Sal shows the polar arc length formula, and explains why it is true.
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- Why is x= rcos(theta) and why is y=rsin(theta)?(17 votes)
- It is derived from the trig functions in a polar graph.
cos Θ = adjacent/hypotenuse = x/r
sin Θ = opposite/hypotenuse = y/r(29 votes)
- Why are we converting Polar to Cartesian form (3:08) ? Y can't we just use ds (arc length) = rdθ and integrate it?(11 votes)
r=f(θ), the length of the arc is not given by
r*θ, (that is only valid on segments of circles).(8 votes)
- I think the question people are asking, which has still not been properly answered, relates to the difference between the Area derivation and the Arc Length derivation. In both, r = f(t), so radius can change over the function of the angle, but when integrating to find the Area this doesn't matter, but for the Arc length it does. Why?(12 votes)
- Why doesn't using the circumference of a infinitesimally small sector of a circle and summing the circles up work?
integral[a,b][(2*pi*r *d(theta)/(2*pi)]=integral [a,b][r*d(theta)]?(9 votes)
- since s=rΘ why isnt or why can't arc length be ds=rdΘ in the same way for how area enclosed in polar functions is dA=(r^2)/2 dΘ(9 votes)
- Because the radius changes over a function of the angle, hence r= f(Θ). When the radius is constant, like a circle fixed at the origin for example, the formula reduces down to the S=Θr.(0 votes)
- I was wondering how the calculus equation for arc length relates to the algebra version: s=r(theta) ? I have attempted to find a connection but cannot seem to find one.(3 votes)
- In case of common equation, not polar curves, can we also figure out the length of line between two points using the integral calculus?
ex) y=f(x) x=[a, b](1 vote)
- This polar graph stuff seems to have popped up out of nowhere for me along this integration playlist. Is there like an introduction to polar graphs or the like somewhere in KA?(1 vote)
- Why can't I integrate "arc length = f(theta)* (d theta) " to find arc length?
I think it's a good approximation that arc length = f(theta)* (d theta)
Also, when we calculate the area of the polar graph, we use "(1/2)(f(theta)^2)(d theta)" to approximate the area of the curve. I think this two are similar, but why arc length can't be found by similar method but area can(1 vote)
- arc length = Integral( r *d(theta)) is valid only when r is a constant over the limits of integration, as you can test by reducing the general formula from this video when dr/d(theta) =0. In general r can change with theta. In Sal's video he could have constructed a different right angled triangle with ds as the hypotenuse and the other two sides of lengths dr and r*d(theta). I will leave the construction of this triangle as an intellectual exercise :-) Hint: use polar coordinates.(1 vote)
- In the process of adding dx^2 and dy^2 you have two dtheta^2 to add together. It makes sense that the result of this addition should be 2dtheta^2. Wouldn't it be like adding x^2 and x^2? The result of that addition is 2x^2. So why wouldn't dtheta^2 plus dtheta^2 equal 2dtheta^2?(1 vote)
- What you say is correct. dtheta^2 is just a variable that can be added like any other.
In the present context of the video though, sal never added two dtheta^2 's together.
Notice that the dtheta^2 was factored out and had the other things as its coefficients.(1 vote)
- [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.