Question

Explain why the point $(r,\theta )=\left(8,-\frac{\pi }{3}\right)$is not a polar representation of the point with rectangular coordinates $(x,y)=(-4,4\sqrt{3})$, even though these values of $r,\theta ,x$, and $y$satisfy the formulas $r^2=x^2+y^2$and $\tan \theta =\frac{y}{x}$. Include a picture with your explanation.

Short answer

The required answer is $(-4,4\sqrt{3})$is in the second quadrant where as point localid="1652685710587" $\left(8,-\frac{\pi }{3}\right)$is in the fourth quadrant

Step-by-step solution

Given information

The rectangular coordinate $(x,y)=(-4,4\sqrt{3})$

Simplification

Consider the rectangular coordinate $(x,y)=(-4,4\sqrt{3})$

The objective is to find a polar representation.

In the coordinate $(-4,4\sqrt{3}),x=-4$and $y=4\sqrt{3}$.

To calculate $\theta$use the formula $\theta ={\tan }^{-1}\left(\frac{y}{x}\right)$.

Then

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{4\sqrt{3}}{-4}\right)[\text{since}x=-4,y=4\sqrt{3}] \\ \theta ={\tan }^{-1}(-\sqrt{3}) \\ \theta =-\frac{\pi }{3}\left[\text{since}\tan \left(-\frac{\pi }{3}\right)=-\tan \frac{\pi }{3}=-\sqrt{3}\right] \end{gathered}$$

To find the value of $r$use the equation $r=\sqrt{x^2+y^2}$.

Then,

$$\begin{gathered} r=\sqrt{(-4{)}^2+(4\sqrt{3}{)}^2}[\text{since}x=-4,y=4\sqrt{3}] \\ r=\sqrt{16+48} \\ r=\sqrt{64} \\ r=\pm 8 \end{gathered}$$

Then $(r,\theta )=\left(8,-\frac{\pi }{3}\right)$.

But the point $\left(8,-\frac{\pi }{3}\right)$is in the fourth quadrant. whereas $(-4,4\sqrt{3})$is in the second quadrant.

The correct representation of the point $(-4,4\sqrt{3})$is $\left(-8,\frac{5\pi }{3}\right)$.

Since every rectangular point will have infinitely many representations in polar plane.

Thus, the coordinate $(-4,4\sqrt{3})$is in the second quadrant $\left(8,-\frac{\pi }{3}\right)$is in the fourth quadrant. That is why the point $\left(8,-\frac{\pi }{3}\right)$is not a polar representation for the point $(-4,4\sqrt{3})$.

Therefore, the required answer is $(-4,4\sqrt{3})$is in second quadrant where as point $(8,-\sqrt{3})$is in the fourth quadrant