## Sketching Polar Curves – 2 Examples (KristaKingMath)

Today we’re going to be talking about how

to sketch polar curves and in this particular video we’re going to be talking about two

different polar equations. The first one which we’ll call equation number one is r equals

seven times sine of two theta and the second one is r equals six plus four sine theta.

And, the method that I use for sketching polar curves is really common but it works really

well. The first thing that we’re going to do whenever

we’re asked to sketch a polar curve is we’re going to take whatever is inside our trigonometric

identity and we’re going to set that equal to pie over two. So in the case of equation

number one, we’re going to set two theta equal to pie over two and then we’re going

to solve for theta and we’ll get theta equals pie over four when we divide both sides by

two. For equation number two, we’ll take, again, whatever is inside the trigonometric

identity, in this case, theta, and we’ll set that equal to pie over two, of course,

we’ve already solved for theta so we don’t have a second step. The reason we do that

is because we’re going to be marking off our x axis when we pot… plot this on a Cartesian

coordinate system, we’re going to be marking off our x axis in increments of whatever we

got here for theta. So, in the case of equation number one, we’re going to be marking off

the x axis in increments of pie over four. So, we’ll get pie over four here, then we’ll

get pie over two, we’ll get three pie over four, and then we’ll get pie, and then we

would get five pie over four, right, and we could keep going, we mark off the x axis in

the increments of pie over four. For equation number two, we’re going to mark off the

x axis in increments of pie over two. So, we’ll have pie over two, then we’ll have

pie, then three pie over two, and then two pie, right, and we could keep going, we would

get five pie over two, etc. So we’re just going to mark off x axis on those increments,

that’s the only reason we do it and it… it ends up being convenient for us because

when we try to evaluate each of our polar equations at the points that we’ve marked

off here on the x axis we’re going to be able to plug those in and use the unit circle

which I’ve drawn a basic unit circle out here. But, all the points that we’re going

to plug in are the four points here on our unit circle that I’ve… that I’ve identified,

the point here, (one, zero), (zero, one), (negative one, zero), and (zero, negative

one). And, those points are obviously really easy to evaluate as oppose to something out

here that’s like an angle pie over four and the coordinate is square root of three

over two or something like that. So, seeking to these four corners here on the unit circle

makes things a lot easier which is why we do it this way.

So now that we’ve marked off our x axis and increments of theta that we solved for

down here, we’re going to go ahead and plug each of those points into our polar equation

and just start plotting on the Cartesian coordinate system or the x-y coordinate graph. We’re

going to plot them here on the x-y coordinate plane and then we’re going to translate

that to the polar coordinate system. So, let’s go ahead here and take equation

number one. So we’ve got r equals seven sine of two theta. And now we’re going to

go ahead and figure out the value of the graph at each of the points we’ve marked off here

on the x axis. So we’re going to start with zero, we’re going to start with the point

here, zero, and then we’re going to find pie over four, pie over two, etc. So when

we plug in zero for theta, right, we’ll be plugging in zero at or, sorry, we’ll

be plugging in for theta and solving for r, so when we plug in zero, we’ll get seven

sine of two times zero which will give us seven sine of zero. Well, sine of zero, remember,

when we go back to our unit circle, all of our coordinates on the unit circle, you can

consider the x coordinate as cosine and the y coordinate as the sine, so when we say sine

of zero we know that we’re taking the y coordinate, sine, of the angle zero which

we know to be here. So, we want this y coordinate right here which is zero. So we get seven

times zero which of course is just equal to zero so that tells us that our point here

is at zero. Now let’s go ahead and look at pie over four, our next increment, right.

We’ll get r equals seven sine of two times pie over four, well, we know that two times

pie over four is pie over two, and now, that takes us to this point here on the unit circle.

Pie over two, we’re looking at sine again so we want the y coordinate there, which we

know is one, so we get seven times one which of course is just equal to seven. So then,

here at pie over four, let’s go ahead and call this seven here, at pie over four we’re

going to get the point seven. If we went ahead and plugged in pie over two, let’s just

give ourselves some more space, if we’d plugged in pie over two, we’d get, oops…

r equals seven sine of two times pie over two, our two’s will cancel and we’ll be

left with seven sine of pie. Well, sine of pie, here we are at the angle pie and we know

sine to be the y coordinate, so that’s the y coordinate there, which is zero, seven times

zero gives us zero. So, again, coming back over to our graph here, our y value is then

zero. Let’s go ahead and plot just one more point to see what this is going to look like.

If we plot three pie over four, we’ll get seven sine of two times three pie over four

which is going to be seven times sine, we’ll get three pie over two. And, three pie over

two, here’s our angle, three pie over two, our sine coordinate there or y coordinate

there is negative one, so we’ll get seven times negative one which is negative seven.

So we can see here then that our graph is going to be down here at negative seven. So

what’s this tells us if we kept going is that our graph would look something like this,

it would come up this way, it would come down in this parabola like shape, it would come

down here and meet negative seven, it would come back up here to meet pie, and then here

at five pie over four, we’d be back up here at positive seven, and that would just repeat

itself, it would come up and down and up and down and repeat itself in that way.

So, now that we have the graph of the function on an x-y coordinate plane, we want to go

ahead and translate this to the polar plane, and let’s actually do ourselves a favor

here and mark off these different sections in different colors. So this, right here,

will be orange, this first section, the second section here will be green, and we will use

this color coding to help us translate, we’ll get this awkward teal color here, and then

blue, and then purple. So hopefully this will help us make a translation here. So now, we

want to take each of these points and plot them on a polar coordinate system as well

and it makes it kind of easy now that we’ve plotted it on x-y coordinate system.

So what we’re going to do is we’re going to start with this point here, well, that’s

at the origin, (zero, zero). So when that point is at the origin, the polar coordinate

system is the same thing. We have this point here at (zero, zero), the origin, so that’s

our starting point. Then, as we move between zero here and pie over four, we’re going

to move up here to seven. So if you think about it on a polar coordinate system, pie

over four and all of these… these increments here on the x axis are angles in the polar

coordinate system, so the angle pie over four is right here, right, this is the angle zero,

and as I move up toward this direction, this right here is the angle pie over four. So

as I move towards pie over four from zero, I come out a distance of seven. So what I

end up doing here is I wrap around up here and, as I approach pie over four, I come out

a distance of seven from the origin, right, so we can call this line here seven, this

is a distance of seven from the origin. So that’s how we plot that and that’s going

to be the orange section of our graph, so we’ll go ahead and… we’ll go ahead and

draw that. Then, you notice, the second section here,

as we come… as we come… let me grab a different color… as we come back from pie

over four down to pie over two. Well, the angle pie over two on our polar system here

is here, pie over two, and we can actually mark this off here. This will be pie, this

will be three pie over two, just like the unit circle, this comes back to two pie, and

then, you know, this line here would be… pie over four… this would be three pie over

four, so these are our angles, right, this will help us a little bit. So now, this green

section of the graph, right, we’re here at this point, here, and as we come back toward

the angle pie over two, so as we move from here to here, toward pie over two, notice

that our x-y coordinate system graph comes back toward zero, right, we head back toward

the x axis or zero, which means that our polar graph is going to curve back in here toward

the origin zero, right, so we curve back toward zero.

Now, notice that we are at this point crossing the x axis, right, we were above the x axis

for this whole range here and now at this point, here at pie over four, or sorry, pie

over two, we’re going to dip below the x axis. Well, what that means is that we’re

going to cross the graph, right; we started here at this point, we’re going to cross

over the graph. And as we move from the angle pie over two to the angle three pie over four,

we know this here is the angle three pie over four, on the angle three pie over four, we’re

going to be at negative seven. So, if we we’re at positive seven, we would be out in this

direction, positive seven. But, because we’re here at negative seven, right here, we need

to move out this way, the opposite way, a distance of seven. So, this section of the

graph would be this teal color, as we move toward three pie over four, we’re going

to be out negative seven which will be something like this, right, and we’ll end up at this

point here. Then, for the blue section of the graph, as

we come back here, we come from negative seven back to zero as we approach the angle pie,

well, here’s the angle pie, we’re coming in this direction so we come back toward zero

here, and we wrap back in, we get to the origin. If we kept going with purple, right, here

at five pie over four, the angle five pie over four is right here, and notice that we

are at positive seven here, which means we are moving in this direction, so, that means

that our point here is going to be in this direction of… of the angle five pie over

four and we come out here, our graph would continue like this. And if we kept going,

we’d wrap back in toward the origin like this, back out here, and then back in, and

we would realize that we have a four leafed… a four leafed graph here on the… on the

polar graph. So, contrast that with the equation r equals

six plus four sine theta. If we plug in each of these points, right, we start with zero,

and we plug that in to our equation, and we’ll move through this one a little bit more quickly,

but if we plug in zero, we get r equals six plus four sine of zero. We know that sine

of zero, right, were here at the angle zero on our unit circle, the y coordinate there

is zero, so this whole thing is going to be zero, we’re just going to be left with six,

so that’s means we’re out… we’ll call this here six, so we start here at six. Then,

when we plug in pie over two, we get r equals six plus four sine of pie over two. We know

that sine of pie over two is one so we get six plus four times one, that’s going to

give us ten so we’re out here at ten, we’ll call this here ten, we’re out here at ten.

Then, if we plug in pie, we’ll get r equals six plus four sine of pie. We know that sine

of pie is this number right here, zero, so, again, that will go away and we’ll come

back to six. If we plug in three pie over two, we’ll get r equals six plus four sine

of three pie over two. We know that sine of three pie over two is this negative one number

here so we’ll get r equals six plus four times negative one, that’s going to give

us six minus four which is equal to two, so we come down here, we’ll call this two,

we come down here to two. And if we kept going, this is what we get, we’d come up here to

six from six to ten like this, then down from ten back to six, six down to two, then here

at two pie, we’re back up to six, and then here at two pie, we come back up to ten as

we approach the angle five pie over two, and we could keep going like that, so we’ll

go ahead and mark these points, so we keep going.

Well, same thing, when we translate this on to the polar curve here, at the angle zero,

right, this here is the angle zero, we are out a distance of six. So, here’s the angle

zero, we move out a distance of six, we’ll call this right here six. So this will be

first here, our orange point, so we come out a distance of six. Then, as we move toward

the angle pie over two, right, we move toward the angle pie over two and let’s just go

ahead and mark this off, we’ve got pie over two here, pie, three pie over two, and back

to two pie here, so, as we move here from the angle zero to the angle pie over two,

our graph goes from six to ten. So here on the… on the axis where at six, when we come

out here, we’re going to be at ten. So we come all the way up to ten, this ten, six,

and that will get us here to ten. Then, as we move from pie over two to pie, here, pie

over two to pie, we come down from ten back to six. So if we plot a point here, again,

we’ll call it six, we’ll be there at that distance and we’ll connect these in a second.

Then, as we move from pie to three pie over two, or pie to three pie over two, we move

from six down to positive two, so, we’ll call this here positive two in that direction,

this is two. And if we connect these, right, we’ll get orange here, which comes up to

ten, then we’ll get green that comes from ten down to six, we get teal that comes from

six to two, and this is what’s going to end up looking like, and then from two, here,

we come back to six, and then we will repeat, purple would wrap around again with orange,

and we could keep going, and it will just be this continuous loop around the graph over

and over again. So, what’s interesting in one of the…

the conclusions I want to draw from comparing these two graphs is? You notice that this

equation, when we graph it on an x-y coordinate plane, it’s above the x axis then its below

the x axis then it’s above the x axis again, and what we’d realize is that that graph

gives us this petaled graph on the polar coordinate system where we cross over the graph, right,

we get this petaled approach and the graph crosses itself. Contrast that with this graph

over here, six plus four sine theta, which is always above the x axis, right, this whole

graph is above the x axis, which means that when we translate this into a polar system

we’re… the graph is never going to cross itself, its’ going to just come in this…

in this loop here but we’re never going to end up with this petaled approach, right,

which we get when we cross above and below the x axis. So, just kind of interesting to

look at how this things graph out differently, given what they look like in an x-y coordinate

plane. So that’s it. I hope this video helped you

guys and I will see in the next one.

11:09 pm

what is 7??

11:10 pm

the x or y axis??

8:59 am

So you know some math, AND you're kinda cute? OK, let's get married… I'll hand you the papers as soon as my lawyer's done with them. I'm well-versed in relativistic theoretical physics myself, so maybe we'll make a good couple. See you soon.

1:56 am

So on r=6+4sin(θ), how do you know that for the angle between π and 3π/2 and 3π/2 to 2π the line curves like that, rather than just being a half of a parabola like the other two?

12:53 am

Awesome help!

12:40 am

😀

6:47 pm

THANKS

5:07 pm

You're welcome!

3:28 pm

Thank you!! Thank you!! This makes so much sense! I did not understand what my teacher was saying.

5:00 am

Thank you very much for this video. I reviewed the book several times and studied my notes and could not understand it. Your video cleared it all up perfectly. Please continue doing what you do, your videos are the best.

5:18 am

why the dip? why not a straight curve from 6 to 2? in the 3rd quadrant.

10:36 pm

you're really awesome !

10:37 pm

very nice. thanks

7:13 pm

you awesome

7:13 pm

i love you

9:41 am

I could listen to you teach calculus all day. You'r much prettier than the mean old man at school!

10:51 am

super explanation,better than class teacher,do upload more!!!!

12:22 pm

sexy and smart

12:51 pm

You have just saved my life… Thumbs up!!!

1:34 am

Can u comment upon how x(n)=e^i((π/2)n + (π/2)) is a power signal,

PLZ…

7:47 pm

Again, you are an awesome teacher!

6:12 pm

Wow, you are an awesome teacher indeed!! You made it so easy to understand

5:17 pm

You are such a life saver! i've tried to understand this stuff for days and in 17 mins you explained everything perfectly! Definitely subscribed! 🙂

3:33 am

will you marry me??? jk great teacher thanks for these videos….not jk…lol

5:26 am

Beautifully drawn with nice smooth curves.

3:43 am

Very helpful! I just needed to brush up on polar sketching for my Calc II final tomorrow and this was perfect. Thanks!

9:34 pm

Watching as many of these polar sketching vids as I can, precalculus honors final will probably have a ton of these problems and I never took notes on the basic patterns to do quick sketches

7:21 pm

That was very helpful…..Thanx Krista you are awesome…….

1:01 am

Very nifty way of doing this stuff. Thanks for sharing

12:23 am

best explanation!

8:27 pm

Thank you so much! this was very helpfull!!

4:54 pm

awesome video, thanks a lot 🙂

1:56 pm

Thanks. Very helpful 🙂

1:59 am

So u set = to pi/2 when is sin, do you work it the same way when is cosine?

2:28 am

Thank u profesor, but I still get very confused about finding the angles qhen plotting this type of graphs, I'd seen some people making the equation equal to zero and then they add a constant Is teally frustrating , because I get diferent values..worst is that my test is next monday.

12:12 am

gracias Profesora King!

12:22 am

quick question though, 11:47 for the purple line is positive 7 right ? should it be on top left side instead?

8:30 pm

I was bashing my head against a wall after missing 10 straight graphing polar problems in my assigned homework until I saw this just now. You're technique is amazing with setting the argument of 2Theta = pi/2, then finding it in its actual xy graph, then translating into polar. Very helpful. Subscribed!

1:35 am

great!

7:53 pm

ı need your blackboard program. how can ı find. can u write here a link.

6:24 am

Hi. Thanks a lot. I have the same question as Justin Jackson: "So on r=6+4sin(θ), how do you know that for the angle between π and 3π/2 and 3π/2 to 2π the line curves like that, rather than just being a half of a parabola like the other two?"

7:17 am

thank u very much,nice lecture

6:20 pm

very interesting. thanks

4:53 pm

how when we go from 0 to pi/2 it was towards positive 7 and also when we go opposite to it from pi to 5pi/2 it is also towards 7 not negative 7

1:32 pm

you should have used a polar graph

12:05 pm

Simple and to the point… great for revision (actually better then the lecture).. thanks..

2:04 pm

Excellent video. Much simpler explanation than my textbook!

7:34 pm

Great explanation on graphing polar equations the whole strategy of equaling theta to pi/2 is not taught in my book which is just an amazing tool to know, thank you so much your are the best !

6:18 pm

Thank you so much about that video, it is really help. by the way you scared me at the last second! lol

12:53 pm

U R AN ANGEL

4:52 pm

thanks so much! your video helps me a lot !

11:24 pm

I wish my professor took time to explain it like you did 😩

11:20 pm

You make me love math thanks again

9:48 am

U r a great teacher !!! even better than my lecturers. Excellent work and keep it up. oh and yeah, thank you so much

2:00 pm

Thank you very much for great tutorials.. 🙂

4:39 am

what board do you use

5:34 pm

This should work for non linear equations also right?? like r=sin^2(theta)+ sin(theta)

2:53 am

Grade saver

6:23 pm

I love your teaching, inspiring and colorful

11:28 am

Polar coordinates make me wonder if flowers know trigonometry.

6:15 am

damn, the bum shaped graph makes a guest appearance

1:43 am

Took me quite some time to try and figure it out, almost gave up and then stumbled on this video. Thanks and keep up the good work!

8:58 pm

omg thank you so much for this video my calculus teacher didn't teach us this and just gave use homework.

8:46 pm

Ive been watching your videos for some time now, really amazing work. everything I have a calc question I come to this channel .

10:05 pm

You definitely helped! Now to calculate areas and lengths of polar curves…

11:29 am

Beyond AMAZING. How come I've just found out about u ?!

6:48 am

This was so helpful!! Thank you so much!!

7:25 pm

You're a life saver.

9:14 pm

in second question why we didnt start to draw the line from 6 as we plug 0 in function we get 6

10:17 pm

Thank you so much! I feel a lot more comfortable with sketching polar curves now 🙂

4:54 pm

For the region from pi/2 to 3pi/4, r = -7 is in the 4th quadrant however why u choose to draw the region below the line(green line region)? why not above the line(blue line region?

10:31 am

thank you Krista

7:12 am

your advice for setting up the increment value is really handy, thx

2:10 pm

Thank you so much! You're AMAZING!!!!

5:44 am

This was very helpful. I forgot how to work with these. Thank you so much.

9:23 pm

Thank Thank Thank You so much, I don't know where I would be without the help of your videos! Your work is amazing

11:52 pm

This was 100% understandable. Appreciate it.

12:30 am

Thanks 😀

6:31 am

You are literally the reason I passed Calculus II hahaha. Well, my final is in 8 hours but I am gonna kill it because of you. So thanks!

3:50 am

Thank you!!!

6:23 am

well presented and wonderfully drawn/annotated

8:33 pm

You have the twist to teach that kind of thing thanks!

7:34 am

I've been watching your videos since I took AP calc my junior yr in high school. I'm now taking vector calc in college and here I am again. This really helped. Thank you!

6:29 pm

@Krista King How do you make these videos?

5:40 am

Well explained.

7:12 pm

if

r=cos theta/2

theta/2 =π/2

theta =π

will you please sketch these …….?

2:03 am

when you graph 2 sin(5theta) i am getting a circle. i set 5theta=pi/2. is seting pi/2 to the inside angle works only for even numbers?

12:54 am

You are great! Thank you

10:59 pm

I DID IT!!!!!

Carlton danceThanks so much Krista!3:22 pm

very clear explain and very useful.Knowing how to draw is essentially for calculating the problem about polar coordinate

11:48 pm

you killed this one… thanks so much~!!!!!

3:28 pm

Amazing! Life saver !

8:47 pm

how did you know that you need to set theta equal to pi/2 whats the reasoning behind it?

3:54 am

Wow

6:54 pm

thanks a lot!!!! this video is very helpful for me !!

2:05 pm

Thank you very much!

3:23 pm

thank you, very well taught.

7:38 am

Can i just say how flippin cute you are! Also thank you so much! I’ve been trying to work on a problem like the whole day

2:44 pm

That was amazing. Thank you, Krista!

10:42 pm

Too Great