Polar Parametric Curve - Arc Length

Hi, everyone. Welcome back to integralcalc.com. Today, we’re going to be talking about how
to find the arc length of a polar parametric curve. So in this case, we have a polar parametric
curve defined as r equals e to the theta over 2 or 1/2 theta and we’ve been asked to find
the arc length associated with that parametric curve when theta is between zero and 4pi.
So in order to find arc length, we’re going to use this formula over here for ds. Essentially,
it’s going to give us arc length. So let’s just go ahead and start writing
our formula and we’ll talk about what we get as we go. So to find arc length ds, we’re
going to take the integral on the range a to b. Well, a and b are defined by the range
here, zero and 4pi so we’re going to be evaluating from zero to 4pi. And then we’re
going to be taking the square root here of dr over d theta which is the derivative of
the function r, r equals e to the theta over 2. So to take the derivative of that function,
remember, we’ll just call it r’. When you take the derivative of something here
with the exponential number e, we just have to multiply the coefficient on the variable
in the exponent out in front. So in other words, because this is e to the theta over
2 which is the same thing as e to 1/2 theta, we bring that 1/2 down in front for the derivative.
So we get 1/2 e to the theta over 2. And that’s our derivative.
 So let’s go ahead and plug that in here. So 1/2 e to the theta over 2 and according
to our formula, we’re going to square that and the4n we add to that r squared. And r
of course, we’re going to define that again by our original equation for r. So we’ll
end up with e to the theta over 2. And we’re going to square that as well. These are all
going to be under our square root sign and then we’re going to put d theta out here.
So now, this is just a matter of simplifying what’s inside our integral and then evaluating
on the range zero to 4pi. So to simplify what’s inside the integral,
first we’re going to square what’s under the square root sign because we have two things
that are squared. So to evaluate this part here, to square that, we’ll get 1/4. That
1/2 squared is going to be 1/4. And then e to the theta over 2. Remember that e to the
theta over 2 is the same thing as e to the 1/2 theta. When we square that, e to the 1/2
theta times e to the 1/2 theta, because the bases of these exponential terms here are
the same, they’re both e, that means we can add the exponents. So 1/2 theta plus 1/2
theta gives us 1 theta which obviously we can drop the 1. It’s redundant. We just
get theta. So e to the theta over 2 squared is just going to be e to the theta. Then we
have this e to the theta over 2 squared again so that we already know is also going to be
e to the theta and we’ve got our d theta out here in the end.
So now, we can go ahead and combine these two terms into one fraction underneath the
square root sign. Let’s go ahead and make the second term. We’ll put the 4/4 in front
of it to get a common denominator with our first fraction. So because our first fraction
was over 4, we want to put the second term over 4 but without changing this integral
at all. So in order to do that, we just need to put 4/4 which if course is just equal to
1. So now, we end up with the square root of 5/4 e to the theta d theta. Now, we can
take these separately. Because the 5/4 is out in front of the e to the theta, we can
take the square root of that separately from taking the square root of e to the theta and
we can move that out in front since it’s a constant and it’s a coefficient on this
e to the theta. So we’ll end up with the square root of 5 because we can’t take the
square root of 5 but the square root of 4 is 2 so we end up with the square root of
5 over 2. That’s this part here and it moves out in front of the integral. Then we’re
going to have the integral from zero to 4pi and all that’s left now is the square root
of e to the theta d theta. Remember we found earlier e to the theta by multiplyi8ng e to
the 1/2 theta and e to the 1/2 theta together so we know that the square root of e to the
theta is e to the 1/2 theta. So we’ll end up with the square root of 5 over 2 times
what we get when we take the integral. The square root of e to the theta is e to the
1/2 theta. And we’re going to be evaluating this on the range zero to 4pi.
So now to evaluate, we’re going to plug in our upper limits of integration first,
4. Then we’ll subtract and then plug in the lower limit of integration so we’ll
get ds equals the square root of 5 over 2 times e to the 4pi times 1/2 is going to be
2pi. Then we’ll subtract and put in our lower limit of integration, zero. We’ll
end up with the square root of 5 over 2. Zero times 1/2 is just zero. e to the zero is 1.
So we end up with 1 there. So now, to get our final answer for arc length, we can factor
out square root of 5 over 2 and we’ll simply end up with e to the 2pi minus 1. And that’s
it. That’s our final answer. So that answer represents the arc length of the polar parametric
curve e to the pi/2 on the range zero to 4pi when theta is between zero and 4pi. So that’s
it. I hope this video helped you guys and I’ll see you in the next one. Bye!