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Arc length

# Arc length intro

AP.CALC:
CHA‑6 (EU)
,
CHA‑6.A (LO)
,
CHA‑6.A.1 (EK)
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.

## Want to join the conversation?

• If `√(dx)² = dx`, then `dx` is a quantity, right? , 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 `dx` thing is still bugging me. Am I missing something from the start?
• 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).

so:

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
• So could we factor out (dy)^2 instead of (dx)^2 and compute the arc length as ∫ sqrt(1+(dx/dy)^2) dy ?
• 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."
• At around Sal 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?
• 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.
• What would be a real-world, practical use of finding an arc length? i.e. physics
• 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).
(1 vote)
• Why the integral of "ds" would be the arc length?
• 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.
• At it 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?
• 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
• 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.
• 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.
• how does the squar root of dx^2 + dy^2=the squar root of dy^2 (1+dy/dx)^2
(1 vote)
• at how is he saying its the integral ?
(1 vote)
• 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!
• 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
``∫ f(x) dx``
, 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
``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!