# Calculus BC 2008 2 a

## Video transcript

Let's keep going with our
2008 Calculus BC free answer questions. So we're on problem number 2. And I cut and pasted the table
that they gave for the problem. And I'll read the
rest of the problem. And let's see what we can do. So it says concert tickets went
on sale at noon, t equals 0, and were sold out
within 9 hours. The number of people waiting in
line to purchase tickets at time t, is modeled by a
twice differentiable function L, L of t. So this is the number of people
waiting in line at anytime. And this is noon, 12:00
noon, 1:00 pm, 9:00 pm. All the tickets are
sold by 9:00 pm. And they tell us it's
twice differentiable. So that means that whatever
function we're modeling this, this L of t, that it's
continuous because it's differentiable. And since it's twice
differentiable, we also know that its derivative is
continuous because the second derivative exists
at all points. So it says part a-- let me make
sure I'm not writing too thickly-- use the data in the
table to estimate the rate-- let me actually just copy
and paste this, I've just figured out how to do this. So I figure it doesn't hurt. There you go. You probably can't read it. But it says use the data in the
table to estimate the rate at which the number of people
waiting in line was changing at 5:30 pm. So they don't give us
any data on 5:30 pm. They give us 4:00 pm and
7:00 pm, or t equals 5.5. Show the computations that
lead to your answer. Indicate the units of measure. So what do they want to know? They want to know an estimate
of the rate at which the number of people waiting
in line was changing. So they don't give us a
continuous function definition. They just give us a bunch
of sample points of this function L of t. So the best estimate I can do,
of the rate at which this L of t is changing at time 5.5-- and
5.5 is in between these two-- is to just figure out the
average rate of change between time 4 and time 7. So how do we figure that out? Well the average rate of
change is just the slope. So let's write it. So we could write average
rate of change or delta L over delta t at 5.5. You can write it however you
want, however you think that the exam graders would
best like to see it. We could say at time 5.5. We could say approximately
equals, or whatever. But it's just going to be the
slope between these two points. So it's L of 7 minus L of 4,
all of that over 7 over 4. Rise over run or change in
the value of the function divided by change in the
independent variable. L of 7 is 154. They give us that. L of 4 is 126. And we divide it by 7 minus 4. So that equals 54 minus
126 is 24 plus 26 is 50. So it's equal to 28. Right, if this was 2
less, it would be 30. And 7 minus 4 is 3. So you could say that
the average rate of change is 28 over 3. Or you could write
that as 9 1/3. And they want us to use
the units of measure. Indicate units of measure. So the numerator,
this is people. And what's the denominator? It's hours, people per hour. So my best estimate, or our
best estimate of the rate at which the number of people
waiting in line was changing at 5:30 pm, which is between these
two points, is the average slope between these two points. Which is 9 1/3 people per hour. That's it. Let's do part b. Let me clear all of this so we
have enough space for part b. And I'll copy and don't
know if you can read it. But maybe you can. So doesn't hurt to
just copy and paste. OK part b says, use a
trapezoidal sum with three subintervals. Maybe I can make this
a little bit bigger. Let me see if I can grow
that a little bit. I don't want to take up
all the space though. No that doesn't look good. OK well I'll read it out loud
in case you can't see it. Use a trapezoidal sum with
three subintervals to estimate the average number of people
waiting in line during the first 4 hours that the
tickets were on sale. If you want to do this
really fast, you wouldn't have to graph it. But I'm about to graph it
because I want you to understand how to
do this problem. If you understand it, you
don't have to graph it. So let's graph some
of these points. And we really just have
to do the first 4 hours. Actually let's graph all
of them, just because it might be useful in future
parts of the problem. If you weren't graphing it
while recording a video, it probably would be a
little bit faster. That should be good enough. What are our data points? We have 1, 2, 3,
4, 5, 6, 7, 8, 9. You don't have do it so neatly. Well it's good to do it neatly
when you're doing the exam, because otherwise you'll
confuse yourself. And let's see, how high
should my L values? This is time right here. This is my L of t axis. It goes up to about, 176
is a high point, at least on the data we have. Let's say this is 200. So halfway up would be 100. This would be 150. This would be 50. I don't know what
that sound was from. But let's keep going forward. Let's plot the points. At t equals 0 there are
120 people in line. That's about right there. At t equals 1, 156. I'm just approximating. It's about right there. At t equals 3-- they skip
2-- at t equals 3 it's 176 people in line. So that's going to be
right about there, just approximating it. At t equals 4, we have
126 people in line. It's a little bit more
than at t equals 0. At t equals 7, we have
150 people in line. It's right about there. At t equals 8 we have 80. That's about right there. And t equal 9, the
line is gone. Everyone has gotten
their tickets, or maybe they got sold out. Let's connect those points. Connect the dots. So from here to there. And from there to there. There to there. There to there, almost done. OK so we've at least plotted
the sample points and connected them with lines. We know that the real L of t,
whatever we used to approximate it, it's not going to have
these sharp edges because it's differentiable. It's actually twice
differentiable. So it's actually going to be
a smoother curve, right? Because we can take the
derivative of any point. If this was the actual
function, you wouldn't be able to take a derivative
at this point. Because there's a positive
slope here and then it immediately switches to a
negative slope like the absolute value functions. You actually wouldn't
be able to take the derivative at that point. But anyway back to the problem. Use a trapezoidal sum with
three subintervals to estimate the average number of people
waiting in line during the first 4 hours. So the first 4 hours,
that's right here. Now this might seem
kind of daunting. The trapezoidal sum
for the fourth. Actually I just realized. For some reason, YouTube used
to let me do longer videos because I thought
I was a partner. But now for some reason it's
been limiting me again. So I will continue part
b in the next video. See you soon.