Converting Rectangular coordinates to polar coordinates

[lennonct]

I understand how to do this, my only problem is that I do not understand how my professor got the end result for the y-value coordinate at the end.

it ended up being tan(theta) = 1 / 6 over sqrt 3 / 6 (only the 3 has the sqrt sign not the 6).

after that she wrote theta = 30 degrees in Q2 (I do not understand where this came from)

the final answer was (1/3, 150 degrees). I understand how to get the 1/3, not the 150 degree.

any help would be appreciated


original problem is convert (-sqrt3 / 6 , 1 / 6) ---- (- radical 3 over 6) , ( one sixth)

 

[ksdhart2]

The formulas for converting from rectangular coordinates to polar coordinates are:

\(\displaystyle r^2=x^2+y^2\) and \(\displaystyle \theta=tan^{-1} \left( \dfrac{y}{x} \right)\)

Applying those to the point \(\displaystyle \left( \dfrac{\sqrt{3}}{6}, \: \dfrac{1}{6} \right)\), we get:

\(\displaystyle r^2=\left(\dfrac{\sqrt{3}}{6}\right)^2+\left( \dfrac{1}{6}\right)^2=\dfrac{3}{36}+\dfrac{1}{36}= \dfrac{4}{36}=\dfrac{1}{9}\)

\(\displaystyle r=\dfrac{1}{3}\)

That's is the part you say you understand, but you say you're lost how your instructor found theta... well, let's apply the usual formula and see where that leads:

\(\displaystyle \theta =tan^{-1}\left(\dfrac{ \dfrac{1}{6}}{ \dfrac{\sqrt{3}}{6}}\right)=tan^{-1}\left( \dfrac{1}{\sqrt{3}}\right)=tan^{-1}\left( \dfrac{\sqrt{3}}{3}\right)\)

This value should look familiar. Recall that chart of angles you memorized. What is \(\displaystyle tan(30^{\circ})\) or \(\displaystyle tan \left( \dfrac{\pi}{6}) \right)\)? If, for some reason, you don't have the basic table of trig functions for 30, 45, 60, 90, 120, 150, and 180 degrees memorized, you really ought to do so - it will prove very helpful.

 

[lennonct]

She is having us use: Tan(theta) = y/x

you are showing me inverse tan, what is the difference?

I do not know the chart you are speaking of and I also do not know how this ended up at 150 degrees

Your input was slightly helpful as you showed that 1 over radical 3 equals radical 3 over 3 - - assuming you multiplied by radical 3 over radical 3.

Still confused how did you go from that to 30 degrees? I am not a trigonometry expert as I took that class 5 years ago

 

[ksdhart2]

She is having us use: Tan(theta) = y/x

you are showing me inverse tan, what is the difference?

I do not know the chart you are speaking of and I also do not know how this ended up at 150 degrees

Your input was slightly helpful as you showed that 1 over radical 3 equals radical 3 over 3 - - assuming you multiplied by radical 3 over radical 3.

Still confused how did you go from that to 30 degrees? I am not a trigonometry expert as I took that class 5 years ago

Okay, now I understand a bit better where you're stuck. Your teachers uses the form \(\displaystyle tan(\theta)=\dfrac{y}{x}\) but that's actually expressing the same idea as the form I showed, just in a different way. Are you familiar with inverse functions at all? If you are, what happens when you take the inverse of an already inverted function? Well, you end up with just the argument. That is to say:

\(\displaystyle f^{-1}\left(f\left(x\right)\right)=x\)

Applying that logic, we can see that:

\(\displaystyle tan^{-1}\left(tan\left(\theta \right)\right)=\theta \)

And, as you should recall from your early days of algebra, when you do something to one side of the equation, you must do the same to the other. So, if we start with the original function:

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

And then take the inverse tangent of both sides, we get:

\(\displaystyle tan^{-1}\left(tan\left(\theta \right)\right)=tan^{-1}\left(\dfrac{y}{x}\right)\) or \(\displaystyle \theta =tan^{-1}\left(\dfrac{y}{x}\right)\)

Once you get to this point (and, really, once you're comfortable enough with this process, there's no reason you can't just start here), you can plug it into your calculator. Or, with some practice at recognizing common, "nice" values for angles in trig functions, you can just "eyeball" them sometimes. One such chart you might look at is here.

Finally, as for how to arrive at the given answer of 150 degrees, that's actually a screw up on my part. I missed the initial minus sign. If the original point is:

\(\displaystyle \left(\dfrac{\sqrt{3}}{6},\:\dfrac{1}{6}\right)\)

as I said in my first post, then the value of theta will be 30 degrees. On the other hand, if it's really what you indicated in your post, and the x-coordinate has a minus sign:

\(\displaystyle \left(-\dfrac{\sqrt{3}}{6},\:\dfrac{1}{6}\right)\)

Then you'll arrive at:

\(\displaystyle \theta =tan^{-1}\left(-\dfrac{\sqrt{3}}{3}\right)\) and thus \(\displaystyle \theta =150^{\circ }=\dfrac{5\pi }{6}\)

You can verify this by either punching the inverse tangent into your calculator or consulting the lookup table and noting the value of tan(150 degrees).

 

[HallsofIvy]

I understand how to do this, my only problem is that I do not understand how my professor got the end result for the y-value coordinate at the end.

If I am reading this correctly, you are given the x and y coordinates and your difficulty is finding the angle, \(\displaystyle \theta\). not the y-value. Looking at your answer it appears that you are given \(\displaystyle (x, y)= \left(-\frac{\sqrt{3}}{6}, \frac{1}{6}\right)\). It would have been helpful if you had said that, instead of starting with "it ended up being"!

it ended up being tan(theta) = 1 / 6 over sqrt 3 / 6 (only the 3 has the sqrt sign not the 6).

So \(\displaystyle \frac{\frac{1}{6}}{\frac{\sqrt{3}}{6}}= \frac{1}{6}\frac{6}{\sqrt{3}}= \frac{1}{\sqrt{3}}= \frac{\sqrt{3}}{3}\)

That looks "kinda" familiar- I remember that the sine of 30 degrees is sin(30)= 1/2 and the cosine is \(\displaystyle cos(30)= \frac{\sqrt{3}}{2}\) so that tangent= sine/cosine is \(\displaystyle \frac{1}{2}\frac{2}{\sqrt{3}}= \frac{1}{\sqrt{3}}= \frac{\sqrt{3}}{3}\).

after that she wrote theta = 30 degrees in Q2 (I do not understand where this came from)

Then you need to brush up on your trigonometry. You should have recognized, as I said above, that the tangent of 30 degrees is \(\displaystyle \frac{\sqrt{3}}{3}\). Further, since the x coordinate is negative and the y coordinate positive, the point is in the second quadrant, Q2.

the final answer was (1/3, 150 degrees). I understand how to get the 1/3, not the 150 degree.

Do you know what the graph of y= cos(x) looks like? It starts at 1 when x= 0, goes down to 0 at x= 90 degrees, down to -1 at 180 degrees. There is a symmetry about x= 90 degrees. That's how we know that angle 30 degrees gives the positive value while angle 180- 30= 150 give the negative value.

any help would be appreciated


original problem is convert (-sqrt3 / 6 , 1 / 6) ---- (- radical 3 over 6) , ( one sixth)

Ah! Now, you tell us!

Do you know what an "equilateral triangle" is? It has all sides of equal length and all angles the same. Since the angles in any triangle add to 180 degrees, each angle in an equilateral triangle has measure 180/3= 60 degrees. Further, drawing a line from 1 vertex perpendicular to the opposite side also bisects the opposite side and bisects the angle at the vertex, dividing the equilateral triangle into two right angles with angles 60 and 30 degrees, hypotenuse of length 1, and one leg of length 1/2. By the Pythagorean theorem \(\displaystyle x^2+ \frac{1}{4}= 1\) where x is the length of the other leg (the line bisecting the vertex angle). So \(\displaystyle x= \sqrt{1- \frac{1}{4}}= \frac{\sqrt{3}}{2}\). From that, \(\displaystyle sin(60)= \frac{\frac{\sqrt{3}}{2}}{1}= \frac{\sqrt{3}}{2}\) and \(\displaystyle cos(60)= \frac{\frac{1}{2}}{1}= \frac{1}{2}\).

Your professor expects that you have already learned trigonometry (it is typically a prerequisite for a course that includes "Cartesian Coordinates" which is what you are doing here). You really should know that
\(\displaystyle sin(0)= 0\)
\(\displaystyle cos(0)= 1\)
\(\displaystyle sin(30)= \frac{1}{2}\)
\(\displaystyle cos(30)= \frac{\sqrt{3}}{2}\)
\(\displaystyle sin(45)= \frac{\sqrt{2}}{2}\)
\(\displaystyle cos(45)= \frac{\sqrt{2}}{2}\)
\(\displaystyle sin(60)= \frac{\sqrt{3}}{2}\)
\(\displaystyle cos(60)= \frac{1}{2}\)
\(\displaystyle sin(90)= 1\)
\(\displaystyle cos(90)= 0\)
and the extensions of those beyond 90 degrees.

And you should be ready to learn those in radians as well.