Calculus: Calculus I in 20 Minutes

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Introduction to Calculus II
Introduction
Calculus I in 20 Minutes Page [1 of 5]
So before we go on to the adventure of Calculus II, it's not a bad idea to think about what went on last semester or last year or whenever or 20 years ago when you thought about Calculus for the first time. So I thought it would be fun to quickly run through all of Calculus, and in fact, I actually produced Calculus in under 20 minutes. I know you're saying, "Wait a minute, if you can do all of Calculus in under 20 minutes, well why did I spend like a whole year of my life - it seemed like eternity - what are you going to do in 20 minutes?" Well, you should have watched the video. What can I tell you? Anyway, so we'll review all the big highlights like with respect to derivatives, how to take derivatives of exotic functions using the chain rule, product rule, quotient rule, all the facts about derivatives, and then all the applications, and then also take a look at the integral, take a look at some very basic techniques to actually integrate functions. So watch this clip, it only takes about 18 minutes. I'll see you at the end.
All right. So now we're going to try to do the impossible. We're going to try to do all of Calculus in under 20 minutes, so we have to work really, really fast, go through the whole course. Let's begin. The first two basic issues of Calculus - two big questions, what are they? How do you find instantaneous rate of change? How do you find how things are changing instantly? And then, completely separate from that, how do you find areas under curves? Two completely different questions. It turns out answers are completely related. Let's see how they go. Question 1, instantaneous rate of change. What are we going to do? Well, what you do there is you first remember what does rate mean? Well, rate is just change in distance over change in time. It's as easy as that. Distance equals rates times time. Not a big deal.
Okay, now what do you do with that? Well, if you would graph a function that represents distance against time, what do you notice? If you want to look at the change in time and the change in distance, what do you got? You actually got a slope of a line. It's a slope - rise over run, change in distance over change in time, distance over time, so you've got a slope. So automatically we see a really neat thing. We see that the average rate of change between two points is equal to the slope of the line connecting them. Well, that's really cool.
By the way, what about lines? Maybe you forgot about lines. I'm going to remind you about lines. Okay, no problem. We can do lines. y - y^1 = m(x - x^1). This is the point slope form. All you needed to give me is a point on the line, x^1 y^1, and a slope, m. You give me those two pieces of information, I can always write down the line uniquely. Always, always, always. Never forget it.
Okay, fine. Now we're back to here. Now we can find average rate. And what do we see? Well, in fact, a line that touches the curve at two points is sometimes called a secant line. So, in fact, we've just discovered that the average rate, change in distance over change in time, is equal to the slope of the secant line. Cool. So all you have to do if you want to find average rate, connect the two points with the line, find the slope, you've got the average rate. No biggie. That's not what we want. We want instantaneous rate. So how do you do that? Well, what would that be? Well, if I want to find instantaneous rate here, what would I do? I'd make that secant line closer and closer and closer. I'd bring those points together, and look what I'm converging to. I'm coming to a tangent line. Wow! Instantaneous rate of change equals the slope of the tangent line. So we want a slope of tangent line. Well, what's the slope of the tangent line? It's change in y over change in x, change in distance over change in time, but now, the change in time, from that point to itself, is zero. Zero over zero. So I'm getting zero over zero, which is complete garbage, and that is big problem. Our first major problem of the course.
Okay, so what do we do? Well, we can that. This is equal to rate times time. We've got to can that. Instantaneous rate of change; I really want that. So what do we do? Well, how do we get that zero over zero problem to go away? The answer is, we inch up to it. We just approach the zero point. And what would that look like? Well, let me remind you what you've done when you were a little kid. When you were a little teeny kid this is what you were looking at. You were looking at a value of function at a point - value of function at a point. Okay, not a big deal. There it is - f(a), it's that point. But you don't now anything else about the function. You don't know what's going round and round there, because all you're looking at is f(a). You open it up, the function will be quite interesting - who knows?
But now what I invite us to do and what Calculus invites us to do is to look at the function this way. Cover up that point and look at everything else. Open the window, look outside, that's where Calculus is. And what you see here is we can see when things are approaching and we can actually determine the idea of a limit. The limit is what things are approaching. We don't care about what actually happens at that point, only what things approaching. On with the idea of a limit. What can we do?
Well, now we can return to the question and figure out - let's take the limit as DeltaT goes to zero. What do we get? We get zero over zero. That is called an "indeterminant form" when you get zero over zero. So what do you do? You've got to do some algebraic gymnastics. You've got to factor the top and cancel with the bottom. You can try to multiply by the conjugant, you can try to combine the fractions. There's all these tricks of the trade to actually reduce this to something that you can actually find. So you find the limit. Okay, so once you find the limit and you take the limit as DeltaT goes to zero, what do you get? You get the answer to the question, how do you find instantaneous rate of change? The answer is what we call "the derivative." And what's the derivative? It's the limit as DeltaX goes to zero. Well, . Looks pretty confusing, doesn't it? It's just rate. It's distance over time, but now I'm letting the time go to zero. Not a big deal. So that's the derivative. Bingo. We're done with the first question.
And so now what do we see? Well, we see now that it's come back to here and the derivative, in fact, gives us the slope of a tangent line. Well, that's really cool. If you want to find the slope of a tangent line ever in life, you just take the derivative, and that gives you the slope of the tangent once you evaluate it at the point you want. Great, but for free we answer the first question, because remember, the derivative also represents the instantaneous rate of change. So you want to find out how things are changing? No big deal. You take the derivative, plug it in, and that will tell you how things are changing at that instant.
Okay, great. So now we know all about how to graph these things, how to look at these things, the derivative ... We've got this all out of the way. Now, let's take a look at some applications. What can we do with this? Well, how would you take derivatives of complicated functions? Well, if you've got a product, use the product rule. Remember, the product rule - don't memorize the formula, memorize the chant. First times the derivative of the second, plus the second times derivative of the first. So the derivative of a product is this. It's not the product of the derivatives. You've got to use the product rule. We've got five minutes, folks. I've got to move faster.
What if you have a quotient? Well, then you use the quotient rule. And what's the quotient rule? If you've got a quotient you take the bottom times the derivative of the top, minus the top times the derivative of the bottom, all over the bottom squared. That's the quotient rule. That's what you use when you have a derivative of a quotient. Okay, great. No problem.
Now, what if I have a really complicated function? What if you've got a function that looks like this? It's got insides; it's got guts right in there. See, you've got something like this, and you want to take the derivative of that. What do you do? Well, you've got to use the chain rule, folks. This is a thing that you can chain together. There's an inside; there's a block right here. There's a whole big block there, and then you've got an outside. So take the derivative of the outside, so the derivative of sin of blop is actually cosine of blop. So it's cosine of blop, and what's the blop? The blop is going to be 3x^3 + 1, and then you multiply that by the derivative of the inside, and the derivative of that turns out to be just, let's see, 9x^2 + 0. And so there's the derivative using the chain rule. The idea is to peel off, like an onion. Just peel off, keep peeling off the outside until you get to the inside. Always remember though, when you take the derivative, if you have sin of blop, the derivative is cosine of the blop. Don't put the derivative in there. Put the blop and multiply by the derivative of the inside. That's the key to the chain rule.
Well, what about when you have functions that aren't functions? Like what if you have things that are relations, like x^2 + y^2 = 1, like a circle? How do you differentiate that? How do you find there? Well, the answer is, we use something called implicit differentiation. Implicit differentiation, how does that work? Well, you've got to remember that , that's an object, that's a noun, and is a verb. It's a commandment. Take the derivative with respect to x. So you differentiate this with respect to x and what do you see? Well, you see something that looks like this. What you do is you say, "Okay, I'll take (x^2 + y^2)= 1. The derivative of x^2 with respect to x is just 2x, not a big deal. The derivative of y^2 - remember how I think about this. I think about this as clumping all this together. And I see this as a blop^2, so I actually use the chain rule, which we've just developed, and the chain rule says the derivative of blop^2 is 2blop, and I multiply that by the derivative of the blop, which is the derivative of y with respect to x. That's called , folks. The derivative of 1 is zero. And now you can actually solve this for by bringing this to the other side. That would be a -2x. You divide by the 2y, and you see = -, and there's the answer. So that's implicit differentiation. Just go right through and differentiate implicitly. When you have a relationship you can still find the derivative. We're making progress here folks.
Well, now that you have derivatives, what can you do with that? Well, If you think about it as a velocity, you have instantaneous velocity. If you take the derivative of velocity you actually get the change in velocity, which is acceleration. So we get acceleration now, we get velocity. Acceleration is just the second derivative of the position. So we can actually take derivatives upon derivatives upon derivatives, as many derivatives as you want. So it's great; so higher-order derivatives - no problem, we can do that.
What can you use these things for? Okay, we know it's true for velocity. We can use it for velocity. What else can we use it for? Well, it turns out you can use it for linear approximation. Suppose you've got some wacko function like this, and you actually want to figure out the value right here. Right here, and you don't know what that value is. But you know nearby there's a point that you can actually compute. So what do you do? You find a tangent line approximation. If you remember, the tangent line closely emulates the activity of the function. The tangent line closely emulates the activity of what the function's doing. They look the same there. So you find the equation of the tangent line and then plug in the mysterious point, then you're all set, because you can approximate the value by plugging in the tangent line.
So how would that look? Well, that's called linear approximation. Here's the formula, but don't bother memorizing it, just think about it. What you've got to do is find the equation of the line that's tangent at the known point, which is x here in this case, and this is going to be x + DeltaX, the known point, plus a little, teeny offset, is going to be approximately equal to the derivative times the change in x plus the function. So that's it. That's the whole thing right there - linear approximation, allows you to actually compute things. Computers know Calculus; everyone knows Calculus; you've got to know Calculus.
Now, what else can you do? Well, suppose that a derivative were to be zero? Well, how could that possibly happen? That could possibly happen because maybe the function goes like this and I see the tangent over here has slope zero. Or maybe the function goes like this, and I see the tangent has slope zero. In particular, if the tangent equals zero, maybe we have a max or a min. Also, maybe the slope or the tangent doesn't exist. Like if we have a wave kind of thing, a cusp. Very pretty. Cusp like the waves. Well, that might be a max, that might be a min. So, in particular you can find out when objects are maximized or minimized. You can find the maxima or the minima very easily by using Calculus. What do you do? You take a derivative and you see where it equals zero or where the derivative doesn't exist, but the function does. Those give you candidates for possible max and min. And what can you use that for? Well, you can do all sorts of max/min problems. You want to maximize profits; you want to minimize costs? There are 10 minutes left. They're trying to pull me here, folks, but I'm going to go for it.
So you want to minimize costs, you want to maximize profits? You want to maximize area; you want to minimize volume? Whatever it is, set up the problem really carefully, figure out exactly what you want to optimize, take the derivative, set it equal to zero, solve, find out where the derivative's undefined, and you've got it made. Really not, not, not a big deal. However, you should always remember and never forget the fundamental method of solving problems. So remember yow you solve all of life's problems. The first thing you have to do is understand what you're being asked. You can't answer a question if you don't understand.
The next thing you do after you understand what you're supposed to find is figure out what you know. List every single thing that you know, every single fact. It may be superfluous, maybe you don't use it, who cares. Write it down, understand it, make it your own. And then the last thing is take the information that you know and see a relationship between that and the thing that you seek. Try to find a connection. Once you've got the connection, then you're on the road to actually finding the solutions. That is the simple method for finding every single answer to any single problem.
Another application, if you think about derivative as a rate is looking at related rates. Suppose, for example, that you actually have a ladder that's falling down. The ladder is falling, and you don't want to be sued, but the only thing you do know is how fast the bottom is falling. You want to know how fast the top is falling. What you need to there is if you know this rate you can find that rate by linking them up with a connection. In this case the connection would be the Pythagorean theorem. You can take the derivative with respect to time, because here you see the variable, the thing that's independent, that's always changing, is time. Time keeps on ticking into the future. So you keep going like this, you actually solve this, take the derivative using implicit differentiation, differentiate with respect to time, and plug in what you know, how this is changing and that ought to tell you how this is changing. Pretty cool. That's called related rates.
Suppose, for example, you drop a stone into a very, very still pool. You have a ripple effect. Those concentric circles are getting larger. If you know how fast the radius is changing, you can find how fast the area is changing, because you have a connection between area and rate. Area = \prodr^2. So there you go, related rates. You know how one rate is changing, you can find out how a related rate is changing.
What else can you use a derivative for? Well, the other thing you can do is actually graph really, really accurate pictures of functions. Finally, you can figure out that a parabola looks really pretty and bowl-like like this. And it's not something really exotic, but it's just a nice, pretty bowl. How do you do it? Well, you just start taking derivatives and analyzing things. First you find the critical points. Those are points where the derivative either equals zero, or the derivative is undefined, but the function is defined. So, for example, in these examples right here you'll notice that the derivative equals zero here, the tangent is horizontal, right there. The tangent is horizontal here. In this example the tangent is horizontal here, and then where does the derivative not exist? The derivative doesn't exist here, and the derivative doesn't exist there. Those are candidates for max and min. Those are called "critical points." Then what do you do with those things? Well, you set them up on a little number line on the x-axis, and you look at the integrals all around it, and you see whether the derivative is positive or negative. If the derivative is positive, that means slopes are positive, so the function must be increasing. If the derivative is negative, then that means the function must be decreasing. So you can see that the function, for example, here is decreasing, decreasing, decreasing, then increasing. Here you can see it's increasing, increasing, increasing, then decreasing, then increasing. So you can see where it's going up and it's going down by the sign of the first derivative. That also determines whether you have maxes or mins anywhere. And that's called the "first derivative test."
Now, how do you figure out the curvature? The curvature is given by the rate of change of the derivative, how the derivative is changing. So what you do there is you take the derivative of the derivative. So you look at the second derivative. When the second derivative is zero are potential points of inflection. Points where the concavity changes. This is concave up, it's curving upwards. This is concave down. This is concave up. So the cup is sitting up, the cup is sitting down, concave down. So here you would see the second derivative is positive, then the second derivative is negative. It changes here. Now, it's sitting up. It's positive, positive, we're concave up. Here, we're concave up, second derivative is positive. Now, here we see it's concave down, second derivative is negative, and second derivative here is also negative. This is a cusp point, the derivative doesn't exist there. This is a point of inflection, this is a point of inflection, this is a point of inflection, and that's a minimum.
So just by taking derivative and then second derivatives, you can actually figure out and graph a very accurate sketch of even very complicated, almost scary-looking functions. By the way, got fractional exponents? Expect cusps. That's my word of warning for the day.
Okay, now, if you've got really exotic functions that have denominators, then actually you may have asymptotes. So don't forget that a vertical asymptote is where the function, after you simplify it, the bottom equals zero. So wherever the bottom equals zero, after you simplify and reduce, those are going to give you your vertical asymptotes. Horizontal asymptotes, you take the limit as x goes off to infinity, as you go off to the horizon, and you see what y-value you're trying to land to. If you're landing somewhere, then you know you've got a horizontal asymptote and it's at y equals that value. So you put in the asymptotes, and you can do all the other Calculus, get the curvature; see exactly what the beautiful picture looks like.
And that was the end of differential Calculus. Great, no problem. Now what? Well, now we move on and look at the exact same thing we just did backwards. So we look at Math Jeopardy - Oh, my goodness, I have only five minutes left.
So Math Jeopardy, here we go. So the idea is, if I tell you what the derivative is, how can you find the function whose derivative is that? Well, this is the notion of an anti-derivative. So how can you find the anti-derivative? Well, we saw the formulas for that. If you want to find the anti-derivative of x^n, it would be x^n+1 divided by n+1, take the derivative of that and see you get x^n, unless n = -1, then you're looking at the integral of , and what's the integral of ? Well, it's the natural log of the absolute value of x, because the derivative of a natural log, we already saw, was a 1/x. So great.
Now, how can you find exotic integrals? Well, remember that represents differentiating with respect to x, so therefore, the integral with respect to x represents to integrate with respect to x. Now, if you have a very complicated thing there with an inside and outside, you might be able to untangle that, which potentially was made by the chain rule by using substitution. Let u equal some big blop, and the big blop's derivative should appear somewhere else in your integral. If you've got that, it sounds like a good candidate for u to u substitution, and then you've got to change the dx to du by taking derivatives and seeing what du equals in terms of dx. So that's the u to u substitution. You've got that going on here.
And now what can you do? You can take that and look at studies in motion again. Now I can give you acceleration; if you integrate, you get velocity. If you integrate again, you get position. So vertical motion, not a big deal anymore. Lands, no problem. Things moving into space, like this thing here. You can get that. Anything that moves, anything at all that moves, we can now analyze whether you have crabs or whether you have a bike. It's not a problem anymore, folks. If it's movement we can antidifferentiate it and figure out what it is. Not a problem.
Now, where does this lead us? Well, it brings us back to the very first question of the course, which was how do you find areas under curves? We have to answer that. We haven't done it yet. It turns out the surprising answer is that it's the fundamental theorem of Calculus, and the idea is if you want to find the area under this curve, from A to B, right here, if you want to find that area, then how do you do it? It turns out that if that function is called, let's say, f(x), then all you do is integrate from A to B, f(x) dx, because you're summing up little rectangles in here that are based, small change in x, base times height, which is a function, you go from A to B, and this equals F(b) -F(a). Well, what's F? That's the anti-derivative. So if you take the derivative of F, you actually get f(x). So just find the anti-derivative, plug in the big point, plug in the small points, subtract, that will always give you the area under the curve.
You can look at areas as more exotic things. For example, if a thing actually goes like this, then actually, this is actually not very x easy. The rectangles aren't very clear, the rectangles change from going from green to green, over to green to orange, and to orange to orange. It's not very uniform. However, if you put the rectangles in this way and stack them this way, now you're summing with respect to y. And so here you would actually sum this with respect to y, so you would integrate this, dy, and put the rectangles in this way, and you'd put the rectangles in that way and stack. You'd stack from low to high and you'd stack the rectangles like this.
Okay, that is all of Calculus. I did it in under 20 minutes. That's what it is. Go back, think about it. Have fun with it. Congratulations, folks, you've just finished Calculus I.
Whew! Okay, I'm ready now. Now that we have a little background under our belts, let's take a look at Calculus II. I'll see you there.