Converting Cartesian Coordinates to Polar

We can place a point in a plane by the Cartesian coordinates $(x, \ y),$ a pair of distances from two perpendicular lines: the vertical line ($y$-axis) and the horizontal line ($x$-axis). Descartes made it possible to study geometry that employs algebra, by adopting the Cartesian coordinates. Other than the Cartesian coordinates, we have another representation of a point in a plane called the polar coordinates. To get some intuition why it was named like this, consider the globe having two poles: Arctic and Antarctic. To find the location of a place on the globe, we may use a compass. Actually, the globe is not a plane, but the angle read from the compass, especially when standing on a pole, gives fine information to reach the place. In a plane, we use the pair $(r, \ \theta)$ as the polar coordinates, where $r$ is the distance from a point (called the pole) and $\theta$ is the angle from a directed line . Usually, the origin and the $x$-axis are chosen as the pole and the directed line, respectively, when converting the coordinates.

Then, how can we convert Cartesian coordinates to polar coordinates? We can employ the Pythagorean theorem and a trigonometric function. That is, $$r = \sqrt{x^2 + y^2}$$ and $$\theta = \arctan{\frac{y}{x}}.$$ However, it is not enough because the tangent function does not have its inverse. When calculating $\arctan,$ the domain of angle is $\left(-\frac{\pi}{2}, \ \frac{\pi}{2}\right).$ Therefore, we must adjust for outside the domain, i.e. we must add $\pi.$ Then we have the full range of angle $\left(-\frac{\pi}{2}, \ \frac{3\pi}{2}\right).$ In conclusion, we have $$r = \sqrt{x^2 + y^2}$$ and $$\theta = \cases{\arctan{\frac{y}{x}} &\text{ if } x > 0, \\ \\ \arctan{\frac{y}{x}}+\pi &\text{ if } x < 0.}$$

Convert the Cartesian coordinates $(3, 4)$ to polar coordinates $(r, \theta),$ where $\theta$ is in radians and approximated up to two digits below the decimal point.

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By the Pythagorean theorem, we have $$r = \sqrt{3^2 +4^2} = 5.$$ By the tangent function, we can calculate $\theta$ as follows: $$\theta = \arctan{\frac{4}{3}} = 0.93 \text{ (rad)}.$$

Thus, our answer is $(5, \ 0.93).$ $_\square$

Convert the Cartesian coordinates $(-12, -5)$ to polar coordinates $(r, \theta),$ where $\theta$ is in radians and approximated up to two digits below the decimal point.

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By the Pythagorean theorem, we have $$r = \sqrt{(-12)^2 +(-5)^2} = 13.$$ By the tangent function, we can calculate $\theta$ as follows: $$\theta = \arctan{\frac{-5}{-12}} + \pi = 3.54 \text{ (rad)}.$$

Thus, our answer is $(13, \ 3.54).$ $_\square$

Convert the locus $(x(t), \ y(t))$ in Cartesian coordinates to the locus in polar coordinates $(r(t), \theta(t)),$ where $x = \sqrt{t}\cos{t}$ and $y = \sqrt{t}\sin{t}.$

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By the Pythagorean theorem, we have $$\begin{aligned} r(t) =& \sqrt{x^2 +y^2} \\ =& \sqrt{\left(\sqrt{t}\cos{t}\right)^2 +\left(\sqrt{t}\sin{t}\right)^2} \\ =& \sqrt{t}\left((\cos{t})^2 + (\sin{t})^2\right) \\ =& \sqrt{t}. \end{aligned}$$ By the tangent function, we can calculate $\theta(t)$ for the case $\cos{t} > 0:$ $$\begin{aligned} \theta =& \arctan{\frac{\sqrt{t}\sin{t}}{\sqrt{t}\cos{t}}} \\ =& \arctan{\frac{\sin{t}}{\cos{t}}} \\ =& \arctan{\tan{t}} \\ =& t, \end{aligned}$$

and for the case $\cos{t} < 0:$ $$\begin{aligned} \theta =& \arctan{\frac{\sqrt{t}\sin{t}}{\sqrt{t}\cos{t}}} + \pi\\ =& \arctan{\frac{\sin{t}}{\cos{t}}} + \pi \\ =& \arctan{\tan{t}} + \pi \\ =& (t - \pi) + \pi \\ =& t. \end{aligned}$$

Thus, our answer is $(r(t), \theta(t)) = (\sqrt{t}, \ t).$ $_\square$