4

$$r=\frac { 4 }{ 1+2\sin\theta } $$

Steps I took:

$$(1+2\sin\theta )r=\frac { 4 }{ 1+2\sin\theta } (1+2\sin\theta )$$

$$r+2r\sin\theta =4$$

$$r+2y=4$$

$$(r+2y)^2=16$$

$$(r+2y)(r+2y)=r^2+4yr+4y^2$$

$$r^2+4yr+4y^2-16=0$$

My outcome doesn't seem to match the correct answer. Where did I go wrong?

Vincenzo Tibullo
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Cherry_Developer
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2 Answers2

3

In going from $r+2y=4$ to $(r+2y)^2=16$ you squared both sides, which adds extraneous solutions. After all, you added $r+2y=-4$ to the solutions. To be precise, you squared at the wrong time. It is OK to square $r$, since it can be positive or negative, but you did not know about $r+2y$.

Continue in this way: $$r+2y=4$$ $$r=4-2y$$ $$r^2=16-16y+4y^2$$ $$x^2+y^2=16-16y+4y^2$$

A quick use of a graphing program shows this to have the same graph as your original equation.

Rory Daulton
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  • I never thought of doing it that way. I guess it takes time and practice to see alternate routes to arrive at the correct solution? – Cherry_Developer Dec 14 '14 at 19:50
  • @Cherry_Developer: I suppose. In this kind of problem, look for ways to get $r\sin\theta$, $r\cos\theta$, and $r^2$ out of your polar equation, since they are can be directly changed to $y$, $x$, and $x^2+y^2$ respectively. – Rory Daulton Dec 14 '14 at 19:52
0

use that $$r=\sqrt{x^2+y^2}$$ and $$\sin(\theta)=\frac{y}{\sqrt{x^2+y^2}}$$ and you will get $$\sqrt{x^2+y^2}=\frac{4}{1+\frac{2y}{\sqrt{x^2+y^2}}}$$

Dr. Sonnhard Graubner
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