+2\frac{y}{\sqrt{x^2+y^2}}\\x^2+y^2= \sqrt{x^2+y^2}+2y\\x^2+y^2-2y= \sqrt{x^2+y^2}\\Squaring\\(x^2+y^2-2y)^{2}=x^2+y^2$

Answer 2

Answer:

Step-by-step explanation:

r = 1+2sinФ

multiplying both sides by r ,we get

$r^2= r+2rsin$Ф

$x^2+y^2= r + 2y$    [ since rsinФ = y ]

$x^2+y^2-2y= r$

$(x^2+y^2-2y)^2 = r^2\\x^4+y^4+4y^2+2x^2y^2-4y^3-4x^2y=x^2+y^2\\x^4+y^4+3y^2+2x^2y^2-4y^3-4x^2y-x^2 =0$