[tex]\bf (\stackrel{a}{2}~,~\stackrel{b}{-1})\qquad \begin{cases} r=\sqrt{a^2+b^2}\\\\ \theta =tan^{-1}\left( \frac{b}{a} \right) \end{cases} \\\\[-0.35em] ~\dotfill\\\\ r=\sqrt{2^2+(-1)^2}\implies r=\sqrt{5} \\\\\\ \theta =tan^{-1}\left( \cfrac{-1}{2} \right)\implies \theta \approx -26.57^o\implies \theta \approx 333.43^o \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill (\sqrt{5}~~,~~333.43^o)~\hfill[/tex]
Answer:
[tex](r,\theta); (\sqrt{5} , tan^{-1}(\frac{x}{y}))\\(r,\theta); (-\sqrt{5} , -tan^{-1}(\frac{x}{y}))[/tex]
Step-by-step explanation:
Here we are given our rectangular coordinates as (2,-1) . We have to convert this into polar coordinates. The formula for conversion into polar form is
[tex]r=\sqrt{x^2+y^2}[/tex]
[tex]\theta=tan^{-1}(\frac{x}{y})[/tex]
Substituting the values of x and y in the above formulas we get
[tex]r=\sqrt{2^2+(-1)^2}\\r=\sqrt{4+1}\\r=\sqrt{5}\\r=-\sqrt{5}\\[/tex]
[tex]\theta=tan^{-1}(\frac{-1}{2})[/tex]
Hence our polar coordinates are
[tex]r=(\sqrt{5},tan^{-1}(\frac{-1}{2}) )\\r=(-\sqrt{5},tan^{-1}(\frac{-1}{2}) )\\[/tex]
