BAM! Mr. Tarrou In this video we're going to be taking a look at a scenario here
of a region R that is enclosed by two functions and we're gonna find arc
length we are going to also find the area between these two curves in the
description you'll find time stamps to skip ahead to the second part of this
video if you still like it is it going to be a calculator based question and I
also have tons of calculus lessons so I'll put a link to the all the videos
I've done for calc a B or calc one all the videos i've done for calc BC or
count two and I'm building up a set of videos for AP Calculus free response
style questions to help you prepare for the AP test I'll also put a link to the
AP central website if you want to download a PDF of old AP questions the
solutions guide marking schemes so on and so on but for now let's get going we
have our we're gonna let R be the region and close by the graphs of y is equal to
the natural log of x squared plus one and Y is equal to two times the cosine
of X what is the length of the boundary of this region R and I went ahead and
color-coded it we've got a blue function here which is going to be the natural
log function of course we understand that the domain is basically all real
numbers you can't take the natural log of the negative number but x squared
plus 1 is always going to be positive and we have an orange function here for
y is equal to 2 cosine of X well if you want to find the length of the boundary
basically the sum of two arc lengths an arc length for the cosine function y is
equal to cosine of X and the arc length of the natural log function we're going
to have to find out where do we start and end our what our integration right
we need to know the formula of course two so in case you forgot BAM pitcher
and pizza arc length of a function which is an you know regular function in
rectangular format there's another equation for arc length when your
function is described as parametric functions but we have arc length is
equal to the DEF and integral from A to B of the square root of 1 plus f prime
of X squared DX so this is what we're going
to be setting up for these two functions and when you I like to just put these
into our calculator graph maybe this y sub 1 Y sub 2 get the graphed up there
on our calculator and just hit second calculate and find this intersection
point which is going to be equal to X is right off the top of my head not really
one point one four zero eight seven dot dot dot and one of my in my last AP
calculus training we were told to tell our students not to round off in the
middle of the work but only round off their final answer to three decimal
places now if you were doing this question by hand you probably maybe
wouldn't take the time to graph it course always help sort comprehension of
the problem if we can see what's going on the cosine function is an even
function in other words F of negative x is equal to f of X if you plug in
opposite numbers and get the same answer from your function then it is an even
function which means it's symmetric to the y axis same here if you plug in
negative x negative x squared becomes positive x plus one so both of these
functions would pass this test both of these functions are even functions so
when we go to find this answer this is a calculator based question I'm not going
to beacon to concerned about making my work more efficient I'm just going to
type it into the calculator but if it was a problem that I have to do by hand
I would definitely want to integrate from zero to make some integrals from
zero to one point one four oh eight seven dot and then multiply my answer by
two because the length of this arc is going to be length is gonna be equal to
the length of this art enough and I can use zero in mind Gration process that's
probably gonna make the math easier or quicker both and then just simply double
my answer but we're going to go ahead and we're going to see that this
intersection point is that value and whether we ask the count
then find the second intersection point which is a little bit of work a little
bit of time or we see that both these functions are even this other value is
going to be X is equal to negative one point one four zero eight seven so not
taking advantage of the even nature of these functions just letting the
calculator do all the work that total length of the boundary is going to be
equal to the definite integral from negative one point one four zero eight
seven to one point one four zero eight seven of the square root of one plus the
square root of one plus f prime of x squared not f of x squared so we're
going to take the derivative of the natural log of x squared plus one which
we of course know is two x over x squared plus one derivative of natural
log of U is U prime over u and we're going to add that to the definite
integral from negative one point one four zero eight seven one point one four
zero eight seven now this is the arc length for the natural log between these
two values basically the lower boundary of this region now we're going to add it
with the arc length of the upper boundary which is going to be the square
root of one plus F prime so the derivative of two cosine of X is
negative 2 sine of x squared DX now make sure that I think it's a good idea for
the grading of the AP test that you write each of these integrals
individually one for each of the arc lengths we do have the same lower and
upper limits on both of these definite integrals so you can say technically as
you type this in the calculator you can even write it this way lower and upper
limit of this square root plus this square root with your DX
there was a previous one of the old AP questions where they were asking us to
find the area between two curves were those two curves intersected in a couple
of places and they were looking on the marking scheme for you to set up a
definite integral from A to B correctly with the upper function minus the lower
function so you get that positive area and then from B to C where again you
have to do the proper upper function minus the lower function so you get that
correct area now if you're really handy with your calculator you could type in
the DEF and end a rule from A to C and just wrap the difference of those two
functions and absolute values and get the answer really more efficiently but
you didn't get all the points on the AP marking scheme because you didn't
recognize that you really had to look for each of those intersection points
and truly unless you want to take advantage of that apps of absolute value
function of your calculator recognize that you needed multiple integrals to
find the total area bound between those two curves that's why I think it's I
think it's a good idea to write it so you have the arc length given by the
natural log function and the arc length given by the cosine y equals 2 cosine of
X even though their upper and lower limits are the same now right off the
top of my head that total length of the boundary around the area is
approximately equal to off your calculator 6.28 1/8 and if you stone the
calculator yes if you want to go ahead and find the area of this region as a
second bonus part of this video it's a really a calculus a B concept near the
end of calc a B here we can take advantage let's go ahead and take
advantage what we could before to but let's go ahead and a kadai functions
pass that that even test again the f of op you know F of negative x is equal to
f of X you plug in the opposite of X you plug in the opposite of actual get the
same answer from both of these functions the area is going to do this in green
the area of that region is equal to 2 times the definite integral from 0 to 1
point 4:08 seven dot dot Oh in case I did
forget to mention I'm actually really shooting this scene because I transposed
my numbers here I wrote 807 so apologize if this is a repeat ap does not want you
this what I was told to my last training they do not want you to round off these
numbers to three decimal places after the decimal one thousands place until
you're done with your work don't round off in the middle so include a couple of
dot dot dots there to indicate you're truly not rounding course not gonna put
in an infinite number of decimals but anyway upper boundary of this region is
defined by the cosine function two cosine of X and the lower boundary of
this region is by defined by the natural log function so the natural log of x
squared plus 1 DX course your lower boundary is zero because even functions
are symmetric to the y axis and in this case since we're gonna gradient along
the x axis that means when x value of 0 let's go ahead and with our calculator
find out that that area is approximately two point eight nine three let's get our
calculator out if you liked this video please hit the like button to help the
YouTube algorithms for more people to see this video and get to grab that
calculator and give us some practice and we'll finish right up we're going to get
these definite integrals typed into the calculator so of course you're gonna hit
your second math go down to where it says the numerical integration function
you're going to type these integrals in I already have these in here because
well you probably already know to do this but we have a lower limit and upper
limit we're going to type in that integral just the way it looks and you
can put in the two separate integrals again one for the arc length of natural
log function and one more for the length of the function 2y is equal to the two
cosine of X and get your answer of six point approximately six point two eight
two but since these integrals do have the
same upper and lower limit you can just put in that square root of 1 plus f
prime of x squared for the natural log function and add that with the same 1
plus f prime of x squared for the cosine function but I do show you that it's
going to give you the same answer in your calculator if you do recognize that
please make sure that you realize that is dangerous they now for the area
between those two curves we're going to go ahead and do two times
math numerical integration the definite integral from zero to one point one four
zero eight seven of two cosine of X minus the natural log off of y equals
fraction and we had well I'm thinking of the other ones that natural log of x
squared plus one and we're getting integrating here with respect to X so
there you go I'm mr. Tarrou BAM go to your homework
you