Convert the rectangular coordinates to polar coordinates with $r 0 \text { and } 0 \leq \theta

Question

Convert the rectangular coordinates to polar coordinates with $r > 0 \text { and } 0 \leq \theta < 2 \pi$ $( - 2 \sqrt { 3 } , - 2 )$
A) $( 4 , \pi / 6 )$
B) $( 4,5 \pi / 3 )$
C) $( 2,11 \pi / 6 )$
D) $( 4,7 \pi / 6 )$
E)none of these

Quizplus · Accepted Answer

We use the formulas $r = \sqrt{ x^2 + y^2}$ and $	an 	heta = \frac{y}{x}$ to find $r$ and $	heta$. First, we have $r = \sqrt{ (-2 \sqrt{3})^2 + (-2)^2} = 4$. Next, we can see from the signs of $x$ and $y$ that the point is in the third quadrant, so we need to add $\pi$ to our answer for $	heta$ if we use the formula $	an 	heta = \frac{y}{x}$. However, we can also see that the angle the point makes with the positive $x$-axis is $60^\circ$ less than $\frac{5\pi}{6} = 150^\circ$, which is equivalent to $\frac{7\pi}{6}$. Adding $\pi$ to account for the point being in the third quadrant gives $	heta = \frac{7\pi}{6}$, so the polar coordinates are $(r,	heta) = \boxed{(4,\frac{7\pi}{6})}$.

Convert the Rectangular Coordinates to Polar Coordinates With r>0 and 0≤θ<2πr > 0 \text { and } 0 \leq \theta < 2 \pir>0 and 0≤θ<2π

Convert the Rectangular Coordinates to Polar Coordinates With $r > 0 \text { and } 0 \leq \theta < 2 \pi$