Calculus BC 2008 2 b &c

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.