7.3 Polar Coordinates

[STARTS AT 1:12]

Professor: So first, the point (1,1)-- this is in our xy plane. We can begin by saying r squared equals 1 squared plus 1 squared, which is 2. So that r is the square root of 2. Simple enough. So this point is going to be converted into square root of 2. 

And then we want our angle theta. Now the easiest of these three to use is going to be tangent, if it applies. We'll see a case where it doesn't. Tangent of theta is going to be y divided by x, which is 1. And then that implies that theta is pi over 4. Square root of 2 over square root of 2, 1. So theta is pi over four. So our point is square root of 2 pi over 4. 

This is actually an alternate way to view the xy plane. That's why it's-- and it's actually called polar coordinates instead of Cartesian like we deal with usually. We're looking at polar coordinates. We'll see this in just a minute, a lot clearer, hopefully. 

Next we take the point (-3,4). Well r squared equals negative 3 squared plus 4 squared, which is 9 and 16, 25, which means that r equals 5. So in our point 5 comma. And now we'll take tangent theta equals y divided by x. That will be negative 4/3, which would mean that theta is actually negative 0.927. 

Now that does not agree with (r,theta) being between 0 and 2 pi. So if we alter that by pi, if we add pi to that, (r,theta) is roughly 2.21. So our point then is (5,2.21). 

C-- we'll take (0,3). r squared equals 0 squared plus 3 squared, which is going to be 9, which makes r 3. And then we'll take our tangent theta equals 3 divided by zero. And this is where we cannot actually use our tangent. So let's take what we know about r. And go and take the equation sine theta. Sine theta equals y over r. 

So our y coordinate is 3. Our r we found is 3. Which makes theta equal to pi over 2. So our point is 3 pi over 2. 

Last one of these, r squared equals 5 square root of 3 squared plus 5 squared. That is going to add up to be 100, which makes r equal to 10. So our point is going to be 10 comma something. 

Now our tangent theta is going to be equal to negative 5 over 5 root 3. Which is negative 1 over root 3. And so theta is negative pi over 6. Now that is not between-- it's not between 0 and 2 pi like we want. So we will add pi to that. So that would be 5 pi over 6. So our point is 10, 5 pi over 6. 

Example two convert-- actually no, still example one. It's a long one. Converge to these into rectangular coordinates. All right this is going to be a whole lot quicker, because we have these equations. We know what r and theta are. And we want x and y. We can plug those into r cosine theta and r sine theta. 

So for the first, our x is going to be 3 cosine pi over 3, which is 3/2. So we have the point-- let's put it over here, 3 over 2. And then y is going to be r times sine of pi over 3. So that will be 3 root 3 over 2. So 3 root 3 over 2. And there are rectangular coordinate from our polar. 

Same idea here. x is r times cosine 3 pi over 2. And y will be 2 sine of 3pi over 2. You were likely told at some point that sine is a y value. That's the way I like to think of it. And our cosine is our x value. That actually corresponds to this way of thinking. So that means x is 0, because our cosine is zero there. And our y at 3 pi over 2, that is negative 1. 

So our point is 0, negative 2. 

G-- x equals 6 cosine of negative 5 pi over 6. And y is 6 times sine of negative 5 pi over 6. So for the first, x is going to be negative 3 root 3. Negative 3 root 3. And our y value is going to be negative 3, which means our point is negative 3 root 3, negative 3. 

[ENDS AT 8:30]