Question

Convert the rectangle coordinates to polar coordinates where r > 0
and 0 ≤ 2 . and Convert the polar coordinates to rectangular coordinates.

please list steps on how to solve so I can solve these in the future

Answer 1

Answer:

(1) - C, $(3,\frac{\pi}{6})$

(2) - B, $(0,-2)$

Step-by-step explanation:

$\text{Using the following conversions:}\\\boxed{\left\begin{array}{ccc}\text{\underline{Rect. to Polar:}}\\(x,y)\rightarrow(r, \theta)\\\\r=\sqrt{x^2+y^2}\\\\\theta=\tan^{-1}(\frac{y}{x})\end{array}\right} \ \ \boxed{\left\begin{array}{ccc}\text{\underline{Polar to Rect.:}}\\(r, \theta)\rightarrow(x,y)\\\\x=r\cos \theta\\\\y=r \sin\theta\end{array}\right}$

Note that when doing these calculations, make sure your calculator is in radians mode.

Question #2:

Given:

$(\frac{3\sqrt{3} }{2} ,\frac{3}{2} ) \ \text{in} \ (x,y)$

$(\frac{3\sqrt{3} }{2} ,\frac{3}{2} )\\\\\Rightarrow r=\sqrt{(\frac{2\sqrt{2} }{2})^2+(\frac{3}{2})^2} \\\\\Longrightarrow r=\sqrt{\frac{27}{4} +\frac{9}{4} } \\\\\Longrightarrow r=\sqrt{9}\\\\\therefore \boxed{ r=3}\\\\\Rightarrow \theta=\tan^{-1}(\frac{\frac{3}{2}}{\frac{3\sqrt{3} }{2}})\\\\\Longrightarrow \theta=\tan^{-1}(\frac{\sqrt{3} }{3})\\\therefore \boxed{\theta=\frac{\pi}{6} }$

Thus, the correct option is C.

$\boxed{\boxed{(3,\frac{\pi}{6})}}$

Question #3:

Given:

$(2,\frac{3\pi}{2} ) \ \text{in} \ (r,\theta)$

$(2,\frac{3\pi}{2} ) \\\\\Rightarrow x=(2)\cos(\frac{3\pi}{2})\\\\\Longrightarrow x=(2)(0)\\\\\therefore \boxed{x=0}\\\\\Rightarrow y=(2)\sin(\frac{3\pi}{2})\\\\Longrightarrow y=(2)(-1)\\\\\therefore \boxed{y=-2}$

Thus, the correct option is B.

$\boxed{\boxed{(0,-2)}}$