CONVERTING b/w POLAR equations & RECTANGULAR equations

**Crysland** · April 23rd 2012, 08:09 PM

I 'd like to see the steps it takes to solve each of these problems. Thanks!
Polar to Rectangular:

1.) r² = sin2θ

2.) r = 2secθ ( i got x=2, not sure if it's right)

3.) r = 6/(2cosθ - 3sinθ)

Rectangular to Polar:

1.) 2xy=1

2.) y²- 8y - 16 = 0

3.) x² + y²- 2ay = 0

4.) x² = y³ ( my final answer is r= cot²θcscθ )

**Soroban** · April 24th 2012, 06:02 AM

Hello, Crysland!

Polar to Rectangular:

$(1)\; r^2\:=\:\sin2\theta$

We have: . . . . . . . $r^2 \:=\:2\sin\theta\cos\theta$

Multiply by $r^2\!:\qquad\;\; r^4 \:=\:2r^2\sin\theta\cos\theta$

. . . . . . . . . . . . . $(r^2)^2 \:=\:2(r\cos\theta)(r\sin\theta)$

Substitute: . $(x^2+y^2)^2 \:=\:2xy$

$(2)\; r \:=\:2\sec\theta$
( I got $x=2.$ ) . Right!

$\text{We have: }\:r \:=\:\dfrac{2}{\cos\theta} \quad\Rightarrow\quad r\cos\theta \:=\:2 \quad\Rightarrow\quad x \:=\:2$

$(3)\; r \:=\:\dfrac{6}{2\cos\theta - 3\sin\theta}$

$\begin{array}{ccc}\text{We have:} & r(2\cos\theta - 3\sin\theta) \:=\:6 \\ \\ & 2r\cos\theta - 3r\sin\theta \:=\:6 \\ \\ & 2x - 3y \:=\:6 \end{array}$

Rectangular to Polar:

$(1)\; 2xy\:=\:1$

$2(r\cos\theta)(r\sin\theta) \:=\:1 \quad\Rightarrow\quad 2r^2 \:=\:\frac{1}{\sin\theta\cos\theta} \quad\Rightarrow\quad r^2 \:=\:\frac{1}{2\sin\theta\cos\theta}$

. . . . . . . . . . . . . . . . . $\Rightarrow\quad r^2 \:=\:\frac{1}{\sin2\theta} \quad\Rightarrow\quad r^2 \:=\:\csc2\theta$

$(2)\; y^2- 8y - 16 \:=\: 0$

$(r\sin\theta)^2 - 8(r\sin\theta) - 16 \:=\:0 \quad\Rightarrow\quad r^2\sin^2\theta - 8r\sin\theta - 16 \:=\:0$

Quadratic Formula:

. . $r \;=\;\dfrac{8\sin\theta \pm \sqrt{64\sin^2\theta + 64\sin^2\theta}}{2\sin^2\theta} \;=\;\dfrac{8\sin\theta \pm\sqrt{128\sin^2\theta}}{2\sin^2\theta}$

. . $r \;=\;\dfrac{8\sin\theta \pm 8\sqrt{2}\sin\theta}{2\sin^2\theta} \;=\;\dfrac{8\sin\theta(1 \pm\sqrt{2})}{2\sin^2\theta} \;=\;\dfrac{4(1\pm\sqrt{2})}{\sin\theta}$

. . $r \;=\;4(1\pm\sqrt{2})\csc\theta$

$(3)\; x^2 + y^2 - 2ay \:=\: 0$

We have: . $r^2 - 2ar\sin\theta \:=\:0 \quad\Rightarrow\quad r(r - 2a\sin\theta) \:=\:0$

Then: . $r-2a\sin\theta \:=\:0 \quad\Rightarrow\quad r \;=\;2a\sin\theta$

(We can disregard $r = 0.$ )

$(4)\;x^2 \:=\: y^3$
(My final answer is: . $r\:=\:\cot^2\theta\csc\theta$ . Yes!

We have: . $(r\cos\theta)^2 \;=\;(r\sin\theta)^3 \quad\Rightarrow\quad r^2\cos^2\theta \;=\;r^3\sin^3\theta$

. . . . . . . . . $\cos^2\theta \;=\;r\sin^3\theta \quad\Rightarrow\quad \dfrac{\cos^2\theta}{\sin^3\theta} \;=\;r$

. . . . . . . . . . $r \;=\;\frac{\cos^2\theta}{\sin^2\theta}\cdot\frac{1 }{\sin\theta} \quad\Rightarrow\quad r \;=\;\cot^2\theta\csc\theta$

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