Q. 17 - Calculus

Question

The polar coordinates for several sets of points are given. Find the rectangular coordinates for each of the points, and then plot and label the points in the same polar coordinate system. 

$\left(3,\frac{\pi }{6}\right),\left(-3,\frac{\pi }{6}\right),\left(3,-\frac{\pi }{6}\right)$ and $\left(-3,-\frac{\pi }{6}\right)$

Step-by-Step Solution

Answer

The required graph is 

Step 1: Given information

The polar coordinates are $\left(3,\frac{\pi }{6}\right),\left(-3,\frac{\pi }{6}\right),\left(3,-\frac{\pi }{6}\right)$ and $\left(-3,-\frac{\pi }{6}\right)$

Step 2: Find the value of x

Let us consider the polar coordinates$\left(3,\frac{\pi }{6}\right),\left(-3,\frac{\pi }{6}\right),\left(3,-\frac{\pi }{6}\right)$ and $\left(-3,-\frac{\pi }{6}\right)$.

The objective is to convert the polar coordinates to the rectangular coordinates.

In the point $\left(3,\frac{\pi }{6}\right)$, then $r=3,\theta =\frac{\pi }{6}$.

Use the equations $x=r\cos \theta ,y=r\sin \theta$

Take $x=r\cos \theta$ and substitute $r=3,\theta =\frac{\pi }{6}$ then

$x=3\cos \frac{\pi }{6}$$x=3\times \frac{\sqrt{3}}{2}\left[Since \left.\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}\right] \right]$

Then the value is $x=\frac{3\sqrt{3}}{2}$.

Step 3: Find the rectangular coordinates

Now take $y=r\sin \theta$ and substitute $r=3,\theta =\frac{\pi }{6}$then

$$\begin{gathered} y=3\sin \frac{\pi }{6} \\ y=3\times \frac{1}{2}\left[since\sin \frac{\pi }{6}=\frac{1}{2}\right] \end{gathered}$$

Thus the rectangular coordinate $(x,y)=\left(\frac{3\sqrt{3}}{2},\frac{3}{2}\right)$.

Therefore, the rectangular coordinates are$(x,y)=\left(\frac{3\sqrt{3}}{2},\frac{3}{2}\right)$.

for the point $\left(-3,\frac{\pi }{6}\right)$ then $r=-3,\theta =\frac{\pi }{6}$.

Use the equations,

Take $x=r\cos \theta$ and substitute $r=-3,\theta =\frac{\pi }{6}$ then

$$\begin{gathered} x=-3\cos \frac{\pi }{6} \\ x=-3\times \frac{\sqrt{3}}{2}\left[\text{ since }\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}\right] \end{gathered}$$

Then the value is

$x=\frac{-3\sqrt{3}}{2}$

Now take $y=r\sin \theta$ and substitute $r=-3,\theta =\frac{\pi }{6}$ then

$y=-3\sin \frac{\pi }{6}$

Thus,

$y=-3\times \frac{1}{2}\left[since\sin \frac{\pi }{6}=\frac{1}{2}\right]$

the rectangular coordinate $(x,y)=\left(\frac{-3\sqrt{3}}{2},\frac{-3}{2}\right)$.

Therefore, the rectangular coordinates are $(x,y)=\left(\frac{-3\sqrt{3}}{2},\frac{-3}{2}\right)$

Step 4: Simplification

For the point $\left(3,-\frac{\pi }{6}\right)$ then $r=3,\theta =-\frac{\pi }{6}$.

Use the equations,

Take $x=r\cos \theta$ and substitute $r=3,\theta =-\frac{\pi }{6}$ then

$$\begin{gathered} x=3\cos \left(-\frac{\pi }{6}\right) \\ x=3\times \frac{\sqrt{3}}{2}\left[\text{ since }\cos \left(-\frac{\pi }{6}\right)=\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}\right] \end{gathered}$$

Then the value is $x=\frac{3\sqrt{3}}{2}$

Now take $y=r\sin \theta$ and substitute $r=3,\theta =-\frac{\pi }{6}$ then

$y=3\sin \left(-\frac{\pi }{6}\right)$

Thus,

$y=-3·\frac{1}{2}\left[since\sin \frac{\pi }{6}=\frac{1}{2}\right]$

The rectangular coordinate $(x,y)=\left(\frac{3\sqrt{3}}{2},\frac{-3}{2}\right)$.

Therefore, the rectangular coordinates are $(x,y)=\left(\frac{3\sqrt{3}}{2},\frac{-3}{2}\right)$.

for the point $\left(-3,-\frac{\pi }{6}\right)$ then $r=-3,\theta =-\frac{\pi }{6}$.

Use the equations,

Take $x=r\cos \theta$ and substitute $r=-3,\theta =-\frac{\pi }{6}$ then

width="344" style="max-width: none; vertical-align: -46px;" $$\begin{gathered} x=-3\cos \left(-\frac{\pi }{6}\right) \\ x=-3\times \frac{\sqrt{3}}{2}\left[\text{ since }\cos \left(-\frac{\pi }{6}\right)=\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}\right] \end{gathered}$$

Step 5: Further simplification

Then the value is

$x=\frac{-3\sqrt{3}}{2}$

Now take $y=r\sin \theta$ and substitute $r=-3,\theta =-\frac{\pi }{6}$ then

$y=-3\sin \left(-\frac{\pi }{6}\right)$

Thus,

$y=-3·-\frac{1}{2}\left[\sin ce\right.$$\left.\sin \left(-\frac{\pi }{6}\right)=-\sin \frac{\pi }{6}=-\frac{\sqrt{3}}{2}\right]$

$y=\frac{3}{2}$

The rectangular coordinate $(x,y)=\left(\frac{-3\sqrt{3}}{2},\frac{3}{2}\right)$.

Thus, the rectangular coordinates are $(x,y)=\left(\frac{-3\sqrt{3}}{2},\frac{3}{2}\right)$.

Therefore, the answer is $\left(\frac{3\sqrt{3}}{2},\frac{3}{2}\right),\left(\frac{-3\sqrt{3}}{2},\frac{-3}{2}\right),\left(\frac{3\sqrt{3}}{2},\frac{-3}{2}\right),\left(\frac{-3\sqrt{3}}{2},\frac{3}{2}\right)$

Step 6: Plot the graph

The graphical representation is as follows,