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Worked example: arc length

AP.CALC:
CHA‑6 (EU)
,
CHA‑6.A (LO)
,
CHA‑6.A.1 (EK)
A finely tuned example demonstrating how the arc length formula works.

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  • blobby green style avatar for user Alan.Rodriguez67
    Why was the bounds of the integral changed on ? Would keeping the original bounds still work when you replace u back after taking the integral of u with respect to the original bounds?
    (17 votes)
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    • leaf blue style avatar for user Stefen
      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.
      (33 votes)
  • male robot hal style avatar for user Konrad Taube
    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.
    (8 votes)
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    • leafers ultimate style avatar for user TripleB
      Good question.
      You'll notice that around Sal 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.
      (16 votes)
  • leafers ultimate style avatar for user Quintin Lenti
    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)
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  • blobby green style avatar for user jdbennett16
    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)
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    • blobby green style avatar for user robshowsides
      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)
  • purple pi pink style avatar for user Carolyn Dewey
    Does the arc length formula s = rθ come from applying this arc length formula to circles?

    P. S. At , Sal draws a gorgeous integral sign :)
    (5 votes)
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  • piceratops seed style avatar for user 489.albert
    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)
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    • leafers seed style avatar for user Travis Bartholome
      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)
  • starky sapling style avatar for user 20leunge
    At , 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)
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  • aqualine ultimate style avatar for user Infinite pi π
    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)
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    • aqualine ultimate style avatar for user Andrew Moore
      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)
  • blobby green style avatar for user Danson Ndungu
    Show that the curved surface area of a cone of radius r and height l is πrl
    (1 vote)
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  • duskpin ultimate style avatar for user Daniel Kassem
    How would you do this is the question stated that y=4/5·(x)5/4 for 0<--x<--9 and dy/dx=x^1/4
    (1 vote)
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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.