University of North Dakota
From over 16 years of teaching experience, he has philosophy that it takes humor, patience and understanding when teaching tough subjects.
University of North Dakota
From over 16 years of teaching experience, he has philosophy that it takes humor, patience and understanding when teaching tough subjects.
AP tricks 5, this episode is going to show some of the more specialized kind of problems that are intended to test your knowledge of calculus theory. We?re going to look at problems involving slope fields, problems involving testing whether or not you understand when the chain and product rules need to be used. Problems that involve the first fundamental theorem of calculus, and problems that involve the second fundamental theorem of calculus. But before we start, I?ve got to show you this. This is the best and the only calculus trick on earth.
This is a joke. To understand the joke though, you have to be able to do calculus. I?ll give you a couple of hints on it, first of all you?re going to have to integrate. Second each of these symbols represents some words or just a word. And finally, once you?ve integrated it, you've figured out the words it?s a common phrase. Have a look in the bonus materials of the ones you think you?ve got it, you?ll get the solution in there.
Differentiating with tables, you?re really likely to run to cross puzzles of this kind in the non calculator multiple choice section of the test. Now they?re intended to find out if you really understand when you need to use the chain, or product, or quotient rules. They don?t give you the function though. So you can?t just say hey I know how to this, and just start working away on the function. Find this derivative of ways in a chain of product rules.
No. What you need to do is realize what it is, right out the chain of product rules, and then just substituting values from the table. They?re always going to ask you to do this at a specific location, because you can?t do it for as general function without knowing what the function is. So here is a problem they?ll usually define a function in terms of other functions. In this case h(x) is defined as f(x) times the quantity g(x) plus 8. And then next you find h'(2). In other words the derivative of h at a specific location, slope at a specific location.
Step back for a second and just realize what this thing is. If we?re going to do this derivative, we have to do the derivative of everything over here. And everything over here has multiplication in it. Multiplication, what rule does that take? Product rule. Because these aren?t just plain old numbers, these are functions. The functions with x in them. If they're function with x in them, that means that the derivative is going to require the product rule.
To review the product rule we?re going to have to do the derivative of the first part, times the non derivative of the second. Plus the non derivative of the first part times the derivative of the second. As I write this out, I?m going to be getting rid of the x?s and replacing 2?s. Because we?re going to be finding this at a particular spot.
So we need to find h'(2), and that will have to equal the derivative of the first part. That?s f' specifically at 2 times the non derivative of the second part. Which would be g(2) at that location, plus 8.
Now notice I didn?t do anything with the 8, that?s because I didn?t do the derivative of that section. So I just wrote it down the way it was. Next part of the product rule, plus the non derivative of the first part, that will be f(2) times the derivative of the second part. The second part is the quantity g(x) plus 8. And a derivative of that is g'. The 8 went away, because the derivative of 8 is 0.
So we have g'(2), if you had decided to distribute this f(x) to both things in there. And then do the derivative great you would have gotten the same answer that we?re going to get, with the method I?m showing you. Now it?s time for the table, the table tells you what you can substitute. So for this particular value of x, it gives you the function value for f, derivative value for f' etcetera. So fill in the numbers, we?re almost done.
F'(2) is -3, -3 times the quantity of g(2) is 4 plus the 8. We don?t need to substitute anything there. Plus 1/2 of 2 is 5 times g'(2). g'(2) is 7. Great, we are finished. Just do a little bit of adding; multiplying that kind of stuff and what you get is h'(2) equals -1. Now we?re going to do one more of these, this next one is going to require the chain rule.
The reason I know this one is going to take the chain rule, is because we?re doing a function of a function. If you were to actually say evaluate one of this. What you do is you put a number in for x and then evaluate that with function g. Then you'd use that answer to go into a function f, to find your final result. Notice the table is a little bigger on this one, because it?s going to take some more parts. Remember the chain rule is going to have you do a derivative of the outer part. And then multiplied by the derivative of the inner part, the inner part is g(x). So let write it out. Again we?re doing it a specific location this time it?s at 5.
So h'(5) is going to require us to do the derivative of the outer part that?s f. And the derivative of the outer part with the interior. Remember we don?t do anything to it. For this derivative part you leave this part alone, so I have to have g(5) in there. Times the derivative of what you would get from the inner part, so that?s g'(5). We?re almost done. I know it looks confusing but it really is the chain rule.
This one is going to take 2 steps unlike the last one, because I?m going to have to substitute in for g(5) and g'(5). And then do one more step of substitution. So the f' doesn?t go away yet, but g(5) according to the table is 1. No equal sign there, times g'(5) and g'(5) is 8.
Here is why you needed the second row on the table. We have to do f'(1) not 5. Because 1 is the result that you get when you put 5 into g(x). So f'(1) is -1, just about done. We have -1 times the 8 from before and a final result is just -8.
Now notice there?s a lot of numbers in the table that we didn?t use, you just have to deal with that. They give you the full table so that they?re not giving you too many hints as to what to do.
The thing to remember about slope fields is that, they show you all the possible places where a function could go. They start off by giving you a differential equation and a differential equation is one that includes locations and slope. You can use this differential equation to show what the slope is, at lots of different places on the graph. Now when you start a slope field, it?s not going to show any particular solution to a differential equation. It will show the direction that all the possible solutions could go. Let?s have a look.
This problem asked you to sketch the slope field and to find the particular solution where f(0) equals 7. This is the differential equation here. It?s the one that says that the slope is always just twice as big as the x value. Sometimes these differential equations very often, they include x and y. This time though just the x. If you?re asked to actually draw the slope field yourself, instead of looking at a set of slope fields and deciding which one fits the differential equation, you?ll be just given a few spot at which you?re supposed to find that slope field.
So let?s see, we have this location (1, 1), (0, 1), (0, -1).
We?ll just do a few more; (0, 2), (-1, 2), you won?t usually have to do very many of this, I?m just going to fill in, 9 of them. We can?t go below the axis as well let's do a few more. Once you start then they really go pretty quickly. I?m going to organize this as a data chart, not that you necessarily have to do this, because you can usually do them in your head. But I want to show you specifically what I?m going.
This formula calls on you to finds slopes by knowing x. So this data chart is going to have x and dy/dx in it. You may need to have a data chart that has x and y in it, if both of them are need for finding dy/dx. When x is 0, 2 times x is also 0. That gives me some slopes that I can put on the graph. What you do with these slopes is draw little sections of straight line, to represent the direction that the curve will go. So when x is zero and that happens at all 3 of these spots, the little section of curve that we draw is horizontal. Got a slope of 0.
Let?s pick another value where also is supposed do this when x is -1. And when x is -1, 2 times x is -2. So at all the places where x is -1, I had to draw a little section of slope that?s our -2 slope, fairly steep. Well let?s see a -1 slope goes like this and -2 is probably about like that. Let?s draw them in, -2 slope.
When x is positive 1, 2 times positive 1 is 2, I get a positive 2 slope. Now let?s draw those in, a little section of positive 2 slope, positive 2. I can just fill in a bunch more of these now. Especially in this case where we didn?t have to use the y?s to calculate the slope. If you have to also use x and y to calculate the slope, it won?t be as easy as just replicating the pattern up and down. You?d actually have to use both coordinates of a particular spot to find the slope. But not for this one, this one is a little simpler
You?ll find another one of this of course in the bonus materials. One that involves using both x and y coordinates. So there are the sketches of the slope field. Now let?s go on to do the differential equation.
We covered how to solve differential equations in another episode. So I?m going to this fairly quickly. The basic steps though remember are to separate and integrate. By separate I mean that you want to have everything involving x and dx on one side, everything involving y and dy on the other side.
So I?m going to multiply both sides by dx and we get dy equals 2xdx. Now, with them separated, I can integrate both sides. Integral of dy is just y equals, integral of 2x is x², that?s easy enough. Plus the constant of integration. The constant of integration is why you need the initial condition. So this differential equation gives you a family of parabolas, which can be either higher or lower. When I put this on the slope field, in a minute you?ll see what I mean.
It?s easy enough to do the one f(0) equals 7 you can solve this in your head. When f(0) equals 7, you put 7 in for y, 0 in for x the constant has to be 7, simple. So our differential equation for that particular solution is y equals x² plus 7. So it?s a basic parabola, its vertex is it going to be a 0, 7. Let?s go back and put it on the graph.
Cool thing about slope fields is that, once you actually have your location, where your initial condition, you didn?t even have to find the differential equation to sketch the function. 7 is about here and let me put in some more of these slope fields. We?re running out of room here. Now you?ll see the shape over this, it has to go through the initial condition of 7. And then it follows the curves. Follows along where those little slope section are. I?ll draw another one here down here.
If the initial condition was 2 it would have gone like that. You?ll often be asked to that as well to put a little sketch of it on the slope field.
You?ll use the first fundamental theorem of calculus all the time. It?s the one that justifies doing an integral and then substituting in specific values, to find the value of that integral between those locations. Or, to find the area underneath a curve between those locations. Well, even though you use it all the time, you?re likely to face some problems where you?re asked to recognize what the first fundamental theorem is. And then apply it in a way that you?re not used to, we?ll try one.
Before we do, we?ll do a little bit of review. The first fundamental theorem, now here it is. The first fundamental theorem says again that, after you take a function, and find its integral, then you can substitute in the limits to find the specific value of that integral. Now this capital F means integral of function f. Sometimes you?ll see it written like this. Sometimes you see it written like this. In this case it really is the same thing. This one says that the integral of the derivative of the function is the function with those values substituted in. You could see either one just be prepared for it.
I?m going to do a little bit of a derivation that will set us up for the problem that we?re going to do. I wrote down this version of the first fundamental theorem. Where the capital F represents the integral of the regular function, the antiderivative. Now to make this so that I can produce a formula, I put t here instead, so that I could use x for the upper limit. I?m going to zero for the lower limit.
So if you do that the a?s get replaced with zero?s, the b?s get replaced with x?s. Last thing to do, is take this equation and solve it for f(x). Get f(x) by itself by adding f(0) to both sides. Now if you pause and think for just a second, this is actually just a variation on something we did in AP tricks 4. Which was defining a function as an integral.
Remember when we did that, what you did was, you looked at a graph to figure out what the integral was up to a certain point, and then you add it in the initial condition. Here it is. This is the part that justifies doing the area of a graph between 0 and x and then adding in the initial condition.
The difference now is that instead of doing it with a graph, we?re going to do it with a natural function. So this one asked you to write the antiderivative, just another way of saying the integral, of capital F. So the, antiderivative is going to be called capital F of that function and there?s the initial condition. I copied down the formula, you might as well memories this thing. You don?t what have to do the derivation of this every time, it?s just memorize. It will be on the flash cards too.
Once you have it though, this is pretty quick. I?ll just set up we have he integral from zero to x. Now I wrote this as f(x) equals ?sinX I?m temporary going to replace the x?s with t?s so that this can be integrated, over a different variables than the one that the final formula is going to have. So I?ll write in the ?sin of t, dt, plus. I know the know the initial condition so I might as well put it in right now it?s 3. And that equals capital F(x) which is the antiderivative of little f.
That done, the integral of ?sin is cosine. So we have cosine of t from 0 to x, plus the 3 from before. Equals capital of F(x), now I have to substitute in the 0 and the x. So this becomes cosine of x minus cosine of 0. Don?t make the assumption that the 0 just goes away. This is one where it doesn?t.
Plus the 3 equals capital F(x), just about done. Cosine of x is a variable can?t do anything with it, but cosine of 0 that?s constant. The cosine of 0 is 1. So we?ve got our final result, capital F(x) equals cosine of x minus 1 plus 3, which is +2 and there you have it.
The second fundamental theorem of calculus, is the one that says that a derivative is the inverse of an integral. It says that they undo each other. So the problems that involve the second fundamental theorem, are typically ones where you're going to have to do some substitutions. Let?s look.
This one is asking you to take your function, which is defined in terms of an integral and do its derivative. Well, hey cool, derivative is just undo integrals, so you'd think hey great it just undoes it and my answer is 3t to the 5th plus 9. But it isn?t actually. There is another thing that you need to be aware of, the chain rule.
Now, I?m going to show you how this one could be done the long way, and then I?m going to do it the way you should really do it, the short way. The long way is, you take care of this integral and the integral would be let?s see. Integral of 3t to the 6th would be t to the 6, or in other words ½t to the 6th.
Integral of 9 of is 9t, not 90. You can tell I love pants and they?re always bad. Well then you?d go ahead and substitute this. You?d substitute in the 4x, so that gives ½ of 4x to the 6th plus 9 times 4x minus, you got to substitute the 8 in now. So I?d have ½ times 8 to the 6th plus 9 times 8. Cool, got the integral. I wrote that behind the derivative, but really this is the integral that I just did. Capital F(t).
Well we we?re supposed to find out what f' is, in other words doing the derivative of this integral. Now here is where the sneaky stuff happens. So let?s say you did this process, you did the integral and now you kind of undo the integral by doing the derivative of this thing. 4x to the 6th power is not just x underneath that power. In other words it takes the chain rule. Yeah, it actually changes. Wow, that?s where the short cut comes in.
If you do this without doing all these steps, you have to realize that you have to multiply your answer by a chain rule factor. As for that 8, now look ½ times 8 to the 6th plus 9t is 8, that?s just a number. It?s a constant and when you do its derivative, that just goes away. So the 8 doesn?t do anything, just the 4x.
Now the fast way. This is the way you really want to do it. And it?s the way you need to know how to do it. Because when you see one of these on the test, there is a solid chance that they?re not going to give you one that can be integrated as easily as this. They?ll give you something that you can?t integrate. In other words you can?t do the problem without understanding that integral and derivative reverse each other, and you have to deal with the chain rule.
So, all you really have to do to get my answer now, is say it?s f'(x). I know it?s f'(x) instead of t because, there?s an x here as one of the limits. And I just have to go and take the 3 and substitute 4x for the t. There is still a 5th power up there, plus 9, so I did my substitution. But that quantity needs to be multiplied by the chain rule factor. The chain rule factor is the derivative of 4x which is 4.
Now this is pretty big, so I worked out in advance. We have to do 4x to the 5th power and then times 3 etcetera. So I found the result of this one is 12,288x to the 5th plus 36. There is one more variation possible on this. For this problem the variable was the upper limit of integration. If the variable was the lower limit of integration, if the 4x had been here, then if you did the integral, that would be the part.
The variable part would be the part that would be negative. And you?d have to take this result and just make it the negative of what it is. That, a drum roll please, is all for AP tricks 5. Before I go, I've got one more beautiful joke. What?s the proper response when someone tells you a really lame calculus joke? Let me show you, rdr. Yeah get it? Yeah.