Give all the polar coordinates corresponding the rectangular point

[shamieh]

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

Am i setting this up right?

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

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

\(\displaystyle r^2 = x^2 + y^2\)

so:

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

 

[HallsofIvy]

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

Am i setting this up right?

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

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

\(\displaystyle r^2 = x^2 + y^2\)

Yes, that is correct.

so:

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

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

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

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

 

[shamieh]

How about:

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

\(\displaystyle r^2 = 4\)

\(\displaystyle r = 2 \) or \(\displaystyle - 2\)... I understand all of this...

But then..What is going on here?

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

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. [tex]\left(-\frac{2\pi}{3}\right)^2= \frac{4\pi}{9}[/tex] and [tex]\left(\frac{-11\pi}{6}\right)^2= \frac{121\pi^2}{36}[/tex]. Those do NOT add to 1.

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

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

\(\displaystyle

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:

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

So what must you conclude regarding $\theta$?

 

[shamieh]

I got

\(\displaystyle cos\theta = x/r = -1/2\)

\(\displaystyle sin\theta = y/r = \sqrt{3}/2\)

thus \(\displaystyle \theta = 2\pi/3\) in Quadrant II

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

 

[soroban]

Hello, shamieh!

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

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

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

. . Hence: .[tex]r \,=\,\pm2[/tex][tex]\tan\theta \,=\,\frac{\sqrt{3}}{\text{-}1} \,=\,\text{-}\sqrt{3}[/tex]

. . Hence: .[tex]\theta \:=\: \frac{2\pi}{3} + \pi n [/tex]

Therefore: .[tex](\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}[/tex]

 

[shamieh]

Hello, shamieh!


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

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

. . Hence: .[tex]r \,=\,\pm2[/tex][tex]\tan\theta \,=\,\frac{\sqrt{3}}{\text{-}1} \,=\,\text{-}\sqrt{3}[/tex]

. . Hence: .[tex]\theta \:=\: \frac{2\pi}{3} + \pi n [/tex]

Therefore: .[tex](\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}[/tex]

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

 

[shamieh]

Hello, shamieh!


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

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

. . Hence: .[tex]r \,=\,\pm2[/tex][tex]\tan\theta \,=\,\frac{\sqrt{3}}{\text{-}1} \,=\,\text{-}\sqrt{3}[/tex]

. . Hence: .[tex]\theta \:=\: \frac{2\pi}{3} + \pi n [/tex]

Therefore: .[tex](\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}[/tex]

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

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

Also this may sound like a dumb question but how come we obtained just + in our final solution where as my teacher has \(\displaystyle \pm\) 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 [tex]\pi[/tex] to the angle. That is [tex](-2, 5\pi/3)[/tex], which, technically, means "go to the direction that make angle [tex]5\pi/3[/tex] with the positive x-axis, then go in the opposite direction" is the same as [tex](2, 5\pi/3- \pi)= (2, 2\pi/3)[/tex].

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