Mengubah Koordinat Kartesius Menjadi Koordinat Polar
Titik \((x, y)\) dapat diubah menjadi koordinat polar \((r, \alpha)\)
\(r = \sqrt{x^2 + y^2}\)
\(\tan \alpha = \dfrac{y}{x}\)
\(\alpha = \tan^{-1} \left(\dfrac{y}{x}\right)\dotso\dotso\color{red} \text{ cek kuadran titik}\)
Titik \(\text{P(4, 4)}\) dapat diubah menjadi koordinat polar \((r, \alpha)\)
\(r = \sqrt{4^2 + 4^2} = \sqrt{2\cdot 4^2} = 4\sqrt{2}\)
\(\tan \alpha = \dfrac{4}{4} = 1\)
\(\alpha = 45^{\circ}\dotso\dotso \text{ titik (4, 4) di kuadran I}\)
Jadi \(\text{P(4, 4)} = \text{P}(4\sqrt{2}, 45^{\circ})\)
\(r = \sqrt{(-\sqrt{3})^2 + 1^2}\)
\(r = \sqrt{3 + 1} = \sqrt{4} = 2\)
\(\tan \alpha = -\dfrac{1}{\sqrt{3}} = -\dfrac{1}{3}\sqrt{3}\)
Karena titik \(\text{P}(-\sqrt{3}, 1)\) berada di kuadran II, maka kita pilih sudut untuk tangen di kuadran II
\(\alpha = 180^{\circ}\:-\:30^{\circ} = 150^{\circ}\)
Jadi \(\text{P}(-\sqrt{3}, 1)= \text{P}(2, 150^{\circ})\)
\(x = r\cdot \cos \alpha\)
\(x = 5\cdot \cos 300^{\circ}\)
\(x = 5\cdot \frac{1}{2} = \frac{5}{2}\)
\(y = r\cdot \sin \alpha\)
\(y = 5\cdot \sin 300^{\circ}\)
\(y = 5\cdot (-\frac{1}{2}\sqrt{3}) = -\frac{5}{2}\sqrt{3}\)
Jadi \(\text{P}(5, 300^{\circ}) = \text{P}(\frac{5}{2}, -\frac{5}{2}\sqrt{3})\)
