MHB 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?

- - - Updated - - -

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.