Question

# The Cartesian coordinates of a point are given. a) (2,-2) b) (-1,\sqrt{3}) Find the polar coordinates (r,\theta) of the point, where r is greater than

Alternate coordinate systems
The Cartesian coordinates of a point are given.
a) $$(2,-2)$$
b) $$(-1,\sqrt{3})$$
Find the polar coordinates $$(r,\theta)$$ of the point, where r is greater than 0 and 0 is less than or equal to $$\theta$$, which is less than $$2\pi$$
Find the polar coordinates $$(r,\theta)$$ of the point, where r is less than 0 and 0 is less than or equal to $$\theta$$, which is less than $$2\pi$$

2021-06-06
Consider the Cartesian coordinates be (x,y).
The polar coordinates $$(r,\theta)$$ is defined as,
$$r^2=x^2+y^2$$
$$x=r\cos\theta$$
$$y=r\sin\theta$$
This implies that, $$\tan\theta=\frac{y}{x}$$
a) The Cartesian coordinates be $$(2,-2)$$.
Calculate r and $$\theta$$ as follows:
$$r^2=2^2+(-2)^2$$
$$=8$$
$$r=\pm2\sqrt{2}$$
$$\tan\theta=\frac{y}{x}$$
$$=\frac{-2}{2}$$
$$=-1$$
Thus, the angle $$\theta$$ is $$\frac{3\pi}{4},\frac{7\pi}{4}$$ as the angle $$0\leq\theta2\pi$$
The polar coordinates of (2,-2) are,
$$(2\sqrt{2},\frac{7\pi}{4})$$, where r>0
$$(-2\sqrt{2},\frac{3\pi}{4})$$, where r
b) The Cartesian coordinates be $$(-1,\sqrt{3})$$.
Calculate r and $$\theta$$ as follows:
$$r^2=(-1)^2+(\sqrt{3})^2$$
$$=4$$
$$r=\pm\sqrt{2}$$
$$\tan\theta=\frac{y}{x}$$
$$=\frac{\sqrt{3}}{-1}$$
$$=-\sqrt{3}$$
Thus, the angle $$\theta$$ is $$\frac{2\pi}{3},\frac{5\pi}{3}$$ as the angle $$0\leq\theta\leq2\pi$$
The polar coordinates of $$(-1,\sqrt{3})$$ are,
$$(2,\frac{2\pi}{3})$$, where r>0
$$(-2,\frac{5\pi}{3})$$, where r<0