Question

what does the graph of r = theta, theta >= 0 looks like?

Answer 1

you cant graph on this website

Answer 2

\[\theta=n \pi, n \in Z:intersection.with.polar.axis\]
\[(1/2)n \pi, n \in odd Z: intersection.with.(1/2) \pi.axis.\]
\[when.r=0, \theta=0\]
counterclockwise spiral.

Answer 3

thanks :)) nice :))

Answer 4

\[r^2=x^2+y^2, x=r \cos(\theta), y=r \sin(\theta)\]
\[x=r \cos(\theta)=> \frac{x}{r}=\cos(\theta)=> \cos^{-1}(\frac{x}{r})=\theta\]
\[r=\theta=>\sqrt{x^2+y^2}=\cos^{-1}(\frac{x}{r})=>\sqrt{x^2+y^2}=\cos^{-1}(\frac{x}{\sqrt{x^2+y^2}})\]
\[x^2+y^2=[\cos^{-1}(\frac{x}{\sqrt{x^2+y^2}})]^2\]
and if for some reason you also want to write as a Cartesian equation here you go

Answer 5

thanks guys :))