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Licemer1 [7]
3 years ago
13

Convert from rectangular to polar coordinates: note: choose rr and θθ such that rr is nonnegative and 0≤θ<2π0≤θ<2π (a)(9,0

)⇒(r,θ)(a)(9,0)⇒(r,θ)( 9 equation editorequation editor , tan(0) equation editorequation editor ) (b)(18,183√)⇒(r,θ)(b)(18,183)⇒(r,θ)( 20.7846 equation editorequation editor , tan(18/sqrt(3)/18) equation editorequation editor ) (c)(−5,5)⇒(r,θ)(c)(−5,5)⇒(r,θ)( 10 equation editorequation editor , equation editorequation editor ) (d)(−1,3‾√)⇒(r,θ)(d)(−1,3)⇒(r,θ)( 4 equation editorequation editor ,
Mathematics
1 answer:
DochEvi [55]3 years ago
6 0

Answer:

Step-by-step explanation:

Convert rectangle (x , y) to polar coordinates ( r , θ)

x=r  \cos \theta, y= r \sin \theta

r=\sqrt{x^2+y^2} , \theta =tan^-^1 (\frac{y}{x} )

a) converts (9, 0) to polar coordinates  ( r , θ)

r=\sqrt{x^2+y^2} \\\\=\sqrt{9^2+0} \\\\=9

\theta= \tan^-^1 (\frac{0}{9} )\\\\=0

b) Convert (18,\frac{18}{\sqrt{3} } ) to polar coordinates ( r, θ)

r = \sqrt{18^2+(\frac{18}{\sqrt{3} })^2 } \\\\=\sqrt{324+108} \\\\=\sqrt{432}

\frac{x}{y} \theta = \tan^-^1(\frac{\frac{18}{\sqrt{3} } }{18} )\\\\= \tan ^-^1(\frac{1}{\sqrt{3} } )\\\\= \frac{\pi}{6}

c)  converts (-5, 5) to polar coordinates  ( r , θ)

r =\sqrt{(-5)^2+(5)^2} \\\\=\sqrt{50} \\\\=5\sqrt{2}

\theta=\tan^-^1(\frac{5}{-5} )\\\\= \tan^-^1(-1)\\\\=\frac{3\pi}{4}

d)  converts (-1, √3) to polar coordinates  ( r , θ)

r=\sqrt{(-1)^2+(\sqrt{3})^2 } \\\\= \sqrt{4} \\\\=2

\theta=\tan^-^1(\frac{\sqrt{3} }{-1} )\\\=\tan^-^1(-\sqrt{3} )\\\\=\frac{2\pi}{3}

= \frac{2\pi}{\sqrt{3} }

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