We can use definite integrals to find the length of a curve. See how it's done and get some intuition into why the formula works.
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√(dx)² = dx, then
dxis a quantity, right?6:03, how can it be treated as notation? In some vids in Differential calculus subject, Sal said something like "dy/dx is not a fraction*. Although I clearly understand this video, the
dxthing is still bugging me. Am I missing something from the start?(42 votes)
- I would be better to say the change in x is delta x, and the change in y is delta x times the slope in that point. (slope is change in y per change in x. say slope*delta x is delta y)
Then let delta x get to zero, so delta x becomes dx. (take the limes).
sqrt( (delta x)^2 + (slope*delta x)) is changing to
sqrt (dx^2+(f ' (x)*dx )^2)
now, factor out the dx^2, which is just the distance moved in the x direction.
and you get ds = sqrt (1+f'(x)^2) * dx(7 votes)
- So could we factor out (dy)^2 instead of (dx)^2 and compute the arc length as ∫ sqrt(1+(dx/dy)^2) dy ?(9 votes)
- Yes! You can. And it's not more complicated to find dy/dx than dx/dy. It all depends upon what you're given. If you're given y = mx + b, d(y)/dx = m. If you're given x = my + b d(x)/dy = m. So it only really matters psychologically what you name, the mathematics doesn't change. As Shakespeare wrote "A rose by any other name smells just as sweet."(20 votes)
- At around2:00Sal writes the sum of all ds as int(ds). int(ds) indicates area under curve f(s)=1, ie, int(1.ds). But on the curve, int(ds) is supposed to denote the arc length. When you set both of them to be equal to each other, isn't that dimensionally incorrect?(13 votes)
- Excellent question, it shows you're watching the video thoughtfully. While it may not have been entirely clear in this informal discussion, the infinitesimal quantities Sal is calling ds are independent of how high the curve is with respect to the x-axis. We're measuring tiny linear distances without multiplying them by another dimension, so integrating ds gives us length, not area.
When we integrate f(x)dx we're actually working with height times width: f(x) is the height of the rectangle and dx is the width element (an infinitesimal distance along the x-axis). That's how we get area: multiplying height times width. When we're working with ds, we don't have height or width, only length.(15 votes)
- Why the integral of "ds" would be the arc length?(5 votes)
- Earth is almost a sphere but you hardly recognize the curvature of the earth, right, that's because you are too small compared to the earth so the curvature turns to almost perfectly straight. Come back to this, if you break down a curve into small enough parts, those part would be indistinguishable from a straight line, that's why he use ds, just like people perceive Earth is flat in the old time because the curvature is extremely small.(8 votes)
- What would be a real-world, practical use of finding an arc length? i.e. physics(7 votes)
- To design a bridge, for example, you would want to choose tension ropes in accordance to their elasticity as a ratio of length. If you do not know an accurate measure of the length, the tension ropes can be too short(bridge collapses due to too much tension), or too long (bridge collapses due to too little tension, i.e. no support).(4 votes)
- At4:10it doesn't seem right that Sal can factor out (dx)^2 so that it creates (dy/dx)^2, which is a derivative squared. Why is it OK to do that?(4 votes)
- It is okay to do that since he is really just rewriting the (dx)^2 + (dy)^2 part. Note that he is not changing the value of the original expression. He is just rewriting it in a different way so that we can algebraically manipulate it to be useful to us. If you were to distribute the (dx)^2 part to (1 + (dy/dx)^2), you would just get the same thing under the original square root, which is
(dx)^2 + (dy)^2. All we are doing is manipulating it algebraically to make it so that it can be of some use(7 votes)
- Sal said the fact the derivative of the e^x is e^x would make me stay awake at night, but I found difficulty sleeping just when started learning about integrals and derivatives, in all calculus we are reasoning and deriving formulas from things we don't have any idea of, I mean no one is able to find the smallest change possible of a function yet you're using it to derive formulas about lengths, areas and volumes..... amazing.(3 votes)
- Make no mistake, we perfectly understand the underpinnings of calculus. Any second-year college math student should be able to trace any basic fact from calculus all the way back to the definition of the real numbers.
The concept we use in calculus is not 'the smallest possible change' in the function, in fact we can prove that such a thing doesn't exist. We use an arbitrary or general amount of change in a function, and relate changes between a functions input and output.(4 votes)
- at1:56how is he saying its the integral ?(2 votes)
- Up until that point, did you understand everything regarding the process of integration (of area) as has already been covered in previous videos, most importantly, the concept of summation of infinitely small regions? If yes, then your question is answered as the video progresses. If no, you may want to review the process of integration, starting here: https://www.khanacademy.org/math/integral-calculus/indefinite-definite-integrals
Keep Studying!(4 votes)
- I think Sal is making a huge jump here, skipping over things we should have learned. Together with the Q/A in this forum I think I've been able to tie the pieces together a bit. Can someone confirm the following things?
1) In the integral notation
, dx is not just a notation, but an actual quantity that gets multiplied while summing the Riemann Sum. It represents the infinitely small width of the rectangle. It is in fact
∫ f(x) dx
lim dx->0 dx
2) dx represents a fixed quantity, as in: the width of the rectangles are constant for the integral. Or can it vary? Should we read the Riemann sum as a sum of f(x)dx, or can we read it as the sum of f(x), multiplied by dx?
3) Even though people in the comments say this is not about area anymore, I like to still see it that way: Sal seems to transform this function from something that is expressed in terms of arc length into a different function that is expressed as area. Put in other words, the new function that is created is actually a function where we evaluate the area from. The area of that function represent the arc length. It seems to me the only way that the Fundamental theorem of calculus holds. If that is true, it seems that, as long as you want to measure some property of the curve, if you can express it in dx, you're good to go. Is all of this correct?
Any help is greatly appreciated!(3 votes)
- [Voiceover] We've used the definite integral to find areas. What I want to do now is to see if we can use the definitive role to find an arc length. What do I mean by that? Well, if I start at this point on the graph of a function, and if I were to go at this point right over here, not a straight line, we know already how to find the distance in the straight line but instead we want to find the distance along the curve. If we lay a string along the curve, what would be the distance right over here? That's what I'm talking about by arc length. What we could think about it is okay, that's going to be from x equals a to x equals b along this curve. So how can we do it? Well, the one thing that integration, integral calculus is teaching us is that when we see something that's changing like this, what we could do is we can break it up into infinitely small parts. Infinitely small parts that we can approximately with things like lines and rectangles, and then we could take the infinite sum of those infinitely small parts. So let me break up my arc length. Let me break it up into infinitely small sections of arc length. Let me call each of those infinitely small sections of my arc length a, I guess I could say a length to differential, an arc length to differential, I call it ds. I'll draw it a much bigger that when I at least I conceptualize what a differential is, just so that we could see it. What do I mean by breaking it up into these ds's? Well, if that's the ds, and then let me do the others in other colors, that's another infinitely small change in my arc length. Another infinitely small change in my arc length. If I summed all of these ds's together, I'm going to get the arc length. The arc length, if I take is going to be the integral of all of these ds's sum together over this integral so we can denote it like this. But this doesn't help me right now. This is in terms of this arc length that's differential. We do know how to do things in terms of dx's and dy's. Let's see if we can re-express this in terms of dx's and dy's. If we go on a really, really small scale, once again, we can approximate. This is going to be a line. We just the way that we approximated area with rectangles at first. But if we have an infinite number of infinite small rectangles, we're actually approximating a non-rectangular region. The area on a non-rectangular region. Similarly, we're approximating with lines with the infinitely small and there's infinite number of them, you are actually finding the length of the curve. Well, just focusing on this is a line for now. This distance right over here, I'm just going to try express in terms of dx's and dy's. So this distance right over here, that's dx. You can do this as infinitely small change in x, and this distance right over here, this is a dy. Once again, I'm being loosey-goosey with differentials. Really giving you conceptual understanding, not a reader's proof, but it'll give you a sense of where the formula for arc length is actually coming from. Based on this, you can see the ds could be expressed based on the Pythagorean Theorem as equal to dx squared plus dy squared, or you could rewrite this as square root of dx squared plus dy squared. So we could rewrite this. We could say this is the same thing as the integral of, instead of writing ds, I'm going to write it as the square root of dx squared plus dy squared. Once again, this is straight out of the Pythagorean Theorem. Now this is starting to get interesting. I've written in terms of dx's and dy's but they're getting squared. They're under radical sign. What can I do to simplify this? Or at least write it in a way that I know how to integrate. Well, I could factor out a dx squared, so let me just rewrite it. This is going to be the same thing as the integral of the square root. I'm going to factor out the dx squared, dx squared times one plus dy over dx squared. Notice this, and this is exact same quantity. If I distribute this dx squared, I'm going to get this right up here. Now I can take the dx squared out of the radical. So this is going to be the integral of, let me write that in the white color, the integral of one plus dy, dx squared. This is interesting because we know what dy, dx is. This is the derivative of our function, dy, dx squared. If you take the dx squared out of the radical, the square root of dx squared is just going to be dx. This is just going to be dx. This is just going to be dx. Now this is really interesting because we know how to find this between two bounds. We can now take the definitive role from a to b. This is now we are integrating a bunch of dx's or we're integrating with respect to x. We could say, "Okay, x equals a to x equals b." Let's take the sum of the product of this expression and dx, and this is essential. This is the formula for arc length. The formula for arc length. This looks complicated. In the next video, we'll see there's actually fairly straight forward to apply although sometimes in math gets airy. If you wanted to write this in slightly different notation, you could write this as equal to the integral from a to b, x equals a to x equals b of the square root of one plus. Instead of dy, dx, I could write it as f prime of x squared, dx. So if you know the function, if you know what f of x is, take the derivative of it with respect to x squared added to one, take the square root, and then multiply, and then take the definite integral of that with respect to x from a to b. We'll do that in the next video.