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Studying for a test? Prepare with these 10 lessons on Applications of definite integrals.

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# Worked example: arc length

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.