With this problem it ask to convert r=-3 cos(theta) from polar form to cartesian

This is what I get

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

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- Aug 20th 2008, 04:00 AMCrakaconverting 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

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

- Aug 20th 2008, 04:21 AMwisterville
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 AMCraka
Sorry could someone clarify my answer please.

- Aug 21st 2008, 06:21 AMwisterville
Hello,

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 AMSoroban
Hello, Craka!

Quote:

Convert to cartesian form.

. . . . . . . . .

Then: .

. .

. . . .

This is a circle with center and radius