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# Calculus BC 2008 2 b &c

Video transcript

Welcome back. We're doing part b and part
of problem number two. And it says, use a trapezoidal
sum with 3 subintervals to estimate the average number
of people waiting in line during the first 4 hours. And so I just took the data
they gave, graphed it. I did that in the last video. And so, use trapezoidal sum
with 3 subintervals to estimate the average number of people--
that sounds very complicated. Like boy, I don't-- that
sounds-- yeah, you know, I didn't learn trapezoidal
sums, or however you want to think about it. It could be very confusing. And frankly, I haven't taken
calculus in 20 years. But you really just have to
think about what they're saying, and it's
not that daunting. So what's the average
value of a function? And you learned
this in calculus. You might not remember it. So let's say the average value
of f is equal to essentially-- let's say, between 2 points, b
and a-- it's the difference between the 2 points divided by
the area under the curve of f of x. From a to b. Right? That's, I think they call
that the second fundamental theorem of calculus. Or corollary to the fundamental
theorem of calculus, whatever. But it makes sense. Right? The average value of a function
is equal to the area under the curve divided by, you
could kind of say, the base of the figure. Right? Because another way to think
about it, if you just took the average value of the function--
so if the function was, say, this, and you just multiplied
that average value times the base, you would get the
same area as the integral. That's another way to view it. You could rewrite
this as b minus a. The base times the average
value is the same thing as the area under the curve. So that's another
way to view it. So let's just apply that here. Let's use the trapezoidal sum
to figure out the area under this curve, or an estimate of
the area under l of t, and then we can divide by b minus a. Or in that case,
b is 4, a is 0. And we'll have the average
value of-- over the first 4 hours-- of the people in line. So let's do that. Let me erase it. We're actually not going
to use an interval. We're going to use a
trapezoidal sum to estimate the integral. So it says, 3 intervals. It says a trapezoidal sum
with 3 subintervals. Well, I see, there's 3 very
natural subintervals here. And let me draw them. So this is one. Actually let me-- I'll
draw the dotted lines. This is one subinterval. This is another subinterval. And there we have it. We have three trapezoids. We have this one, we
have this one, and we have this trapezoid. So we essentially just have
to figure out the area of each of these trapezoids. Let me see if I can draw that
a little bit-- I drew them in different colors, just so
you can see the trapezoids. So this trapezoid goes goes
from-- oh, I don't know if I'm doing this right. Let's see. I'll use a line tool. So 1 trapezoid here, and
another trapezoid there. And then another
trapezoid there. And I just need to figure out
the areas of each of these trapezoids, and that's my
estimate of the area under the curve of l of t, from 0 to 4. And then I just divide
it by 4, and I'm done. Part b. So what's the area of
this first trapezoid? Well, what do you do to find
an area of a trapezoid? It's the base. So this area of the trapezoid
right here, this first one, is the base. Which is just 1. 1 times the average
of the two heights. Right? So it's the average. Right here it's 120,
and this is 156. So what's the average
of 120 and 156? Well, the average of 120
and 156-- let's see. We get 276 divided by 2 is 138. So it's 1 times 138
or it's just 138. That's the area of
that trapezoid. What's the area of this
trapezoid right here? This bigger one? Well, its base is 2, right? 1, 2. Its base is 2. And what's the average value? Well, it's going to
be the average value between 156 and 176. Or the average value between
this height and this height. That's an easier one, right? 166 is right in between
those two, and we don't have to do much. So it's 2 times 166. Sorry, 166, not 1.66. And finally, what's the average
value of this trap-- or what's the area in this trapezoid? Well, the base is 1, right? It's going from 3 to 4. 1 times the average value. So 176 and 126. So that is 1/2
times-- what's 176? Let me use a calculator, just
so I don't make any mistakes. Trusty TI-85 emulator. Let's see. 176 plus 126. Enter. Divided by 2, is equal to 151. So this area right here
is 1 times 151, or 151. So what's the total area? It's just going to be the
sum of these 3 trapezoids. So it's the sum of
those 3 trapezoids. So it's going to be 138
plus 2 times 166 plus 151. So that's the total area
under those trapezoids. But then they ask us,
well, what's the average value of the function? Well, the average value is just
the area divided by the base. The base is 4, right? From noon to 4:00 p.m. or from time equals
0 to time equals 4. So we just divide
this value by 4. So divided by 4, we get 155.25. So they say, use a trapezoidal
sum with 3 subintervals to estimate the average number of
people waiting in line during the first 4 hours that the
tickets were on sale. So the answer to
part b is 155.25. They don't tell us to round it
or anything, so I'm assuming that's, you know, if we're
saying the average number of people, I guess we can
have a fraction there. So that's reasonable. Let's see, what part are we? We're on part c. OK. So part c. I think that's going to
use this graph, too. So let me erase at
least this top part. The top part I can erase. I can erase this,
I can erase that. I want to erase all the
unnecessary things. Let me see if I can-- let's
see, I can erase all of that. Just want to make it
as neat as possible. I think this graph is going
to come in useful in part c, by just glancing at it. OK. That's good enough, I think. OK, let's do part c. Let me copy and paste it. Part c. OK, copied it. Edit, paste. OK. For t is greater than 0,
is less than 9, right? What is the fewest number
of times at which l prime of t must be equal to 0? Give a reason for your answer. So l prime of t, that's
the derivative of this. When I connected these dots,
I just kind of wanted to get a shape of the curve. But it's really a curve. It's not like this sharp
edged line, right? And how do we know
it's a curve? Because it's differentiable. It's continuous. It's actually twice
differentiable. Which tells us the
derivative is continuous. So the real graph of this is
going to look something-- you know, let me do it in
a bold color like magenta. You know, it might look
something like this. I don't know, a curve and
then it maxes out at some point, comes down. Then it comes up like this. And then it, maybe it
maxes out, and it comes down like that. It'll be a curve, right? Anyway, they say, what is the
fewest number of times that l prime of t must be equal to 0? Well, what happens when l
prime of t is equal to 0? That means that we are at
a local minima or a local maxima, or sometimes
an inflection point. But we don't know
that for sure. But at minima and maxima point,
we know for sure that l prime of t is equal to 0. And so we can just eyeball it. We could look at this graph and
we could say, well, you know, there has to be a minima point
someplace in this range. Right? Between here and here. So, you know, the way I
drew it, it looks like the slope is 0 there. And between here and here, it
looks like the slope is also going to have to bottom
out at some point. Like there. And then there's a maximum
point right-- someplace in this range, and it's
going to max out. So when I just eyeball it, I
say, well, I have 2 maximum points, and one minima point,
so the slope is going to equal 0 at at least 3 points, right? But they tell us, give us
a reason for your answer. Give us a reason. And my reason-- and actually,
you could use Rolle's Theorem or the mean value theorem, but
the simplest reason is that here we have 3 sign changes
in the average velocity. Or the average, not the
average velocity, the rate-- the l of t. We're not doing velocity
in this problem. So from here-- let me
use a thicker line. Use yellow. So from here to here,
we have a sign change. From here to here, we
have a sign change. And from here to here, we
could have a sign change. So what I would write down at
the AP exam, is, I was like, well, the function is twice
differentiable, which means its derivative is continuous. So let me write down
the important points. So derivative is continuous. And we have 3 sign changes
in the derivative, right? The derivative is positive over
this interval, or at least its average is positive,
then it goes negative. So at some point, since the
derivative is continuous-- it doesn't jump around-- it had
to be positive someplace in this range. And you can be more formal
when you write it. It had to be negative
someplace in this range. So there had to be a sign
change, and so it had to pass through 0. And that same argument could
be used between here and here, and here and here. And that's my answer. And I will continue part d in
the next video, because I'm already over 10 minutes.