AP Tricks, Part 4 views
Guten tag, my journal, and halo once again. Today we are going to be covering functions of integrals and average rates of change. We've thrown these two topics together in AP Tricks 4, as we've often done with the AP Tricks episodes. Combining a few different topics that you are very likely to see as tricky problems on the AP test.
Now, both of these topics are integral related. You've seen average rates of change before. We did them in AP Tricks 2. But I think it's worth looking at some more of them. Before we start though, got the bad joke. You knew it was coming, you were dreading it too. So you heard about the mother dog. She was ready to give birth to her puppies, so she had to find a nice nest for them to hang out in. Dogs usually look for a nice secluded place to have their puppies. She found a beautiful spot, right by a tree but really near the road too. Well the poor dog, after she had the puppies, the police came by, found the puppies. They weren't amused at all. They gave her a ticket for littering.
Functions of integrals. Well, the AP test very often includes questions where a function is defined in terms of being an integral of some other function. They look scary. They look absolutely horrible. And yet, in a lot of ways they are the easiest questions on the test. So let's look. The graph of the function f is defined by two line segments as shown in the graph.
You are usually presented with a graph on these. And what's more, you are usually given some coordinates as well. The coordinates are part of getting the answer. Let me fill those in. Let's see what we have. This coordinate right here is (4,4). That's when you can see a (0,0).
This one over here is (-3,3). So what we are supposed to do is look at this thing that says, h is the function defined by h(x) as the integral of f(t)dt. Oh no, there's no t. Don't worry, you don't need it. It's a technicality which is necessary to make the problem correct.
You could do this. You could say, that's a line I know I can make that line equation. And then you can put the line equation for f(t). And then you can integrate the line function. And then you can substitute in x and substitute in 0. And beautiful you would have h(x). And that took probably took you about ten minutes that you could have spent on doing something that's harder.
All you are being asked to do here is remember that integrals are area. That's it. If you are being asked to find h(4). What they are saying is, hey put 4 in front of the x there, and there. So you are expected to do the integral between 0 and 4, which is just the area between 0 and 4. Man, this is the easiest thing on earth. Look, wow triangle. That was tough. You won't have a complicated shape to do. Maybe you'll have a trapezoid to do, or a combination of some rectangles and triangles. But it really doesn't get harder than that.
So on this triangle, let's see the base is 4. My scale on the graph is not equal left to right, and top and down. That's why the triangle looks different than it's real shape. Because the height is also four. That's it. So even if you see one of these on the free response section, you won't be expected to do the integral. I'll show you all you will be expected to do.
It's just to set up the problem. All you have to do is write out the technicality. It's what you are doing.
Substituting in this case 4 in, for that. Again f(t)dt. The t is just a working variable, which is necessary so that you can integrate it. And then you'll be substituting in numbers other than what t were. And then over on the side, you can just work out area, equals 1/2 base times height. I've been doing this since, probably fourth grade.
1/2 base times height. That's 1/2 4 times 4 which is 8. Now before you relax and decide you have the answer, one other thing to watch out for, the direction of integration. You did this when you started integrals. You did it for about a week. And then you forgot it because you never saw it again. Correction matters, we are going from left to right. So the integral result is going to be positive. And it's above the x axis. So the result is going to be positive. So just leave that area alone. The answer is 8. Let's do one more on this.
This time we'll do h(-3). The co-ordinate that would be given to you on this problem is (-3,3). The problem is about the same as the other one. We are going to find h(-3). And that's what we get when we integrate from 0 to -3. F(t) dt. Not f(d)dt. Which is a poison which was banned years ago because it was really hard on eagles.
Let's draw in our triangle, easy enough. The base is 3, the height is 3. You know you could actually use the negative in there. But we'll throw that back in a second though. We'll do the area.
Area is 1/2 base times height. 1/2 3 times 3 equals 4.5. But that's not our finished answer. Our finished answer for this one is -4.5. And I'll give you two reasons why it has to be -4.5. The first one is, we are integrating from right to left. We are going this way, building up integral area, the reverse of the way that we should. That makes the result negative. Or think of it in another way, you can call this a height of +3 and call this base -3. In which case the answer would have come out at negative anyhow.
Let's go onto one that's another step harder. This is getting to be more like the scale of thing that you might see on the AP test. So I've got something defined by two line sections again. And we have to find h(3). So I'm going to film the coordinates. Again, you are typically going to be given the coordinates. This one is (0,5). And this one here is (3,-2.5). Last one is at (6,5).
So we had to find h(3}. So 3 is right here. If I draw it straight down. I have to actually find two areas; that one and that one. So let's work out those out. It's easy enough to do. If you are not given this point, you actually are going to have to make the line function to find out where it is. It might also be given very clearly on the graph too. It probably would be.
This location right here is at 2. So the base of this triangle is 2, the height of that triangle is 5. That's easy enough. 1/2 base times height gives us an area of 5. Then down here. From 2 to a coordinate of 3, that's a base of 1. A base of 1 and a height of 3. So 1/2 times 1 times 3 is 1Â². The area of that section is 1.5. That section though, it's below the axis.
So even though the area of that triangle is 1.5, we have to say that the integral, the area, is negative. And you know, actually if you use this negative coordinate in there, it would have come out that way. We are almost done with this.
So again write out the technicalities to show that you know what you are doing. Doesn't take much to do that. f(3)dt. I think on the last problem I didn't substitute the value. That should probably be in there. Easy enough. So we've got our areas of 5 for the triangle on the left, plus the area on the next section which is -1.5. And it looks like we have 3.5 for h(3).
We'll move on. On the next part we are going to do the same one, the same function. But we'll find out something else about the function. This time we are supposed to find out what h'(3) is. Then again, you can say, okay, h'(3).
Well there it is. And you might go down here and say, "Oh! h'(3) oh no, we can't do that can we?." Because it's a sharp pointed that spot. Actually we can. We can do that because we don't have to come in from both sides.
This is asking you to recognise that this function right here is an integral of this. The picture is f(t). The integral of that gives you the area. But derivatives reverse integrals. So really, this derivative just is that function.
So, h'(3) is just f(3). That's all. All you have to do is go and read the value off the graph. So f(3), reading this spot off the graph is -2.5. One more thing we can do about this graph. Find the second derivative of 4. Well if we did the first derivative of this, remember that it undoes the function. First derivative undoes the integral function. Second derivative, it's the derivative of f(t). What's a derivative? It's a slope. This problem is just asking us to find the slope at 4. That's quick.
So the slope, we need this location and that location. This one again was at (6,5). This location was at (3,-2.5). Just use old fashioned rise over run on that.
So, the slope is 5/6 minus minus. -2.5 over 3. And we are just about done with this. Let's see 5 minus -2.5 that's 7.5 on the top. 7.5 over 6 minus 3, easy enough. That slope is 7.5, so that's it. h"(4) is 7.5. Now you might be thinking, wait a minute, on the last two parts past of it said to put 3's in there. We did h(3) and we did h'(3). Why did he suddenly switch to h'(4)? Look at what happens at 3.
Remember, this double derivative is the slope of the original function. And the original function has a sharp little point there. It's not differentiable at that point. So if I had asked you for h"(3), you would have had to say, 'doesn't exist'.
Average rate of change. Do you remember dealing with the mean value theorem for integrals back in the Ap Tricks 2 episode? The problem we are going to do now, is just another variation on that. It's phrased differently, but surprisingly the problem is much the same.
Now this one tells you, Find the average rate of change of the function. There's the function on interval from 0 to 5. So you are thinking, "Rates of change. Cool, I know rates of change, those are just slopes. And I'm going to do the derivative." And you are not completely wrong. Because a derivative is involved in this.
But if you'd follow your instinct of saying, you know what I'm just going to do the derivative, and then substitute in 0 and 5, and then average those two results, you'll get the wrong answer. Because the rate of change isn't necessarily linear. You could have a function that looks like this, getting steeper and steeper and steeper. So across that interval from 0 to 5, it's getting steeper and steeper and steeper. But that does not mean that the average rate of change over that time spent, is going to be the same as just averaging the slopes at the ends.
No, we've got to do something a little bit more complicated that involves the mean value theorem for integrals. So here we go. We have to find this average value, average rate again, of this function on that interval. So here the mean value theorem. I'm going to start setting this up, and then there's one thing we are going to have to modify on this.
So we want the average rate. It's going to be based on the mean value theorem. Now if it is the mean value theorem, what I'm going to find out is the quantity of stuff that got accumulated over the period from 0 to 5. If I divide that by how long it took, then I know the average amount of stuff that you have happening per second in this interval. Well, we are going to set this up, but there's one more thing to realise.
Rates of change have to take the derivative of the function. So we really do have to do the derivative of this function. But then we have to integrate it as well. So here's the weird thing. I've got to use this formulae. I have to use 1 over b minus a. And I have to use the integral from a to b of the function.
But since we want the average rate of change, not the average of the function, we have to do a derivative of that. Well, remember, what do derivatives do to integrals? What do integrals do to derivatives? They're inverses, they get rid of each other. So really I can just replace this entire unit with the function, and evaluate it over the 0 to 5 span. So here's the set up. I have 1 over b minus a. Those are the endpoints of the interval, 5 to 0. And f(x)dx, just gets written down the way it is. Because remember, if you did the integral, you could get a different function, but then you are doing the derivative of it right away. Which undoes what the integral does and all I have left is xÂ³ minus xÂ². xÂ³ minus xÂ² and then we'll evaluate that from 0 to 5 as well.
That's easy enough to do. You'll put 5s in for the X's. I should have done that. I'll do that on the next step. So I'll have 1/5. Parenthesis 5 cubed. This is the correct time to do it, after I've written out the evaluation form. And when we are all done, what you have is 20.
One more thing before we close up for the day. At this point, you've probably watched the previous AP Tricks episodes. So you've seen all sorts of odd problems solved. Usually two or three different types in every AP trick episode.
Well, in the course of these episodes, I've gone through pretty much everything I could find for strange and unusual problems in the official AP guides. The official guide that you can download from their website. Problem is, it's very hard to predict what you are going to have on the free response section. Because the AP people change them every year. And then as soon as they've issued the test, they give them y for free. Which means you can look at them.
And again the problem is, they don't really want to find out if you can just do certain problems. They want to find out if you really understand what is behind the calculus. So they throw in really surprising variations on the problems, which are really difficult to predict. What you need to know in order to do well on the free response section, is general strategies for how you can tackle a problem, if it's not straight forward. Let's look at a few.
Strategies for puzzling problems. Well, we've gone over this one a lot of times. And you'll see it very frequently. A problem where integrals are depicted as areas. So often times you'll have something with an area drawn. And the problem will be asking you about an integral. And you need to realise they're really the same thing. Co-related to that is that areas can be found geometrically. So there are a lot of times when you don't actually have to do any integrals or derivatives, any calculus. Really all you have to do, is realise what the calculus is finding. And then just use a geometric formula to get the answer. Like, trapezoid formula or the triangle area formula.
Also along those lines, any time you are asked for a Riemann Sum or a trapezoidal sum, realise it's just geometric.
You are breaking this section, you are breaking the shape down into sections which are approximated by geometric shapes. Lots of them. And the more of those you make, the closer you are to the real answer. Well, there's another one. Integrating a rate gives you a quantity. So they undo each other and you are able to figure out how much is built up of some material, like a tank filling up or, cars going through an intersection over period of time.
Derivatives and integrals. They are inverses of each other. They undo each other. Now when they undo each other, there are times when you can't just simply write down the original formula the way it was. Sometimes you have to use the chain rule on those. We'll get into problems like that in AP Tricks 5. So derivatives and integrals are inverses. Does that mean they can really sing? Does that mean they know their poetry? No.
Derivatives; derivatives are the slopes of functions. Got a function, derivative tells you the steepness. And as far as that goes, you can take it a step further if you have the slope of a function. The second derivative is the slope of the slope. In other words it tells you how fast the slope is changing.
Well, rates can be read off of graphs of derivatives. So for example, if you have a graph that shows you x values versus functions or derivative values. If you want to know the function slope, really all you do is read off an (x,y) coordinate from that x derivative graph.
And finally, functions of functions. They take chain rule. Remember this one when we are looking at the AP Tricks 5 episode.
Once upon a time there was this guy who was a real genius. He's an all round guy, he could do everything. He was fantastic at math. He'd go out and chop down a tree and turn it into kindling just like that. And besides that, he was an operatic tenor. He could really sing. Do you know what they said about this mathematical lumberjack? They said that he had great log rhythm. Logarithm, yeah, okay. Well, anyhow, in this episode, what we've done is started off by talking about how you can take a function and define it as being the integral of some other function.
We went on to talk about problems where you are looking at the average rates of change, where you had to use the mean value theorem. But in addition to using the mean value theorem, you had to realise that since you had a rate of change, you were being asked to do a derivative of the integral that was in the mean value theorem. Derivatives and integrals undo each other, leaving you with just the function to evaluate.
We finished off by going over some general strategies. The things that you can do to help calm and orient yourself. If you run across a problem in the free response section of the test, where it is not clear what you are supposed to do. Well, again thanks for tuning in.