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Lecture Comments (9)

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

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
    =


     

    x2 + y2
     
    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

A horizontal force of 60N is applied to a box on a 45° ramp. Find the components of the force parallel to the ramp and perpendicular to the ramp.
  • Use SOHCAHTOA to solve this problem
  • cos45° = [x/60], sin45° = [y/60]
  • x = 60cos45°, y = 60sin45°
Parallel force is 30√2 N; Perpendicular force is 30√2 N
A ball is thrown with an initial velocity of 70 feet per second, at an angle of 50° with the horizontal. Find the vertical and horizontal components of the velocity.
  • Horizontal component of velocity: x = rcosθ
  • x = 70cos50°
  • Vertical component of velocity: y = rsinθ
  • y = 70sin50°
x = 45.0 ft/sec, y = 53.6 ft/sec
A ball is thrown with an initial velocity of 50 feet per second, at an angle of 70° with the horizontal. Find the vertical and horizontal components of the velocity.
  • Horizontal component of velocity: x = rcosθ
  • x = 50cos70°
  • Vertical component of velocity: y = rsinθ
  • y = 50sin70°
x = 17.1 ft/sec, y = 47.0 ft/sec
A heavy object is pulled 35 feet across the floor, using a force of 70 pounds, find the work done if the direction of the force is 45° above the horizontal. (Use the work formula W = FD where F is the component of the force in the direction of motion; D is the distance)
  • F = 70cos45°, D = 35
  • W = (70cos45°)(35)
W ≈ 1732.4 ft/lb
A heavy object is pulled 10 feet across the floor, using a force of 65 pounds, find the work done if the direction of the force is 30° above the horizontal. (Use the work formula W = FD where F is the component of the force in the direction of motion; D is the distance)
  • F = 65cos30°, D = 10
  • W = (65cos30°)(10)
W ≈ 562.9 ft/lb
Convert the following point in polar coordinates to standard form (with r ≥ 0 and 0 ≤ θ ≤ 2π), and then convert them to rectangular coordinates: (2, − [(4π)/3])
  • 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
Rectangular Coordinates:( − 1, √3 )
Convert the following point in polar coordinates to standard form (with r ≥ 0 and 0 ≤ θ ≤ 2π), and then convert them to rectangular coordinates: ( − 4, − [(π)/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
Rectangular Coordinates:( − 2, 2√3 )
Convert the following point in polar coordinates to standard form (with r ≥ 0 and 0 ≤ θ ≤ 2π), and then convert them to rectangular coordinates: ( − 2, [(3π)/4])
  • 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
Rectangular Coordinates: (√2 , − √2 )
Convert the following point in polar coordinates to standard form (with r ≥ 0 and 0 ≤ θ ≤ 2π), and then convert them to rectangular coordinates: ( − 1, − [(3π)/4])
  • 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]
Rectangular Coordinates: ([(√2 )/2], [(√2 )/2])
Graph the polar equation: r = 3sin2θ
  • Create a table of values
  • Plot those values on a polar plane
Graph the polar equation: r = 3cos2θ
  • 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 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

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

I’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 I’m going to add another 2pi to that and that gives me 11pi/6, that is good.0039

I’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 –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 –root 3 and 1.0149

That is the first point, the second one here 6 (–pi/3), let me graph that one.0166

I think I’m going to add 2pi to that right away to get it in the proper range, add 2 pi to that, 2pi – 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 ½.0215

(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 –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 I’m going to add 2pi to that right away and get 3pi/4.0371

I’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

I’m going to try and graph that, I’m going to start by graphing this using x and y.0007

I’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

I’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 I’m going to use this to make a chart of values for (theta) and then finally I’m going to make a polar graph.0129

Let me make a chart of values for (theta), I’m not going to fill in every (theta) I know.0134

I’m just going to fill in probably the multiples of pi/4.0141

Each one I’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

I’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, I’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 I’m going to fill in my dots and I’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 I’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 I’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 I’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, I’m going to use that as a graph to fill in all the (theta) I’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 x2+y2.0149

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=x2+y2.0160

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 x2+y2.0465

Let me remind you of those formulas up here because we're going to use them quite a bit, x2+y2.0472

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 42+32.0628

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 22+52, 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 x2+y2.0995

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 r2=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 x2+y2, r2=x2+y2.2468

With this nice new version of the equation r2=2rsin(θ), I can write that as x2+y2.2480

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 x2+y2-2y=0.2503

I like to simplify that y2-2y.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 y2-22.2535

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 -12=1.2550

Add that to both sides.2555

The point of that is that now we have a perfect square.2558

We have y2-2y+1, that's (y-1)2=1.2562

This is actually an equation I recognize.2571

Remember that if you have x2+y2 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 r2, on the right, we get 2rsin(θ).2755

The point of that is that the rsin(θ) converts into y, and the x2+y2 is what you get from r2.2760

That of course comes from this, r2=x2+y2.2770

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