Main content

## Calculus, all content (2017 edition)

### Course: Calculus, all content (2017 edition) > Unit 8

Lesson 2: AP Calculus BC questions- 2015 AP Calculus BC 2a
- 2015 AP Calculus BC 2b
- 2015 AP Calculus BC 2c
- 2015 AP Calculus BC 2d
- 2015 AP Calculus BC 5a
- 2015 AP Calculus BC 5b
- 2015 AP Calculus BC 5c
- 2015 AP Calculus BC 5d
- 2015 AP Calculus BC 6a
- 2015 AP Calculus BC 6b
- 2015 AP Calculus BC 6c
- AP Calculus BC exams: 2008 1 a
- AP Calculus BC exams: 2008 1 b&c
- AP Calculus BC exams: 2008 1 c&d
- AP Calculus BC exams: 2008 1 d
- Calculus BC 2008 2 a
- Calculus BC 2008 2 b &c
- Calculus BC 2008 2d
- 2011 Calculus BC free response #1a
- 2011 Calculus BC free response #1 (b & c)
- 2011 Calculus BC free response #1d
- 2011 Calculus BC free response #3a
- 2011 Calculus BC free response #3 (b & c)
- 2011 Calculus BC free response #6a
- 2011 Calculus BC free response #6b
- 2011 Calculus BC free response #6c
- 2011 Calculus BC free response #6d

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# 2011 Calculus BC free response #3a

Arc length of a function. Created by Sal Khan.

## Want to join the conversation?

- At3:36, Sal uses the arc length formula with 1 + dy^2 whereas in the previous video for #1d, he used dy^2 + dx^2 at the end. What determines which to use?(1 vote)
- Nevermind. If anyone wants the answer, when it is parametric or x and y each receive their own equation, use this equation and for f(x)=y, use the 1 +...(2 votes)

- Surface area of a parametric equation?(1 vote)
- integrate from a≤t≤b 2π*y(t)√(x'(t)^2+y'(t)^2)(1 vote)

- Hi, where is the video sal did on finding the arc length?(1 vote)
- I think you may find the preceding video helpful:

https://www.khanacademy.org/math/calculus/ap_calc_topic/bc_sample_questions/v/2011-calculus-bc-free-response--1d(1 vote)

## Video transcript

Problem three. Let f of x is equal
to e to the 2x. Let R be the region in
the first quadrant bounded by the graph of f,
the coordinate axes, and the vertical
line, x is equal to k. So they drew that
right over here. And it's in the figure
above in the actual AP exam. But I put it below
here so I didn't have to waste the screen
real estate above that. And you see right here,
this is our f of x, this is the line
x is equal to k, and this region R
is between those, and it's in the first
quadrant right over here. So this right over
here is our region R. So what are they
asking us to do? So k is greater
than 0, the region R shown the figure above. And this is the figure. Part a. Write, but do not evaluate
an expression involving an integral that gives the
perimeter of R in terms of k. So for the perimeter
of R, we're going to have to find the
length of the sides of R, and the hardest of
these is to find the length of the actual
curve between this point and this point right over here. And I like to always rederive
the arc length formula, although if you're about
to take the AP exam, it probably doesn't hurt
to have it memorized just so you could save some time. But you should
always know how it's derived so that when
you're 35 years old and you don't have a
calculus book as a reference, you can come up
with it yourself. And so you know where
the formula comes from. So let's rederive it. So let's say we were to zoom
in on a little bit of the curve right over there. So let's say let's zoom in
on that part of the curve right over there. And what we could do is,
same way we derived an arc length when the function
was defined parametrically, which we did, I think, in
part one of this AP exam. This time it's not
defined parametrically, so we can do it in
terms of just x and y. So if you viewed this
distance right over here, this is going to be a
very small change in x. Let's call that dx. And then this right
over here is going to be a very small
change in y, dy. And we know that dy, the
derivative of y with respect to x, or dy/dx, this is
equal to f prime of x. Or we could say that if we
multiply both sides by dx, we know that dy is equal
to f prime of x dx. So this is equal
to f prime of x dx. And then for a very small change
in x and very small change in y, we can approximate
this curve length with the Pythagorean theorem. And if we get small enough,
it'll probably directly kind of measure
that curve length. So we use the Pythagorean
theorem right here. This length is going to be
the square root of dx squared plus dy squared. And if we write dy this
way, that arc length right over here, this
little small arc length, is going to be equal
to, right over here, the square root of dx squared
plus-- this is dy-- so plus f prime of x
squared times dx squared. All I did is I squared
this. dy squared is this thing right over here. And let me expand the
radical sign a little bit. And just the same
exact way we did it when we had the function
defined parametrically, we can factor out
a dx squared here. So this is going to be equal
to the square root of-- if you factor out a dx squared
of this, you get a 1. You factor out a dx
squared out of this part, plus f prime of x squared. And we factor a dx squared
out of the radical sign, it just becomes a dx. And so that would be the length
of this little small arc here, and it wouldn't
even be this big. It would be a super
infinitesimally small piece of arc. We could call it d
arc right over there. And essentially want to sum up
all of these little segments. So we want to sum them up from x
is equal to 0, all the way to x is equal to k, because we're
integrating with respect to x right over here. So this arc length
right over here-- so let me write what they
want us to figure out; we're trying to figure
out the perimeter of R-- is equal to first, this part. So this is the integral
from 0 to k. x is equal to 0 to x is equal to k of the
square root of 1 plus-- and f prime of x. f of x is e to the 2x. So f prime of x is equal to the
derivative of this with respect to x, which is 2e to the 2x. The derivative of 2x is 2. Derivative of e to the 2x with
respect to 2x is e to the 2x. So 2e to the 2x, we
want to square that, because this is f
prime of x, but we want to square f prime of x. So that's 4e to the 4x. And of course, we have a dx. So this is the length of the
arc, and that's the hard part. And now we just have to get
the rest of the perimeter. So you have this little
segment right over here. That's length 1. We're going from
1 to 0 or 0 to 1. So plus 1. Then you have this part
right along the x-axis. Well, that's going
to be of length k. Plus k. And then finally, you have
this height right over here. And that's going to be of length
of f of k, or e to the 2k. Plus e to the 2k. And we're done. We found the perimeter of R, and
we don't have to evaluate it. So we're done with this part.