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## Calculus 2

### Course: Calculus 2 > Unit 4

Lesson 13: Arc length# Worked example: arc length (advanced)

A worked example of finding arc length using definite integrals. This example involves some challenging algebra.

## Want to join the conversation?

- Here's the thing, I figured out this problem on my own without help, but it took me an hour. It took me 3 pages of re-writing the problem to figure out that I can take the problem out of the square root, then I watch the video and a couple minutes into the problem he's just like "oh btw this can be taken out of the square root" So naturally my question is: Is there some sort of pattern I'm not seeing or does this just come from practice and experience, because I'm starting to get a little discouraged(20 votes)
- Don't be discouraged. What you just did, figuring it out on your own, is worth 10 videos. You can watch as much as you want, but without PRACTICE, you won't get anywhere. I wish more students would spend that kind of time trying to figure things out - that is how your mathematical intuition and situation recognition grows. I prefer to use the phrase "situation recognition" rather than "pattern recognition", because patterns are too specific - if the pattern does not match exactly to what you have seen, you may pass over the similarity of the current situation to a previously encountered situation and abandon your investigation prematurely.

So keep watching videos, do the exercises, prove results to yourself, then, if all that fails, ask for help, everyone here has asked for help in math at one time or another. Just remember to be prepared to explain what you have tried when you ask for help - that way we can see you have been honest in your attempt to figure it out on your own.(31 votes)

- What is the practical use of calculating arc lengths?(13 votes)
- My professor said that he was an electrical engineer and used it while working on the cable harness in cars. He had to calculate how long a wire had to be, and know the exact length, so that if anyone else needed to look at his measurements, they would be precise. So essentially, you can find the exact length of arcs or even curved objects in the real world.(39 votes)

- I dont understand how he got the square of the derivative. I understand that the 1 + f'(x) is from the formula, but how is (1/2 x^2 - 1/2 x^-2) squared equal to (1/4 x^4 - 1/2 + 1/4 x^-4) ?(8 votes)
- Will calculating arc length prove useful in calculating surface area of objects?(4 votes)
- Yes! The arclength formula is part of the surface area formula for a solid of revolution.(5 votes)

- How did Sal know what x^8+2x^4+1 = (x^4+1)^2?

Is there a shortcut involved, or maybe a trick?

Is there a formula to find out how x^8+2x^4+1 = (x^4+1)^2 ?(3 votes)- The easiest way to see this is by using a u-substitution.

Let u=x⁴

Thus,

x⁸+2x⁴+1

= u²+ 2u + 1

And from algebra you should know that is a perfect square quadratic:

= (u+1)²

Now back-substituting gives:

= (x⁴+1)²

You should have learned this technique in Algebra II. If you didn't or if you have forgotten it, you might need to review this material.(5 votes)

- at2:56sal finished up the equation. the next equation he left out the 1 on the right side of the equation? I thought he canceled out the 1's on both sides, but the 1 on the left hand side appears at5:57when he brings us up to speed with what he's done so far. what happened to the 1 on the right hand side of the equation at2:56?(3 votes)
- Find the area of the subtending arc when () = 30° and r = 6.

can we do this using integration?if possible,how?(1 vote) - Why is the beginning function, f(x)=x^3/6+1/2x, almost identical to x^3/6-1/2x, the equation Sal finds at7:45?(2 votes)
- It is just a coincidence - nothing more.(2 votes)

- would this have been possible with u substitution?(2 votes)
- Before or after getting rid of the square root sign?

Because integrating square roots can be very difficult in the case the derivate is not is not also outside the square root. Then you need trig substitutions. Think the difficult part in this video is to see how you can complete the square. Otherwise it's very hard.(1 vote)

- can I use arc formula on this function f(x)= 0.26x^2-0.06x-0.05 where the minimum is not on the origin(0,0) ?(2 votes)

## Video transcript

- [Voiceover] This right here is the graph of Y is equal to X to the third over six plus one over two X. And what I want to do in this video is to figure out the arc
length along this curve between X equals one and X equals two. And so we've already highlighted that in this purple-ish color. So I encourage you to pause this video and try it out on your own. And I'll give you one hint. Assuming that you apply
the arc length formula correctly, it'll just be
a bit of power algebra that you'll have to do to
actually find the arc length. So I'm assuming you've had a go at it. Let's work through it together. So a few videos ago,
we got a justification for the formula of arc length. We got arc length, arc length is equal to the integral from the
lower boundary in X to the upper boundary in X,
and this is the arc length, if we're dealing in terms
of X we could actually deal in terms of other variables. And so it's going to be
one plus F prime of X squared D X. So for this particular F of
X we just need to figure out what F prime of X is, we
need to square it, we need to add one to it, and then we
have to take the square root of that. So let's do all of that step-by-step. So what is F prime of X going to be? Let's see, X to the third,
the derivative of that is three X squared. So three X squared, three
over six is going to be X squared over two, and
this is one half X to the negative one is one way to think about it. And so that's going to
be negative one half X to the negative two. Negative one half X to
the negative two power. Now, what is F prime of X squared? So it's going to be, Actually let's just write
out what is one plus F prime of X squared? So it's going to be, So one, one plus F prime of X, F prime of X squared is going to be equal to it's gonna be one plus this thing squared. And so, let's see, this
term right here squared is going to be X to the fourth over four. It's gonna be X to the fourth over four. Now the product of
these times two is going to get us negative, Let's see X squared time
X to the negative two which is going to be one. And so it's going to be negative one half. Right, I just took the
product of these two and then multiplied that by two, yep, it's just gonna be negative one half. And then this term right over here squared is going to be positive one fourth X to the negative four. One fourth X to the negative four. Now let's see we have
one and a minus one half so we can simplify them a little bit. This is going to be
equal to X to the fourth over four. One minus one half is plus
one half plus one fourth, plus one fourth X to
the negative four power. Let me make that a little bit clearer. X to the negative four power. So this seems a little bit strange. And we're going to have to
take the square root of it. But maybe we can set this
up so that it's a product of perfect squares, because it does look, you know X to fourth
and let's see if we can write this in a way
that we could recognize how to factor it a little bit better. So let's factor out a one fourth X to the negative fourth power. So this is going to be equal to one fourth X to the negative four times, let's see I factored that out, so if you factor one
fourth X to the negative fourth out of this first term, and I could color-code 'em a little bit. This first term, when you factor this out is going to be X to the eighth power. Now this term right over
here, if I factored out a one fourth, it's going to be equal to two X to the fourth. So this term right over here, once again, this powering
through the algebra part. So plus two X to the fourth, and then this one is
pretty straight forward. This is just going to be equal to one. So plus one. Plus one. And now this is looking interesting. Because this one, this right over here we could rewrite as one half
X to the negative two squared. This is equal to, that same color, one half X to the negative two squared. And this over here we
could rewrite this as X to the fourth plus, so we write it this way, this is going to be X to the fourth plus one squared. Alright, X to the fourth
times X to the fourth is X to the eighth. One squared is one. The product of them times
two is two X to the fourth. So this over here is
the same thing as that. And so now if we wanted
to take the square root of both sides. So if we wanted to say the
square root of one prime plus X squared. Notice I'm just focusing on the algebra. So what we've done so far
this is one plus F prime of X squared. Now we want to take the
square root of that. So that's going to be
the square root of this. And so that's going to be equal to square root of this is,
let me just color code it, this is going to be one half
X to the negative two power times, times X to the fourth plus one. Times X to the fourth plus one. X to the fourth plus one. I did that right, yep. Or I could actually, now that I did all this to put this in a form
that I could actually recognize, now I could
distribute things back. This is going to be equal to one half, X to the negative two
times X to the fourth is X squared. One half X squared plus one
half X to the negative two. And if we're gonna take,
if we're gonna take, Let me rewrite this. That's one plus F prime of X squared. Now let's take the definite integral. I'm going to give myself
some space to write my D X. So we're gonna take the definite integral, in this case, from X equals one X equals one to X equals two. So this is the definite
integral from X equals one to X equals two of this D X. So it's going to be the
definite integral of this from X equals one to X equals two D X. And so this is fairly straight forward. The anti-derivative of one half X squared that's going to be, what? That's going to be one,
let's see it's X to the third and then we divide by three. So one half divided by three is one sixth. One sixth X to the third. And then this is going to be, we're going to increment, this is gonna be X to the negative one. We're gonna divide by
that so minus one half X to the negative one power. Is that right? Did I do that? Yep, negative one, when you
take it, yep negative two and then this one, yep, that looks good. And we're in the homestretch. We're gonna evaluate it. At two and at one. And so we get, when you evaluate it at two, you get two to the third, which is eight over six minus one half times one half. So this is minus one fourth. And when you evaluate it at one, you're gonna have minus
so we're gonna subtract evaluating it at one, it's gonna be one sixth minus one half) And now we just have to
evaluate these fractions. So this is going to be, let's see This is, if we divide,
this is four thirds, four thirds minus one
fourth minus one sixth and then we have plus plus one half. Now let's see the common
denominator here would be 12. So this is be four thirds over 12 is 16 over 12. Alright multiply the numerator
and denominator by four. Minus one fourth, the same
thing as three over 12. Now this is minus two over 12. And then we have plus six over 12. And so this is going to be equal to, I think we deserve at least
a little mini-drum roll right over here. So 16 minus three is going
to be 13 minus two is 11 plus six is 17. So there we have it. The length of that arc along this curve. Between X equals one and X equals two. That length right over
there is 17, 17 twelfths. Were done.