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# Justification for polar arc length formula

Video transcript

- [Voiceover] What I want to do with this video is come up with
the formula for the arc length of a curve that's
defined in polar coordinates. So, if this curve right over here is r is equal to F of theta, how do we figure out the length of this curve between two thetas, say between theta is equal to, well let's say, in this case, it looks like between theta zero radius and say, pi over two radius, but between
any two bounds for r theta. So the way we're going to do this, an if at any point you get excited or inspired, you definitely should pause the video and see if you can run with the formula for arc length when you're dealing with something in polar form. But the way that we're going to tackle it is the exact same way that we tackled arc length when we were dealing with standard rectangular coordinates. So, let's take a little, small section of the arc length, let's take a little small section, I'm going to blow it up. So, let's call this right over here, this is our infinitesitely, this is our infinitely smaller, infinitesimal sized, our piece of our arc length,
I'm going to call it DS. And obviously, this is
a lot bigger than maybe you would imagine when you think of infinitesimal, but then if you integrate together all of the DSes, if you integrate together all of the DSes, then you're going to have, you're going to have the length of the actual
curve that you care about. So we can say that the length is going to be all of the DSes integrated, all of the infinitely, the infinite sum of all these infinitely small, all of
these infinitely small DSes. Now, to actually put this in terms we can relate to in terms of Rs and thetas, I'm first going to relate this to Xs and Ys and then relate the Rs and thetas to Xs and Ys, which we have seen before when we have converted between polar
form and rectangular form. So we know that this DS is going to be equal to our infinitely
small change in X squared. So if this is our,
going from this point to that point, that's our
change in our arc length. But this distance, right over here would be our change in our X, I'll write that as DX and I'm writing everything as differentials, which is a little bit mathematically, I guess we could say, hand-wavy or loosey
goosey, but it gives you a good conceptual understanding
of where this comes from. You could, if we were a little bit more precise, we could take DX, we could do delta Xs and then eventually take limits and all the rest, but I'll just go with this because it makes a little bit, at least for my brain, more conceptual sense. So that's, that's our change in X when we go from that point to that point. Then this is our change in Y when we go from this point to this point, DY. And we've seen this before when we got our justification for the arc length formula rectangular coordinates, we could say that DS is going to be equal to the square root of DX squared, DX squared, plus DY squared and this just comes straight out of
the Pythagorean Theorum. Plus DY squared, and then if we can integrate these, then we're
kind of in the same place. But how do we get these
in terms of Rs and thetas? Well, to do that, we just have to remind ourselves what X is in terms of r and theta, what Y is in terms of r and theta. So X, we know, is going to be equal to r cosine of theta and we first saw this when we just first were going back and forth between polar and rectangular coordinates. And Y is going to be r sine of theta. And now we can use this to say what DX and what DY are going to be. DX is then going to be equal to, and we have to remember that r is going to be a function of theta, so actually, let me write it this way, let me just rewrite it. So X, we could also
write it as, F of theta times cosine theta and Y is equal to F of theta times sine of theta. So now, what's a DX? DX is going to be, this is just, we're just going to apply the product rule here. It's going to be F prime of theta, derivative of the first expression times the second one, times cosine to theta plus the derivative of the second one. Well, the derivative of cosine of theta is negative sine of theta, so
we'll say minus sine theta. Minus sine theta times the first expressions, so F of theta, that was just the product rule, that's our DX. And then, of course, D theta, D theta. Another way you could have said if you treated these differentials like numbers, you could divide both sides by D theta, you would have the derivative of X with respect to theta is this business right over here, so those
are equivalent statements. And also do the same thing for DY. So DY, same again, by the product rule is going to be F prime, F prime of theta times sine of theta plus F of theta times the derivative of sine of
theta, which is cosine of theta. Cosine of theta, and now if we want to figure out what DS is,
we're going to have to take the sum of DX squared and
DY squared, so let's do that. So, DX, DX squared is going to be equal to, we just need to square all of this business, so I'm just going to square this and then multiply that times D theta squared, so that's going to be equal to, this is going to be F prime of theta squared, cosine squared theta, minus 2 times the product of
these, minus 2 times F prime of theta, F of theta, cosine theta, sine theta, and then this one squared, so negative times negative is a positive, so plus F of theta squared, sine
squared, sine squared theta. So that's DX squared, and then, of course, we have the D theta, well not done yet, then we have the D theta squared and now let's figure out what DY squared is. So, DY squared is going to be equal to... Well, this term, squared, oh I have to forget, this DY is going to have a D theta at the end, don't want to forget that. And so over here, this is going to be F prime of theta squared, sine squared, sine squared theta, and then two times the products of these, so plus two times F prime of, let me, F prime of theta, F of theta, it's a little bit hairy, but we'll see in a few it's going
to clean up nicely. F of theta, cosine sine, cosine theta, sine theta and then we just want to square this plus F of theta
squared cosine, cosine squared theta, and then D theta squared. D theta squared, now let's
add these two together. So let's add them together,
and what are we going to get? So, if we add DX squared and DY squared, we're going to get, so DX squared plus DY squared plus DY squared is equal to. So over here we have cosine squared theta times F prime of theta squared and then sine squared there times
F prime of theta squared. So we can factor out an F prime of theta squared, so it's going to be equal to, so if we factor these characters out, it's going to be F prime of theta squared times, times cosine squared theta, cosine squared theta plus sine squared theta. Plus sine squared theta and we see that that's going to simplify nicely, that this is just going to be equal to one, that's just one basic trig identity. And then, these middle two terms actually cancel out, this is negative of this, so these two
cancel out and then over here, we can factor out
an F of theta squared. So, we could factor out an F of theta squared, so it becomes plus F of theta squared times sine squared theta, sine squared theta plus cosine squared theta, plus cosine squared theta. Well, that simplifies nicely. This is just going to be equal to one. And then, we have these and then this D, D theta squared is multiplied
times everything. So, so everything right over here is going to be, so times D theta, D theta squared. You can almost view these as the coefficients on D theta squared and we added those two coefficients. So, this is going to clean up nicely now, so this simplifies as DX squared plus DY squared plus DY squared is equal to, is equal to F prime of
theta squared plus F of theta plus F of theta squared. And then all of that
times, I'm going to do this in a new color, actually. All of that times D theta squared. Actually, that, I've
already used that color. I'll use the magenta. All of that times D theta
squared, D theta squared. Now we know that DS is going to be the square root of this, so let's write that, so DS is equal to the square root of this, which is equal to the square root of this. Which is going to be equal to, well we can factor out a, the square root of D theta squared is just going to be D theta, so we can just take that out. And we are left with, we are left with F prime of theta squared. F prime of theta squared plus F of theta squared,
plus F of theta squared. And now we take a D theta out of the radical, if you put it
in as D theta squared, you take it out, it's going
to be D theta, D theta. So this is interesting, so if we wanted to integrate them, so if you wanted to integrate this, if you want to integrate this, if you want to integrate this, you just want to integrate
this right over here. And you would integrate it from your starting theta, maybe we could call that alpha, to your ending theta, beta. And just like that, we have given ourselves a reasonable justification, or hopefully a conceptual understanding, for the formula for arc length when we're dealing with something in polar form. If you have r is equal to F of theta, you find what F prime of theta is, or you could think of it as the derivative of r with respect to theta, square that, add that to F of theta squared, take the square root and then integrate with respect to theta from alpha to beta. And so this right over
here, our arc length, is going to be equal to
this right over here. And the next few videos, we
will actually apply this.