Chapter 9.1, Problem 48AYU

To determine

The rectangular coordinates of polar coordinates of a point $\left(-6,-\frac{\pi }{4}\right)$.

Answer to Problem 48AYU

Solution:

The rectangular coordinates of a point $\left(-6,-\frac{\pi }{4}\right)$ are $\left(-3\sqrt{2},3\sqrt{2}\right)$.

Explanation of Solution

Given information:

The polar point $\left(-6,-\frac{\pi }{4}\right)$.

Explanation:

The point $P$ in a polar coordinate system is represented by an ordered pair $\left(r,\theta \right)$.

Here the point is $\left(-6,-\frac{\pi }{4}\right)$.

$\Rightarrow r=-6,\theta =-\frac{\pi }{4}$.

If $P$ is a point with polar coordinates $\left(r,\theta \right)$, then the rectangular coordinates $\left(x,y\right)$ of $P$ are $x=r\cos \theta ,y=r\sin \theta$.

Therefore, the $x$-coordinate is $x=-6\cos \left(-\frac{\pi }{4}\right)$

By using the rule: $\cos \left(-\theta \right)=\cos \left(\theta \right)$.

$\Rightarrow x=-6\times \left(\frac{\sqrt{2}}{2}\right)$

By simplifying,

$\Rightarrow x=-3\sqrt{2}=-4.24264$.

Now, $y=-6\sin \left(-\frac{\pi }{4}\right)$

By using the rule: $\sin \left(-\theta \right)=-\sin \left(\theta \right)$

$y=-6\left(-\frac{\sqrt{2}}{2}\right)$.

By simplifying,

$\Rightarrow y=3\sqrt{2}=4.24264$.

The rectangular coordinates of the point $\left(-6,-\frac{\pi }{4}\right)$ are $\left(-3\sqrt{2},3\sqrt{2}\right)$, which lies in quadrant $\text{II}$.