If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content
Current time:0:00Total duration:7:03
AP.CALC:
CHA‑6 (EU)
,
CHA‑6.A (LO)
,
CHA‑6.A.1 (EK)

Video transcript

so right over here we have graph we have a graph the graph of the function y is equal to X to the three halves power and what I want to do is find the arc length of this curve but when from when x equals zero to when X is equal 2 and I'm going to pick a strange number here and I picked the strange number because it makes the numbers work out very well 2x is equal to 32 over 9 32 over 9 is let's see that's a it's a three and three and five ninths so it's going to be right around so that's three and a half so it's going to be a little bit past three and a half so it's going to be like right over there so we want to find we want to find this this arc length right over here this thing that I have depicted in yellow so it's from 0 to 32 over nine 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 if 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 0 to 32 over 9 of the square root actually let me let me just write it in general terms first so that you can kind of see the formula then how we apply it so it's the square root of 1 plus f prime of x squared DX and in this case that's going to be it's going to be the definite integral from 0 to 32 over 9 ty of I should say the square root of 1 plus now what's the derivative if f of X if f of X is x to the three-halves then f prime of x is going to be 3 has X to the 1/2 and we picked this this particular function because it simplifies quite well when we put it under the radical it's it's fairly straightforward to find the antiderivative 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 over 4 X to the 1/2 squared is X so 1 plus 9/4 X DX and so now we just have a definite integral that we have we know how to solve this type of thing and you might be able to even do this in your head essentially you do the u substitution and say ok I have 1 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 use u substitution so use substitution so if i say u is equal to 1 plus 9 over 4x and then we know let's see d u DX would be d u DX is going to be equal to 9/4 or we could say D U is equal to 9/4 DX or we could say DX or we could say let me scroll down a little bit we could say DX is equal to I'm just going to multiply both times both sides times 4/9 is equal to 4/9 D U and then we just have to change the bounds of integration when X is equal to when X is equal to 0 then U is going to be equal to 9 4 times 0 is 2 0 so U is going to be equal to 1 and when X is equal to 32 over 9 32 over 9 and this is why that number was picked what's you going to be equal to 32 over 9 times 9 4 is going to be 32 over 4 which is going to be 8 plus 1 so that worked out very nicely imagine that so there you have it so this is going to be equal to this is going to be equal to the definite integral actually let me make it clear this is what is equal to this the definite integral from U is equal to U is equal to 1/2 U is equal to 9 to make it very explicit that I'm dealing with you now of the square root of u instead of DX we have DX is 4/9 D you let me do it this way square root whoops that's not the right color square root of U instead of DX we have times 4/9 D U and I'm just going to take the 4/9 and stick it out here 4/9 D u and we know how to we know how to apply the fundamental or the SEC is the second fundamental theorem of calculus here to to evaluate this definite integral this is going to be four ninths times the antiderivative of U of the square root of u which is the same thing as u to the one-half is going to be u to the three-halves and then we divide by three halves which is the same thing as multiplying by two thirds and we're going to evaluate that at U equals nine and U is equal to one and so we're in the homestretch here this is going to be equal to 4/9 times times let's see two thirds times 9 to the three-halves minus 2/3 times 1 to the 3 half so 9 to the three-halves that is let's see the square root of 9 is 3 to the third power is 27 and this of course is 1 so we are left with 2/3 of actually let's just factor out the 2/3 this makes it easier so we're going to have to be left with this is going to be equal to 2/3 times 4/9 is is equal to 8 over 27 I've just factored out the 2/3 and then we're gonna have 27 minus 1 inside I guess you could say the brackets now so 27 minus 1 is just going to be 26 it's going to be 26 times 26 and we could obviously we could simplify this more if we want we could live whatever 8 times 26 is so actually let's just let's just figure that out just for fun so eight times 26 is going to be so this is going to be 160 plus 8 times 6 is 48 so it's 208 208 over 27 and we are and we are done
AP® is a registered trademark of the College Board, which has not reviewed this resource.