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## Calculus 2

### Unit 5: Lesson 9

Arc length: polar curves# Worked example: Arc length of polar curves

What is the arc length of the polar curve r = 4sin(theta)?

## Want to join the conversation?

- For Pi/2<Theta<Pi and 3Pi/2<Theta<2Pi, the value of r is negative. Do you mean r=Absolute value(4*sin(2Theta))?(1 vote)
- It does not matter if it is negative or not, because the values will get squared, which means that it will always be positive.

And another fact about r, is that when you are sketching the graph, the negative r values will transform into positive values, but in the opposite direction. This is why polar graphs look so pretty and why r(theta) is a function, even though it looks like it will not pass a vertical line test. If you look at the function definition, it's clear that for every theta there is only one r, so r(theta) is indeed a function.(4 votes)

- We can’t use calculators on tests :-( (sad face)(1 vote)
- Mostly because teachers don't want you to cheat. ;-) (winking face)(1 vote)

- The expression does convert into sqrt(24cos(4theta)+40) using the double-angle identity, but that is still troublesome to do out.(0 votes)
- Why cant we do this for arc's lenght...

∫ dr for some range of theta ?(0 votes) - Can you please make a video explaining how to do perimeters of regions between 2 polar curves? These problems show up on the practice but there's no video explaining them :((0 votes)
- I tried to solve this with u-sub: u=2theta, du=2 dtheta. This clears the radical nicely

but gives the wrong answer. Anyone explain this?(0 votes)

## Video transcript

- What I want to do in this video is find the arc length of one petal, I guess we could call it, of the graph of r is equal
to four sine of two theta. So I want to find the length
of this portion of the curve that is in red right over here. We'll do this in two phases. First of all I want to set
up the definite integral for finding that arc length,
and then we'll evaluate it. I'll actually use a
calculator to evaluate it because it'll be a little
bit more straightforward at least for this integral. Let's just start with that. I encourage you to pause the video. Knowing what we know about
the formula for arc length, when we have it in polar form, see if you can apply it to
figure out this arc length right over here. I'm assuming you've had a go at it, so let's remind ourselves
that the arc length is going to be the integral
from our starting angle to our ending angle, we'll
call it from alpha to beta, of the square root of the
derivative of our function. Which respect to theta squared,
plus our function squared. Plus our function squared d theta. Let's figure out what r prime of theta and what r of theta actually are. I'll color code it. R prime of theta, we
know what r of theta is let me just rewrite it. We know r of theta is equal
to four sine of two theta. Our prime of theta is just
going to be derivative of two theta with respect
to theta is two times four is eight, derivative of sine
is cosine, cosine of two theta. If we wanted to write the
definite integral for arc length it's going to be equal
to the definite integral. What's our starting bound? For the petal that we care about, we're going to start at
an angle of zero radians. When we have an angle of
zero radians, r is zero, that's this point right over here. We're going to start at zero radians. And we're going to go all
the way up to pi over two. When sine of two times pi
over two is sine of pi, which is zero, so we'll
get back to this point right over here. We're going to go from
zero to pi over two, to pi over two radians. It's going to be the square root of, give myself some real estate here, r prime of theta squared is
going to be this thing squared. It's going to be 60-- Let me do it in that same blue color, it's going to be 64 cosine
squared of two theta, then plus this squared
which is going to be 16 sine squared of two theta, then of course we have our d theta. You can attempt to solve this
or evaluate this analytically, it's not too straightforward so I'm going to use a calculator. It just gives us a
practice with knowing that hey, there are tools out
there that can help us do these types of things. I'll go to the definite integral, I'll go to second calculus on my t 85, I picked this choice, which
is for the definite integral. Let me express when I'm taking
the definite integral of, it's going to be the square root of 64 times cosine of two theta squared. I'm going to use x instead of theta just because it's an easier variable to use on my calculator. Cosine of two-- Actually let me make sure
my parenthesis are good. Cosine of two x, alright I'll close that. Now I want to square that. I did that first part. Plus 16 times plus 16 times 16 times sine of two x. Close that, close that. Squared. I'm not sure if the calculator
knows to interpret that as multiplication, so let me
insert a times right over here. 64 times cosine of two
x, that thing squared. Plus 16 times sine of two
x, that thing squared. Let me go to the end. Let me close my radical, so that's going to close
that right over there, that closes what I'm
taking the square root of. Then comma, the variable
that I'm integrating with respect to is x, in this case. Everywhere I saw theta here I'm just replacing with an x there. I want to do it from,
instead of theta equals zero to theta equals pi over two, we'll say x is equal to zero
to x is equal to pi over two. Hopefully I have not made a
mistake when I've typed this in, And I get... It's chugging on a little bit. And we get... Will it actually come up? Oh finally! If we were to round it
to the nearest thousandth it'll be nine point six eight eight. Approximately nine point six eight eight. Let's just see, does that
make intuitive sense? It goes as far as four away, so if you just went out
to four and then back down that would be about eight. Of course we have actually gone out some, so it actually makes intuitive
sense that this arc length would be nine point six eight eight. Hopefully you enjoyed that.