converting polar equation to cartesian equation

• Aug 20th 2008, 04:00 AM
Craka
converting polar equation to cartesian equation
With this problem it ask to convert r=-3 cos(theta) from polar form to cartesian

This is what I get
$

\begin{array}{l}
r = - 3\sin (\theta ) \\
(x - \frac{3}{2})(x - \frac{3}{2}) + (\frac{{ - 3}}{2})^2 + y^2 = 0 + (\frac{{ - 3}}{2})^2 \\
(x - \frac{3}{2})^2 + (\frac{{ - 3}}{2})^2 + y^2 = (\frac{{ - 3}}{2})^2 \\
\end{array}

$

However the answer given is as below. What have I done wrong?
$

(x - \frac{3}{2})^2 + y^2 = (\frac{{ - 3}}{2})^2 \\

$
• Aug 20th 2008, 04:21 AM
wisterville
Hello,

If the polar coordinate is defined as:
x=r cos(theta)
y=r sin(theta)

then,
r=\sqrt{x^2+y^2},
cos(theta)=x/r,
sin(theta)=y/r.

By substituton,
r=-3 cos(theta) becomes
r=-3x/r that is, r^2=-3x
so that x^2+y^2=-3x.
((x+3/2)^2+y^2=(3/2)^2.)

Please adjust the above argument to the appropriate polar coordinate you are using...

Bye.
• Aug 21st 2008, 05:32 AM
Craka
• Aug 21st 2008, 06:21 AM
wisterville
Hello,

Quote:

Originally Posted by Craka

Sorry I couldn't figure out what you had written.
Would you explain how you got these?
(If you are asked to convert r=-3 cos(theta), why begin with r=-3sin(theta)? Where does the second line come from?)
http://www.mathhelpforum.com/math-he...a556f5bc-1.gif

Besides, there are several ways to define coordinates.
What is the definition (in your book) of the polar coordinates?

Bye.
• Aug 21st 2008, 07:02 AM
Soroban
Hello, Craka!

Quote:

Convert $r\:=\:-3\cos\theta$ to cartesian form.

$\text{Multiply by }r\!:\;\;\underbrace{r^2} \;\;=\;-3\underbrace{r\cos\theta}$

. . . . . . . . . $x^2+y^2 \;=\quad\; -3x$

Then: . $x^2 + 3x + y^2 \;=\;0$

. . $x^2 + 3x + {\color{blue}\frac{9}{4}} + y^2 \;=\;{\color{blue}\frac{9}{4}}$

. . . . $\boxed{\left(x + \frac{3}{2}\right)^2 + y^2 \;=\;\frac{9}{4}}$

This is a circle with center $\left(-\frac{3}{2},\:0\right)$ and radius $\frac{3}{2}$