For more information, please see full course syllabus of Trigonometry
For more information, please see full course syllabus of Trigonometry
Related Articles:
Polar Coordinates
Main definition and formulas:
- Rectangular or Cartesian coordinates (x,y) describe a point based on the distances from the x- and y-axes.
- Polar coordinates (r,θ ) describe a point based on the distance from the origin and the angle it makes with the positive x-axis.
- Conversions:
r =
√
x^{2} + y^{2}θ = arctan y x, if x > 0 (Quadrants I and IV) π + arctan y x, if x < 0 (Quadrants II and III) x = r cosθ y = r sinθ - Usually, r ≥ 0, but not necessarily.
- Usually, 0 ≤ θ < 2π , but not necessarily.
Example 1:
Convert the following points from rectangular coordinates to polar coordinates: (3√ 2, − 3√ 2), (− 4,− 3), (− √ 3,1), (− 2,5).Example 2:
Convert the following points in polar coordinates to standard form (with r ≥ 0, 0 ≤ θ < 2π ), and then convert them to rectangular coordinates: (8,− [3π /4]), (− 6,− [11π /6]), (− 2,[11π /3]), (3,− [2π /3]).Example 3:
Graph the polar equation r = 2sinθ . Check your answer by converting the equation to rectangular coordinates and solving it algebraically.Example 4:
Convert the following points in polar coordinates to standard form (with r ≥ 0, 0 ≤ θ < 2π ), and then convert them to rectangular coordinates: (− 2,− [13π /6]), (6,− [π /3]), (− 5,[7π /4]), (− 4,− [5π /4]).Example 5:
Graph the polar equation r = sin2θ .Polar Coordinates
- Use SOHCAHTOA to solve this problem
- cos45^{°} = [x/60], sin45^{°} = [y/60]
- x = 60cos45^{°}, y = 60sin45^{°}
- Horizontal component of velocity: x = rcosθ
- x = 70cos50^{°}
- Vertical component of velocity: y = rsinθ
- y = 70sin50^{°}
- Horizontal component of velocity: x = rcosθ
- x = 50cos70^{°}
- Vertical component of velocity: y = rsinθ
- y = 50sin70^{°}
- F = 70cos45^{°}, D = 35
- W = (70cos45^{°})(35)
- F = 65cos30^{°}, D = 10
- W = (65cos30^{°})(10)
- x = rcosθ, and y = rsinθ
- Standard form: − [(4π)/3] + 2π = [(2π)/3] ⇒ (2,[(2π)/3])
- x = 2cos[(2π)/3] = − 1
- y = 2sin[(2π)/3] = √3
- x = rcosθ, and y = rsinθ
- Standard form: − [(π)/3] + 2π = [(5π)/3] − π = [(2π)/3] had to subtract π to get a positive r value ⇒ (4,[(2π)/3])
- x = 4cos[(2π)/3] = − 2
- y = 4sin[(2π)/3] = 2√3
- x = rcosθ, and y = rsinθ
- Standard form: [(3π)/4] + π = [(7π)/4] had to add π to get a positive r value ⇒ (2,[(7π)/4])
- x = 2cos[(7π)/4] = √2
- y = 2sin[(7π)/4] = − √2
- x = rcosθ, and y = rsinθ
- Standard form: − [(3π)/4] + 2π = [(5π)/4] − π = [(π)/4] had to subtract π to get a positive r value ⇒ (1,[(π)/4])
- x = 1cos[(π)/4] = [(√2 )/2]
- y = 1sin[(π)/4] = [(√2 )/2]
- Create a table of values
- Plot those values on a polar plane
- Create a table of values
- Plot those values on a polar plane
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
Answer
Polar Coordinates
Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
- Intro
- Polar Coordinates vs Rectangular/Cartesian Coordinates
- Converting Between Polar and Rectangular Coordinates
- Example 1: Convert Rectangular to Polar Coordinates
- Example 2: Convert Polar to Rectangular Coordinates
- Example 3: Graph the Polar Equation
- Extra Example 1: Convert Polar to Rectangular Coordinates
- Extra Example 2: Graph the Polar Equation
- Intro 0:00
- Polar Coordinates vs Rectangular/Cartesian Coordinates 0:12
- Rectangular Coordinates, Cartesian Coordinates
- Polar Coordinates
- Converting Between Polar and Rectangular Coordinates 2:06
- R
- Theta
- Example 1: Convert Rectangular to Polar Coordinates 6:53
- Example 2: Convert Polar to Rectangular Coordinates 17:28
- Example 3: Graph the Polar Equation 28:00
- Extra Example 1: Convert Polar to Rectangular Coordinates
- Extra Example 2: Graph the Polar Equation
Trigonometry Online Course
I. Trigonometric Functions | ||
---|---|---|
Angles | 39:05 | |
Sine and Cosine Functions | 43:16 | |
Sine and Cosine Values of Special Angles | 33:05 | |
Modified Sine Waves: Asin(Bx+C)+D and Acos(Bx+C)+D | 52:03 | |
Tangent and Cotangent Functions | 36:04 | |
Secant and Cosecant Functions | 27:18 | |
Inverse Trigonometric Functions | 32:58 | |
Computations of Inverse Trigonometric Functions | 31:08 | |
II. Trigonometric Identities | ||
Pythagorean Identity | 19:11 | |
Identity Tan(squared)x+1=Sec(squared)x | 23:16 | |
Addition and Subtraction Formulas | 52:52 | |
Double Angle Formulas | 29:05 | |
Half-Angle Formulas | 43:55 | |
III. Applications of Trigonometry | ||
Trigonometry in Right Angles | 25:43 | |
Law of Sines | 56:40 | |
Law of Cosines | 49:05 | |
Finding the Area of a Triangle | 27:37 | |
Word Problems and Applications of Trigonometry | 34:25 | |
Vectors | 46:42 | |
IV. Complex Numbers and Polar Coordinates | ||
Polar Coordinates | 1:07:35 | |
Complex Numbers | 35:59 | |
Polar Form of Complex Numbers | 40:43 | |
DeMoivre's Theorem | 57:37 |
Transcription: Polar Coordinates
We are learning about polar coordinates today and we got some points here in polar coordinates which we have to convert to standard from.0000
The (r) should be positive and the (theta) should be between 0 and 2pi and then we have to convert them to rectangular coordinates.0008
Let me get started graphing this first point, we are given -13pi/6, I do not like that already.0018
Iâ€™m going to add 2pi to that and that gives me -13pi/6 + 12pi/6 = -pi/6.0026
I still do not like that, that is still not in standard range so Iâ€™m going to add another 2pi to that and that gives me 11pi/6, that is good.0039
Iâ€™m going to graph that now, 11pi/6 is just short of 2pi, over there 11pi/6.0051
But we are supposed to go negative two units in this direction, that means I want to end up in the opposite direction from 11pi/6.0063
Maybe you can tell from the graph that is 5pi/6.0074
If you can not tell that immediately from the graph, just subtract pi from 11pi/6 and you will get 5pi/6.0078
The answer there is that we want to go positive 2 units in the direction of 5pi/6.0088
Now we got to find my rectangular coordinates, remember the master formula we are using here x=arcos(theta), (y)=arcsin(theta), (x)=(2)cos(2pi/6) which is (2 x â€“root3)/2, that one I got memorized.0101
It is negative because the (x) value is negative there, (y) =(2)sin(5pi/6) on both of those which is (2)x(1/2) because the (y) is positive.0126
My (xy) collectively, the rectangular coordinates are â€“root 3 and 1.0149
That is the first point, the second one here 6 (â€“pi/3), let me graph that one.0166
I think Iâ€™m going to add 2pi to that right away to get it in the proper range, add 2 pi to that, 2pi â€“ pi/3= 5pi/3.0172
If I graph that, 5pi/3 is down here in the fourth quadrant and our radius is 6.0188
Our standard form there is 6 (5pi/3).0205
I will find (x) and (y) using arcos(theta) and arcsin(theta), so 6 x ((cos(5pi/3)) which is 6, now the cos of 5pi/3 is positive because the (x) value is positive Â½.0215
(y)=6sin(5pi/3) negative because the (y) coordinate is negative so 6 x(- root 3)/2, these are common values that I have memorized.0234
The (xy) collectively can be written and simplified down to 3 and â€“square root of 3.0247
The third one here is 7pi/4, let me find that, this is number three now, that was number two.0267
Number three, 7pi/4, that is all the way around in the fourth quadrant, just short of 2pi.0275
7pi/4 but I want to go -5 units in that direction, that takes me -5 units in the opposite direction.0283
I can tell from the graph that is 3pi/4 but if you are concerned about working that out graphically, just subtract pi from 7pi/4 and you will get 3pi/4.0292
Our standard form of polar coordinates would be (phi)(3pi/4).0305
Let me use that to find x and y, x=arcos(theta) that is (phi)cos(3pi/4) which is (5) x (-square root of 2)/2 cos is negative.0317
Y=(5) sin(3/4) which is (5) x (square root of 2)/2.0337
The x and y collectively give you the coordinates -5(root 2)/2 and 5(root 2)/2.0353
Finally, the fourth point here, we go -5pi/4, I do not like that already and Iâ€™m going to add 2pi to that right away and get 3pi/4.0371
Iâ€™m going to go in the 3pi/4 direction, there is 3pi/4 but I have to go -4 units in that direction.0389
That means I actually go in the opposite direction 4 units and to find that angle I add pi and that puts me back down to 7pi/4.0401
7pi/4 is my reference angle and I want to go 4 units in that direction so my standard polar coordinates there are 4 and 7pi/4.0415
My x and y, I find using arcos(theta) and arcsin(theta), x=4cos(7pi/4) now the cos is positive because we are in the fourth quadrant so that is (root 2)/2.0429
(y)=(4)sin(7pi/4) which is 4 x (- root 2)/2, negative because we are in the fourth quadrant, the( y) coordinate is negative.0451
The x and y collectively are 2(root 2) and -2(root 2).0465
There are several steps involved in that problem, we are given these polar coordinates but they are not necessarily in standard form.0483
They all might not be positive and the (theta) might not be between 0 and 2pi.0491
The first thing you do in all of these is to draw a graph, figure out where your angle is and if you do not like the angle, if it is not between 0 and 2pi you can add or subtract multiples of 2 pi to get it in between 0 and 2pi.0497
What we did we a lot of this is we added and subtracted multiples of 2 pi to get the angle in between 0 and 2 pi. 0513
Then we plot the radius and for some of these, the radius turned out to be negative which means we are actually walking in the opposite direction from the angle that we expected.0521
Example on this first one, we thought we are going down to 11 pi/6, in fact we are going in the opposite direction.0531
To figure out what the opposite direction was, I subtracted pi and that is how I got 5 pi/6 as my answer.0540
That happened on several of those, we ended up walking in the negative direction and so we have to subtract pi or add pi to get our final answers for the reference angle.0548
Once we add or subtract pi, we can make the radius positive and that is how we got the radius and reference angle for each of these.0559
Once we find the radius and the reference angle, it was a matter of using the standard component formulas arcos(theta) and arcsin(theta).0568
In each case we used x=arcos(theta) and y=arcsin(theta), plug those in, these were all common values.0580
I knew the cos and sin for all of them without looking them up on the calculator because they were all multiples of pi/3 and pi/4.0588
I was able to just write down the coordinates for x and y on all of those.0596
For our last example here of polar coordinates, we have to graph the polar equation r=sin(2 theta)0000
Iâ€™m going to try and graph that, Iâ€™m going to start by graphing this using x and y.0007
Iâ€™m going to start by graphing y=sin(x) that is a very familiar one, I know that has period of 2pi.0014
I graph that, now we practice modifying these equations from the basic sin way to more complicated ones in the earlier lecture.0031
You might want to review that if this is not making sense to you.0045
Iâ€™m going to try and graph y=sin(2x), what that does the way that two changes it, if you remember back from the earlier lecture on www.educator.com that changes the period. 0049
The period is now 2pi/2, so it is pi, which means this thing has a whole period just between 0 and pi.0066
I will do a second period between pi and 2pi.0078
Let me draw that a little lighter, it is going to go up and down, come back down on pi/2, go down to -1, come back and finish a whole period at pi.0086
Then from pi to 3pi/2, it goes up to 1 again, goes down to -1, and comes back at 2 pi.0109
There is 1, -1, and this is pi/2, this is 3pi/2.0117
Now Iâ€™m going to use this to make a chart of values for (theta) and then finally Iâ€™m going to make a polar graph.0129
Let me make a chart of values for (theta), Iâ€™m not going to fill in every (theta) I know.0134
Iâ€™m just going to fill in probably the multiples of pi/4.0141
Each one Iâ€™m going to write down what sin(2) (theta) is, I stated sin(theta)=0, sin(2x0)=sin0 so that is 0.0151
Now when (theta) is pi/4, I can remember that sin(2)( theta) is sin(pi/2) which is 1.0163
Or I can just look at my graph here, that was the point of making this graph.0170
I see that sin goes up to 1 at pi/4, when (theta)=pi/2, I see that it will back at 0, sin(2 theta)=sin(pi), so it is 0.0176
(theta)=3pi/4, I see them down at -1, (theta)=pi we are back to 0, (theta)=5pi/4 that is right here so we are back up at 1.0189
3pi/2 we are back down to 0, 5pi/4 we are back down to -1, and 2pi we are back up to 0.0216
You can work all those out, you can plug each one into sin(2)( theta) and do the calculations or you can just check this graph, it is really helpful if you just check the graph.0236
Iâ€™m going to graph that as a polar equation and let me fill in my key angles here, so there is 0, pi/2, pi, and 3pi/2.0248
We are back to 0 at 2pi, now I made some values for all the multiples of pi/4, Iâ€™m going to fill those in.0268
There is pi/4, 3pi/4, 5pi/4, and 7pi/4, I just labeled my 7pi/4 in the chart, I labeled it as 5 pi/4 but that is 7pi/4.0287
Just right here 7/pi and that is where 5pi/4 is.0310
Now Iâ€™m going to fill in my dots and Iâ€™m going to do this in blue.0319
At 0 it is just 0, at pi/4 I have grown out to 1, pi/2 Iâ€™m back at 0, 3pi/4 I met -1, but -1 in the 3pi/4 direction means you are actually walking in the opposite direction down to -1.0325
That is really 3pi/4, -1 in the 3pi/4 direction, let me write that down.0352
-1 (3pi/4) that is why I ended up there, at pi I ended up at 0.0362
Let me graph what I have so far, I started at 0, I grew out to 1 at pi/4, by the time I got to pi/2 I was back to 0, and then I went to -1 in the 3pi/4 direction.0371
And then by pi I was back to 0, at 5pi/4 Iâ€™m back at +1, I go +1 in the 5pi/4 direction, let me write that down 1 (5pi/4) so I come down here.0395
3pi/2 Iâ€™m back to 0, 7pi/4 I met -1 again, now 7pi/4 is down here but I want to walk -1 unit in that direction.0425
I end up walking out to the 1 unit in the 3pi/4 direction.0447
By 2pi I have come back to 0. 0457
That last point, even though it looks like it is on the 3pi/4 axis, it is actually you should think of it as -1 on the 7pi/4 direction.0466
We have this interesting four leaf clover comes out as our graph as r=(2)sin(theta).0478
Let me recap the steps that I followed there.0486
The first thing I wanted to do was really graph r=(2)sin(theta), think about it as a rectangular equation and graph it as a rectangular equation.0490
To warm up , I graphed y=sin(x) and then I graphed y=(2)sin(x).0499
I remember back from the previous lecture on www.educator.com, that, that two changes the period.0507
That takes y=sin(x) and it shrinks the period down to be pi, the thing oscillates up and down twice as fast.0514
I get this faster oscillating graph, Iâ€™m going to use that as a graph to fill in all the (theta) Iâ€™m interested in.0526
I just took multiples of pi/4, and I looked at my sin(2)(theta) and I got these 0, 1, -1.0536
I plotted each one of those on the axis over here, but the important thing to remember is when you have a negative number you are walking in the opposite direction from what you expected.0544
Let me put a little timeline here if you are trying to follow us later on.0556
The first thing we did was we graphed 1 in the pi/4 direction that really represented going here, that was the first loop we drew.0561
The second thing was going down to -1 in the 3pi/4 direction, -1 in the 3pi/4 direction brought us out in to the 7pi/4 direction that was down there.0573
The third thing we did is we went back to 1 in the 5pi/4 direction that brought us down to quadrant 3.0590
We got to -1 in the 7pi/4 direction but 7pi/4 is in the fourth quadrant walking -1 in that direction actually puts you up on the second quadrant.0599
That is why the fourth loop that we graphed was up over here in the second quadrant even though it came from looking at the values in the fourth quadrant.0619
It was a pretty complicated example there but it is a matter of plotting your values and then plotting them on each of this axis and keeping track of one thing is positive and one is negative.0628
If they are negative remember to walk in the opposite direction going to the opposite quadrant from what you expected.0641
That is the end of these lectures on polar coordinates.0648
Thanks for watching, these are the trigonometry lectures on www.educator.com.0651
Hi these are the trigonometry lectures for educator.com, and today we're going to learn about polar coordinates which is really a kind of a new way of looking at the coordinate plane.0000
It's going to be a lot of fun to check this out.0010
I want to remind you how you have been keeping track of points in the plane.0012
You've been using ...0019
That's a really ugly access.0021
You've been using x and y coordinates which are known as rectangular coordinates or Cartesian coordinates.0024
The way that works is you graph a point based on its distance from the x-axis and its rectangular distance from the y-axis.0033
You look at that distance, that's the y-coordinate, and that distance is the x-coordinate.0044
That's all based on rectangles, it's also called Cartesian coordinates because it comes from Descartes, and you've done that before.0051
Polar coordinates are the new idea here, and it's really quite different.0059
If you have a point somewhere in the plane, instead of trying to orient that point based on rectangular distances from the x and y axis, what you do is you draw a line straight out from the origin to that point.0066
You describe that point in terms of how long that line is, r, it's the distance from the origin, in terms of the angle that line makes with the positive x-axis.0089
There's the positive x-axis.0098
You measure what that angle is from the positive x-axis, and you measure the distance along that line, then give the coordinates of that point in terms of r and θ.0100
That's really a new idea.0113
We're going to figure out how.0115
If you know what the point is in terms of x and y, how do you figure out what the r and θ are and vice versa.0117
Let's check that out now.0124
If you know the x and y coordinates of a point, the rectangular coordinates of a point, then you can figure out r and θ based on the Pythagorean theorem.0126
There's x and there's y, there's θ, and there's r.0141
r, just based on the Pythagorean theorem, is the square root of x^{2}+y^{2}.0149
That's the same as the magnitude of a vector that we learned in the lecture last time about vectors.0153
You don't really have to remember any new formulas here, it's the same formula r=x^{2}+y^{2}.0160
Same with θ, we know that the tan(θ)=y/x, that's based on SOH CAH TOA, which means that θ is arctan(y/x).0167
Here is where it gets tricky.0178
Remember that arctan(y/x) will always give you an angle in the fourth quadrant or in the first quadrant.0179
What you do if you have a point that's in the second or third quadrant, or if you have a point in the second or third quadrant, we said before that you add 180 degrees to θ.0189
In polar coordinates, you often use radians instead of degrees.0202
I've changed that formula slightly.0207
It's the same formula except in terms of radians, it's π+arctan(y/x), instead of 180+arctan(y/x).0209
You use this formula when your point is in quadrant 2 or 3, is when you have to add on a π to the arctan(y/x).0217
In other words, that occurs when the x-coordinate is negative, when x is less than 0.0239
This formula, the basic formula arctan(y/x), you use that for quadrants 1 and 4, or when your x-coordinate is positive.0249
That's how if you know the x and the y, you can find the r and the θ.0264
If you know the r and the θ, you can find the x and the y, by taking rcos(θ) and rsin(θ).0271
Those are the same formulas we had when we started with the magnitude and angle of a vector, and we convert it back to find the components of the vector.0279
You don't really have to remember any new formulas here.0289
It's the same formulas as before, x=rcos(θ), y=rsin(θ).0292
Those come from SOH CAH TOA, so it's not very hard to derive those formulas based on the SOH CAH TOA relationships in a right angle.0299
That's θ, that's r, that's x and that's y.0306
You can use SOH CAH TOA to find x and y equals rcos(θ) and rsin(θ).0310
There's some conventions that we tried to follow but it's not absolutely essential.0316
The r is usually assumed to be positive.0322
Certainly, if you used this formula to find r, you get a square root, so it's always positive.0325
But you can also talk about -r's.0331
You think about a -r, if you have an angle θ, what would a -r represent?0336
Well that just represents going r units in the other direction.0343
If r is less than 0, then that just represents going r units on the other side of the origin.0347
That's what a -r would represent, but we try to use positive r's if we can.0355
Similarly, θ, we try to keep in between the range 0 to 2π.0360
Here's 0, π/2, π, 3π/2, and 2π.0367
We try to keep our θ in between 0 and 2π.0373
If it's outside of that range, then what you can do is you could add or subtract multiples of 2π to try to get it back into that range.0377
That's how we try to restrict the values of r and θ.0385
You'll see some examples of this as we practice converting points.0390
It's probably best just to move on to some examples and you'll see how we generally find r's that are positive and θ's between 0 and 2π.0394
But if someone gives us values that are not in those ranges, we can modify them to find different sets of coordinates that do have r's and θ's in those ranges.0401
Let's move on to some examples.0411
First example is to convert the following points from rectangular coordinates to polar coordinates.0414
Remember, we're given here the x and the y, and we want to find the r's and the θ's, because we're given rectangular coordinates that's x and y.0422
We want to find polar coordinates, that's r and θ.0432
I'm going to draw some little graphs of these points to help me keep track of where they are.0435
You can also figure them out just using the formulas that we learned on the previous slide.0440
For the first point here, we've got 3 square root of 2 and -3 square root of 2.0445
That's positive in the x direction, and negative in the y direction.0455
That's a point down there.0460
I want to figure out where that is.0464
I'm going to use the formulas to figure that out, r is equal to the square root of x^{2}+y^{2}.0465
Let me remind you of those formulas up here because we're going to use them quite a bit, x^{2}+y^{2}.0472
θ=arctan(y/x), that's the case if x is greater than 0, or π+arctan(y/x) if the case x is less than 0.0479
In this case, our r, our magnitude is the square root of 3 root 2 squared, plus negative 3 root 2 squared.0504
3 root 2 squared is 9 times 2, that's the square root of 18, plus another 18, which is the square root of 36, which is 6.0521
θ is arctan of negative 3 root 2 over 3 root 2, because it's y/x, which is arctan(-1).0534
Arctan(-1), that's a common value that I remember.0556
That's -π/4, but I would like to get an answer in between 0 and 2π, -π/4 doesn't qualify.0559
I'm going to add +2π, and that will give me, 2π is 8π/4, 7π/4.0570
My polar coordinates for that point (r,θ) are (6,7π/4).0583
That really corresponds with what I can check visually.0597
That angle is -π/4 if you go down south from the x-axis, but if you go all the way around the long way, it's 7π/4.0602
That's our first set of polar coordinates.0613
Moving on to the next one, (-4,-3), that's -4 in the x direction, -3 in the y direction.0616
That's somewhere down there.0625
So, r is the square root of 4^{2}+3^{2}.0628
Since we're squaring them, I'm not going to worry about the negative signs.0634
16+9=25, square root of that is 5.0637
θ=arctan(-4/-3).0644
Now, we know that this angle is in the third quadrant, its x-coordinate is negative, that means we have to add a π to the θ.0655
This is arctan(4/3)+π.0668
If you plug in arctan(4/3), your calculator would give you an angle up here.0671
We have to add π radians to whatever the calculator's answer is.0676
Now, arctan(4/3), one thing that's very important here is that I'm going to switch my calculator over to radian mode, that's because I'm looking for answers in terms of radians now.0681
I don't want to mix up my radians and my degrees.0693
The π is certainly in radians, I didn't say 180 degrees, I said π, I need to make sure that my calculator's in radian mode when I do arctan(4/3).0697
I do arctan(4/3), negative's cancelled, that gives me 0.93+π.0708
That simplifies down to 4.07.0726
My (r,θ) for that second one is (5,4.07).0734
Just a reminder that that 4.07 is a radian measure.0745
It's not so obvious when you don't have the π in there, when you don't have a multiple of π, but it is a radian measure there.0750
That's the radius and the reference angle for that point.0757
The third one here, negative square root of 3 and 1.0762
Let me graph that one.0768
Negative square root of 3 and positive 1, that's the point over there.0773
My r is, root 3 squared is 3, plus 1, so that's 2, θ is arctan(y/x), so 1 over negative root 3.0780
The x is negative, I have to add a π there.0798
That's arctan, if we rationalize that, that's negative root 3/3, plus π.0807
Now, negative root 3 over 3, that's a common value.0818
I recognize that as something that I know the arctangent of, that's -π/6+π, that's 5π/6.0821
My polar coordinates for that point, our (r,θ), is (2,5π/6).0837
Finally, we have -2π.0860
That's -2 in the x direction, π in the y direction.0869
That's a point somewhere up there in the second quadrant, so r is the square root of 2^{2}+5^{2}, which is the square root of 29, nothing very much I can do with that.0871
θ, arctan(5/-2), my x is negative and I'm in the second quadrant so I'm using the other formula for the θ, I have to add on a π there.0888
My calculator's set to radian mode, I'm going to do arctan(-5/2), and I get -1.19+π, which is 1.95.0904
As long as I'm giving decimal approximations, I may as well give a decimal approximation for the r.0939
The square root of 29 is approximately equal to 5.39.0943
My (r,θ) is (5.39,1.95) radians.0955
That's the answer for that one.0978
All of these, it really helped me to draw a picture even though I didn't really need that for the calculations, but it was really useful to kind of check that I was in the right place.0979
In all four, I drew a picture of where the point was, I figured out which quadrant it was in.0989
I worked out the r using the square root of x^{2}+y^{2}.0995
I worked out θ using arctan(y/x), but then I had to check which quadrant the point was in.1001
If it was in the second or third quadrant, then I had to add π to the answer that the calculator gave me for arctan(y/x).1008
In this fourth one, for example, the calculator gave me an answer of -1.19.1020
The calculator's answer was down there, that's why I had to add π to it inorder to get the answer.1027
I came back to these formulas for r and θ every time, but it's still very helpful to draw your coordinate axis and to show where the point is on those axis.1034
For our second example, we have to convert the following points in polar coordinates to standard form.1048
That means that the r should be positive and the θ should be between 0 and 2π, then we'll convert these points to rectangular coordinates.1056
We're going to start by graphing each one of these points.1064
First one, we have -3π/4, that means we're going down 3π/4 in the negative direction.1071
That's the same as going around 5π/4 in the positive direction.1080
We can write that as (8,5π/4).1084
That is legit because it's between 0 and 2π.1091
If you didn't like doing that graphically, you could do -3π/4+2π, and that gives you immediately the 5π/4.1094
That's how you could figure it out using equations instead of doing it graphically.1105
That's the standard form of that point.1111
Now, I have to find the rectangular coordinates, and I'm going to use rcos(θ) for x, and rsin(θ) for y.1113
My x is 8cos(5π/4).1127
The cos(5π/4), that's a common value that I've got it memorized, that's 8 times root 2 over 2, but it's negative because the x is negative there, negative root 2 over 2.1135
y is sin(5π/4), the y is also negative there down on the third quadrant, negative root 2 over 2.1151
This is why it's so useful to draw a graph where the point is.1163
It helps you find the sine and cosine and remember which one is positive or negative.1165
(x,y) together simplify down to (-4 root 2,-4 root 2).1169
That was my first point.1190
The second one, I'm going around 11π/6, that's just short of 2π, but it's in the negative direction, so I go around just short of 2π in the negative direction.1194
That's actually the same as going π/6 in the positive direction.1205
If you don't like doing that graphically, you can add 2π to 11π/6, and you'll get π/6.1211
But then I'm going -6 units in that direction.1218
That really takes me 6 units in the opposite direction.1223
Really, my reference angle is down here in the third quadrant.1229
I can write that as 6, that's π/6, that's an angle of π/6.1234
This is π/6 beyond π.1246
I have here (6,7π/6) would be the standard version of that point in polar coordinates.1251
Now I've got a positive radius and a positive angle between 0 and 2π.1262
Now I've got to find the rectangular coordinates for that point.1269
I'm going to use rcos(θ) and rsin(θ).1272
x=6cos(7π/6), which is 6×cos(7π/6), is negative root 3 over 2.1274
y=6sin(7π/6), these are common values that I've gotten memorized which is 6×(-1/2) down in the third quadrant.1292
The (x,y) collectively, the coordinates, simplify down to -3 root 3, and -3.1302
Okay, next one, -2 and 11π/3.1316
Let's figure out where that one is.1323
11π/3 is pretty big, I'm going to subtract 2π from that right away.1325
11π/3-2π, 2π=6π/3, that's 5π/3.1331
That's up there in the second quadrant, 5π/3.1342
We'll have to go -2 units in that direction, which takes me down in the negative direction.1346
Sorry, I graphed that in the wrong place.1360
I was graphing 5π/6 instead of 5π/3.1363
Let me modify that slightly.1366
5π/3, that's in the fourth quadrant, that's down there.1370
5π/3 is down there, not 5π/6.1376
I want to go -2 units in that direction.1381
That's up there, back in the second quadrant.1384
That's really our reference angle.1389
We want to go 2 units in that direction.1391
Where is that direction?1394
That is 2π/3.1395
Our final answer for the standard polar coordinates there is (2,2π/3).1401
The way we got that just to remind you is we took 11π/3.1413
That was really big so I subtracted 2π right away, I got 5π/3.1415
Then I graphed 5π/3, but because I had to go -2 units in that direction, I subtracted π off from 5π/3, that's how I got the 2π/3.1421
Now, my x and y, rcos(θ) and rsin(θ).1436
2cos(2π/3), cos(2π/3)=-1/2, because the x-coordinate is negative, 2×-1/2.1445
y is 2sin(2π/3) which is 2 times positive root 3/2, because the y-coordinate's positive in the third quadrant.1458
x and y together are, 2×-1/2=-1, 2×root 3/2=root 3.1470
I've got x and y for that third point.1487
Finally, the fourth point here is (3,-2π/3).1489
Let me graph that one.1495
-2π/3, we're going down in the opposite direction from the x-axis, that's 2π/3.1497
If we make that something in the positive direction, that's 4π/3 in the positive direction.1507
If you don't like doing that graphically, just add 2π to -2π/3, and you'll get 4π/3.1516
This is the same as (3,4π/3).1526
3 is already positive, so we don't have to do anything else clever there.1529
That's our standard polar coordinate form.1533
The x and y, x=3cos(4π/3).1537
Cosine is negative there, -1/2, so 3×-1/2.1548
The sine, the y-coordinate, is also negative there.1557
3×sin(4π/3), common value, I remember that one, is 3 times negative root 3 over 2.1558
The x and y collectively are -3/2 and negative 3 root 3/2.1568
This one was a little bit tricky because we've been given r's and θ's that are not in the standard ranges.1587
Some of the r's were less than 0, and some of the θ's were either less than 0, or bigger than 2π.1594
The trick on all four of these points, is first of all, graph the point.1602
Once you graph the point, if it's not in the standard range of 0 to 2π, add or subtract multiples of 2π until you get it into that standard range.1607
That's what we did at first, we added or subtracted multiples of 2π to each of those to get it into the standard range.1615
After we did that, we looked at the r's.1625
If the r's were negative, then we had to go in the opposite direction from the direction we expected.1628
That's what happened in this second and third problem with -r's.1635
We had to go in the opposite direction from what we expected.1643
What we did was we added or subtracted π to get the correct reference angle, and to get a positive value of r.1648
That's how we got the r's and θ's into the standard range then.1658
After that, we used x=rcos(θ), y=rsin(θ), in each case to give the x and y the rectangular coordinates of the point.1662
That's how we did those conversions.1677
We'll try some more examples.1679
This one we have to graph the polar equation r=2sin(θ), then we're going to check our answers by converting the equation to rectangular coordinates and solving it algebraically.1681
What I'm going to do here is make a little chart of all my reference angles.1693
I'm going to make a list of all the possible values of θ that I can easily figure out 2sin(θ).1703
Then I'll fill in all my 2sin(θ)'s.1708
I'll try to graph them and see what happens.1713
First of all, let me remind you which angles we know the common values for.1717
I'm going to draw a unit circle here.1720
The angles that I know, first, the easy ones, 0, π/2, π, 3π/2, and then 2π.1727
I also know all the multiples of π/4 and I also know all the multiples of π/3.1740
There's π/6, π/3, 2π/3, 5π/6, 7π/6, 4π/3, 5π/3 and 11π/6.1747
I know every multiple of π/6 and π/4 between 0 and 2π.1764
I'm going to make a chart showing all of these and then we'll try to graph those and see what it turns out to be.1767
2sin(θ), first one is 0, sin(0)=0, that's just 0.1777
π/6, 30-degree angle, the sin is 1/2, and 2 times that is 1.1787
Next one is π/4, 45-degree angle.1799
The sine of square root of 2 over 2, this stuff I have memorized, that's square root of 2, which for future reference is about 1.4.1802
π/3 is the next one.1814
The sine of that is square root of 3 over 2, 2sin(θ) square root of 3, which is about 1.7.1817
Next one is π/2, sine of that is 1, 2 sine of that is 2.1827
Next one is 2π/3, sine of the square root of 3 again, that's 1.7.1835
3π/4, sine is root 2 over 2, twice that is root 2, 1.4.1845
5π/6, sine of that is 1/2, 2 sine of that is 1.1857
Finally, π, sine of that is just 0.1866
Let me go through and figure out the values on the southern half of the unit circle.1871
Then we'll get to put all these together and see what kind of graph we get.1880
Below the unit circle, the first value after π is 7π/6.1887
The sine of that is negative now, it's -1/2, 2 sine of that is -1.1894
Then we get to 5π/4, negative root 2 over 2.1898
We just multiply by 2 and get negative root 2, which is -1.4.1906
Then we get 4π/3, sine of that is negative root 3 over 2, we get -1.7.1913
3π/2, sine of that is -1, because we're down here at the bottom of the unit circle.1927
This is -2.1936
5π/3, the sine of that is negative root 3 over 2, so it's negative root 3 because we're multiplying by 2, about 1.7.1939
7π/4, sine of that is negative root 2 over 2, so I multiply it by 2, you get 1.4.1954
I forgot my negative sign up above.1965
Finally, 11π/6, sine of that is -1/2.1968
All of these are values that you should have memorized to be able to figure out very quickly.1975
2 times the sine of that is -1.1980
Finally, 2π, we're back to 0 again, sin(0).1981
We've made this big chart, now I want to try and graph this thing.1986
Those are my values of multiples of 0, π/2, π, 3π/2, and 2π.2006
Now I'm filling in π/4, 3π/4, 5π/4, and 7π/4.2019
This one is π/6 and 7π/6.2034
That is π/3 and 4π/3.2050
That's 2π/3 and 5π/3.2067
Finally, 5π/6 and 11π/6.2081
I want to graph each one of this radii on the spokes of this wheel.2086
Starting at 0 ...2094
I'm at 0, I'll just put a big ...2096
I think I better do this in another color or it's going to get obscured.2101
I'll do my graph in red.2103
I'll put a big 0 on the 0 axis there.2110
At π/6, I'm 1 unit out.2114
Let's say, that's about 1 unit at π/6.2116
At π/4, I'm 1.4 units out.2122
It's a little bit farther out, about there.2126
At π/3, I'm 1.7 units out, a little farther out.2130
At π/2, I'm 2 units out in the π/2 direction.2138
2π/3 is 1.7.2148
1.7 in the 2π/3 direction.2155
1.4 in the 3π/4 direction.2157
1 in the 5π/6 direction, it's shrinking back down.2163
When we get to π, it's 0.2169
If I just connect up what I've got so far, I've got something that looks fairly circular.2173
It's a little hard to tell, my graph isn't perfect.2190
It's also kind of an optical illusion here because of the spokes on the wheel here, but it looks kind of circular.2197
Remember that that was two units out.2201
This is one unit out.2205
I want to see what happens when I fill in the values from π to 2π.2207
I'm going to fill those in in blue.2211
It's 7π/6, I'm in the -1 direction.2215
I look at the 7π/6 spoke, but then I go -1 in that direction.2220
That means I go in the opposite direction for 1 unit.2225
I actually end up back in the opposite direction, I end up back here.2228
At 5π/4, that's down here, but I go -1.4 in that direction which puts me back there.2237
At 4π/3, I go -1.7 which ends me back there.2247
3π/2, -2 puts me up there.2254
I start out in the 3π/2 direction but I go -2 units, that takes me back up here.2259
5π/3 puts me at -1.7.2267
7π/4 puts me at -1.4.2272
11π/6 puts me at -1.2277
At 2π, I'm back at 0 again, and it starts to repeat.2280
If you connect up those dots, what you see is another copy of the same graph.2286
I'm drawing them slightly separated so that you can see both of them, but it really is a graph just retracing itself again.2295
You might have thought that you might get some action down here in the south side of the coordinate plane, but you don't.2304
You just get a graph that traces itself over the top side of the coordinate plane, and then retraces itself even when the angles are down on the south side of the coordinate plane.2315
Let me recap what happened there.2327
We started out with an equation r=2sin(θ).2331
First thing I did was I made a big chart of all the θ's that I know the common values of.2334
I made a big chart of all my possible θ's.2340
Then I filled in what 2sin(θ) is for each one.2343
I drew my axis and I drew my spokes with all the angles listed here.2347
I plotted the points for each one of those spokes.2352
I went out the correct direction for each one of those spokes.2358
Except that when I got down to these angles below the horizontal axis, all my signs were negative, which means I'm going in the opposite direction, which actually landed me up on the top side of the axis.2363
That's why you have this blue graph that retraces the original red graph.2378
That's how we did that one.2386
That's really all we need to do to graph that equation, but the problem also asked us to check our answers by converting the equation to rectangular coordinates and solving it algebraically.2387
Let's just remember here what this graph look like because I'm going to have to go to a new slide to check it.2400
We have what looks like a circle sitting on top of the x-axis.2407
It peaks at y=2, and it looks like its center is y=1.2412
In terms of x and y coordinates, it looks like we have that circle peaking at y=2 and centered at y=1.2419
We're going to check that by working with the equation algebraically.2426
Now we're going to work with this equation algebraically, r=2sin(θ).2434
What I'm going to do there is I'm going to multiply both sides by r.2439
The reason I'm going to do that is because that will give me r^{2}=2rsin(θ).2444
I recognized because I remember my transformations of coordinates from rectangular to polar.2452
I remember that x=rsin(θ), and y=rcos(θ).2460
I also remember that r is equal to the square root of x^{2}+y^{2}, r^{2}=x^{2}+y^{2}.2468
With this nice new version of the equation r^{2}=2rsin(θ), I can write that as x^{2}+y^{2}.2480
rsin(θ), that's exactly my y.2490
I've converted that equation into rectangular coordinates.2494
I'm going to solve that algebraically and see if the graph checks out to what I found on the previous slide.2496
What I'm going to do is I'm going to write this as x^{2}+y^{2}-2y=0.2503
I like to simplify that y^{2}-2y.2511
I'm going to complete the square there.2515
We're going to complete the square.2521
That's an old algebraic technique, hopefully, you'll learn about that in the algebra lectures on educator.com.2529
This is y^{2}-2^{2}.2535
You take the middle number and you divide it in half and square it.2538
Minus 2 divide that by 2, and you square it.2545
That's -1^{2}=1.2550
Add that to both sides.2555
The point of that is that now we have a perfect square.2558
We have y^{2}-2y+1, that's (y-1)^{2}=1.2562
This is actually an equation I recognize.2571
Remember that if you have x^{2}+y^{2} equals a number, let's say 1, that's a circle of radius 1 centered at the origin.2575
That's the unit circle.2601
This y-1, what that does is, it just takes the original graph of the unit circle and raises it up in the y direction by 1 unit.2604
This is a circle of radius 1, that's because of that 1 right there, centered at x=0, that's because we think of that as (x-0)^{2}, and y=1, that's because of the y-1.2614
If we graph that, there's 1, there's 2, I'm going to graph a circle of radius 1 centered at the point (1,0).2644
There's (1,0).2661
There's my circle of radius 1 centered at the point (0,1).2670
If you flip back to the previous slide, you'll see that that is exactly the same graph that we got on the previous slide.2680
We got it by using polar coordinates and a lot of trigonometry, but the graph that we drew actually really resembled a circle of radius 1 centered at (0,1).2687
Let's recap what we did there.2698
We started by graphing this thing in polar coordinates.2700
I made a big chart of θ's and 2sin(θ)'s, then I graph the r's, their radius, at each one of those values including when the radius turn out to be negative.2704
I ended up graphing this.2715
I connected them up into a circle.2718
That's one way to solve that problem.2720
The second way to check our answer was to start with the equation, I see r=2sin(θ).2722
Remembering my conversion formulas x=rcos(θ) and y=rsin(θ).2730
That's very important not to mix up.2745
I multiplied both sides of the equation by r.2748
On the left, we get r^{2}, on the right, we get 2rsin(θ).2755
The point of that is that the rsin(θ) converts into y, and the x^{2}+y^{2} is what you get from r^{2}.2760
That of course comes from this, r^{2}=x^{2}+y^{2}.2770
Then it was a matter of doing some algebra to realize that that's the equation of a circle that involve completing a square.2775
We figured out that it was equation of a circle, that we can rid of the radius of the circle is 1, the coordinates of the center of the circle are (0,1), so we can graph that equation algebraically.2783
It checks out with what we got from the polar coordinates.2795
We'll try some more examples later.2798
Try them out yourself and then we'll work them out together.2800
1 answer
Last reply by: Dr. William Murray
Fri Oct 19, 2012 3:32 PM
Post by joey cal on October 17, 2012
Mr. Murray is Awesome. He simplifies what seems to appear as so complex thank you sir.
1 answer
Last reply by: Dr. William Murray
Thu May 30, 2013 3:53 PM
Post by Garth Anderson on July 9, 2011
In Example 1, Problem 2 (-4,-3):
Theta should be arctan of -3/-4 plus pi...not -4/-3 plus pi. Mr. Murray put x over y by accident.
So, Theta is 3.79 radians, not 4.07 radians.
And I agree...Mr. Murray is an excellent Trig teacher!
2 answers
Last reply by: Dr. William Murray
Thu May 30, 2013 3:49 PM
Post by varsha sharma on June 7, 2011
Instead of 5pi/4, it has to be 7pi/4.
Thank you.
Varsha
1 answer
Last reply by: Dr. William Murray
Thu May 30, 2013 3:49 PM
Post by varsha sharma on June 7, 2011
you are a terrific instructor !
Nobody ever explained me the concepts of trigonometry the way you do. I have made used of all of your lectures completely and want to thank you for it. I cannot think of being successful without these lectures and notes.
Varsha