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# AP Tricks, Part 2

###### John Postovit

###### John Postovit

**University of North Dakota**

M.Ed.,Stanford University

From over 16 years of teaching experience, he has philosophy that it takes humor, patience and understanding when teaching tough subjects.

[0:00:00]

To tell you the truth, nothing we're doing today is really a trick. I mean it's real Math, it's good stuff. But we're going to show you some stuff today that you might have seen too much even in your regular calculus course. Now we're going to begin with differential equations, you've definitely done those. Differential equations are ones that tend to really slope on x,y coordinates on the graph. Integrating rates though, maybe different. Integrating rate problems, usually not covered that thoroughly in calculus textbooks. And if you haven't started the AP review for portion of your class yet, you might not have never seen any of these. So that's what we'll be doing in this episode.

Before we start though, I've got a real fine joke for you here. If you were studying political science, you know a little bit of feudalism and kings and democracies. But the thing to remember is that, in democracy it's your vote that counts but in feudalism, it's your count that votes.

Basically differential equations is one that includes a slope, a derivative, as part of the equation. Let's have a look at one. So you can see you've got an equation that's not just y equals, it's a y' equals. So what this is doing is it's relating an x coordinate to a slope. Let's look at a particular example here.

It says to find the particular solution, that's a technicality to y=f(x) to y'=3xÂ² with the initial condition y(1)=10. That initial condition is pretty important actually. Because it's going to turn out there is an infinite number of possible solutions and they're only going to want just one. Let me show you where the infinite solutions come in.

[0:02:00]

Now remember that y' is a derivative, to start you have to get rid of the primes, and replace it with Leibniz notation. That's what has been done here, y' is the same as dy/dx. Now, we've got to get rid of the derivative. What gets rid of the derivatives? Integrals, we have to do an integrals. I'm going to start by separating the parts. These are called separable differential equations because we're going to get it into a point easily where it's separated to do an integral. I'm going to multiply both sides by dx and I have this, I've got dy=3xÂ²dx.

Now, I want to get rid of the differentials, so I'm going to integrate both sides. This side is integrating with respect to y, this one is integrating with respect to x. The integral of dy is just y, and the integral of 3xÂ² that's nice and neat. That is just xÂ³. But remember, whenever you do an integral and it's not a definite integral, you have your old friend plus C, constant of integration. That's where the initial condition comes in. Because, no matter what that constant is, if we did the derivative of this, it would just go away and the equation would look like that. That's why there are so many solutions to this. Because of the constant can be anything you want, it's called a family of equations.

We're almost done. All I need to do is use this bit of data to find out what C is. So I'll substitute, I'll put a 10 in for the y, and I've got a 1 for the x. So that's 10 equals 1Â³ plus C. I just went back and used this equation and that data point. That's easy enough, C works out to be 9 and we've got our solution.

[0:04:00]

This particular solution is y= xÂ³+9. Let's try something a little more difficult now. This problem is asking us to find a particular solution again to y=f(x) to y' times y-2x=0, that's the initial condition again. Well think back to our steps.

We have to undo a derivative by doing an integral but before we do that, we've got to separate. We have to replace the y' with dy/dx. So this differential equation will have dy/dx times y-2x=0. Now by separate, I mean that you have to get the x and dx on one side, the y and dy on the other side.

So I'm going to start by adding 2x to both sides. And there's the 2x and here I've got dy/dx times y. I'm going to reverse the order of that, you'll see why in a second. So y turns dy/dx.

Now we've got to separate. So I'm going to multiply both sides with dx. Now we're separated. We've got ydy on one side and 2xdx on the other. Time to integrate. So we're going to do the integral of ydy equals the integral of 2xdx.

[0:06:00]

The integral of y, well one more power then y to the first is y to the second. We've got y to the second and we're going to need a correction factor here, because when I do the derivative of yÂ², I get 2y. So I need that correction factor of 1/2 over here. The correction factor of 1/2 is just going to undo that too and I'll have xÂ².

Now I did some indefinite integrals, so I need the constant of integration. You don't really need them on both sides because constants just add up to form other constants. And as long as we have a lot done to our constant it is yet, it's okay to just forget about the one on the other side. Now last time that we went and we solved this for y before we put in the initial condition. This time is going to be a little easier actually if we put the initial condition in right now, before we do the square root.

I'm going to do another little thing here. Let's get rid of that half. I'm going to multiply both sides by 2. This is another little trick you can do with constant of integration. So 2 times 1/2 times yÂ² is yÂ². And that equals 2xÂ² plus, it's two constants of integration. Because we don't know what the constant is yet, I can say all this is C1 and 2 times C1 is constant 2. I'm going to put the 4 in for the y, and that equals 2 times -1Â² plus the constant.

[0:08:00]

So we'll have 2 times 1, that's 2, that's 16, take that from both sides, looks like our constant is going to be 14. We are almost done. Substituting into here we have yÂ² equals 2xÂ² plus constant of 14. I have to do the square root of both sides, so that gives me y equals +/- the square root of 2xÂ²+14. Plus or minus there is two solutions to these. How do you pick which one?

Go and look at the initial condition again. Look f(-1) is +4, we have a positive result. If I had a negative in front of this square root, we could never get the positive result. So because of that, we're going to ignore the positive or minus, and get rid of it. The positive solution is the one that works.

Now let's go and take a look at one more little detail. You might see something like this on the AP test and you think, "Oh my gosh! This looks horrible." No, it's actually one of the easiest problems in the world. It's asking me to find the slope and the differential equation. The slope is y'. You're really just being asked to substitute in to it to find out what y' is at that particular spot.

Now if I solve this for y' I would have y'=2x/y. Look at that, just put the numbers in. So y' at that particular location, has to be 2 times x is 4, at that location and y is 3 at that location.

[0:10:00]

Let's look, it can depend on x and y, it's possible. The result is 8/3.

I'm moving upwards, I want to change the gears right now to something unrelated to differential equations. Integrating Rate Practice. Now you think you see this word rate and you think, "My gosh! Rates I know what rates are." You look at this problem, and say; A water tank started out holding 1000 gallons at time t=0. The rate at which the tank is filled is given by such and such. So you think, "Oh men! This is a related rates problem." But it's not.

This is a kind of problem which again, isn't covered that often in calculus textbooks, but they're not that hard to do once you realise what's going on. Now, bear in mind what a rate is. A rate is the speed at which something is being accumulated. How do you accumulate something? With an integral. A rate is a slope, a slope is a derivative, and an integral undoes a derivative.

The first thing we need to do though, is to set up our complete rate equation, because they only gave us part of this. They told us the rate at which this liquid is being added. So the rate at which it's being added is 40t, 40 times the time, times the cosine of 5tÂ²/500. So the liquid is being added at that rate. It's a rate that changes. But at the same time, the drain is open on the bottom and it's losing 120 gallons per minute.

So we have to subtract 120, and that rate is just a constant rate. It's always 120. Now, we need to set up our integral. When we do that, we're going to integrate the this rate.

[0:12:00]

And we have to integrate it over the time span, and which we're accumulating liquid. That is going to be from 0 to 10. So now let's have a look at this integral. You could probably do this integral by hand. I mean, yeah you have the 5tÂ², there is a 40t there, you could use integration by substitution. But you're much more likely to see this kind of problem on the part AP test where you're allowed to use a calculator. If you can use a calculator do it. If you're allowed to use it, it's going to increase your rate of getting the problems right. Just use it.

Now I can't go through the steps on this because there are so many different calculators out there. With my calculator, what I do is I type in this equation in a graph, and then use an integrate function with 0 and 10 as the limits. You have to explore your calculator to find out how it works for this kind of a thing.

When I did it what I found was that, we accumulated 483 gallons. Let's go back for just a second, there is one last thing we have to look at.

This tank started holding out at a 1000 gallons and notice I didn't use that. I didn't use that yet. The reason I didn't was because, the rate at which fluid is being added, the amount that's accumulated, has nothing to do with what it started with. Nothing, so it isn't included in the rate. I used it at the end. We just added 483 gallons, started off with a 1000. So there is 1483 gallons in the tank at the end of 10 minutes.

If you look at the graph of this one in detail, you will notice that, for the first couples of minutes or so, actually the fluid is going downwards. We started out with a 1000 gallons, and for a while is being drained faster than it's being added.

[0:14:00]

That could cause you a complication if there wasn't enough fluid in there to begin with. Say you started off at 20 gallons, and it drained down to 0 before it filled up. But we don't have to worry about that complication on the AP test, they won't give anything that tricky. You won't have to worry about drain past the bottom.

Now we'll try a second problem that has an extra elaboration. This is integrating rates with the mean value theorem. There is a couple of key words to watch out for in the problems. Before we go into the problem though, let's have a look at the mean value theorem again.

The mean value theorem, it is probably worth memorising, because you might see it in the multiple choice section of the test. Basically what it says is, if you integrate up an area, and divide it by how wide the area is, you're going to find out how the mean value is. Now, myself, I'm terrible at memorising. I've always remembered this one in terms of what it does.

What it does is, it basically builds up a rectangle that has the same area as the integral. So let's say that a is right here. Let's put it there for convenience. And let's put B right here. If you do this integral what you'll be finding is the area that's underneath that curve and the x axis. The mean value is really just the height of a rectangle that has the same width as that base. (b-a) that's just the distance between those, it's the length of that base. And the rectangle that has the same area as the integral, is probably a little bit bigger than that, probably about that high. So that rectangle has the same area as the integral. And the height of that rectangle that's the mean value.

[0:16:00]

Now on to the problem. It says the rate at which water flows in a pipe is given by gallons per minute for time can't equals 0. You can't have time equals 0 on this because it would be undefined. The question is what's the average rate of flow between t=1 and t=5.

You might really be tempted to say, "You know I'm just going to put 1 in here and find out the rate, put 5 in there to find out the rate, and we're in business. Just add and divide by 2." If you do that you'll get the wrong answer, and if this is the multiple choice question, that answer will probably be waiting for you to trap you. They always put in the answers for the common mistakes. No, instead we still have to integrate.

Key words here are the rate and average. Now remember in the last problem, what we did was, we did an accumulation of rates to find out how much got added to the tank. If we had divided that by how long it took for that to get into the tank, we would have the average rate of flow. That's what we're going to do.

So for this one, we are going to do the integral from 1 to 5, that is the time span and which we're looking of 10/tÂ². I'm just going to write this out as the quantity, that is the amount that gets added. And at the end we'll just divide it by how long it took to happen.

You could put the 1 over (b-a) factor here, so that it looks like the mean value theorem. But you don't really need to if you understand what's going on. I'm going to rewrite this integral a little bit before I integrate it. You might be able to do this at the top of your head. But if you can't, remember tÂ² of the denominator is t of the negative second of the numerator.

[0:18:00]

And so we have got 10, t to the -2, dt and one more power then -2 is -1. So I'll have -10t to the -1. That negative came from the correction factor, because if you did the derivative of t to the -1 you get t to the -2. That negative there corrects for it. And the limits of integration are from 1 to 5. Let's go ahead and finish this, we're almost done.

So I'm going to take this -10 out front as a factor, it's going to save me a little bit of time. And t to the -1 it's 1/t. Substitute in the 5, so that gives you 1/t being 1/5 minus 1/t, when t=1 is 1/1. So 1/5 minus 1 is going to be -4/5 times the 10, we have +8. So 8 gallons got added, one last step though.

It took from 1 to 5 minutes for that to be added, so the average rate is going to be the 8 gallons that got added, divided by how long that took which is 4 minutes. So the rate, 8 divided by 4 is 2 gallons per minute.

This episode covered two very different topics, differential equations and integrating rates problems. You'll very likely to see them on the AP test, make sure you study them. Now there is one other find of integrating rates problem, it's called total distance problems. But we're going those in yet another episode.[0:20:00]

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###### John Postovit

University of North Dakota

M.Ed.,Stanford University

From over 16 years of teaching experience, he has philosophy that it takes humor, patience and understanding when teaching tough subjects.

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