The Arc Length of a Smooth Curve (in Detail)

Video Series: Applications of Integration (Complete In-Depth Tutorials For Calculus)

(D4M) — Here is the video transcript for this video.

00:00
in this episode you’ll learn how to find
the arc length of a smooth curve without using a tape measure let’s do some math
[Music] hi everyone um we’re going to begin by
asking uh the question what is the arc length of a smooth curve
and in fact let’s start off by trying to understand what a smooth curve is
so we’re going to say um we’re going to take a plane curve just a curve given by
y equals f of x and we’re going to be starting with an interval here so let’s
you know think about um here is this f is a plane curve it’s a function of x so
that’s out so let’s look at something like this and we got some a and we got
some b and we’re going to talk about um this being smooth
so whenever the derivative is continuous so we know what that means the

00:01
derivative and what it means to be continuous right
so the derivative has to be continuous then we know it’s a smooth curve
so um for example um the absolute value function is not differentiable at zero
there’s a sharp corner so we want it to uh not be interested in those uh prop um
type of problems right now so we’re gonna have a nice smooth curve and
what we’re going to say is that if we have a smooth curve
then we can talk about the arc length of the curve and it’s going to be given by
this formula right here so this is what the this video is about this episode is
about so obviously we need to know what um derivative is what continuous means
and and integrate right so i just before we get started i want to mention that
this episode is part of the series applications of integration complete

00:02
in-depth tutorials for calculus um and so this is um something that
we’re going to be talking about in this episode right now now
um if we have here um the the arc length is what we’re going
to be interested in so the arc length is i’ll just shade it in this
this would be the arc length and here would be f of a and here would be f of b
and we could talk about finding the arc length right there
so we’re going to need to we’re going to use definite integrals to find the arc
length and we’re going to do approximation by straight lines
so we’ll be able to take a straight line as an approximation right
here and then do another one and then do another straight line and then choose
another point and another straight line so we can approximate the arc length
by uh doing that by choosing some x values down here and building the um

00:03
so we’re gonna we’re gonna rely upon a partition and the distance formula so
let’s just recall the distance formula right here so the distance between two
points here we’re going to have the square root and
we’re going to have x2 minus x1 squared plus y2 minus y1 squared
where a point here is given say is x1 y1 and then x2 y2
and the y’s we’re gonna you know be be using the function to
find the heights there so um a rectifiable curve is
let’s let’s spell this here rectifiable so this means
that the arc length is finite in other words if you find the arc length of it
it’s finite and so sufficient condition for a function to have
a finite arc length or to be rectifiable between a and f a and

00:04
the other point will be b and f b to have finite arc length between them is for
for our function to be have a smooth uh sorry have a continuous
derivative in other words it’s smooth so
a continuous derivative is the condition that we’re going to be working under
here and um you know just as a quick thing here if your derivative is continuous
then this whole integral here this integrand here is continuous so the integral
we’ll be able to find the integral right there a definite integral there right
there uh assuming we can find an anti-derivative um
yeah so let’s get uh started here let’s look at sort of a
motivating example right here if we have say y equals f of x here and

00:05
let’s go with something more general than that one right there
so let’s say here we have y equals f of x and we’re going to
just put down a curve right here say something like this and
and we’re going to have an a and we’re going to have a b
and so we’re looking at finding this um distance right here this arc length
what is the distance along this path here
and so to do that what we’re going to do is to partition here we’re going to
partition this and we’re going to do it in the abstract
now instead of doing just four or whatever we did in the
uh so let’s just put down a partition here and you know we’re going to have
say a representative here because we’re going to have a whole bunch of them
let’s just put a whole bunch of them here and
we’re going to say that the the width of each one delta x i

00:06
is going to be x i minus x i minus 1. so
that’s the width of each one right there and and the delta y
is the change in height so that’ll be y i minus y i minus 1. so here’s the
change in x and here’s the change in y and the reason why we need both of these
is we’re going to be finding the distance we’re going to do a finite
number of these um little little edges here these little
pieces you know if you just did two it wouldn’t be very accurate of
approximation we could find the distance and add it to this distance here but
that’s not that’s not going to be very accurate so let’s do this n times so we
got these little tiny little line segments going in here and the smaller
the the part the the finer the partition then the finer the then the smaller the
you know the smaller those little distance are but they keep adding up right
so we add in all these little tiny little distances for each for each of

00:07
these sub intervals here we have a little line segment here all right and so
how would we find uh the the distance right here so the arc
length which i’m going to denote by s is going to be approximately equal to
the sum of all the distances so we’ll go from 1 to n and we’re going to add up
all the distances and so we’re going to say x i minus x i minus 1 squared
plus and then y i minus y i minus 1 squared and so and so like if we have a
representative right here this will be x i minus 1 and then this one right here
will be x i and so that little distance right there that little change
and we’re going to come up here and get a y i and a y i minus 1 using the
function right there and we’ll get a little change in the y right there
so we get a little change in the y we’ll get a little change in the and the

00:08
little change in the y little change in the x and you know [Music]
uh sorry let’s put it that way and so the distance right there that we’re
approximating is using that little triangle that little right triangle in
there so ultimately we’re using pythagorean theorem here but
but we’ll just say distance formula okay so
now we can manipulate this a little bit here so let’s go from one to n
and what i’m going to just do is say this is delta x i
and then square that and then delta y i and square that
because that’s the notation we can use it’s it’s a lot nicer notation
um and so what i’m going to do is i’m going to factor out this delta x i
squared part right here and so what we’re going to have is the sum from 1 to n
and if we factor this out here what we’re going to get is i’ll just leave
the square root here first and say delta x i squared plus and

00:09
let’s say delta y i over delta x i and then all that squared
and then times the delta x i squared right so before we factor it out let’s
just go ahead and get uh you know delta x i squared over delta
x i squared and then now we’ll just go ahead and factor this out right here we
factor out the delta x i in each one of these right here
and so when we do that we’re just going to get a 1 plus and then we get a delta
y i over delta x i here and yes we’re factoring out this delta x
i squared when it factors out we’re going to take square root of that
squared so we just get delta x i here and so this right here is going to be
the motivating formula right there let’s
see if i’m out of the way like that yeah
so this is delta x i hanging out here we have the square root of 1 plus
and then this is the delta y i over the delta x i squares

00:10
all right so this is just a manipulation right here so all of these are equal in
here except for the fact that the arc length is approximately equal to all of
these so the uh arc length so just to like rehash here this is the arc length is
approximately equal to this so this is the formula right here that i like to
keep here before we move on to [Music] this next part here
so yeah um what we’re going to do now is keep refining the partition here and so
we may take the norm of the partition which means the largest piece but
because if we use a a regular partition then you could just
say n goes to infinity so however you want to think about it you want to think
about if you want to choose arbitrarily representatives in an arbitrary
partition then you want to say that your norm is going to zero in other words
your largest of these sub intervals is is always going to zero or you want to

00:11
say n goes to infinity so whichever route you want to take we want to pass a
limit and then we’ll get the actual arc length let’s put that right here
so the arc length of the actual arc length will be equal to the limit of
this right here so we’ll say the limit as n goes to infinity
of all this right here i equals 1 to n and then we have the square root 1 plus
delta y delta x all that squared and then times delta x
and so this right here will be the arc length now this is a limit of a sum
though and that’s not exactly what we want this is certainly possible to work
out especially if you’re going to use a computer you could use this uh to do
some approximations which is just basically tracing this up here
but the fact of the matter is is that we can do better than this right here this
is not what the goal was that i showed you on
the first um at the first part of the video so what we want to do is get

00:12
better than this so let me erase this up here
and what we can do is we can bring in the derivative we can bring in the
derivative and i’m going to show you how right now
so um first off what we’re going to say is that
the derivative exists so this right here exists our remember our function has a
continuous derivative so in particular this exists
and it and it does so in the interval x i minus 1 to x i
and so we’re going to use so by the mean value theorem
we’re going to get an existence of a c so there exists c i in this interval here
x i to x i minus one so for each one of the partitions right there’s a ci in it
and such that so if you don’t remember the mean value theorem here’s what it

00:13
says it says that f of x i minus f of x i minus one all over x i minus x i
minus 1 is equal to the derivative at c i and so what we have going on here is
this is the actual delta y i’s right because that’s how we’re finding
the y’s you plug in the x’s right so this is delta y i over delta x i
and that’s equal to f prime at c i and so now we can piece it all together
now we can rewrite this limit right here and instead of using the delta y i over
the delta x i we can write the derivative right here so let’s do that right here
let’s just come back up here and say s is remember s is with the limit so

00:14
we’re going to have the limit as n goes to infinity of the sum from 1 to n
of 1 plus and now instead of delta y i over delta x i we can get f prime at ci
and the c’s are coming because the mean value theorem holds the mean value mean
value theorem gives us the existence of the c’s that you can plug into the
derivative to get this right here so this will be cis here and then we have
delta x i here and so yeah there we go there is the
limit of the sum of this right here which by definition here is going to be
the integral here from a to b of 1 plus the derivative times dx
let’s put the square root here there we go so there’s how we get the
arc length formula right there the arc length s is the integral from a
to b of square root of 1 plus the derivative at x times dx right there

00:15
um and so yeah if the derivative is continuous
then this whole thing here is continuous and then this integral here will exist
and and perhaps we’ll be able to find it and
so in some cases we’ll be able to find this right here it just depends if we
are able to find an anti-derivative of this integrand here and be able to
integrate that here so this is the arc length here all right so
let’s do some examples yeah so here’s our first example here
let me get this stuff out of the way so in our first example we’re going to
look at this function right here and x is six plus eight over 16x squared
over the interval two three now um we motivated all that and i wanna
you know not lose that understanding because you know we can just integrate
um and be done with it and find the exact value so but what i actually am more

00:16
interested in is trying to actually understand the process and make sure
that we that we’re understanding it so to help us with that we’re going to
go to some python now so we’re going to open up a python notebook
and so let’s make this bigger so we can see this here
now if you’ve never used python before i recommend checking out the link below
in the description where you can open up your own python
notebook and you can follow along you can type up everything that we have in
here and it’ll work perfectly so yeah follow along now if you don’t want
to look at python you don’t have to you can just follow along but what we’re
going to do is we’re going to import some packages into our python notebook here
and this will make it so this will allow
us to do a lot of work very easily we’re
going to import numpy and we’re going to make some plots here so we’re going to
import those two packages here so we execute the cell by hitting shift enter
now i like to make up this function right here to customize my axes on my plots

00:17
so i’m going to execute that cell also and then i’m going to make up this plot
function right here to help me make some plots
all right so again we executed those cells
and then one last thing is we’re going to use the distance formula so here’s
just a quick quick function that gives us the
distance so here x1 y1 is a point and here x2 y2 is a point so make sure you
put them in that order uh if you switch orders it won’t have the same meaning
and so we’re going to have the square root of x2 minus x1 squared plus y2
minus y1 squared all right perfect so let’s execute that
all right so there’s our setup and so we’re just going to have some plots
and some pretty plots and and do a distance formula here so now let’s look
at some examples so i want to look at this first example uh that we’re working
on where it says find the exact here the
word exact is unnecessary because we use the word the

00:18
but i just put it in here for emphasis our our problem is to find the exact arc
length but i’m actually going to approximate it first
so i’m going to type in a function x to the sixth plus eight
so x to the six plus eight and i’m going to divide by x squared right here 16 x
squared and i’m looking on the interval 2 3.
so i’m going to execute the cell here and now my function is defined
and i’m going to go and plot it so plot function
and then plot the function f and then work on two three and so here’s what our
function looks like on two three and so we’re trying to find the arc length
now as in a first approximation i’ll just say what is the distance between
two and f of two in other words this point right here two and f of two
and this point right here three and three at f of three
so what’s that distance so that distance is straight just a straight line so
that’s obviously not going to be the same thing as the arc length so if i

00:19
execute this cell right here hit shift enter it’s 4.1
you know and that’s too much that is going to be too big
sorry that’s going to be too small right the distance between two points is
always the shortest shortest distance but anyways
that’s going to be too small so what we really are interested is the arc length
so um that’s the first approximation now let’s do another approximation so now
let’s look at two f of two this one right here
and then let’s look at uh so this point to 2.5 so 2.5 is about right here
and then the next point we’re going to look at is 3 f of 3.
so this first one right here this first function d right here finds the distance
right here and the second one right here the
distance between 2.5 and f of 2.5 in this one so this one right here is
so that will give us a better approximation

00:20
of the arc length and so if i execute this cell right here
we’ll get a better approximation so this one right here is going to get to be f
of 1.27 let’s execute that all right and then i did a whole bunch
of them right so a whole bunch of them so let’s see here can we see
how many i did i try to zoom in the right amount
so i did 2 to 2.5 2.25 2.25 to 2.5 2.5 to 2.75
and then to finally the three right there so that would be
an even better approximation right there but you know what rather than type all
that up let’s just go to the next one right here
um sorry let’s go to the next one right here
and i’m just going to make up an arc length formula right now so this will
allow us to do this for not only two points or
you know a finite number of points any finite number that we want and n so i’m

00:21
going to input a function into this function called approximate
arc length i want to input a function an interval a b
and how many uh like line segments do i want to add up the distance of the line
segments right so i’m going to set my initial distance to
be zero i’m going to loop through however many line segments i’m going to have
and so i’m going to start off with zero of them
and then i’m going to make a bunch of x so this is like putting down a partition
so we have a to b and depending upon what n is we’re going to
make a partition here it’s going to be a
regular partition it’s going to be equal with right here
and then i’m going to loop through the um all the way to n i’m going to loop
through them so the first one second one third one fourth one fifth one and i’m
going to add up all the little tiny distances so this is our distance formula d
and we’re going to choose an x and then this will be the y right you

00:22
you take the x and you put it into the function
and then this will be the next one so the x i and then the x i plus 1. so
this will be the next x and then i’ll take that next x and plug it into y
right here so this is the distance formula and what this right here means
is that i’m just adding what the previous distance is and i’m going to
add on the new distance so i’m just adding up all the little distances
all right and so this will give us the approximate arc length here so let’s
execute that cell there and then now i’m going to take the function on the
interval 2 3 and i’m going to do with two line segments there and then three
and then we can just jump down here and do as many as we’d like
and each time we do this we’re getting closer and closer to the arc length if i
do a big one like this let’s see how long that will take my computer to go a
little bit of a pause here and then it comes out so that one is pretty close
now we actually know using calculus so this is all just approximating just

00:23
adding up just kind of a brute force method of adding up all the little
distances but the actual exact value is 595 over 144 and there’s an
approximation to that number but this right here is the number that’s the
exact so remember the problem was to find the exact arc length and this is
the exact arc length right here and i’m going to show you how to find that right
now of course using calculus using integrals and derivatives
so what we’re going to say is the arc length here is given as
right so the arc length was given as the integral from a to b so two to three
and then one plus and then we have the derivative
and then we have d of x right so we need to know what the derivative of this
right here is so let’s go over here and do that so first thing i’m going to do
is say what’s a better way to think of the y so the y is like uh

00:24
1 over 16 and then x to the fourth and then plus
and then we have 8 over 16 so one half and then x to the minus 2
and so what is the derivative here and since i’m using f notation over here
um i’ll just say f of x right here and so what is y prime
so it’s just going to be 1 4 x to the minus 3 sorry and then minus and then x
to the minus 3 right here and so there’s our derivative right there
and now we want to square our derivative don’t forget that right there
we want to square our derivative and so now we know what it is we can go
in here and do some work so this will be integral 2 to 3 and then 1 plus
and then we’re going to get 1 4 x to the third plus x to the minus 3

00:25
and then we’re going to square all of that so let’s say here one plus and
dx there all right and so what we’re going to do
is we’re going to expand this out and regroup and when we do that
so we’re going to expand this out here and
yeah so when we do that we’re going to get here square root of
1 4 x to the third plus x to the minus 3 dx
and yeah then if we integrate this part right here we’re going to get the 595
over the 144 here all right so yeah it’s just a matter of
expanding this out right here adding you’re going to get a constant
term in there because you’re going to get a number um
you know this times this you know times two and the exponents are going to

00:26
add up and get x to the zero so you’re going to get a constant then you’re
going to add the one to this and then you’re going to be able to refactor it
out like that all right so um this will be a squared here
okay so yeah we’re going to get 1 plus and then if we square this out we’re
going to get what 1 over 16 x to the sixth
and then two times this one times this one so two times one half so it’ll be
one half and then x to the third times x to the
minus three right and so then here we’re going to get here um x to the minus um

00:27
sixth six so let’s see here think of this as three halves does this
so it’s going to be um 16 and then okay so this will be minus right here
and so this will be that’s a minus right there so this will be minus right here
so this will be positive one-half now um and so then we can factor this
into that right there all right perfect great
all right so i hope that helps um and then
let’s see so let’s do another one that might be um even more challenging
or well we’ll see if it is or not and let’s go and do the same thing before
let’s before we find the actual arc length and do the calculus part
let’s look at how you would approximate such things

00:28
um and then i’ll talk about why why this actual part right here is
actually important so find the arc length of the natural log of cosine x on 0 4
so we’re going to define our function g right here let’s execute that
and then we’re going to work on zero to pi over four
and i thought it would be good idea to just graph the function right there so
we can see what we’re trying to find the arc length of and i did some
approximations so here’s the point right here
zero g of zero and then pi over four which is about right there to about
right there so my initial estimate just going straight here just straight there
that’s the approximate arc length there and so then we did some more i did some
more i took in another point here at 0.2 so 0.2 is about right here so now
we can find the arc length here and then the arc length all the way to there

00:29
and so that’s the this distance right there so the distance and then plus and
then the second distance um and then so let’s just go ahead and
use our approximate arc length function that we defined above and so we
can find the approximate approximation there and we can keep doing them until
we feel satisfied that we found a good approximation right here
so let’s see if we can do 100 000 right here and a little bit of a
pause but not much um and then i actually want to
um yeah so there’s the approximate arc length there
and so let’s go find the exact value now let’s go back here and find the
um exact arc length or find the arc length of the graph right here
all right so first off we’re going to find that the derivative is equal to

00:30
let’s move that down a little bit the derivative is equal to minus tangent
and we’re going to use the arc length formula so i’m going to use an l this
time l for arc length and so we’re going to integrate from 0 to pi over 4
and we’re going to have 1 plus and then minus tangent x and then all that’s
going to be squared and then dx and so let’s go ahead and integrate this
so we’re going to get 1 plus tangent squared
and we’re going to square root on that so let’s just
take the square root of the squared so this will just be secant then and so now
to integrate the integral of secant well maybe one way is to
uh use a conjugate so i’m going to say secant x plus tangent x

00:31
and i think that might help us so i’m going to say u is is secant tangent
and then d u will be secant tangent plus secant squared
and then all that times dx all right so good so let’s go here and
find the arc length here that’s a zero so the arc length is equal to that is
equal to that is equal to this so now let’s use our u substitution here
and so what we’re going to have here is um the secant tangent here is a u
down here but here i have secant times secant that’s the secant squared and
secant times secant tangent right so all this numerator right here is the du

00:32
so this will be one over udu and so now when x is zero
so when x is zero what do we get out here we get a
zero here and we get a one here so this will be one so when x is zero
then u is one when x is pi over four now we’re gonna get
1 plus square root of 2 right 1 plus square root of 2. so 1 plus square root
of 2. so we’re integrating from 1 to 1 plus square root of 2 of 1 over udu
and so this will just be natural log of 1 plus square root of 2.
right we just have natural log um maybe that looks a little funny
so i’ll just say natural log absolute value of u 1 to 1 plus square
root of 2 over 2 the natural log of 1 all right and so
then we’re just going to get parentheses here first
1 plus square root of 2. there we go all right so that gives us the arc
length that gives us the exact value of the arc length and just to finish up

00:33
what we talked about a minute ago is that’s why i approximated this number
right here natural log of 1 plus square root of 2 over 2.
this is an approximation of the abs of the exact value and you’ve seen here
when we did the um when we did the approximation with 100 000 right here we
got pretty close to the approximation to the exact value so
that’s definitely the exact value here’s one way to approximate made it using
these um [Music] little pieces here and we were able to do this because
this part right here we’re able to integrate so right so not
always are we able to find the integral even though its value may exist
because because the integrand it will be continuous but
you know in this example we were able to find some

00:34
substitution or some method so that we’re able to find the exact value
but the approximate value is something that we can always do we can always
approximate it as much as we want all right so now let’s look at
what if we don’t have a function of x so let me get rid of this real quick
all right so let’s switch the roles of x
and y and obtain arc length for a smooth curve having the form now we have x
equals g of y so right so g here has to have a derivative and it has to
be continuous and that’s the condition that we’re
going to work on so if this is a smooth curve on an interval remember this is a
y interval now then the arc length can be done using the same methodology we’re
just going to keep adding up line segments
and um so we’re going to just look at an example of how to do this now i’m not
going to go and do the do the python or the approximation anymore let’s just go

00:35
ahead and find the exact value here now so we’re going to say that
g of y is right here and we need to find g of y prime
so g of y prime is one half y to the third and then minus one half
and then y to the minus three all right um
and so we’re good to go now let’s find the arc length so the arc length is from
one to four so we’re going to go from one to four
we’re integrating along the y axis now so we’re going to have square root of 1
plus and then we’re going to work all this out here one half y to the third
minus one half y to the minus 3. we’re going to square all that and then this
is d y here all right and so we can integrate this right here one to four and um
yeah so this right here is the derivative right here squared that’s good

00:36
and so let’s go right here and say 1 plus 1 4 y to the 6 right and then minus 1
half and then plus 1 4 and then y to the minus 6 and then dy
so just squaring that out right there and then now we’re going to get the
integral 1 to 4 and we’re going to have the square of the
square root i’ll put it in but then i’ll erase it y to the third
plus one half y to the minus three and all that’s a squared so we’re going to
add those together and then factor it d y um yeah i guess i’ll leave it and so
integral one to four and we have square square root so it’s going to be one half
y to the third plus one half y to the minus three all right so there we go

00:37
and so yeah if we integrate this out and substitute in the four and the one
we’re going to get 20 05 over 64. and this is going to be the exact
arc length the exact value of the arc length all right so good
let’s look at another example so i promised you that you know maybe
we’ll look at one that was uh challenging that maybe we couldn’t solve um
find the arc length what about if we have this curve right here
will we be able to find the derivative and integrate so
if you try that you’re going to run into problems so i’m going to show you
something that will will help so i’m going to say e to the y we have
natural log in all this right here right so i’m going to say i’m going to bring
in e to the y here and take that approach so i’m going to say e to the y

00:38
is this inside part right here so x minus
all right so that’s not a 4. it’s a y so let’s make that a y so
x minus and then square root of x squared minus 1.
so what would be useful in doing that right
so we can also say what e to the minus y is so e to the minus y what is that
well let’s say here e to the minus y i’ll come and work out out over here and
then i’ll come and put it over here and save it so e to the minus y
is right so if we have a minus on here so let’s don’t let’s don’t
start off with either the minus y let’s just put a minus on here so minus y
minus natural log of x minus x squared plus one [Music]
so just multiplying by minus on both sides but now what can log do with a

00:39
minus sign so we can bring this up as a power and say minus one
and then we in fact can just say that this is one over and so
e to the y e to the minus y if i now take e to the minus y it’ll be
e to the ln and that will all cancel out so we’re going to get 1 over x minus
so that’s what e to the y is but actually we can do even better than this
i’m going to multiply by the conjugate and now
what we’re going to get here in this numerator is just x plus
and what happens down here in the denominator we’re going to get x times x
which is x squared and then we’re going to get minus x squared plus 1
and the x squares cancel out we’re going to get a minus 1 so it’s going to be a

00:40
minus 1 down here um so let’s see here this is plus plus
and this is going to be x squared and then minus
and then this here so that’s going to it seems like we get a minus 1 down here
and so excuse me um yeah so we’re gonna get a

00:41
minus one it sure does seem like we’re going to get a minus one here but um
this seems all good and x minus oh i see what it is that’s a minus one there
that’s minus one and then over here i put plus one okay so minus one
and then minus one because this was not going to um [Music] this is minus 1 here
and so this is minus minus so this is plus so
this is just a 1 down here underneath all right so e to the minus y is x plus
and then square root of x squared plus 1 right because
sorry actually this is still a minus 1 here and a minus 1 here

00:42
because i had made that one mistake up here i brought it all the way down so
when i multiply by negative 1 here it’s negative
y minus natural log and then this is a minus 1 here minus 1 minus 1 minus 1
minus one minus one minus one and then this is all over one and the reason why
it’s all over one is because you have negative times positive so it’s a
negative and we have the square root with an x squared minus one so this
would be a negative times a negative that’s positive long story short this is
the square root of x plus x squared minus 1. so
even though we’re start we’re given a function of x that doesn’t necessarily
look so friendly we can look at the e to the y’s and they
look much more friendly so let me erase all that right there and now
what we’re going to say here is if we add these up together see if we

00:43
add these up together then what happens is this part right here cancels out so
we’re going to get e to the y plus e to the minus y is just simply 2x
so in other words x equals e to the y plus e to the minus y all over two
and so what we’ve done is is we started with a function of x
and we got something that that may have been very difficult to work with right
because what we’re what we’re going to start off with we’re going to do
the the arc length is integral from 1 to square root of 2 and then square
root 1 plus and then we’re going to take the derivative y prime and then we’re
going to square that y prime squared and then dx
and we’re going to have to do that so we’re going to have to take the
derivative of this and then square that and then plug all that into here and
that was not going to be uh very friendly so
instead what we’re going to do is we’re going to integrate with respect to y

00:44
in order to do that we have to solve for x how do you solve for x
well we’re going to use this method here
we’re going to say e to the y if we just
say e to the y and then e to the natural log so i’m hoping i didn’t lose anyone
when i said e to the natural log of something is just that something so
yeah all right so we solve for x so now we have our g of y
function right here and now our arc length can be given by instead of
integrating with respect to x we can do with respect to y
so now we’re going to integrate and now when we integrate here we need
to change our bounds when x is one then what is the y so when x is one
that’ll be zero we have natural log of one so we’re gonna have um
i see yeah so we’re gonna have a zero and then when x is square root of two

00:45
we’re gonna have a square root of two so it’ll be two minus one so it’ll be one
so would be square root of natural log of square root of 2 minus 1 parentheses
all right and so that right there are the new bounds and we’re going to
have square root of 1 plus and so now we’re going to need to take the
derivative here of g of y here and so what would be the derivative here
so we’re going to take the derivative here g of y prime will be e to the y
and then minus e to the minus y all over 2 right 2 is just a constant there
and so that’ll be y prime g of y prime there
and so now we’ll use all that up in here and square this right here so we’re
going to have e to the y plus e to the sorry minus
minus right here minus e to the minus y all over 2 and then we’re going to

00:46
square that and then say d of d y so this will be the arc length right
here we just need to integrate this right here can we can we do that
and so the answer is yes because um you know we’re just going to get e to the y
minus e to the minus y all over 2 and we’re going to evaluate from natural
log of square root of 2 minus 1 [Music] 0 and for all that work
substituting in zero and then the natural log
and when you substitute the natural log you’re going to get e to the natural log
right so this will be workable and then we’re going to get one for the arc
length that seems like a lot of work just to get the number one all right and so
there’s one way which to do this if you try to
integrate with respect to x you’re going to run into

00:47
a lot more problems if you’re going to integrate with respect to y it’ll be a
challenge to solve for x but if you’re clever enough you can do
that and then we can integrate right there like that all right so um
one more thing before we go this right here is i think important is
if you’re going to find the arc length of a straight line it sure does seem
like the arc length formula should just give you the distance formula
um and so how does that work though because
the arc length formula has derivative in it and the distance formula doesn’t so
how can we reconcile that so let’s find the arc length of one
point to another so let’s say we have this point in that point and we want to
find the arc length and it’s just a straight line m is a
slope and b is a y intercept that’s given now remember the slope is given by y2

00:48
minus y1 over x2 minus x1 and also remember that the slope of the
tangent line in this case is just the derivative so the derivative is just the
slope so let’s find the arc length l we’re going to integrate from x1 to x2
and we have our linear function here so we’re going to have square root of 1
plus and then we’re going to have our derivative squared so what’s our
derivative right so the derivative is just constant m
just take the derivative right and so we have m squared actually
all right and so we have i’ll just put here yeah m squared and then dx
but what is the m squared well the m is just y2 minus y1 x2 minus x1
square all that dx and so now what we’re going to do

00:49
is do some algebra and work with this here a little bit x1 to x2
and so now we’re going to basically undo
the process we did at the very beginning instead of
factoring out the x2 minus x1 now we’re going to
do a common denominator so we’re going to have here square root of we’re going
to put this 1 as an x2 minus x1 squared over x2 minus x1 squared
plus and then y2 minus y1 squared then x2 minus x1 squared dx
so now i’ll go down here and you see that we can factor out here
this x2 minus x1 squared on here and so actually though we’re going to integrate
right we integrate we’re going to get an x and

00:50
then we’re going to do x2 minus x1 so let’s do this right here
let’s say we have square root of x2 minus x1 squared plus y2 minus y1 squared
all that’s a constant and then we have the common denominator
here so i’ll just put this right here as x2 minus x1
and then since we’re integrating with it just a dx and remember that’s just a
constant so like the integral of 2 dx is just 2x and then from
from a to b it’s just you know a a to b right so
if it’s just a constant there right so anyways so this would just be x
i’ll just put an x here and then this will be x1 x2
and so now we’re getting these x2 minus x1 squared plus y2 minus y1 squared

00:51
and here when we have the x2 and then minus the x1 and then that will cancel
with that so this part right here cancels with this part right here and we
in fact just get the distance formula right there
so there we have it yeah the arcane formula of course
should give you the distance formula if you’re just along a straight line
all right so i hope you enjoyed this episode and um
i’ll see you next time if you enjoyed this video please like and subscribe to
my channel and click the bell icon to get new video updates

About The Author
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David A. Smith (Dave)

Mathematics Educator

David A. Smith is the CEO and founder of Dave4Math. His background is in mathematics (B.S. & M.S. in Mathematics), computer science, and undergraduate teaching (15+ years). With extensive experience in higher education and a passion for learning, his professional and academic careers revolve around advancing knowledge for himself and others. His work helps others learn about subjects that can help them in their personal and professional lives.

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