Question

i need your help. cartesian to polar conversion..

Answer 1

\[r^2 \sin^2 \theta = \frac{ r^3\cos^3 \theta }{ 2a - r costheta }\]

Answer 2

the real equation was y^2 = x^3 / 2a - x

Answer 3

and i need to derive the answer which is r= 2a sin theta tan theta

Answer 4

Okay first left side: \[
r^2\sin^2\theta = (r\sin\theta)^2=y^2
\]

Answer 5

There are four formula to remember: \[
r^2=x^2+y^2\\
\theta = \tan \frac yx\\
x=r\cos\theta\\
y=r\sin\theta\\
\]

Answer 6

y^2 = x^3 / 2a - x.. i just substituted the rcos theta and rsin theta hehe :)

Answer 7

okay so square root both sides if you want \(y=f(x)\)

Answer 8

then...

Answer 9

the answer of this is r= 2a sin theta tan theta :) i need the process.. im always lost..

Answer 10

Wait... you started in Cartesian and are going to polar?

Answer 11

yes

Answer 12

\[
\begin{split}
y^2 &= \frac{x^3}{2a - x}\\
2ay^2-xy^2&=x^3\\
2ay^2&=x^3+xy^2\\
2ay^2&=x(x^2+y^2)\\
2ar^2\sin^2\theta &=r^3\cos\theta \\
2a\sin^2\theta &=r\cos\theta \\
r&=2a\tan\theta\sin\theta
\end{split}
\]

Answer 13

another equation.. y^2 = x(x-a)^2/ 2a- x..

Answer 14

the answer is r^2 - 2arsec theta + a^2 = 0.. i dont know how to derive sec