Conversion of polar equations into cartesian equation

• Aug 23rd 2008, 07:46 AM
roshanhero
Conversion of polar equations into cartesian equation
Find the cartesian equations of these polar equations-
1.r=a(theta)/(theta-1)
2.r(1-e^theta)=a
3.r(theta)cos(theta)=acos2(theta)
• Aug 23rd 2008, 09:04 AM
Soroban
Hello, roshanhero!

These are particularly ugly problems!
. . Very little simpliying is possible.

We need: .$\displaystyle \begin{array}{ccc}r^2 &=&x^2+y^2 \\ r\cos\theta &=& x \\ r\sin\theta &=& y \\ \theta &=&\arctan(\frac{y}{x}) \end{array}$

Quote:

Find the cartesian equations of these polar equations:

$\displaystyle 1)\;r\:=\:\frac{a\theta}{\theta-1}$

We have: .$\displaystyle \sqrt{x^2+y^2} \;-\;\frac{a\arctan\left(\frac{y}{x}\right)}{\arctan\ left(\frac{y}{x}\right) - 1}$

Quote:

$\displaystyle 2)\;r\left(1-e^{\theta}\right) \:=\:a$

We have: .$\displaystyle \sqrt{x^2+y^2}\left(1 - e^{\arctan(\frac{y}{x})}\right) \:=\:a$

Quote:

$\displaystyle 3)\; r\!\cdot\!\theta\!\cdot\!\cos\theta \:=\:a\cos2\theta$

Multiply by $\displaystyle r^2\!:\;\;r^3\!\cdot\!\theta\!\cdot\!\cos\theta \;=\;ar^2\!\cdot\!\cos2\theta$

. . . . . . $\displaystyle r^2\!\cdot\!\theta\!\cdot\!r\cos\theta \;=\;ar^2(2\cos^2\!\theta - 1)$

. . . . . . $\displaystyle r^2\!\cdot\!\theta\!\cdot\!r\cos\theta \;= \;2ar^2\cos^2\!\theta - ar^2$

. . . . . . $\displaystyle r^2\!\cdot\!\theta\!\cdot\!r\cos\theta \;=\;2a(r\cos\theta)^2 - ar^2$

. . $\displaystyle (x^2+y^2)\!\cdot\!\arctan(\frac{y}{x})\!\cdot\!x \;=\;2ax^2 - a(x^2+y^2)$

. . $\displaystyle x(x^2+y^2)\arctan\left(\frac{y}{x}\right) \;= \;a(x^2-y^2)$

• Aug 24th 2008, 01:51 AM
roshanhero
Can't we have the form that contains only x and y terms?
• Aug 24th 2008, 12:41 PM
Soroban
Quote:

Originally Posted by roshanhero
Can't we have the form that contains only x and y terms?

Isn't that what I found?