- #1

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If I have the following equation:

x^2 - 2y = 0

I need to convert it into polar coordinants:

r^2 * cos(phi)^2 - 2*r* sin(phi) = 0

Is this correct ?

/Bob

- Thread starter Bob19
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- #1

- 71

- 0

If I have the following equation:

x^2 - 2y = 0

I need to convert it into polar coordinants:

r^2 * cos(phi)^2 - 2*r* sin(phi) = 0

Is this correct ?

/Bob

- #2

arildno

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Yes, it is correct.

- #3

HallsofIvy

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And, of course, as long as r is not 0, you can divide through by r to getBob19 said:

If I have the following equation:

x^2 - 2y = 0

I need to convert it into polar coordinants:

r^2 * cos(phi)^2 - 2*r* sin(phi) = 0

Is this correct ?

/Bob

[tex]r cos^2(\theta)- 2sin(\theta)= 0[/tex]

or

[tex]r= \frac{2sin(\theta)}{cos^2(\theta)}[/tex]

(It is customary to use "theta" in polar coordinates rather than "phi")

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