Main content

## Calculus, all content (2017 edition)

### Course: Calculus, all content (2017 edition) > Unit 8

Lesson 2: AP Calculus BC questions- 2015 AP Calculus BC 2a
- 2015 AP Calculus BC 2b
- 2015 AP Calculus BC 2c
- 2015 AP Calculus BC 2d
- 2015 AP Calculus BC 5a
- 2015 AP Calculus BC 5b
- 2015 AP Calculus BC 5c
- 2015 AP Calculus BC 5d
- 2015 AP Calculus BC 6a
- 2015 AP Calculus BC 6b
- 2015 AP Calculus BC 6c
- AP Calculus BC exams: 2008 1 a
- AP Calculus BC exams: 2008 1 b&c
- AP Calculus BC exams: 2008 1 c&d
- AP Calculus BC exams: 2008 1 d
- Calculus BC 2008 2 a
- Calculus BC 2008 2 b &c
- Calculus BC 2008 2d
- 2011 Calculus BC free response #1a
- 2011 Calculus BC free response #1 (b & c)
- 2011 Calculus BC free response #1d
- 2011 Calculus BC free response #3a
- 2011 Calculus BC free response #3 (b & c)
- 2011 Calculus BC free response #6a
- 2011 Calculus BC free response #6b
- 2011 Calculus BC free response #6c
- 2011 Calculus BC free response #6d

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# 2011 Calculus BC free response #1d

Arc length for a curve. Created by Sal Khan.

## Want to join the conversation?

- It would be awesome if there was a video for just arc length in Calculus.(13 votes)
- Can we get an example of how to find the arc length without using a calculator, (except maybe for basic number crunching?)(4 votes)
- Wouldn't you just evaluate the integral by hand?(3 votes)

- Just to clarify, if they had asked displacement, would one find the distance between points (0, -4) and (21, -3.226)? When I first saw the question, this is what I thought.(2 votes)
- You have to be careful to distinguish between the distance (just the length of the straight line between two points) and the length of a traced curve between two points (arc length)... It also was my first thought!(2 votes)

- I thought arc length was the square root of 1 + (dy/dx)^2 at6:20. Why is it the pythag and not that?(2 votes)
**If**you could convert the parameterized equations he is using into an equation for y in terms of x then that formula would be helpful ...

What we are being shown by Sal is how to derive the 'arc length formula' ... if you watch from2:00to3:00, we are shown that:

`ds = √( (dx)² + (dy)² )`

divide both sides by dx:

`ds/dx = √( (dx)² + (dy)² ) / dx`

move the dx inside the square root, and then distribute it:

`ds/dx = √( (dx)² + (dy)² / (dx)² )`

`ds/dx = √( (dx/dx)² + (dy/dx)² )`

simplify & rearrange:

`ds/dx = √( (1)² + (dy/dx)² )`

`ds/dx = √( 1 + (dy/dx)² )`

`ds = √( 1 + (dy/dx)² ) dx`

... when integrated this gives you the arc length ...

You may also find this helpful:

http://en.wikipedia.org/wiki/Arc_length

HTH(2 votes)

- why is Sal finding the arc length (the small piece) when he doesn't know how the actual graph looks like? The actual might not even have the arc shape, rather it can be a straight line connecting from (0,=4) to (21,-3.226)(2 votes)
- Why is the distance the arc length? I haven't worked too much with parametric equations and usually the total distance is the absolute value of the integral of the velocity function with respects to x.(1 vote)
- But isn't the distance almost the same as the displacement from 0 to 21

that is the same value ?(1 vote) - I thought the length formula was the square root of 1+dx/dt^2 + dydt^2(0 votes)
- Maybe it makes more sense if you think of the nonparametric equation as a special case of the parametric one.

This is the parametric version.

sqrt(dx/dt^2 + dy/dt^2)dt

If you parametrise in terms of x instead of t, as the nonparametric, regular version does, you get

sqrt(dx/dx^2 + dy/dx^2)dx

which of course simplifies to

sqrt(1 + dy/dx^2)dx(2 votes)

## Video transcript

Part d: Find the total distance traveled by the particle over the time interval between time t = 0 and t = 3, or 0 <= t <= 3. So let's draw some axes here to just make sure and you wouldn't obviously have to do this if you were doing it under time pressure during the actual AP exam but my point here to make sure we all are understanding what's going on. So when t = 0, where are we? Well they tell us; x(0) is 0 so x is 0 and y(0) is -4 so we're at the point (0, -4). That's when t = 0 and in the last problem we figured out what happens when t = 3; we figured out that x is at 21. I'll just say that's 21 right over there. x(3) is 21 and y(3) was like -3 point something, so that puts us right over here. So this is 21 and we figured this out in the last problem, I think it was -3.226, and so this is what happens when t = 3. And between these point who knows what the path is. We could plot it if we wanted but it might look something like this, so who knows what it does, and we go like that. And so in are part d right here they're asking us "What is the total distance traveled?", or another way to think about it is "What is the length of this path?" And we could, there is a formula for arc length and if you know it, you could just apply it. It's not a bad thing to know going into the AP exam, especially under time pressure, but I always forget what it is, I'm almost 35 now, so I always like to re-derive it and there's always something a little bit satisfying about that too because we remind ourselves why the formula works. Well how do we figure a little bit of that arc length? So actually let me do a little other part; I like this part more. So let's say, "How do we figure out a little bit of that arc length right over there?" So let's say you have a little bit of that arc length that... I'm going to zoom in on that. Well you have a little bit of a small change in x over that arc length that I've blown up, so call that dx, and you have a small change in y as well, dy, and we know from the Pythagorean theorem, if we get small enough you can approximate this as just the hypotenuse as this is the base, this is the height, and this is the hypotenuse. And so this is going to be, especially if you get these small enough, a line is going to be a pretty - you could really approximate this arc by the hypotenuse right over here. We know what this is from the Pythagorean theorem. This is going to be the square root of dx squared plus dy squared, straight up from the Pythagorean theorem. Now, how do we write these as functions of t? Well we know that dx/dt is equal to x'(t) or if we treat differentials kind of like numbers, and you can most of the time, this is not very rigorous, but you that that dx is equal to x'(t) dt and we know that dy of t or dy/dt, we know this is the same thing at the derivative of y with respect to t; y'(t) multiply both sides by dt you get dy = y'(t) dt. And all I'm doing here is essentially reproving or re-deriving the arc length formula. So we know that this right over here is just a small little arc length so we could call that dL or da for small arc length. Or actually we don't have to worry about what we call that right now. But this expression for that very little small arc length right over there; if I rewrite it in terms of this and this, what I have on the right-hand side, and I want to do that so that I can get everything in terms of t, I get it is equal to the square root of dx squared... Well dx squared is the same thing as (I'll do it in that magenta color) x'(t) dt squared and then you have plus dy squared, and dy squared is just this stuff: y'(t) dt squared. And so this little arc length is this right over here, and then you can actually factor out a dt squared, so this is going to be equal to dt squared times x'(t) squared plus (I'll do this in green) y'(t) squared. And then of course you can factor out the dt out of the radical sign now. The square root of dt squared is just dt so all of this simplifies to (I'll write it in yellow) this part right over here is x'(t) squared and this part right over here is y'(t) squared, and then we factor out the dt. Now this is all just a way to derive what this little small d-arc or this little small arc length is but we don't want to find just a small arc length, we want to sum over all of them, so what we want to do is integrate the dt's, take the infinite sum of these infinitely small pieces right over here, these infinitely small arc lengths from t = 0 to t = 3. And so now we just do it. And so they tell us what x'(t) is and y'(t) is. Let me just rewrite the expression. This is going to be equal to the integral from 0 to 3 of the square root of x'(t) squared; the problem they give us what x'(t) is, its 4t + 1. So its 4t + 1 squared plus y'(t) squared... the derivative of y with respect to t is sine of t squared and we need to square that as well and then dt. And this is not a simple thing to find the antiderivative of, but lucky for us we are allowed to use our calculators in this part of the AP exam. And so all you have to do at this point, we've already done really the hard work, is use this definite integral function. It's in our catalog right over here and you can go straight to the 'f's and go a page down. We need to find the definite integral, we have to punch this into our calculator, of the square root of, (and I'll write it as x's just because the x button is much easier to get to than trying to put a variable t in there), so the square root of 4x plus 1 squared plus sine of x squared (instead of writing a t here, I'm just writing an x here, just to clarify what I'm doing) and then we want to square that, let's see, that closes that parentheses, and then I need to close the radical parentheses right over there. I want to say that my variable of integration is x I could have put t's here and set up my variable of integration as t, there's not any difference there and then I'm going between 0 and 3. And so, let's let the calculator do all the work here and we get... its taking a little time. We get 21.091. So this is equal to 21.091 and that's the entire arc length. That's the distance that this particle traveled.