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### Course: Calculus 2 > Unit 4

Lesson 13: Arc length# Worked example: arc length

A finely tuned example demonstrating how the arc length formula works.

## Want to join the conversation?

- Why was the bounds of the integral changed on3:40? Would keeping the original bounds still work when you replace u back after taking the integral of u with respect to the original bounds?(18 votes)
- The original bounds were in terms of x. Once we did the u-sub, we are now in terms of u, so to find the new bounds in terms of u we did the process you referenced.

In answer to your second question, YES, if you want to convert back to terms of x after you do the integration in terms of u, then you can use the original bounds.

Do what works best for you.(35 votes)

- Why did Sal not revert from the u-sub back to the original values in terms of x? The problem began in terms of x, it should end in terms of x also I feel.(9 votes)
- Good question.

You'll notice that around4:00Sal changes the interval of the integral into values of u.

This part was sort of "optional". He could just as well have left the integral boundaries in terms of x and back substituted at the end of the problem. The important thing is that this has to be consistent or you will get the wrong answer.

Also this situation is specific to the definite integral. The answer is just a number so there is no x or u value in the final answer. If this were an indefinite integral and you left the answer in terms of u then that would be pretty much wrong. For those problems you should back substitute and leave the answer in terms of x.(17 votes)

- I wanted to play around with this method for calculating the arc length of a simple
`y=x^2`

parabola and chose the boundaries of 0 and 2...

So first step, you know the derivative of`x^2`

is`2x`

and you have to square that derivative in the formula, so you get`4x^2`

.

Plug in the interval and that derivative squared, and you have the`integral from 0 to 2 of √(4x^2+1)`

.

After this, I've tried integration by parts and some u substitution, maybe I missed learning how to solve integrals like this, but I don't know how to solve the`integral from 0 to 2 of √(4x^2+1)`

. I'd really love some help!(7 votes)- This turns out to be a pretty tricky integral. It
*can*be done via trig substitution, but you need to know how to find the integral of the secant function and what to do when the original integral reappears after using integration by parts .

There's a nice step-by-step explanation of how to find the integral of sqrt(1+x^2), which is basically the same problem, here:

https://in.answers.yahoo.com/question/index?qid=20100303103406AAzKlz5(7 votes)

- how would you find the arc length, when you have two variables for example if you have x=t^3, and y=4t ?(7 votes)
- The arclength in the x-y plane is ALWAYS ∫ √( dx² + dy²). Thus, if you are given x(t) and y(t) (we say "parametric" equations for x and y), then we can write this as:

L = ∫ √( (dx/dt)² + (dy/dt)² ) dt

Basically, we have "divided" everything inside the radical by dt², and so we then multiply on the outside of the radical simply by dt. Since x and y are functions of t, this will be a purely "t" integral.(3 votes)

- Does the arc length formula s = rθ come from applying this arc length formula to circles?

P. S. At4:17, Sal draws a gorgeous integral sign :)(6 votes)- Yes, it does! If you use this formula on
`f(x) = sqrt(r^2 - x^2)`

over a certain interval of the curve, you will get just this result. Very interesting how everything connects together. :)(5 votes)

- If the equation had been y^2=x^3 and then you were asked to find the integral from point a to b. Would you have to consider both curves? Since when you take the square root you would have y=x^3/2 and y=-x^3/2.

So would we have to consider both curves? Thus, doubling our answer for the top portion since they are symmetric?(3 votes)- It would depend on your definition of "point A to point B" and on your variable of integration. Knowing that y^2=x^3 is not a one-to-one function in terms of y=f(x), you would need to decide whether you were looking for the length of both halves of the curve or just one between two given x-values. (That is, if you're integrating with respect to x.) If you were trying to find the length of the curve between two specific points (x,y), then yes, you would need to account for both parts of the curve.

If you're integrating with respect to**y**, on the other hand, your equation would give you the length of both parts of the curve without any extra trouble (given a lower and upper limit for y) because this*can*be expressed as a one-to-one function in terms of x=f(y). So in this case, integration with respect to y would probably be a better choice for representing the true length of the curve between any two given points.

Hope that helps.(3 votes)

- Why did Sal not revert from the u-sub back to the original values in terms of x?(1 vote)
- Observe that he changed the bounds from 0 to 32/9 --> 1 to 9. So, once you change bounds, you can remain in the same variable. If you decide to use the original bounds, you
**must**revert from "u" back to "x"

It honestly doesn't matter if you wanna stay in "x" or "u", but be cautious of what bounds you're using, as failure to change bounds will give you the wrong answer.(3 votes)

- why does the sqrt of u = u^3/2?(1 vote)
- The square root of u is u^1/2. The derivative of u^3/2 is (3/2)u^1/2, so the anti-derivative of u^1/2 is (u^3/2)/(3/2) or (2/3)u^3/2.(2 votes)

- At2:38, shouldn't it be the absolute value of x? Squaring the square root should produce an absolute value right? Or at least restrict the domain to non-negative x-values.(1 vote)
- In this case, x is positive, so abs(x) = x.(2 votes)

- I was thinking about this integrals when we derive int ds we get a int b sqrt(1+(f'(x))^2)dx isn't dx suppose to multiply rather than being left out like that? Or calculus is made for almost an obvious answers?(1 vote)
- When you are writing and rearranging an integral, you want to get a differential (such as dx) multiplied out and placed at the end of the integral. This is the standard form of an integral, as it shows what you are integrating with respect to, and then you can start to find the value of the integral.

Hope this helps!(2 votes)

## Video transcript

- [Voiceover] So, right
over here, we have the graph of the function y is equal
to x to the 3/2 power. And what I wanna do is find
the arc length of this curve, from when x equals zero
to when x is equal to-- and I'm gonna pick a strange number here, and I picked this strange number 'cause it makes the numbers
work out very well-- to x is equal to 32/9. 32/9 is, let's see... That's three and 5/9, so
it's gonna be right around... So that's three and 1/2, so
it's gonna be a little bit past three and 1/2, so it's
gonna be right over there. So we wanna find this arc
length right over here, this thing that I have depicted in yellow. So it's from zero to 32/9. And I encourage you to pause the video and try this out on your own. So I'm assuming you've had a go at it. And if at any point while
I'm working through it, you feel inspired, always
feel free to pause the video and continue working with it. So let's just apply the arc length formula that we got kind of a conceptual proof for in the previous video. So we know that the arc length... Let me write this. The arc length is going to be
equal to the definite integral from zero to 32/9 of the square root... Actually, let me just write
it in general terms first, so that you can kinda see the
formula and then how we apply it. So, it's the square root of one
plus f-prime of x squared d x. And in this case, it's going
to be the definite integral from zero to 32/9 of the
square root of one plus... Now, what's the derivative? If f of x is x to the 3/2,
then f-prime of x is going to be 3/2 x to the 1/2. And we picked this particular function because it simplifies
quite well when we put it under the radical, and
it's fairly straightforward to find the anti-derivative. So we've done a lot of
engineering of this problem to make the numbers work out well, but let's just go through it. So this is f-prime of x;
f-prime of x squared is going to be this quantity squared. It's going to be 9/4 x
to the 1/2 squared is x. So, one plus 9/4 x d x. And so, now, we just
have a definite integral that we know how to
solve this type of thing. And you might be able to
even do this in your head, essentially, do the u-substitution: say I have one plus 9/4 x. Its derivative is 9/4. I can kind of engineer that if I want, but instead, I'm just going to
do straight up u-substitution. So, if I say u is equal to one plus 9/4 x, then we know... Let's see. D u d x is going to be equal to 9/4. Or, we could say d u is equal to 9/4 d x. Or, we could say d x... Let me scroll down a little bit. We could say d x is equal to... I'm just gonna multiply
both sides times 4/9. Let's go to 4/9 d u. And then we just have to change
the bounds of integration. When x is equal to zero, then
u is going to be equal to... 9/4 times zero is just zero, so u is going to be equal to one. And when x is equal to 32/9-- and this is why that number was picked-- what's u going to be equal to? 32/9 times 9/4 is gonna be 32/4, which is going to be eight plus one. So that worked out very
nicely; imagine that. So there we have it. So, this is going to be equal
to the definite integral... Actually, let me make
it clear that this is what is equal to this. The definite integral
from u is equal to one, to u is equal to 9-- I'm gonna make it very
explicit that I'm dealing with u now-- of the square root of u. And instead of d x, we
have d x is 4/9 d u. Let me do it this way: the square root... Whoops, that's not the right color. Square root of u; instead of
d x, we have times 4/9 d u. And I'm just gonna take the
4/9 and stick it out here. And we know how to apply the fundamental, or, I guess, the second fundamental
theorem of calculus here, to evaluate this definite integral. This is going to be 4/9
times the anti-derivative of the square root of u,
which is the same thing as u to the 1/2. It's going to be u to the
3/2, and then we divide by 3/2, which is the same
thing as multiplying by 2/3. We're going to evaluate
that at u equals nine and u is equal to one. And so, we're in the home stretch here. This is going to be equal
to 4/9 times 2/3 times 9 to the 3/2 minus 2/3 times one to the 3/2. So, 9 to the 3/2, that is... Let's see: the square
root of nine is three, to the third power is 27. And this, of course, is one. So we are left with 2/3 of... Actually, let's just factor out the 2/3, that makes it easier. This is going to be equal to
2/3 times 4/9 is equal to 8/27. I've just factored out the 2/3. And then we're gonna have 27 minus one inside, I guess you could
say the brackets now. So 27 minus one is just going to be 26. Times 26. And we could obviously
simplify this more if we want. Eight times 26 is... So, actually, let's just
figure that out, just for fun. So, eight times 26 is going to be 160 plus eight times six
is 48, so it's 208 over 27. And we are done.