WEBVTT mathematics/calculus-ii/murray
00:00:00.000 --> 00:00:06.000
Hi, this is educator.com and we are here to talk about polar coordinates.
00:00:06.000 --> 00:00:14.000
The idea of polar coordinates is that we are not going to keep track of things in terms of x's and y's anymore.
00:00:14.000 --> 00:00:25.000
Instead, we are going to keep track of points in terms of the radius r and the angle θ.
00:00:25.000 --> 00:00:30.000
Every point now will have coordinates in terms of r and θ.
00:00:30.000 --> 00:00:37.000
We will talk about functions r = f(θ).
00:00:37.000 --> 00:00:42.000
There are sort of two places that calculus comes in in polar coordinates.
00:00:42.000 --> 00:00:54.000
If you have a function, here I am talking about plugging in different values of θ and getting different values of r as your output.
00:00:54.000 --> 00:01:00.000
There are two things you might be interested in calculating.
00:01:00.000 --> 00:01:10.000
One is the area, before we calculate under, here it makes more sense to talk about calculating the area inside a curve.
00:01:10.000 --> 00:01:25.000
The equation we have for that is the integral from, these are values of θ, θ = a to θ = b.
00:01:25.000 --> 00:01:30.000
Of f(θ²/2).
00:01:30.000 --> 00:01:38.000
If you are given f(θ), this f(θ) is just whatever the r is, you square that and divide by 2.
00:01:38.000 --> 00:01:43.000
Then you take the integral with respect to θ.
00:01:43.000 --> 00:01:51.000
The second thing we are interested in calculating is the arc length.
00:01:51.000 --> 00:01:54.000
What is the length of that curve?
00:01:54.000 --> 00:02:02.000
The way you figure that out is you find f(θ²), f'(θ²).
00:02:02.000 --> 00:02:07.000
Then you use the Pythagorean theorem, you add them, take their square root,
00:02:07.000 --> 00:02:15.000
And integrate from θ = a to θ = b.
00:02:15.000 --> 00:02:18.000
Let us try those out using some examples.
00:02:18.000 --> 00:02:25.000
The first example is the area inside the graph of r = θ, from θ going to 0 to π/2.
00:02:25.000 --> 00:02:40.000
If r = θ, then when θ = 0, r is just 0 but as θ increase to π/2, r gradually increases.
00:02:40.000 --> 00:02:45.000
We are trying to find that area there.
00:02:45.000 --> 00:02:50.000
Let us work it out.
00:02:50.000 --> 00:03:00.000
Our formula for the area is f(θ²/2) dθ
00:03:00.000 --> 00:03:12.000
In this case, our f(θ) is just θ itself, so this is the integral of θ = 0 to π/2.
00:03:12.000 --> 00:03:14.000
Of θ/2 dθ.
00:03:14.000 --> 00:03:16.000
The integral of θ² is θ³/3
00:03:16.000 --> 00:03:31.000
This whole thing is θ³/3/2, so over 6, evaluated from θ = 0 to θ = π/2.
00:03:31.000 --> 00:03:40.000
That is just π/2³/6.
00:03:40.000 --> 00:03:46.000
Now, 2³ is 8 and 8 × 6 is 48 so our answer is π³/48.
00:03:46.000 --> 00:03:55.000
That represents that area inside that curve.
00:03:55.000 --> 00:04:02.000
To recap here, what we did was look at the function we were given,
00:04:02.000 --> 00:04:09.000
R = f(θ), and then we just plugged it into this integral formula f(θ²)/2.
00:04:09.000 --> 00:04:12.000
Then we worked out the integral.
00:04:12.000 --> 00:04:15.000
Let us try that out with a slightly harder example.
00:04:15.000 --> 00:04:20.000
We want to find the area inside one loop of the graph of r = cos(2θ).
00:04:20.000 --> 00:04:26.000
Perhaps the first thing that makes this example difficult is we have not been told what the boundaries of θ are.
00:04:26.000 --> 00:04:33.000
We really need to look at a graph to figure this out.
00:04:33.000 --> 00:04:36.000
What does a graph of cos(2θ) look like.
00:04:36.000 --> 00:04:50.000
Well if we graph cos(x) to warm up, it starts at 1, it goes down to -1, then it comes back to 1 at 2π.
00:04:50.000 --> 00:05:05.000
If we graph y = cos(2x), we get a graph with the same basic shape, but it oscillates twice as fast.
00:05:05.000 --> 00:05:20.000
If that is π, and that is 2π, it does a complete period in the space of π, and then another complete period by 2π.
00:05:20.000 --> 00:05:30.000
If we graph r = cos(2θ), we take that graph and we wrap it around a circle.
00:05:30.000 --> 00:05:41.000
In the sense that when θ = 0, we start out at 1 at radius 1.
00:05:41.000 --> 00:05:50.000
By the time θ gets to be π/4, it has gone to radius 0.
00:05:50.000 --> 00:05:53.000
This goes down to radius 0.
00:05:53.000 --> 00:06:05.000
At π/2, r = -1 so you know that if π/2 is up here, the radius has gone down to -1.
00:06:05.000 --> 00:06:08.000
It comes back to 0.
00:06:08.000 --> 00:06:10.000
Now we are graphing this part of the graph.
00:06:10.000 --> 00:06:19.000
By the time we get up to π it has come up to 1 again, so here is π in this direction.
00:06:19.000 --> 00:06:27.000
Then between π and 5π/4 it goes back down to 0.
00:06:27.000 --> 00:06:40.000
That is graphing that part of the graph and if we graph this part of the graph going from π/4 to 3π/2,
00:06:40.000 --> 00:06:52.000
It is negative so the graph ends up here and then it spirals back down to 0.
00:06:52.000 --> 00:06:57.000
And back to 1 again when it comes back to 2π.
00:06:57.000 --> 00:07:05.000
We get this interesting 4-leaf clover and we are trying to find the area inside one of those loops.
00:07:05.000 --> 00:07:12.000
A good way to do it might be to find that area inside there, inside half of one of those loops.
00:07:12.000 --> 00:07:21.000
That is between θ = 0 and θ = π/4.
00:07:21.000 --> 00:07:25.000
Then we will multiply our answer by π/2.
00:07:25.000 --> 00:07:34.000
Our area is 2 × the integral from 0 to π/4.
00:07:34.000 --> 00:07:48.000
Remember f(θ²)/2 so our f(θ) = cos(2θ) this is cos²(2θ)/2 dθ
00:07:48.000 --> 00:07:55.000
Our 2's will cancel, that is convenient, so we get the integral from 0 to π/4.
00:07:55.000 --> 00:07:58.000
Remember how to integrate cos² of something.
00:07:58.000 --> 00:08:04.000
You write that as 1+cos(2 × that thing), but the thing is 2θ, so this is actually 1 + cos(4θ).
00:08:04.000 --> 00:08:12.000
All of that over 2.
00:08:12.000 --> 00:08:17.000
So this 2 over here was not the 2 above, that cancelled, but this came from the half angle formula.
00:08:17.000 --> 00:08:20.000
Now we integrate this with respect to θ.
00:08:20.000 --> 00:08:25.000
I am going to pull the 1/2 outside now, we get 1/2.
00:08:25.000 --> 00:08:29.000
Now we have to integrate 1 + cos(4θ).
00:08:29.000 --> 00:08:45.000
The integral of 1 is just θ + the integral of cos(4θ) is just sin(4θ)/4.
00:08:45.000 --> 00:09:00.000
Then we evaluate that from θ = 0 to θ = π/4.
00:09:00.000 --> 00:09:05.000
Then we get 1/2 of π/4.
00:09:05.000 --> 00:09:13.000
Plugging in θ = π/4 + 1/4 sin(4θ).
00:09:13.000 --> 00:09:17.000
Sin(4θ) is sin(π) which is just 0.
00:09:17.000 --> 00:09:26.000
- plug in θ = 0, we get 0 - sin(4×0) is just 0 again.
00:09:26.000 --> 00:09:30.000
Our final answer is just π/8.
00:09:30.000 --> 00:09:40.000
The tricky part there is that we were not given the boundaries of integration.
00:09:40.000 --> 00:09:44.000
We really had to look at the graph of r = cos(2θ)
00:09:44.000 --> 00:09:50.000
Then interpret from the graph what useful boundaries of integration we could use to find the area.
00:09:50.000 --> 00:10:00.000
Once we found the boundaries of integration, we just plugged it into the formula f(θ²)/2.
00:10:00.000 --> 00:10:02.000
Then we worked out the integral and it was not too bad.
00:10:02.000 --> 00:10:10.000
We will try some more examples later, this is educator.com.
00:10:10.000 --> 00:10:18.000
Another example is the length of the polar curve, r = e^5θ as θ goes from 0 to 2π.
00:10:18.000 --> 00:10:23.000
Fortunately we have been given the limits of integration.
00:10:23.000 --> 00:10:29.000
We are going to set up our arc length formula.
00:10:29.000 --> 00:10:40.000
Which remember is f'(θ²) + f(θ)² dθ.
00:10:40.000 --> 00:10:50.000
Our f'(θ) is well, r = e^5θ.
00:10:50.000 --> 00:11:05.000
f'(θ) would be (5e^5θ)² + just f(θ²) so that is (e^5θ)².
00:11:05.000 --> 00:11:10.000
We want to square root that and integrate it.
00:11:10.000 --> 00:11:25.000
5 squared is 25, e^5θ)² is e^10θ.
00:11:25.000 --> 00:11:30.000
Square root that and integrate it.
00:11:30.000 --> 00:11:44.000
This is sqrt(26e^10θ).
00:11:44.000 --> 00:11:50.000
I am going to pull the sqrt of 26 all the way out of the integral.
00:11:50.000 --> 00:11:58.000
Then we have the sqrt(e^10θ) which is just e^5θ.
00:11:58.000 --> 00:12:08.000
We want to integrate this from θ = 0 to 2π.
00:12:08.000 --> 00:12:22.000
Now the integral of e^5θ is just e^5θ/5
00:12:22.000 --> 00:12:30.000
We want to evaluate that from θ = 0 to 2π.
00:12:30.000 --> 00:12:35.000
I can write this as sqrt(26/5) ×
00:12:35.000 --> 00:12:43.000
If you plug 2π into e^5θ you get e^10π.
00:12:43.000 --> 00:12:50.000
If you plug 0 into e^5θ you get e⁰, which is just 1.
00:12:50.000 --> 00:12:56.000
That is our answer for the arc length.
00:12:56.000 --> 00:13:04.000
The key to that problem is recognizing that it is a length problem and then going to the arc length formula.
00:13:04.000 --> 00:13:12.000
Which is this Pythagorean formula f'(θ)² + f(θ)² square of that and then integrate it.
00:13:12.000 --> 00:30:59.000
We use the f(θ) that we are given, work it through, and then plug in the limits that we are given.