My Math Forum Turning Polar Equations to Cartesian Equations

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 June 15th, 2010, 04:00 PM #1 Newbie   Joined: Jun 2010 Posts: 4 Thanks: 0 Turning Polar Equations to Cartesian Equations Just to clarify I understand how to turn easy polar equations to Cartesian ones, such as r=10sin(theta)--> x^2+(y-5)^2=25 However I don't know how to do some problems when the equations are different than the one above such as: 1. r=-2csc(theta) 2. r^(2)sin(2theta)=6 3. 2r^(4)-r^(2)-1=0 4. r=3/(cos(theta)-4sin(theta)) Please explain to me how to do each of these different kinds of "scenarios" thank you.
June 15th, 2010, 04:27 PM   #2
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Re: Turning Polar Equations to Cartesian Equations

Quote:
 Originally Posted by maximade Just to clarify I understand how to turn easy polar equations to Cartesian ones, such as r=10sin(theta)--> x^2+(y-5)^2=25 However I don't know how to do some problems when the equations are different than the one above such as: 1. r=-2csc(theta) 2. r^(2)sin(2theta)=6 3. 2r^(4)-r^(2)-1=0 4. r=3/(cos(theta)-4sin(theta)) Please explain to me how to do each of these different kinds of "scenarios" thank you.

Keep the following relationships in mind:

$x= (r)(cos(\theta))$

$y= (r)(sin(\theta))$

The reason for these relationships become obvious if one draws a right triangle with base on the x-axis. The hypotenuse of this triangle is the radius of the polar coordinates. So to find x simply minuplate the cosine function of x over r to get the above, same with y.

Now we have:

$r=-2csc(\theta)$

To find x and y we simply plug the equations into the formulas:

$x= -2csc(\theta)cos(\theta)$

$y= -2csc(\theta)sin(\theta)$

These both (by simple trigonomic identities) become the following:

$x= -2cotan(\theta)$

$y= -2$

Since y is constant this is simply a horizontal line at y=-2

I assume you can do the other ones from this exapmle. But if you want more help or need better explaination just post here and I'll answer more of the questions or give more explaination if desired.

 June 15th, 2010, 05:20 PM #3 Newbie   Joined: Jun 2010 Posts: 4 Thanks: 0 Re: Turning Polar Equations to Cartesian Equations How would I do something like the third one?
June 15th, 2010, 08:14 PM   #4
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Re: Turning Polar Equations to Cartesian Equations

I assume you know the basic conversion formulas . . .

Quote:
 $(1)\;\;r\:=\:-2\,\csc\theta$

$\text{We have: }\;r \:=\:\frac{-2}{\sin\theta}$

[color=beige]. . . . . [/color]$\underbrace{r\,\sin\theta}_{\text{This is }y} \:=\:-2$

[color=beige]. . . . . . . . . . [/color]$y \:=\:-2$

Quote:
 $(2)\;\; r^2\,\sin2\theta \:=\:6$

$\text{We have: }\:r^2(2\,\sin\theta\,\cos\theta) \:=\:6$

[color=beige]. . . . . . . .[/color]$\underbrace{(r\,\cos\theta)}_{\text{This}\:\text{i s}\,x} \,\cdot\, \underbrace{(r\,\cos\theta)}_{\text{This}\:\text{i s}\,y} \:=\:3$

[color=beige]. . . . . . . . . . . . . . [/color]$xy \:=\:3$

Quote:
 $(3)\;\;2r^4\,-\,r^2\,-\,1 \:=\:0$

$\text{Factor: }\2r^2\,+\,1)(r^2\,-\,1) \;=\;0" />

$2r^2\,+\,1\:=\:0 \;\;\;\Rightarrow\;\;\;r^2 \:=\:-\frac{1}{2}\;\cdots\;\text{ no real roots}$

$r^2\,-\,1 \:=\:0 \;\;\;\Rightarrow\;\;\;r^2 \:=\:1$

[color=beige]. . . . . . . . . . . .[/color]$\overbrace{x^2\,+\,y^2}\:=\:1$

Quote:
 $(4)\;\;r \:=\:\frac{3}{\cos\theta\,-\,4\sin\theta}$

$\text{We have: }\;\underbrace{r\,\cos\theta}_{\text{This is }x}\,-\,4\underbrace{r\,\sin\theta}_{\text{This is }y} \;=\;3$

[color=beige]. . . . . . . . . . . . . . . . . .[/color]$x\,-\,4y \;=\;3$

 June 15th, 2010, 08:37 PM #5 Member   Joined: Feb 2010 Posts: 53 Thanks: 0 Re: Turning Polar Equations to Cartesian Equations Well soroban seems to have the correct method explained. The way I did it was more complicated it seems, and unneccesarily so. Good luck

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# what is the Cartesian equation for the curve r(1-5sin^2theta)=2costheta

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