Give all the polar coordinates corresponding the rectangular point

[shamieh]

Give all the polar coordinates corresponding the rectangular point $$(-1, \sqrt{3})$$

Am i setting this up right?

so would I use $$(r, \theta)$$

so $$x = rcos(\theta)$$
$$y = rsin(\theta)$$

$$r^2 = x^2 + y^2$$

so:

$$(-1)^2 = (-1*\frac{2\pi}{3})^2 + (-1*\frac{11\pi}{6})$$ ?

 

[HallsofIvy]

Give all the polar coordinates corresponding the rectangular point $$(-1, \sqrt{3})$$

Am i setting this up right?

so would I use $$(r, \theta)$$

so $$x = rcos(\theta)$$
$$y = rsin(\theta)$$

$$r^2 = x^2 + y^2$$

Yes, that is correct.

so:

$$(-1)^2 = (-1*\frac{2\pi}{3})^2 + (-1*\frac{11\pi}{6})$$ ?

But I have no idea what is meant by this. But that equation certainly isn't true. $$\left(-\frac{2\pi}{3}\right)^2= \frac{4\pi}{9}$$ and $$\left(\frac{-11\pi}{6}\right)^2= \frac{121\pi^2}{36}$$. Those do NOT add to 1.

You are given that x= -1 and $$y= \sqrt{3}$$ so that $$r^2= (-1)^2+ (\sqrt{3})^2= 1+ 3= 4$$

By dividing [math]y= r sin(\theta)[/tex] by $$x= r cos(\theta)$$ you get
$$tan(\theta)= \frac{y}{x}= \sqrt{3}{-1}= -\sqrt{3}$$.

 

[shamieh]

How about:

$$r^2 = (-1)^2 + (\sqrt{3})^2$$

$$r^2 = 4$$

$$r = 2 $$ or $$ - 2$$... I understand all of this...

But then..What is going on here?

How are we obtaining the points $$(2, \frac{2\pi}{3} +(-) 2n\pi)$$

HOW does he know that we have the angle of 2pi/3 from the work above alone?

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Yes, that is correct.
But I have no idea what is meant by this. But that equation certainly isn't true. $$\left(-\frac{2\pi}{3}\right)^2= \frac{4\pi}{9}$$ and $$\left(\frac{-11\pi}{6}\right)^2= \frac{121\pi^2}{36}$$. Those do NOT add to 1.

You are given that x= -1 and $$y= \sqrt{3}$$ so that $$r^2= (-1)^2+ (\sqrt{3})^2= 1+ 3= 4$$

By dividing [math]y= r sin(\theta)[/tex] by $$x= r cos(\theta)$$ you get
$$tan(\theta)= \frac{y}{x}= \sqrt{3}{-1}= -\sqrt{3}$$.

Also, would that mean that my answer is just $-\sqrt{3}$ as the final answer??

 

[MarkFL]

Look at two things: in which quadrant is $\theta$, and as HallsofIvy mentioned:

$$\tan(\theta)=\frac{y}{x}$$

So what must you conclude regarding $\theta$?

 

[shamieh]

I got

$$cos\theta = x/r = -1/2$$

$$sin\theta = y/r = \sqrt{3}/2$$

thus $$\theta = 2\pi/3$$ in Quadrant II

So would my points be $$(2,2\pi/3)$$ ?

 

[soroban]

Hello, shamieh!

Give all the polar coordinates corresponding to
the rectangular point $$(\text{-}1,\,\sqrt{3})$$

We are given: .$$\begin{Bmatrix}x &=& \text{-}1 \\ y &=& \sqrt{3} \end{Bmatrix}$$

We want: .$$\begin{Bmatrix}r, & \text{where }r^2 \,=\,x^2+y^2 \\ \theta, & \text{ where }\tan\theta \,=\,\frac{y}{x}\quad \end{Bmatrix}$$$$r^2 \:=\:x^2+y^2 \;=\;(\text{-}1)^2 + (\sqrt{3})^2 \:=\:1+3 \:=\:4$$

. . Hence: .$$r \,=\,\pm2$$$$\tan\theta \,=\,\frac{\sqrt{3}}{\text{-}1} \,=\,\text{-}\sqrt{3}$$

. . Hence: .$$\theta \:=\: \frac{2\pi}{3} + \pi n$$

Therefore: .$$(\text{-}1,\,\sqrt{3}) \;=\;\begin{Bmatrix}\left(2,\;\tfrac{2\pi}{3}\!+\! 2\pi n\right) \\ \left(\text{-}2,\;\tfrac{5\pi}{3}\!+\!2\pi n\right) \end{Bmatrix}$$

 

[shamieh]

Hello, shamieh!


We are given: .$$\begin{Bmatrix}x &=& \text{-}1 \\ y &=& \sqrt{3} \end{Bmatrix}$$

We want: .$$\begin{Bmatrix}r, & \text{where }r^2 \,=\,x^2+y^2 \\ \theta, & \text{ where }\tan\theta \,=\,\frac{y}{x}\quad \end{Bmatrix}$$$$r^2 \:=\:x^2+y^2 \;=\;(\text{-}1)^2 + (\sqrt{3})^2 \:=\:1+3 \:=\:4$$

. . Hence: .$$r \,=\,\pm2$$$$\tan\theta \,=\,\frac{\sqrt{3}}{\text{-}1} \,=\,\text{-}\sqrt{3}$$

. . Hence: .$$\theta \:=\: \frac{2\pi}{3} + \pi n$$

Therefore: .$$(\text{-}1,\,\sqrt{3}) \;=\;\begin{Bmatrix}\left(2,\;\tfrac{2\pi}{3}\!+\! 2\pi n\right) \\ \left(\text{-}2,\;\tfrac{5\pi}{3}\!+\!2\pi n\right) \end{Bmatrix}$$

That makes complete sense! But can you explain to me why the answer says:
$$(2, 2\pi/3 $$ +/- $$ 2n\pi)$$ and $$(2, -\pi/3$$ +/- $$ 2n\pi)$$ ?

 

[shamieh]

Hello, shamieh!


We are given: .$$\begin{Bmatrix}x &=& \text{-}1 \\ y &=& \sqrt{3} \end{Bmatrix}$$

We want: .$$\begin{Bmatrix}r, & \text{where }r^2 \,=\,x^2+y^2 \\ \theta, & \text{ where }\tan\theta \,=\,\frac{y}{x}\quad \end{Bmatrix}$$$$r^2 \:=\:x^2+y^2 \;=\;(\text{-}1)^2 + (\sqrt{3})^2 \:=\:1+3 \:=\:4$$

. . Hence: .$$r \,=\,\pm2$$$$\tan\theta \,=\,\frac{\sqrt{3}}{\text{-}1} \,=\,\text{-}\sqrt{3}$$

. . Hence: .$$\theta \:=\: \frac{2\pi}{3} + \pi n$$

Therefore: .$$(\text{-}1,\,\sqrt{3}) \;=\;\begin{Bmatrix}\left(2,\;\tfrac{2\pi}{3}\!+\! 2\pi n\right) \\ \left(\text{-}2,\;\tfrac{5\pi}{3}\!+\!2\pi n\right) \end{Bmatrix}$$

So you are just taking $$0 + \frac{2\pi}{3}$$ to get : $$(2, \frac{2\pi}{3} + 2\pi n)$$

and then you are taking $$2\pi - \frac{2\pi}{3}$$ to get: $$(-2, \frac{5\pi}{3} + 2\pi n)$$Also can you explain how my teacher got for his solution, $$(2, \frac{2\pi}{3} \pm 2\pi n)$$ and $$(2, -\frac{\pi}{3} \pm 2\pi n)$$

Also this may sound like a dumb question but how come we obtained just + in our final solution where as my teacher has $$\pm$$ in his solution...

 

[HallsofIvy]

Although Soroban wrote "-2" for r, in polar coordinates, "r" is usually taken to be positive. Of course, that just reverses the direction which is the same as adding $$\pi$$ to the angle. That is $$(-2, 5\pi/3)$$, which, technically, means "go to the direction that make angle $$5\pi/3$$ with the positive x-axis, then go in the opposite direction" is the same as $$(2, 5\pi/3- \pi)= (2, 2\pi/3)$$.

As for the difference between Soroban's "+" and your teacher's "$$\pm$$", Soroban is allowing n to be any integer, positive or negative while your teacher is, apparently, only allowing n to be non-negative.